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**Unformatted text preview: **1.1. Solve: 1.2. Solve: 1.3. Solve: 1.4. Solve: (a) The basic idea of the particle model is that we will treat an object as if all its mass is concentrated into a single point. The size and shape of the object will not be considered. This is a reasonable approximation of reality if (i) the distance traveled by the object is large in comparison to the size of the object, and (ii) rotations and internal motions are not significant features of the objects motion. The particle model is important in that it allows us to simplify a problem. Complete realitywhich would have to include the motion of every single atom in the objectis too complicated to analyze. By treating an object as a particle, we can focus on the most important aspects of its motion while neglecting minor and unobservable details. (b) The particle model is valid for understanding the motion of a satellite or a car traveling a large distance. (c) The particle model is not valid for understanding how a car engine operates, how a person walks, how a bird flies, or how water flows through a pipe. 1.5. Solve: (a) An operational definition defines a concept or an idea in terms of a procedure, or a set of operations, that is used to identify or measure the concept. (b) The displacement .r.. of an object is a vector found by drawing an arrow from the objects initial location to its final location. Mathematically, f i.r.. = r.. - r... The average velocity v.. of an object is a vector that points in the same direction as the displacement .r.. and has length, or magnitude, .r../.t, where f i .t = t - t is the time interval during which the object moves from its initial location to its final location. 1.6. Solve: The player starts from rest and moves faster and faster (accelerates). 1.7. Solve: The particle starts with an initial velocity but as it slides it moves slower and slower till coming to rest. This is a case of negative acceleration because it is an acceleration opposite to the positive direction of motion. 1.8. Solve: The acceleration of an object is a vector formed by finding the ratio of .v.. , the change in the objects velocity, to .t, the time in which the change occurs. The acceleration vector a.. points in the direction of .v.., which is found by vector subtraction. 1.9. Solve: (a) Acceleration is found by the method of Tactics Box 1.3. Let 0 v.. be the velocity vector between points 0 and 1 and 1 v.. be the velocity vector between points 1 and 2. (b) Speed 1 v is greater than speed 0 v because more distance is covered in the same interval of time. 1.10. Solve: (a) Acceleration is found by the method of Tactics Box 1.3. Let 0 v.. be the velocity vector between points 0 and 1 and 1 v.. be the velocity vector between points 1 and 2. (b) Speed 1 v is greater than speed 0 v because more distance is covered in the same interval of time. 1.11. Solve: (a) (b) 1.12. Solve: (a) (b) 1.13. Model: Represent the car as a particle. Visualize: The dots are equally spaced until brakes are applied to the car. Equidistant dots indicate constant average speed. On braking, the dots get closer as the average speed decreases. 1.14. Model: Represent the (child + sled) system as a particle. Visualize: The dots in the figure are equally spaced until the sled encounters a rocky patch. Equidistant dots indicate constant average speed. On encountering a rocky patch, the average speed decreases and the sled comes to a stop. This part of the motion is indicated by a separation between the dots that becomes smaller and smaller. 1.15. Model: Represent the tile as a particle. Visualize: The tile falls from the roof with an acceleration equal to a = g = 9.8 m/s2. Starting from rest, its velocity increases until the tile hits the water surface. This part of the motion is represented by dots with increasing separation, indicating increasing average velocity. After the tile enters the water, it settles to the bottom at roughly constant speed. 1.16. Model: Represent the tennis ball as a particle. Visualize: The particle falls freely for the three stories under the acceleration of gravity. It strikes the ground and very quickly decelerates to zero (while decompresses) and finally travels upward with negative acceleration under gravity to zero velocity at a height of two stories. The downward and upward motions of the ball are shown in the figure. The increasing length between the dots during downward motion indicates increasing average velocity or downward acceleration. On the other hand, the decreasing length between the dots during upward motion indicates acceleration in a direction opposite to its motion; that is, in the downward direction. Assess: For a free-fall motion, acceleration due to gravity is always vertically downward. 1.17. Model: Represent the toy car as a particle. Visualize: As the toy car rolls down the ramp, its average speed increases. This is indicated by the increasing length of the velocity arrows. That is, motion down the ramp is under an acceleration a... At the bottom of the ramp, the toy car continues with the speed obtained with no change in velocity. 1.18. Solve: Dot Time (s) x (m) 1 0 0 2 2 30 3 4 95 4 6 215 5 8 400 6 10 510 7 12 600 8 14 670 (a) 9 16 720 (b) 1.19. Solve: A forgetful physics professor goes for a walk on a straight country road. Walking at a constant speed, he covers a distance of 300 m in 300 s. He then stops and watches the sunset for 100 s. Finding that it was getting dark, he walks faster back to his house covering the same distance in 200 s. 1.20. Solve: Forty miles into a car trip north from his home in El Dorado, an absent-minded English professor stopped at a rest area one Saturday. After staying there for one hour, he headed back home thinking that he was supposed to go on this trip on Sunday. Absent-mindedly he missed his exit and stopped after one hour of driving at another rest area 20 miles south of El Dorado. After waiting there for one hour, he drove back very slowly, confused and tired as he was, and reached El Dorado in two hours. 1.21. Visualize: The bicycle is moving with an acceleration of 1.5 m/s2. Thus, the velocity will increase by 1.5 m/s each second of motion. 1.22. Visualize: The particle moves upward with a constant acceleration a.. . The final velocity is 200 m/s and is reached at a height of 1000 m. 1.23. Solve: (a) 6 9.12 s (9.12 s) 10 s 9.12 10 6 s 1 s - - . . = . . = . . (b) 3 3.42 km (3.42 km) 10 m 3.42 103 m 1 km . . = . . = . . (c) 2 2 3 44 cm/ms 44 cm 10 m 1 ms 4.4 10 m/s ms 1 cm 10 s - - . .. .. . = . .. .. . = . .. .. . (d) km 103 m 1 hour 80 km/hour 80 22 m/s hour 1 km 3600 s . .. .. . = . .. .. . = . .. .. . 1.24. Solve: (a) 2.54 cm 10 2 m 8.0 inches 8.0 (inch) 0.20 m 1 inch 1 cm . .. - . = . .. . = . .. . (b) 66 feet/s 66 feet 12 inch 1 m 20 m/s s 1 foot 39.37 inch = . .. .. . = . .. .. . . .. .. . (c) miles 1.609 km 103 m 1 hour 60 mph 60 27 m/s hour 1 mile 1 km 3600 s . .. .. .. . = . .. .. .. . = . .. .. .. . (d) 2 14 square inches 14 (inches)2 1 m 9.0 10 3 square meter 39.37 inches = .. .. = - . . 1.25. Solve: (a) 1 hour 1(hour) 3600 s 3600 s 3.60 103 s 1 hour = . . = = . . . . (b) 1 day 1 (day) 24 hours 3600 s 8.64 104 s 1 day 1 hour . .. . = . .. . = . .. . (c) 4 1 year 1 (year) 365.25 days 8.64 10 s 3.16 107 s 1 year 1 day . .. . = . .. . = . .. . (d) 2 2 2 32 ft/s 32 ft 12 inch 1 m 9.75 m/s s 1 ft 39.37 inch = . .. .. . = . .. .. . . .. .. . 1.26. Solve: (a) 20 ft 20(ft) 1 m 7.0 m 3 ft = . . = . . . . (b) 60 miles 60(miles) 1 km 1000 m 1.0 105 m 0.6 miles 1 km = . .. . = . .. . . .. . (c) 60 mph 60(mph) 1 m/s 30 m/s 2 mph . . = . . = . . (d) 1 cm 10 2 m 8 in 8(in) 0.16 m 1/2 in 1 cm . .. - . = . .. . = . .. . 1.27. Solve: (a) (30 cm) 4 in 12 in 10 cm . . = . . . . (b) (25 m/s) 2 mph 50 mph 1 m/s . . = . . . . (c) (5 km) 0.6 mi 3 mi 1 km . . = . . . . (d) 1 in 1 cm 2 1 in 2 1 cm 4 . . . .. . . .. . = . .. . 1.28. Solve: (a) 33.3 25.4 = 846 (b) 33.3 - 25.4 = 7.9 (c) 33.3 = 5.77 (d) 333.3 25.4 =13.1 1.29. Solve: (a) (33.3)2 =1.109103. For numbers starting with 1 an extra digit is kept. (b) 33.3 45.1 =1.50103 Scientific notation is an easy way to establish significance. (c) 22.2 -1.2 = 3.5 (d) 1/ 44.4 = 0.0225 1.30. Solve: The length of a typical car is 15 ft. Or 15(ft) 12 inch 1 m 4.6 m 1 ft 39.37 inch . .. .= . .. . . .. . This length of 15 ft is approximately two-and-a-half times my height. 1.31. Solve: The height of a telephone pole is estimated to be around 50 ft or 15 m. This height is approximately 8 times my height. 1.32. Solve: I typically take 15 minutes in my car to cover a distance of approximately 6 miles from home to campus. My average speed is 6 miles 60 min 24 mph 24(mph) 0.447 m/s 11 m/s 15 min 1 hour 1 mph . . . . . . = = . . = . . . . 1.33. Solve: My barber trims about an inch of hair when I visit him every month for a haircut. The rate of hair growth is 2 9 6 9 1(inch) 2.54 cm 10 m 1 month 1 day 1 h 9.8 10 m/s (month) 1 inch 1 cm 30 days 24 h 3600 s 9.8 10 m 10 m 3600 s 35 m/h s 1 m 1 h - - - . .. .. .. .. .. .. .. .. .. . = . .. .. .. .. . . .. .. . = . .. .. . = . .. .. . 1.34. Model: Represent the Porsche as a particle for the motion diagram. Visualize: 1.35. Model: Represent the watermelon as a particle for the motion diagram. Visualize: 1.36. Model: Represent (Sam + car) as a particle for the motion diagram. Visualize: 1.37. Model: Represent the speed skater as a particle for the motion diagram. Visualize: 1.38. Model: Represent the wad as a particle for the motion diagram. Visualize: 1.39. Model: Represent the ball as a particle for the motion diagram. Visualize: 1.40. Model: Represent the ball as a particle for the motion diagram. Visualize: 1.41. Model: Represent the motorist as a particle for the motion diagram. Visualize: 1.42. Model: Represent Bruce and the puck as particles for the motion diagram. Visualize: 1.43. Model: Represent Fred and yourself as particles for the motion diagram. Visualize: 1.44. Solve: Rahul was coasting on interstate highway I-35 from Wichita to Kansas City at 65 mph. Seeing an accident at a distance of 200 feet in front of him, he braked his car to a stop with steady deceleration. 1.45. Solve: A car starts coasting at an initial speed of 30.0 m/s up a 10 incline. 230 m up the incline the road levels out to a flat road and the car continues coasting at a reduced speed along the road. 1.46. Solve: A skier starts from rest down a 25 slope with very little friction. At the bottom of the 100 m slope the skier moves to a flat area and continues at constant velocity. 1.47. Solve: A ball is dropped from a height to check its rebound properties. It rebounds to 80% of its original height. 1.48. Solve: Two boards lean against each other at equal angles to the vertical direction. A ball rolls up the incline, over the peak, and down the other side. 1.49. Solve: (a) (b) Sue passes 3rd Street doing 40 mph, slows steadily to the stop sign at 4th Street, stops for 1 s, then speeds up and reaches her original speed as she passes 5th Street. If the blocks are 50 m long, how long does it take Sue to drive from 3rd Street to 5th Street? (c) 1.50. Solve: (a) (b) A train moving at 100 km/hour slows down in 10 s to a speed of 60 km/hour as it enters a tunnel. The driver maintains this constant speed for the entire length of the tunnel that takes the train a time of 20 s to traverse. Find the length of the tunnel. (c) 1.51. Solve: (a) (b) Jeremy has perfected the art of steady acceleration and deceleration. From a speed of 60 mph he brakes his car to rest in 10 seconds with a constant deceleration. Then he turns into an adjoining street. Starting from rest, Jeremy accelerates with exactly the same magnitude as his earlier deceleration and reaches the same speed of 60 mph over the same distance in exactly the same time. Find the cars acceleration or deceleration. (c) 1.52. Solve: (a) (b) A coyote (A) sees a rabbit and begins to run toward it with an acceleration of 3.0 m/s2. At the same instant, the rabbit (B) begins to run away from the coyote with an acceleration of 2.0 m/s2. The coyote catches the rabbit after running 40 m. How far away was the rabbit when the coyote first saw it? (c) 1.53. Solve: Since area equals length width, the smallest area will correspond to the smaller length and the smaller width. Similarly, the largest area will correspond to the larger length and the larger width. Therefore, the smallest area is (64 m)(100 m) = 6.4 103 m2 and the largest area is (75 m)(110 m) = 8.3 103 m2. 1.54. Solve: (a) We need kg/m3. There are 100 cm in 1 m. If we multiply by 3 100 cm (1)3 1 m . . = . . . . we do not change the size of the quantity, but only the number in terms of the new unit. Thus, the mass density of aluminum is 3 3 3 3 3 2.7 10 kg 100 cm 2.7 10 kg cm 1 m m - .. .... .. = . .. . (b) Likewise, the mass density of alcohol is 3 3 3 0.81 g 100 cm 1 kg 810 kg cm 1 m 1000 g m . .. . . . . .. . . . = . .. . . . 1.55. Model: The car is represented by the particle model as a dot. Solve: (a) Time t (s) Position x (m) 0 1200 10 975 20 825 30 750 40 700 50 650 60 600 70 500 80 300 90 0 (b) 1.56. Solve: Susan enters a classroom, sees a seat 40 m directly ahead, and begins walking toward it at a constant leisurely pace, covering the first 10 m in 10 seconds. But then Susan notices that Ella is heading toward the same seat, so Susan walks more quickly to cover the remaining 30 m in another 20 seconds, beating Ella to the seat. Susan stands next to the seat for 10 seconds to remove her backpack. 1-1 1.57. Solve: A crane operator holds a ton of bricks 30 m above the ground. Four seconds after he is told to lower the bricks, he takes four seconds to lower them 15 m at a constant rate before stopping the bricks to make an eight-second safety check. He then continues lowering the bricks the remaining 15 m, taking four more seconds. 2.1. Model: We will consider the car to be a particle that occupies a single point in space. Visualize: Solve: Since the velocity is constant, we have xf = xi + vx.t. Using the above values, we get 1 x = 0 m+ (10 m/s)(45 s) = 450 m Assess: 10 m/s 22 mph and implies a speed of 0.4 miles per minute. A displacement of 450 m in 45 s is reasonable and expected. 2.2. Model: We will consider Larry to be a particle. Visualize: Solve: Since Larrys speed is constant, we can use the following equation to calculate the velocities: f i s f i v s s t t - = - (a) For the interval from the house to the lamppost: 1 200 yd 600 yd 200 yd/min 9:07 9:05 v - = =- - For the interval from the lamppost to the tree: 2 1200 yd 200 yd 333 yd/min 9:10 9:07 v - = =+ - (b) For the average velocity for the entire run: avg 1200 yd 600 yd 120 yd/min 9:10 9:05 v - = =+ - 2.3. Model: Cars will be treated by the particle model. Visualize: Solve: Beth and Alan are moving at a constant speed, so we can calculate the time of arrival as follows: 1 0 1 0 1 0 1 0 v x x x t t x x t t t v . - - = = . = + . - Using the known values identified in the pictorial representation, we find: Alan 1 Alan 0 Alan 1 Alan 0 Beth 1 Beth 0 Beth 1 Beth 0 8:00 AM 400 mile 8:00 AM 8 hr 4:00 PM 50 miles/hour 9:00 AM 400 mile 9:00 AM 6.67 hr 3:40 PM 60 miles/hour t t x x v t t x x v - = + = + = + = - = + = + = + = (a) Beth arrives first. (b) Beth has to wait Alan 1 Beth 1 t - t = 20 minutes for Alan. Assess: Times of the order of 7 or 8 hours are reasonable in the present problem. 2.4. Solve: (a) The time for each segment is .t1 = 50 mi/40 mph = 5/4 hr and 2 .t = 50 mi/60 mph = 5/6hr. The average speed to the house is 100 mi 48 mph 5/6 h 5/4 h = + (b) Julie drives the distance 1 .x in time 1 .t at 40 mph. She then drives the distance 2 .x in time 2 .t at 60 mph. She spends the same amount of time at each speed, thus 1 2 1 2 1 2 .t = .t ..x /40 mph = .x /60 mph..x = (2/3).x But 1 2 .x + .x =100 miles, so 2 2 (2/3).x + .x =100 miles. This means 2 .x = 60 miles and 1 .x = 40 miles. Thus, the times spent at each speed are 1 .t = 40 mi/40 mph =1.00 h and 2 .t = 60 mi/60 mph =1.00 h. The total time for her return trip is 1 2 .t + .t = 2.00 h. So, her average speed is 100 mi/2 h = 50 mph. 2.5. Model: The bicyclist is a particle. Visualize: Please refer to Figure EX2.5. Solve: The slope of the position-versus-time graph at every point gives the velocity at that point. The slope at t = 10 s is 100 m 50 m 2.5 m/s 20 s v s t . - = = = . The slope at t = 25 s is 100 m 100 m 0 m/s 10 s v - = = The slope at t = 35 s is 0 m 100 m 10 m/s 10 s v - = =- 2.6. Visualize: Please refer to Figure EX2.6. Solve: (a) We can obtain the values for the velocity-versus-time graph from the equation v = .s/.t. (b) There is only one turning point. At t = 3 s the velocity changes from +5 m/s to -20 m/s, thus reversing the direction of motion. At t =1 s, there is an abrupt change in motion from rest to +5 m/s, but there is no reversal in motion. 2.7. Visualize: Please refer to Figure EX2.7. The particle starts at x0 =10 m at 0 t = 0. Its velocity is initially in the x direction. The speed decreases as time increases during the first second, is zero at t =1 s, and then increases after the particle reverses direction. Solve: (a) The particle reverses direction at t =1 s, when vx changes sign. (b) Using the equation f 0 x = x + area of the velocity graph between 1 t and f t , 2 s 3 s 10 m (area of triangle between 0 s and 1 s) + (area of triangle between 1 s and 2 s) 10 m 1 (4 m/s)(1 s) + 1 (4 m/s)(1 s) 10 m 2 2 10 m area of trapazoid between 2 s and 3 s 10 m 1 (4 m/s + 8 m/s 2 x x = - = - = = + = + 4 s 3 s )(3 s 2 s) 16 m area between 3 s and 4 s 16 m 1 (8 m/s + 12 m/s)(1 s) 26 m 2 x x - = = + = + = 2.8. Visualize: Please refer to Figure EX2.8. Solve: A constant velocity from t = 0 s to t = 2 s means zero acceleration. On the other hand, a linear increase in velocity between t = 2 s and t = 4 s implies a constant positive acceleration. 2.9. Visualize: Please refer to Figure EX2.9. Solve: (a) The acceleration of the train at t = 3.0 s is the slope of the v vs t graph at t = 3 s. Thus a = (2 m/s - ( - 2 m/s)) (8 s) = 0.5 m/s2 . (b) 2.10. Visualize: Please refer to Figure EX2.10. Solve: (a) At t = 2.0 s, the position of the particle is 2 s 2.0 m area under velocity graph from 0 s to 2.0 s 2.0 m 1 (4 m/s)(2.0 s) 6 m 2 x = + t = t = = + = (b) From the graph itself at t = 2.0 s, v = 4 m/s. (c) The acceleration is f i 2 6 m/s 0 m/s 2 m/s 3 s x x x x a v v v t t . - - = = = = . . 2.11. Visualize: Please refer to Figure EX2.11. Solve: (a) Using the equation xf = xi + area under the velocity-versus-time graph between i t and f t we have x(at t =1 s) = x(at t = 0 s) + area between t = 0 s and t =1 s = 2.0 m + (4 m/s)(1 s) = 6 m Reading from the velocity-versus-time graph, (at 1 s) 4 m/s. xv t = = Also, slope / 0 m/s2. xa = = .v .t = (b) x(at t = 3.0 s) = x(at t = 0 s) + area between t = 0 s and t = 3 s = 2.0 m + 4 m/s 2 s + 2 m/s 1 s + (1/2) 2 m/s 1 s =13.0 m Reading from the graph, ( 3 s) 2 m/s. x v t= = The acceleration is ( 3 s) slope (at 4 s) (at 2 s) 2 m/s2 2 s x x x a t v t v t = - = = = = =- 2.12. Model: Represent the jet plane as a particle. Visualize: Solve: (a) Since we dont know the time of acceleration, we will use 2 2 1 0 1 0 2 2 2 2 1 0 2 1 2 ( ) (400 m/s) (300 m/s) 8.75 m/s 2 2(4000 m) v v ax x a v v x = + - - - . = = = (b) The acceleration of the jet is approximately equal to g, the acceleration due to gravity. 2.13. Model: We are using the particle model for the skater and the kinematics model of motion under constant acceleration. Solve: Since we dont know the time of acceleration we will use 2 2 f i f i 2 2 2 2 f i 2 f i 2 ( ) (6.0 m/s) (8.0 m/s) 2.8 m/s 2( ) 2(5.0 m) v v ax x a v v x x = + - - - . = = = - - Assess: A deceleration of 2.8 m/s2 is reasonable. 2.14. Model: We are assuming both cars are particles. Visualize: Solve: The Porsches time to finish the race is determined from the position equation 2 P1 P0 P0 P1 P0 P P1 P0 2 2 P1 P1 ( ) 1 ( ) 2 400 m 0 m 0 m 1 (3.5 m/s )( 0 s) 15.1 s 2 x x v t t a t t t t = + - + - . = + + - . = The Hondas time to finish the race is obtained from Hondas position equation as 2 H1 H0 H0 H1 H0 H0 H1 H0 2 2 H1 H1 ( ) 1 ( ) 2 400 m 50 m 0 m 1 (3.0 m/s )( 0 s) 15.3 s 2 x x v t t a t t t t = + - + - = + + - . = So, the Porsche wins. 2.15. Model: Represent the spherical drop of molten metal as a particle. Visualize: Solve: (a) The shot is in free fall, so we can use free fall kinematics with a = -g. The height must be such that the shot takes 4 s to fall, so we choose 1 t = 4 s. Then, 2 2 2 2 1 0 0 1 0 1 0 0 1 ( ) 1 ( ) 1 1 (9.8 m/s )(4 s) 78.4 m 2 2 2 y = y + v t - t - g t - t . y = gt = = (b) The impact velocity is 1 0 1 0 1 v = v - g(t - t ) = -gt = -39.2 m/s. Assess: Note the minus sign. The question asked for velocity, not speed, and the y-component of v.. is negative because the vector points downward. 2.16. Model: Assume the ball undergoes free-fall acceleration and that the ball is a particle. Visualize: Solve: (a) We will use the kinematic equations 2 0 0 0 0 0 0 ( ) and ( ) 1 ( ) 2 v = v + a t - t y = y + v t - t + a t - t as follows: v(at t =1.0 s) =19.6 m/s + (-9.8 m/s2 )(1.0 s - 0 s) = 9.8 m/s y(at t =1.0 s) = 0 m+ (19.6 m/s)(1.0 s - 0 s) +1/2(-9.8 m/s2 )(1.0 s - 0 s)2 =14.7 m v(at t = 2.0 s) =19.6 m/s + (-9.8 m/s2 )(2.0 s - 0 s) = 0 m/s y(at t = 2.0 s) = 0 m+ (19.6 m/s)(2.0 s - 0 s) +1/2(-9.8 m/s2 )(2.0 s - 0 s)2 =19.6 m v(at t = 3.0 s) =19.6 m/s + (-9.8 m/s2 )(3 s - 0 s) = 9.8 m/s y(at t = 3.0 s) = 0 m+ (19.6 m/s)(3.0 s - 0 s) +1/2(-9.8 m/s2 )(3.0 s - 0 s)2 =14.7 m v(at t = 4.0 s) =19.6 m/s + (-9.8 m/s2 )(4.0 s - 0 s) = -19.6 m/s y(at t = 4.0 s) = 0 m+ (19.6 m/s)(4.0 s - 0 s) +1/2(-9.8 m/s2 )(4.0 s - 0 s)2 = 0 m (b) Assess: (a) A downward acceleration of 9.8 m/s2 on a particle that has been given an initial upward velocity of +19.6 m/s will reduce its speed to 9.8 m/s after 1 s and then to zero after 2 s. The answers obtained in this solution are consistent with the above logic. (b) Velocity changes linearly with a negative uniform acceleration of 9.8 m/s2. The position is symmetrical in time around the highest point which occurs at t = 2 s. 2.17. Model: We represent the ball as a particle. Visualize: Solve: Once the ball leaves the students hand, the ball undergoes free fall and its acceleration is equal to the acceleration due to gravity that always acts vertically downward toward the center of the earth. According to the constant-acceleration kinematic equations of motion 2 1 0 0 1 2 y = y + v .t + a.t Substituting the known values 2 2 1 1 -2 m = 0 m+ (15 m/s)t + (1/2)(-9.8 m/s )t The solution of this quadratic equation gives 1 t = 3.2 s. The other root of this equation yields a negative value for 1t , which is not valid for this problem. Assess: A time of 3.2 s is reasonable. 2.18. Model: We will use the particle model and the constant-acceleration kinematic equations. Visualize: Solve: (a) Substituting the known values into 1 2 1 0 0 2 y = y + v .t + a.t , we get 2 2 1 1 10 m 0 m 20 (m/s) 1 ( 9.8 m/s ) 2 - = + t + - t One of the roots of this equation is negative and is not relevant physically. The other root is 1 t = 4.53 s, which is the answer to part (b). Using 1 0 v = v + a.t , we obtain 2 1 v = 20(m/s) + (-9.8 m/s )(4.53 s) = -24 m/s (b) The time is 4.5 s. Assess: A time of 4.5 s is a reasonable value. The rocks velocity as it hits the bottom of the hole has a negative sign because of its downward direction. The magnitude of 24 m/s compared to 20 m/s, when the rock was tossed up, is consistent with the fact that the rock travels an additional distance of 10 m into the hole. 2.19. Model: We will represent the skier as a particle. Visualize: Note that the skiers motion on the horizontal, frictionless snow is not of any interest to us. Also note that the acceleration parallel to the incline is equal to g sin10. Solve: Using the following constant-acceleration kinematic equations, 2 2 f i f i 2 2 2 1 1 f i f i 2 2 ( ) (15 m/ s) (3.0 m / s) 2(9.8 m / s )sin10 ( 0 m) 64 m ( ) (15 m / s) (3.0 m / s) (9.8 m / s )(sin10 ) 7.1 s x x x x x x v v a x x x x v v a t t t t = + - . = + - . = = + - . = + . = Assess: A time of 7.1 s to cover 64 m is a reasonable value. 2.20. Model: Represent the car as a particle. Visualize: Solve: Note that the problem ends at a turning point, where the car has an instantaneous speed of 0 m/s before rolling back down. The rolling back motion is not part of this problem. If we assume the car rolls without friction, then we have motion on a frictionless inclined plane with an accleration a = -g sin. = -g sin 20 = -3.35 m/s2. Constant acceleration kinematics gives 2 2 2 2 2 2 2 0 1 0 1 0 0 1 1 2 2 ( ) 0 m /s 2 (30 m/s) 134 m 2 2( 3.35 m/s ) v v a x x v ax x v a = + - . = + . = - = - = - Notice how the two negatives canceled to give a positive value for x1. Assess: We must include the minus sign because the a.. vector points down the slope, which is in the negative x-direction. 2.21. Solve: (a) The position t = 2 s is 2 2 s x = [2(2) - 2 +1] m = 7 m. (b) The velocity is the derivative v = dx/dt and the velocity at t = 2 s is calculated as follows: 2 2 s v = (4t -1) m/s.v = [4(2) -1] m/s = 7 m/s (c) The acceleration is the derivative a = dv/dt and the acceleration at t = 2 s is calculated as follows: 2 2 2 s a = (4) m/s .a = 4 m/s 2.22. Solve: The formula for the particles position along the x-axis is given by f i f i t x t x = x + .v dt Using the expression for vx we get 3 3 2 2 f i f i 2 (2 m/s) 4 m/s 3 x x x x t t a dv d t t dt dt = + .. - .. = = = (a) The particles position at t =1 s is 2 5 1 s 3 3 x =1 m+ m = m. (b) The particles speed at t =1 s is 1 s v = 2 m/s. (c) The particles acceleration at t =1 s is 2 1 s a = 4 m/s . 2.23. Solve: The formula for the particles velocity is given by vf = vi + area under the acceleration curve between i t and f t For t = 4 s, we get 2 4 s 8 m/s 1 (4 m/s )4 s 16 m/s 2 v = + = Assess: The acceleration is positive but decreases as a function of time. The initial velocity of 8.0 m/s will therefore increase. A value of 16 m/s is reasonable. 2.24. Solve: (a) Time (s) Position (m) 0 -4 0.5 -2 1.0 0 1.5 1.75 2.0 3 2.5 4 3.0 5 3.5 6 4.0 7 (b) (c) .s = s(at t =1 s) - s(at t = 0 s) = 0 m - (-4 m) = 4 m. (d) .s = s(at t = 4 s) - s(at t = 2 s) = 7 m - 3 m = 4 m. (e) From t = 0 s to t =1 s, s v = .s/.t = 4 m/s. (f) From t = 2 s to t = 4 s, s v = .s/.t = 2 m/s. (g) The average acceleration is 2 m/s 4 m/s 2 m/s2 2 s 1 s a v t . - = = =- . - 2.25. Solve: (a) (b) To be completed by student. (c) 2 4 (at 1 s) [2 m/s2 (1 s) 4 m/s] 2 m/s x x dx v t v t dt = = - . = = - = - (d) There is a turning point at t = 2 s. (e) Using the equation in part (c), 4 m/s (2 4) m/s 4 xv = = t - .t = Since x = (t2 - 4t + 2) m, x = 2 m. (f) 2.26. Visualize: Please refer to Figure P2.26. Solve: The graph for particle A is a straight line from t = 2 s to t = 8 s. The slope of this line is -10 m/s, which is the velocity at t = 7.0 s. The negative sign indicates motion toward lower values on the x-axis. The velocity of particle B at t = 7.0 s can be read directly from its graph. It is -20 m/s, The velocity of particle C can be obtained from the equation f i v = v + area under the acceleration curve between i t and f t This area can be calculated by adding up three sections. The area between t = 0 s and t = 2 s is 40 m/s, the area between t = 2 s and t = 5 s is 45 m/s, and the area between t = 5 s and t = 7 s is -20 m/s.We get (10 m/s) + (40 m/s) + (45 m/s) - (20 m/s) = 75 m/s. 2.27. Visualize: Please refer to Figure P2.27. Solve: (a) We can calculate the position of the particle at every instant with the equation xf = xi + area under the velocity-versus-time graph between i t and f t The particle starts from the origin at t = 0 s, so xi = 0 m. Notice that the each square of the grid in Figure P2.27 has area (5 m/s) (2 s) =10 m. We can find the area under the curve, and thus determine x, by counting squares. You can see that x = 35 m at t = 4 s because there are 3.5 squares under the curve. In addition, x = 35 m at t = 8 s because the 5 m represented by the half square between 4 and 6 s is cancelled by the 5 m represented by the half square between 6 and 8 s. Areas beneath the axis are negative areas. The particle passes through x = 35 m at t = 4 s and again at t = 8 s. (b) The particle moves to the right for 0 s = t = 6 s, where the velocity is positive. It reaches a turning point at x = 40 m at t = 6 s. The motion is to the left for t > 6 s. This is shown in the motion diagram below. 2.28. Visualize: Solve: We will determine the objects velocity using graphical methods first and then using calculus. Graphically, v(t) = v0 + area under the acceleration curve from 0 to t. In this case, 0 v = 0 m/s.The area at each time t requested is a triangle. t = 0 s 0 v(t = 0 s) = v = 0 m/s t = 2 s ( 2 s) 1 (2 s)(5 m/s) 5 m/s 2 v t= = = t = 4 s ( 4 s) 1 (4 s)(10 m/s) 20 m/s 2 v t= = = t = 6 s ( 6 s) 1 (6 s)(10 m/s) 30 m/s 2 v t= = = t = 8 s v(t = 8 s) = v(t = 6 s) = 30 m/s The last result arises because there is no additional area after t = 6 s. Let us now use calculus. The acceleration function a(t) consists of three pieces and can be written: 2.5 0 s 4 s ( ) 5 30 4 s 6 s 0 6 s 8 s t t a t t t t = = .. = - + = = .. . = = These were determined by the slope and the y-intercept of each of the segments of the graph. The velocity function is found by integration as follows: For 0 = t = 4 s, 2 2 0 0 ( ) ( 0 s) ( ) 0 2.5 1.25 2 t v t = v t = + . t a t dt = + t = t This gives t = 0 s v(t = 0 s) = 0 m/ s t = 2 s v(t = 2 s) = 5 m/s t = 4 s v(t = 4 s) = 20 m/ s For 4 s = t = 6 s, 2 2 4 4 ( ) ( 4 s) ( ) 20 m/s 5 30 2.5 30 60 2 t v t v t t a t dt t t t t .- . = = + = + . + . = - + - . . . This gives: t = 6 s v(t = 6 s) = 30 m/s For 6 s = t = 8 s, 6 ( ) ( 6 s) ( ) 30 m/s 0 m/s 30 m/s t v t = v t = + . a t dt = + = This gives: t = 8 s v(t = 8 s) = 30 m/s Assess: The same velocities are found using calculus and graphs, but the graphical method is easier for simple graphs. 2.29. Visualize: Please refer to Figure P2.29. Solve: (a) The velocity-versus-time graph is the derivative with respect to time of the distance-versus-time graph. The velocity is zero when the slope of the position-versus-time graph is zero, the velocity is most positive when the slope is most positive, and the velocity is most negative when the slope is most negative. The slope is zero at t = 0, 1 s, 2 s, 3 s, . . . ; the slope is most positive at t = 0.5 s, 2.5 s, . . ; and the slope is most negative at t = 1.5 s, 3.5 s, . . . (b) 2.30. Visualize: Please refer to Figure P2.30. Solve: (a) We can determine the velocity as follows: vx = vx (at t = ti ) + area under the acceleration-versus-time graph from t = ti to tf (at 4 s) 0 m/s ( 1 m/s2 )(4 s) 4 m/s xv t= = + - =- (at 8 s) 4 m/s (3 m/s2 )(4 s) m/s 8 m/s xv t= =- + = (at 10 s) 8 m/s ( 2 m/s2 )(4 s) 4 m/s xv t= = + - = (b) If (at 0 s) 2.0 m/s, xv t= = the entire velocity-versus-time graph will be displaced upward by 2.0 m/s. 2.31. Solve: (a) The velocity-versus-time graph is given by the derivative with respect to time of the position function: (6 2 18 )m/s x v dx t t dt = = - For 0 m/s, x v = there are two solutions to the quadratic equation: t = 0 s and t = 3 s. (b) At the first of these solutions, x(at t = 0 s) = 2(0 s)3 - 9(0 s)2 +12 =12 m The acceleration is the derivative of the velocity function: x (12 18) m/s2 (at 0 s) 18 m/s2 x a dv t a t dt = = - . = =- At the second solution, x(at t = 3 s) = 2(3 s)3 - 9(3 s)2 +12 = -15 m (at 3 s) 12(3 s) 18 18 m/s2 xa t= = - =+ 2.32. Model: Represent the object as a particle. Solve: (a) Known information: 0 0 1 1 1 0 m, 0 x = v = m/s, x = 40 m, v =11 m/s, t = 5 s. If the acceleration is uniform (constant a), then the motion must satisfy the three equations 2 1 1 1 2 x = at .a = 3.20 m/s2 1 1 v = at .a = 2.20 m/s2 2 1 1 v = 2ax .a =1.51 m/s2 But each equation gives a different value of a. Thus the motion is not uniform acceleration. (b) We know two points on the velocity-versus-time graph, namely at 0 t = 0 and 1 t = 5 s.What shape does the function have between these two points? If the acceleration was uniform, which its not, then the graph would be a straight line. The area under the graph is the displacement .x. From the figure you can see that .x = 27.5 m for a straight-line graph. But we know that, in reality, .x = 40 m. To get a larger .x, the graph must bulge upward above the straight line. Thus the graph is curved, and it is concave downward. 2.33. Solve: The position is the integral of the velocity. 1 1 1 0 2 1 3 1 3 1 0 0 0 0 3 0 0 3 1 t t t t x x = x + . v dt = x + . kt dt = x + kt = x + kt Were given that 0 x = -9.0 m and that the particle is at 1 x = 9.0 m at 1 t = 3.0 s. Thus ( ) 1 3 ( ) ( 3 ) 3 9.0 m = -9.0 m + k(3.0 s) = -9.0 m + k 9.0 s Solving for k gives k = 2.0 m/s3. 2.34. Solve: (a) The velocity is the integral of the acceleration. 1 1 ( ) ( ) 1 0 1 2 1 2 1 0 0 2 0 1 2 1 0 m/s 10 10 10 t t t x x t x v = v + . a dt = + . - t dt = t - t = t - t The velocity is zero when ( 1 2 ) ( 1 ) 1 1 2 1 2 1 1 1 1 0 m/s 10 10 0 s or 20 s xv t t tt t t = = - = - . = = The first solution is the initial condition. Thus the particles velocity is again 0 m/s at 1 t = 20 s. (b) Position is the integral of the velocity. At 1 t = 20 s, and using 0 x = 0 m at 0 t = 0 s, the position is 1 ( ) 0 20 2 2 20 3 20 1 1 1 0 0 2 0 6 0 0 m 10 5 667 m t t x x = x + . v dt = + . t - t dt = t - t = 2.35. Model: Represent the ball as a particle. Visualize: Please refer to Figure P2.35. Solve: In the first and third segments the acceleration as is zero. In the second segment the acceleration is negative and constant. This means the velocity vs will be constant in the first two segments and will decrease linearly in the third segment. Because the velocity is constant in the first and third segments, the position s will increase linearly. In the second segment, the position will increase parabolically rather than linearly because the velocity decreases linearly with time. 2.36. Model: Represent the ball as a particle. Visualize: Please refer to Figure P2.36. The ball rolls down the first short track, then up the second short track, and then down the long track. s is the distance along the track measured from the left end (where s = 0). Label t = 0 at the beginning, that is, when the ball starts to roll down the first short track. Solve: Because the incline angle is the same, the magnitude of the acceleration is the same on all of the tracks. Assess: Note that the derivative of the s versus t graph yields the vs versus t graph. And the derivative of the vs versus t graph gives rise to the as versus t graph. 2.37. Model: Represent the ball as a particle. Visualize: Please refer to Figure P2.37. The ball moves to the right along the first track until it strikes the wall, which causes it to move to the left on a second track. The ball then descends on a third track until it reaches the fourth track, which is horizontal. Solve: Assess: Note that the time derivative of the position graph yields the velocity graph, and the derivative of the velocity graph gives the acceleration graph. 2.38. Visualize: Please refer to Figure P2.38. Solve: 2.39. Visualize: Please refer to Figure P2.39. Solve: 2.40. Visualize: Please refer to Figure P2.40. There are four frictionless tracks. Solve: For the first track, as is negative and constant and vs is decreasing linearly. This is consistent with a ball rolling up a straight track but not so far that vs goes to zero. For the second track, as is zero and vs is constant but greater than zero. This is consistent with a ball rolling on a horizontal track. For the third track, as is positive and constant and vs is increasing linearly. This is consistent with a ball rolling down a straight track. For the fourth track, as is zero and vs is constant. This is again consistent with a ball rolling on a horizontal track. The as on the first track has the same absolute value as the as on the third track. This means the slope of the first track up is the same as the slope of the third track down. 2.41. Model: The plane is a particle and the constant-acceleration kinematic equations hold. Solve: (a) To convert 80 m/s to mph, we calculate 80 m/s 1 mi/1609 m 3600 s/h = 179 mph. (b) Using as = .v/.t , we have, 2 2 s ( 0 to 10 s) 23 m/s 0 m/s 2.3 m/s ( 20 s to 30 s) 69 m/s 46 m/s 2.3 m/s 10 s 0 s 30 s 20 s s a t t a t t - - = = = = = = = = - - For all time intervals a is 2.3 m/s2. (c) Using kinematics as follows: 2 fs is f i f f v = v + a(t - t ).80 m/s = 0 m/s + (2.3 m/s )(t - 0 s).t = 35 s (d) Using the above values, we calculate the takeoff distance as follows: 2 22 f i is f i s f i ( ) 1 ( ) 0 m (0 m/s)(35 s) 1 (2.3 m/s )(35 s) 1410 m 2 2 s = s + v t - t + a t - t = + + = For safety, the runway should be 31410 m= 4230 m or 2.6 mi. This is longer than the 2.4 mi long runway, so the takeoff is not safe. 2.42. Model: The automobile is a particle. Solve: (a) The acceleration is not constant because the velocity-versus-time graph is not a straight line. (b) Acceleration is the slope of the velocity graph. You can use a straightedge to estimate the slope of the graph at t = 2 s and at t = 8 s. Alternatively, we can estimate the slope using the two data points on either side of 2 s and 8 s. Either way, you need the conversion factor 1 mph = 0.447 m/s from Table 1.4. ( ) ( ) ( ) ( ) ( ) ( ) 2 2 at 4 s at 0 s at 2 s 11.5mph 0.447 m/s 5.1 m/s 4 s 0 s s 1 mph at 10 s at 6 s mph 0.447 m/s at 8 s 4.5 2.0 m/s 10 s 6 s s 1 mph x x x x x x v v a v v a - = = - - = = - (c) The displacement, or distance traveled, is ( ) ( ) 10 s 0 s at 10 s at 0 s area under the velocity curve from 0 s to 10 s x .x = x - x = . v dx = We can approximate the area under the curve as the area of five rectangular steps, each with width .t = 2 s and height equal to the average of the velocities at the beginning and end of each step. avg avg avg avg avg 0 s to 2 s 14 mph 6.26 m/s area 12.5 m 2 s to 4 s 37 mph 16.5 m/s area 33.0 m 4 s to 6 s 53 mph 23.7 m/s area 47.4 m 6 s to 8 s 65 mph 29.1 m/s area 58.2 m 8 s to 10 s 74 mph 33.1 m/s a v v v v v = = = = = = = = = = = = = = rea = 66.2 m The total area under the curve is 217 m , so the distance traveled in 10 s is 217 m . 2.43. Model: Represent the car as a particle. Solve: (a) First, we will convert units: 60miles 1 hour 1610 m 27 m/s hour 3600 s 1 mile = The motion is constant acceleration, so 1 0 2 1 0 (27 m/s 0 m/s) 2.7 m/s 10 s v v a t a v v t - - = + . . = = = . (b) The fraction is a/g = 2.7 /9.8 = 0.28. So a is 28% of g. (c) The distance is calculated as follows: 2 2 2 2 1 0 0 1 ( ) 1 ( ) 1.3 10 m 4.3 10 feet 2 2 x = x + v .t + a .t = a .t = = 2.44. Model: Represent the spaceship as a particle. Solve: (a) The known information is: 2 0 0 x = 0 m, v = 0 m/s, t0 = 0 s, a = g = 9.8 m/s , and 8 1 v = 3.010 m/s. Constant acceleration kinematics gives 1 0 7 1 0 1 v v a t t t v v 3.1 10 s a - = + . .. = = = The problem asks for the answer in days, so we need a conversion: 7 2 1 (3.1 10 s) 1 hour 1 day 3.6 10 days 3600 s 24 hour t = = (b) The distance traveled is 2 2 15 1 0 0 1 1 ( ) 1 4.6 10 m 2 2 x - x = v .t + a .t = at = (c) The number of seconds in a year is 1 year 365 days 24 hours 3600 s 3.15 107s 1 day 1 hour = = In one year light travels a distance 1 light year = (3.0108m/s)(3.15107s) = 9.461015m The distance traveled by the spaceship is 4.61015m/9.461015m = 0.49 of a light year. Assess: Note that 1 x gives Where is it? rather than How far has it traveled? How far is represented by 1 0x - x . They happen to be the same number in this problem, but that isnt always the case. 2.45. Model: The car is a particle, and constant-acceleration kinematic equations hold. Visualize: Solve: This is a two-part problem. During the reaction time, we can use 2 1 0 0 1 0 0 1 0 ( ) 1 ( ) 2 0 m (20 m/s)(0.50 s 0 s) 0 m 10 m x = x + v t - t + a t - t = + - + = During deceleration, 2 2 2 1 1 2 1 v = v + 2a (x - x ) 2 2 2 2 0 = (20 m/s) + 2(-6.0 m/s )(x -10 m). x = 43 m She has 50 m to stop, so she can stop in time. 2.46. Model: The car is a particle and constant-acceleration kinematic equations hold. Visualize: Solve: (a) This is a two-part problem. During the reaction time, 2 1 0 0 1 0 0 1 0 ( ) 1/2 ( ) 0 m (20 m/s)(0.50 s 0 s) 0 m 10 m x = x + v t - t + a t - t = + - + = After reacting, 2 1 x - x =110 m-10 m =100 m, that is, you are 100 m away from the intersection. (b) To stop successfully, 2 2 2 2 2 2 1 1 2 1 1 1 v = v + 2a (x - x ).(0 m/s) = (20 m/s) + 2a (100 m).a = -2 m/s (c) The time it takes to stop once the brakes are applied can be obtained as follows: 2 2 1 1 2 1 2 2 v = v + a (t - t ).0 m/s = 20 m/s + (-2 m/s )(t - 0.50 s).t =11 s The total time to stop since the light turned red is 11.5 s. 2.47. Model: We will use the particle model and the constant-acceleration kinematic equations. Visualize: Solve: (a) To find x2 , we first need to determine 1x . Using 1 0 0 1 0 x = x + v (t - t ), we get 1 x = 0 m+ (20 m/s)(0.50 s - 0 s) = 10 m. Now, 2 2 2 2 2 2 2 1 1 2 1 2 2 v = v + 2a (x - x ).0 m /s = (20 m/s) + 2(-10 m/s )(x -10 m). x = 30 m The distance between you and the deer is 3 2 (x - x ) or (35 m - 30 m) = 5 m. (b) Let us find 0 max v such that 2 v = 0 m/s at 2 3 x = x = 35 m. Using the following equation, 2 2 2 2 2 2 2 0 max 1 2 1 0 max 1 v - v = 2a (x - x ).0 m /s - v = 2(-10 m/s )(35 m - x ) Also, 1 0 0 max 1 0 0 max 0 max x = x + v (t - t ) = v (0.50 s - 0 s) = (0.50 s)v . Substituting this expression for x1 in the above equation yields 2 2 2 22 0 max 0 max 0 max 0 max -v = (-20 m/s )[35 m- (0.50 s) v ].v + (10 m/s)v - 700 m /s = 0 The solution of this quadratic equation yields 0 max v = 22 m/s. (The other root is negative and unphysical for the present situation.) Assess: An increase of speed from 20 m/s to 22 m/s is very reasonable for the car to cover an additional distance of 5 m with a reaction time of 0.50 s and a deceleration of 10 m/s2. 2.48. Model: The car is represented as a particle. Visualize: Solve: (a) This is a two-part problem. First, we need to use the information given to determine the acceleration during braking. Second, we need to use that acceleration to find the stopping distance for a different initial velocity. First, the car coasts at constant speed before braking: x1 = x0 + v0 (t1 - t0 ) = v0t1 = (30 m/s)(0.5 s) =15 m Then, the car brakes to a halt. Because we dont know the time interval, use 2 2 2 1 1 2 1 v = 0 = v + 2a (x - x ) 2 2 1 2 1 2 1 (30 m/s) 10 m/s 2( ) 2(60 m 15 m) a v x x . =- = - = - - - We used 1 0 v = v = 30 m/s.Note the minus sign, because 1 a.. points to the left. We can repeat these steps now with v0 = 40 m/s. The coasting distance before braking is 1 01 ( x = v t = 40 m/s)(0.5 s) = 20 m The position 2 x after braking is found using 2 2 2 1 1 2 1 v = 0 = v + 2a (x - x ) 2 2 1 2 1 2 1 20 m (40 m/s) 100 m 2 2(10 m/s) x x v a . = - = - = - (b) The car coasts at a constant speed for 0.5 s, traveling 20 m. The graph will be a straight line with a slope of 40 m/s. For t = 0.5 the graph will be a parabola until the car stops at t2.We can find t2 from 1 2 1 12 1 2 1 1 0 ( ) 4.5 s v v v a t t t ta = = + - . = - = The parabola will reach zero slope (v = 0 m/s) at t = 4.5 s. This is enough information to draw the graph shown in the figure. 2.49. Model: The rocket is represented as a particle. Visualize: Solve: (a) There are three parts to the motion. Both the second and third parts of the motion are free fall, with a = -g. The maximum altitude is y2 .. In the acceleration phase: 2 2 2 2 1 0 0 1 0 1 0 1 2 1 0 1 0 1 ( ) 1 ( ) 1 1 (30 m/s )(30 s) 13,500 m 2 2 2 ( ) (30 m/s )(30 s) 900 m/s y y v t t at t at v v at t at = + - + - = = = = + - = = = In the coasting phase, 2 2 2 2 1 2 1 2 1 2 1 2 0 2 ( ) 13,500 m (900 m/s) 54,800 m 54.8 km 2 2(9.8 m/s) v v v g y y y yg = = - - . = + = + = = The maximum altitude is 54.8 km ( 33 miles). (b) The rocket is in the air until time 3t .We already know 1 t = 30 s. We can find 2 t as follows: 1 2 1 21 2 1 0 m/s ( ) 122 s v v v g t t t tg = = - - . = + = Then t3 is found by considering the time needed to fall 54,800 m: 2 2 2 3 2 23 2 3 2 2 3 2 3 2 0 m ( ) 1 ( ) 1 ( ) 2 228 s 2 2 y y v t t g t t y g t t t t y g = = + - - - = - - . = + = (c) The velocity increases linearly, with a slope of 30 (m/s)/s, for 30 s to a maximum speed of 900 m/s. It then begins to decrease linearly with a slope of -9.8(m/s)/s. The velocity passes through zero (the turning point at 2 y ) at 2 t =122 s. The impact velocity at 3 t = 228 s is calculated to be 3 2 3 2 v = v - g(t - t ) = -1040 m/s. Assess: In reality, friction due to air resistance would prevent the rocket from reaching such high speeds as it falls, and the acceleration upward would not be constant because the mass changes as the fuel is burned, but that is a more complicated problem. 2.50. Model: We will model the rocket as a particle. Air resistance will be neglected. Visualize: Solve: (a) Using the constant-acceleration kinematic equations, 1 0 0 1 0 0 0 2 2 2 1 0 0 1 0 0 1 0 0 0 2 2 1 1 2 1 1 2 1 2 2 2 2 0 0 0 ( ) 0 m/s (16 s 0 s) (16 s) ( ) 1 ( ) 1 (16 s 0 s) (128 s ) 2 2 ( ) 1 ( ) 2 5100 m 128 s 16 s (20 s 16 s) 1 ( 9.8 m/s )(20 s 16 s) 27 m/s 2 v v a t t a a y y v t t a t t a a y y v t t a t t a a a = + - = + - = = + - + - = - = = + - + - . = + - + - - . = (b) The rockets speed as it passes through a cloud 5100 m above the ground can be determined using the kinematic equation: 2 2 2 1 1 2 1 0 v = v + a (t - t ) = (16 s)a + (-9.8 m/s )(4 s) = 4.010 m/s Assess: 400 m/s 900 mph, which would be the final speed of a rocket that has been accelerating for 20 s at a rate of approximately 20 m/s2 or 66 ft/s2. 2.51. Model: We will model the lead ball as a particle and use the constant-acceleration kinematic equations. Visualize: Note that the particle undergoes free fall until it hits the water surface. Solve: The kinematics equation 1 2 1 0 0 1 0 2 0 1 0 y = y + v (t - t ) + a (t - t ) becomes 2 2 1 1 5.0 m 0 m 0 m 1 ( 9.8 m/s )( 0) 1.01 s 2 - = + + - t - .t = Now, once again, 2 2 1 1 2 1 1 2 1 2 1 1 1 ( ) 1 ( ) 2 (3.0 s 1.01 s) 0 m/s 1.99 y y v t t a t t y y v v = + - + - . - = - + = 1 v is easy to determine since the time 1 t has been found. Using 1 0 0 1 0 v = v + a (t - t ) , we get 2 1 v = 0 m/s - (9.8 m/s )(1.01 s - 0 s) = - 9.898 m/s With this value for 1v , we go back to: 2 1 1 y - y =1.99v = (1.99)(-9.898 m/s) = -19.7 m 2 1 y - y is the displacement of the lead ball in the lake and thus corresponds to the depth of the lake. The negative sign shows the direction of the displacement vector. Assess: A depth of about 60 ft for a lake is not unusual. 2.52. Model: The elevator is a particle moving under constant-acceleration kinematic equations. Visualize: Solve: (a) To calculate the distance to accelerate up: 2 2 2 2 2 1 0 0 0 0 1 1 ( ) 2 ( ) (5 m/v = v + a y - y . s) = (0 m/s) + 2(1 m/s )( y - 0 m). y =12.5 m (b) To calculate the time to accelerate up: 2 1 0 0 1 0 1 1 v = v + a (t - t ).5 m/s = 0 m/s + (1 m/s )(t - 0 s).t = 5 s To calculate the distance to decelerate at the top: 2 2 2 2 2 3 2 2 3 2 3 2 3 2 v = v + 2a ( y - y ).(0 m/s) = (5 m/s) + 2(-1 m/s )( y - y ). y - y =12.5 m To calculate the time to decelerate at the top: 2 3 2 2 3 2 3 2 3 2 v = v + a (t - t ).0 m/s = 5 m/s + (-1 m/s )(t - t ).t - t = 5 s The distance moved up at 5 m/s is 2 1 3 0 3 2 1 0 y - y = ( y - y ) - ( y - y ) - ( y - y ) = 200 m-12.5 m-12.5 m =175 m The time to move up 175 m is given by 2 2 1 1 2 1 1 2 1 2 1 2 1 ( ) 1 ( ) 175 m (5 m/s)( ) ( ) 35 s 2 y - y = v t - t + a t - t . = t - t . t - t = To total time to move to the top is 1 0 2 1 3 2 (t - t ) + (t - t ) + (t - t ) = 5 s + 35 s + 5 s = 45 s Assess: To cover a distance of 200 m at 5 m/s (ignoring acceleration and deceleration times) will require a time of 40 s. This is comparable to the time of 45 s for the entire trip as obtained above. 2.53. Model: The car is a particle moving under constant-acceleration kinematic equations. Visualize: Solve: This is a three-part problem. First the car accelerates, then it moves with a constant speed, and then it decelerates. First, the car accelerates: 2 1 0 0 1 0 2 2 2 1 0 0 1 0 0 1 0 ( ) 0 m/s (4.0 m/s )(6 s 0 s) 24 m/s ( ) 1 ( ) 0 m 1 (4.0 m/s )(6 s 0 s) 72 m 2 2 v v a t t x x v t t a t t = + - = + - = = + - + - = + - = Second, the car moves at v1: 2 2 1 1 2 1 1 2 1 ( ) 1 ( ) (24 m/s)(8 s 6 s) 0 m 48 m 2 x - x = v t - t + a t - t = - + = Third, the car decelerates: 2 3 2 2 3 2 3 2 3 2 2 22 3 2 2 3 2 2 3 2 3 2 ( ) 0 m/s 24 m/s ( 3.0 m/s )( ) ( ) 8 s ( ) 1 ( ) (24 m/s)(8 s) 1 ( 3.0 m/s )(8 s) 96 m 2 2 v v a t t t t t t x x v t t a t t x x = + - . = + - - . - = = + - + - . - = + - = Thus, the total distance between stop signs is: 3 0 3 2 2 1 1 0 x - x = (x - x ) + (x - x ) + (x - x ) = 96 m + 48m+ 72 m = 216 m Assess: A distance of approximately 600 ft in a time of around 10 s with an acceleration/deceleration of the order of 7 mph/s is reasonable. 2.54. Model: The car is a particle moving under constant linear acceleration. Visualize: Solve: Using the kinematic equation for position: 2 2 2 2 1 1 2 1 1 2 1 1 1 1 2 2 1 1 ( ) 1 ( ) 30 m (5.0 s 4.0 s) 1 (2 m/s )(5.0 s 4.0 s) 2 2 30 m (1.0 s) 1 (2 m/s )(1.0 s) 29 m/s 2 x x v t t a t t x x v v v = + - + - . + = + - + - . = + . = And 4.0 seconds before: 2 1 0 0 1 0 0 0 v = v + a (t - t ).29 m/s = v + (2 m/s )(4.0 s - 0 s).v = 21 m/s Assess: 21 m/s 47 mph and is a reasonable value. 2.55. Model: Santa is a particle moving under constant-acceleration kinematic equations. Visualize: Note that our x-axis is positioned along the incline. Solve: Using the following kinematic equation, || 2 2 2 2 1 0 1 2 1 2 ( ) (v = v + a x - x = 0 m/s) + 2(4.9 m/s )(10 m- 0 m).v = 9.9 m/s Assess: Santas speed of 20 mph as he reaches the edge is reasonable. 2.56. Model: The cars are represented as particles. Visualize: Solve: (a) Ann and Carol start from different locations at different times and drive at different speeds. But at time 1 t they have the same position. It is important in a problem such as this to express information in terms of positions (that is, coordinates) rather than distances. Each drives at a constant velocity, so using constant velocity kinematics gives xA1 = xA0 + vA (t1 - tA0 ) = vA (t1 - tA0 ) C1 C0 C 1 C0 C0 C 1 x = x + v (t - t ) = x + v t The critical piece of information is that Ann and Carol have the same position at 1t , so A1 C1x = x . Equating these two expressions, we can solve for the time 1 t when Ann passes Carol: A 1 A0 C0 C1 v (t - t ) = x + v t A C 1 C0 AA0 .(v - v )t = x + v t C0 A A0 1 A C 2.4 mi (50 mph)(0.5 h) 2.0 h 50 mph 36 mph t x v t v v + + . = = = - - (b) Their position is 1 A1 C1 C0 C1 x = x = x = x + v t = 74 miles. (c) Note that Anns graph doesnt start until t = 0.5 hours, but her graph has a steeper slope so it intersects Carols graph at t 2.0 hours. 2.57. Model: We will use the particle model for the puck. Visualize: We can view this problem as two one-dimensional motion problems. The horizontal segments do not affect the motion because the speed does not change. So, the problem starts at the bottom of the uphill ramp and ends at the bottom of the downhill ramp. At the top of the ramp the speed does not change along the horizontal section. The final speed from the uphill roll (first problem) becomes the initial speed of the downhill roll (second problem). Because the axes point in different directions, we can avoid possible confusion by calling the downhill axis the z-axis and the downhill velocities u. The uphill axis as usual will be denoted by x and the uphill velocities as v. Note that the height information, h = 1 m, has to be transformed into information about positions along the two axes. Solve: (a) The uphill roll has 2 a0 = -g sin30 = -4.90 m/s . .. The speed at the top is found from 2 2 1 0 0 1 0 v = v + 2a (x - x ) 2 2 2 1 0 01 .v = v + 2a x = (5 m/s) + 2(-4.90 m/s )(2.00 m) = 2.32 m/s (b) The downward roll starts with velocity 1 1 u = v = 2.32 m/s and 2 1 a = +g sin 20.. = 3.35 m/s . Then, 2 2 2 2 2 1 1 2 1 2 u = u + 2a (z - z ) = (2.32 m/s) + 2(3.35 m/s )(2.92 m - 0 m).u = 5.00 m/s (c) The final speed is equal to the initial speed, so the percentage change is zero! Assess: This result may seem surprising, but can be more easily understood after we introduce the concept of energy. For now, imagine this is a one dimensional vertical problem. The total vertical change in height of the puck is zero. We have already seen how an object with an initial velocity upward has the same velocity in the opposite direction as it passes through that height going down. 2.58. Model: We will model the toy train as a particle. Visualize: Solve: Using kinematics, 2 1 0 0 1 0 0 1 0 ( ) 1 ( ) 2 m (2.0 m/s)(2.0 s 0 s) 0 m 6.0 m 2 x = x + v t - t + a t - t = + - + = The acceleration can now be obtained as follows: 2 2 2 2 2 2 2 1 1 2 1 1 1 v = v + 2a (x - x ).0 m /s = (2.0 m/s) + 2a (8.0 m- 6.0 m).a = -1.0 m/s Assess: A deceleration of 1 m/s2 in bringing the toy train to a halt over a distance of 2.0 m is reasonable. 2.59. Model: We will use the particle model and the kinematic equations at constant-acceleration. Visualize: Solve: To find x2 , let us use the kinematic equation 2 2 2 2 2 2 1 1 2 1 2 1 2 1 v = v + 2a (x - x ) = (0 m/s) = (50 m/s) + 2(-10 m/s )(x - x ). x = x +125 m Since the nail strip is at a distance of 150 m from the origin, we need to determine 1 x : 1 0 0 1 0 x = x + v (t - t ) = 0 m+ (50 m/s)(0.60 s - 0.0 s) = 30 m Therefore, we can see that 2 x = (30 +125) m =155 m. That is, he cant stop within a distance of 150 m. He is in jail. Assess: Bob is driving at approximately 100 mph and the stopping distance is of the correct order of magnitude. 2.60. Model: We will use the particle model with constant-acceleration kinematic equations. Visualize: Solve: The acceleration, being the same along the incline, can be found as 2 2 2 2 2 1 0 1 0 2 ( ) (v = v + a x - x . 4.0 m/s) = (5.0 m/s) + 2a(3.0 m- 0 m).a = -1.5 m/s We can also find the total time the puck takes to come to a halt as 2 2 0 2 0 2 2 v = v + a(t - t ).0 m/s = (5.0 m/s) + (-1.5 m/s )t .t = 3.3 s Using the above obtained values of a and 2 t , we can find 2 x as follows: 2 22 2 0 0 2 0 2 0 ( ) 1 ( ) 0 m (5.0 m/s)(3.3 s) 1 ( 1.5 m/s )(3.3 s) 8.3 m 2 2 x = x + v t - t + a t - t = + + - = That is, the puck goes through a displacement of 8.3 m. Since the end of the ramp is 8.5 m from the starting position x0 and the puck stops 0.2 m or 20 cm before the ramp ends, you are not a winner. 2.61. Model: We will use the particle model for the skiers motion ignoring air resistance. Visualize: Solve: (a) As discussed in the text, acceleration along a frictionless incline is a = g sin 25, where g is the acceleration due to gravity. The acceleration of the skier on snow therefore is a = (0.90)g sin 25. Also since 1 h/x = sin 25, x = 200 m/ sin 25. The final velocity can now be determined using kinematics 2 2 2 2 1 0 1 0 1 2 ( ) (0 m/s) 2(0.90)(9.8 m/s )sin 25 200 m 0 m sin 25 59.397 m/s (59.397 m/s) 1 km 3600 s 214 km/h 1000 m 1 h v v ax x v = + - = + . - . . . . . . = = . .. . = . .. . . .. . (b) The speed lost to air resistance is (214 -180)/214100% =16%. Assess: A record of 180 km/h on such a slope in the presence of air resistance makes the obtained speed of 214 km/h (without air resistance) physical and reasonable. 2.62. Model: We will use the particle model for the motion of the rocks, which move according to constantacceleration kinematic equations. Visualize: Solve: (a) For Heather, 2 H1 H0 H0 H1 H0 0 H1 H0 2 2 H1 H1 2 2 H1 H1 ( ) 1 ( ) 2 0 m (50 m) ( 20 m/s)( 0 s) 1 ( 9.8 m/s )( 0 s) 2 4.9 m/s 20 m/s 50 m 0 y y v t t a t t t t t t = + - + - . = + - - + - - . + - = The two mathematical solutions of this equation are -5.83 s and +1.75 s. The first value is not physically acceptable since it represents a rock hitting the water before it was thrown, therefore, H1 t =1.75 s. For Jerry, 2 J1 J0 J0 J1 J0 0 J1 J0 2 2 J1 J1 ( ) 1 ( ) 2 0 m (50 m) ( 20 m/s)( 0 s) 1 ( 9.8 m/s )( 0 s) 2 y y v t t a t t t t = + - + - . = + + - + - - Solving this quadratic equation will yield J1 t = -1.75 s and +5.83 s. Again only the positive root is physically meaningful. The elapsed time between the two splashes is J1 H1 t - t = 5.83 s -1.75 s = 4.08 s 4.1 s. (b) Knowing the times, it is easy to find the impact velocities: H1 H0 0 H1 H0 2 J1 J0 0 J1 J0 ( ) ( 20 m/s) ( 9.8 m/s)(1.75 s 0 s) 37.1 m/s ( ) ( 20 m/s) ( 9.8 m/s )(5.83 s 0 s) 37.1 m/s v v a t t v v a t t = + - = - + - - =- = + - = + + - - =- Assess: The two rocks hit water with equal speeds. This is because Jerrys rock has the same downward speed as Heathers rock when it reaches Heathers starting position during its downward motion. 2.63. Model: The ball is a particle that exhibits freely falling motion according to the constant-acceleration kinematic equations. Visualize: Solve: Using the known values, we have 2 2 2 2 2 1 0 0 1 0 0 0 v = v + 2a ( y - y ).(-10 m/s) = v + 2(-9.8 m/s )(5.0 m- 0 m).v =14 m/s 2.64. Model: The car is a particle that moves with constant linear acceleration. Visualize: Solve: The reaction time is 1.0 s, and the motion during this time is x1 = x0 + v0 (t1 - t0 ) = 0 m+ (20 m/s)(1.0 s) = 20 m During slowing down, 2 2 1 1 2 1 1 2 1 2 2 1 1 ( ) 1 ( ) 200 m 2 20 m (20 m/s)(15 s 1.0 s) 1 (15 s 1.0 s) 1.02 m/s 2 x x v t t a t t a a = + - + - = = + - + - . =- The final speed 2 v can now be obtained as 2 2 1 1 2 1 v = v + a (t - t ) = (20 m/s) + (-1.02 m/s )(15 s -1 s) = 5.7 m/s 2.65. Solve: (a) The quantity 4 2 2(3.6 10 W) 60 m2/s3. 1200 kg P m = = Thus ( 2 ) 3 60m x s v = t At ( 2 ) 3 10 s, 60m (10 s) 24 m/s ( 50 mph), x s t = v = = and at t 20 s, 35 m/s x = v = ( 75 mph). (b) With 1 2 2 , we have x v Pt m = 1 2 2 1 2 2 x x a dv P t P dt m mt - = = = (c) At t =1 s, 4 = (3.6 10 W) 3.9 m/s2. 2 2(1200 kg)(1 s) x a P mt = = Similarly, at 10 s, 1.2 m/s2. x t = a = (d) Consider the limiting case of very short times. Note that as 0. xa .8 t . This is physically impossible for the Alfa Romeo. (e) We can use the relationship that x v dx dt = and integrate to find x(t).We have 1 2 2 x v Pt m = and the initial condition 0 i x = at 0. i t = Thus 1/ 2 0 0 x dx 2P t t dt m . = . 3/ 2 and 2 2 2 3/ 2 3/2 3 x P t Pt m m = = (f) Time to travel a distance x is found by solving the above equation for t. 2 / 3 3 2 2 t m x P . . = . . .. .. For x = 402 m, t =18.2 s. 2.66. Model: Both cars are particles that move according to the constant-acceleration kinematic equations. Visualize: Solve: (a) Davids and Tinas motions are given by the following equations: 2 D1 D0 D0 D1 D0 D D1 D0 D0 D1 2 2 T1 T0 T0 T1 T0 T T1 T0 T T1 ( ) 1 ( ) 2 ( ) 1 ( ) 0 m 0 m 1 2 2 x x v t t a t t v t x x v t t a t t at = + - + - = = + - + - = + + When Tina passes David the distances are equal and D1 T1t = t , so we get 2 D0 D1 T1 D0 D1 T T1 D0 T T1 T1 2 T 1 1 2 2(30 m/s) 30 s 2 2 2.0 m/s x x v t a t v a t t v a = . = . = . = = = Using Tinas position equation, 2 2 2 T1 T T1 1 1(2.0 m/s )(30 s) 900 m 2 2 x = a t = = (b) Tinas speed T1 v can be obtained from 2 T1 T0 T T1 T0 v = v + a (t - t ) = (0 m/s) + (2.0 m/s )(30 s - 0 s) = 60 m/s Assess: This is a high speed for Tina (~134 mph) and so is Davids velocity (~67 mph). Thus the large distance for Tina to catch up with David (~0.6 miles) is reasonable. 2.67. Model: We will represent the dog and the cat in the particle model. Visualize: Solve: We will first calculate the time tC1 the cat takes to reach the window. The dog has exactly the same time to reach the cat (or the window). Let us therefore first calculate C1 t as follows: 2 C1 C0 C0 C1 C0 C C1 C0 2 2 C1 C1 ( ) 1 ( ) 2 3.0 m 1.5 m 0 m 1 (0.85 m/s ) 1.879 s 2 x x v t t a t t t t = + - + - . = + + . = In the time D1 t =1.879 s, the dogs position can be found as follows: 2 D1 D0 D0 D1 D0 D D1 D0 2 2 ( ) 1 ( ) 2 0 m (1.50 m/s)(1.879 s) 1 ( 0.10 m/s )(1.879 s) 2.6 m 2 x = x + v t - t + a t - t = + + - = That is, the dog is shy of reaching the cat by 0.4 m. The cat is safe. 2.68. Model: We use the particle model for the large train and the constant-acceleration equations of motion. Visualize: Solve: Your position after time tY1 is 2 Y1 Y0 Y0 Y1 Y0 Y Y1 Y0 Y1 Y1 ( ) 1 ( ) 2 0 m (8.0 m/s)( 0 s) 0 m 8.0 x x v t t a t t t t = + - + - = + - + = The position of the train, on the other hand, after time tT1 is 2 2 T1 T0 T0 T1 T0 T T1 T0 2 2 2 T1 T1 ( ) 1 ( ) 2 30 m 0 m 1 (1.0 m/s )( ) 30 0.5 2 x x v t t a t t t t = + - + - = + + = + The two positions Y1 x and T1 x will be equal at time Y1 T1 t (= t ) if you are able to jump on the back step of the train. That is, 2 2 2 Y1 Y1 Y1 Y1 Y1 30 + 0.5t = 8.0t .t - (16 s)t + 60 s = 0.t = 6 s and 10 s The first time, 6 s, is when you will overtake the train. If you continue to run alongside, the accelerating train will then pass you at 10 s. Let us now see if the first time tY1 = 6.0 s corresponds to a distance before the barrier. From the position equation for you, Y1 x = (8.0 m/s)(6.0 s) = 48.0 m. The position equation for the train will yield the same number. Since the barrier is at a distance of 50 m from your initial position, you can just catch the train before crashing into the barrier. 2.69. Model: Jill and the grocery cart will be treated as particles that move according to the constantacceleration kinematic equations. Visualize: Solve: The final position of Jill when the cart is caught is given by 2 2 22 J1 J0 J0 J1 J0 J0 J1 J0 J0 J1 J1 ( ) 1 ( ) 0 m 0 m 1 ( 0 s) 1 (2.0 m/s ) 2 2 2 x = x + v t - t + a t - t = + + a t - = t The carts position when it is caught is 2 2 2 C1 C0 C0 C1 C0 C0 C1 C0 C1 2 2 C1 ( ) 1 ( ) 50 m 0 m 1 (0.5 m/s )( 0 s) 2 2 50 m (0.25 m/s ) x x v t t a t t t t = + - + - = + + - = + Since J1 C1 x = x and J1 C1t = t , we get 2 2 2 2 2 J1 C1 C1 C1 2 2 2 2 C1 C1 1 (2.0) 50 s 0.25 0.75 50 s 8.20 s 2 50 m (0.25 m/s ) 50 m (0.25 m/s )(8.2 s) 67.2 m t t t t x t = + . = . = . = + = + = So, the cart has moved 17.2 m. 2.70. Model: The watermelon and Superman will be treated as particles that move according to constantacceleration kinematic equations. Visualize: Solve: The watermelons and Supermans position as they meet each other are 2 W1 W0 W0 W1 W0 W0 W1 W0 2 S1 S0 S0 S1 S0 S0 S1 S0 2 2 W1 W1 S1 S1 ( ) 1 ( ) 2 ( ) 1 ( ) 2 320 m 0 m 1 ( 9.8 m/s )( 0 s) 2 320 m ( 35 m/s)( 0 s) 0 m y y v t t a t t y y v t t a t t y t y t = + - + - = + - + - . = + + - - . = + - - + Because S1 W1t = t , 2 2 W1 W1 S1 W1 y = 320 m- (4.9 m/s ) t y = 320 m - (35 m/s) t Since W1 S1y = y , 2 2 W1 W1 W1 320 m- (4.9 m/s )t = 320 m- (35 m/s)t .t = 0 s and 7.1 s Indeed, W1 t = 0 s corresponds to the situation when Superman arrives just as the watermelon is dropped off the Empire State Building. The other value, W1 t = 7.1 s, is the time when the watermelon will catch up with Superman. The speed of the watermelon as it passes Superman is 2 W1 W0 W0 W1 W0 v = v + a (t - t ) = 0 m/s + (-9.8 m/s )(7.1 s - 0 s) = -70 m/s Note that the negative sign implies a downward velocity. Assess: A speed of 140 mph for the watermelon is understandable in view of the significant distance (250 m) involved in the free fall. 2.71. Model: Treat the car and train in the particle model and use the constant acceleration kinematics equations. Visualize: Solve: In the particle model the car and train have no physical size, so the car has to reach the crossing at an infinitesimally sooner time than the train. Crossing at the same time corresponds to the minimum 1 a necessary to avoid a collision. So the problem is to find a1 such that x2 = 45 m when 2 y = 60m. The time it takes the train to reach the intersection can be found by considering its known constant velocity. 2 0 0 2 2 2 0 2 30 m/s 60 m 2.0 s y y v v y y t t t t - = = = = . = - Now find the distance traveled by the car during the reaction time of the driver. ( ) ( )( ) 1 0 0 1 0 0 20 m/s 0.50 s 10 m x x = x + v t - t = + = The kinematic equation for the final position at the intersection can be solved for the minimum acceleration 1a . ( ) ( ) ( )( ) ( ) 2 2 1 121 12 1 2 1 2 1 45 m 1 2 10 m 20 m/s 1.5 s 1 1.5 s 2 4.4 m/s x x x v t t a t t a a = = + - + - = + + . = Assess: The acceleration of 4.4 m/s2 = 2.0 miles/h/s is reasonable for an automobile to achieve. However, you should not try this yourself! Always pay attention when you drive! Train crossings are dangerous locations, and many people lose their lives at one each year. 2.72. Solve: A comparison of the given equation with the constant-acceleration kinematics equation 2 1 0 0 1 0 x 1 0 ( ) 1 ( ) 2 x = x + v t - t + a t - t yields the following information: 0 1 0 1 x = 0 m, x = 64 m, t = 0, t = 4 s, and 0 v = 32 m/s. (a) After landing on the deck of a ship at sea with a velocity of 32 m/s, a fighter plane is observed to come to a complete stop in 4.0 seconds over a distance of 64 m. Find the planes deceleration. (b) (c) 64 m 0 m (32 m/s)(4 s 0 s) 1 (4 s 0 s)2 64 m 128 m (8 s2 ) 8 m/s2 2x xx = + - + a - = + a .a = - 2.73. Solve: (a) A comparison of this equation with the constant-acceleration kinematic equation 2 2 (v1y ) = v0y + 2(ay )( y1 - y0 ) yields the following information: 2 0 1 0 m, 10 m, 9.8 m/s , y y = y = a = - and 1 10 m/s. y v = It is clearly a problem of free fall. On a romantic Valentines Day, John decided to surprise his girlfriend, Judy, in a special way. As he reached her apartment building, he found her sitting in the balcony of her second floor apartment 10 m above the first floor. John quietly armed his spring-loaded gun with a rose, and launched it straight up to catch her attention. Judy noticed that the flower flew past her at a speed of 10 m/s. Judy is refusing to kiss John until he tells her the initial speed of the rose as it was released by the spring-loaded gun. Can you help John on this Valentines Day? (b) (c) 2 2 2 0y 0y (10 m/s) = v - 2(9.8 m/s )(10 m- 0 m).v =17.2 m/s Assess: The initial velocity of 17.2 m/s, compared to a velocity of 10 m/s at a height of 10 m, is very reasonable. 2.74. Solve: A comparison with the constant-acceleration kinematics equation 2 2 (v1x ) = (v0x ) + 2ax (x1 - x0 ) yields the following quantities: 0 0 1 0 m, 5 m/s, 0 m/s, x x x = v = v = and (9.8 m/s2 )sin10 . x a = - .. (a) A wagon at the bottom of a frictionless 10 incline is moving up at 5 m/s. How far up the incline does it move before reversing direction and then rolling back down? (b) (c) 2 2 2 1 2 2 1 1 (0 m/s) (5 m/s) 2(9.8 m/s )sin10 ( 0 m) 25(m/s) 2(9.8 m/s )(0.174) 7.3 m x x x = - - . = . = 2.75. Solve: (a) From the first equation, the particle starts from rest and accelerates for 5 s. The second equation gives a position consistent with the first equation. The third equation gives a subsequent position following the second equation with zero acceleration. A rocket sled accelerates from rest at 20 m/s2 for 5 sec and then coasts at constant speed for an additional 5 sec. Draw a graph showing the velocity of the sled as a function of time up to t = 10 s. Also, how far does the sled move in 10 s? (b) (c) 2 2 2 1 1 2 1 (20 m/s )(5 s) 250 m 20 m/s (5 s) 100 m/s 250 m (100 m/s)(5 s) 750 m 2 x = = v = = x = + = 2.76. Model: The masses are particles. Visualize: Solve: The rigid rod forms the hypotenuse of a right triangle, which defines a relationship between x2 and y1 : 2 2 2 2 1 x + y = L . Taking the time derivative of both sides yields 2 1 2 1 2x dx 2y dx 0 dt dt + = We can now use 2 2x v dx dt = and 1 1y v dy dt = to write 2 2 1 1 0 x y x v + y v = . Thus 1 2 1 2 x y v y v x . . = -. . . . . But from the figure, 1 2 1 2 tan tan x y y v v x = . . = - . . Assess: As 2 x decreases 2 1 ( 0), xv < y increases 1 ( 0), y v > and vice versa. 2.77. Model: The rocket and the bolt will be represented as particles to investigate their motion. Visualize: The initial velocity of the bolt as it falls off the side of the rocket is the same as that of the rocket, that is, vB0 = vR1 and it is positive since the rocket is moving upward. The bolt continues to move upward with a deceleration equal to g = 9.8 m/s2 before it comes to rest and begins its downward journey. Solve: To find R a we look first at the motion of the rocket: 2 R1 R0 R0 R1 R0 R R1 R0 2 R R ( ) 1 ( ) 2 0 m 0 m/s 1 (4.0 s 0 s) 8 2 y y v t t a t t a a = + - + - = + + - = To find aR we must determine the magnitude of R1 y or B0y . Let us now look at the bolts motion: 2 B1 B0 B0 B1 B0 B B1 B0 2 2 R1 R1 ( ) 1 ( ) 2 0 (6.0 s 0 s) 1 ( 9.8 m/s )(6.0 s 0 s) 2 y y v t t a t t y v = + - + - = + - + - - R1 R1 . y =176.4 m - (6.0 s) v Since R1 R0 R R1 R0 R R v = v + a (t - t ) = 0 m/s + 4 a = 4 a the above equation for R1 y yields R1 R y =176.4 - 6.0(4a ). We know from the first part of the solution that R1 R y = 8a . Therefore, R R 8a =176.4 - 24.0a and hence 2 R a = 5.5 m/s . 2.78. Model: The rocket car is a particle that moves according to the constant-acceleration equations of motion. Visualize: Solve: This is a two-part problem. For the first part, 2 2 2 1 0 0 1 0 0 1 0 0 0 1 0 0 1 0 0 0 ( ) 1 ( ) 0 m 0 m 1 (9.0 s 0 s) 1 (81 s ) 2 2 2 ( ) 0 m/s (9.0 s 0 s) (9.0 s) x x v t t a t t a a v v a t t a a = + - + - = + + - = = + - = + - = During the second part of the problem, 2 2 1 1 2 1 1 2 1 2 2 2 0 0 2 0 ( ) 1 ( ) 2 990 m 1 (81 s ) (9.0 s) (12 s 9.0 s) 1 ( 5.0 m/s )(12 s 9.0 s) 2 2 15 m/s x x v t t a t t a a a = + - + - . = + - + - - . = This leads to: 2 1 0 v = (9.0 s)a = (9.0 s)(15 m/s ) =135 m/s Using this value of 1v , we can now calculate 2 v as follows: 2 2 1 1 2 1 v = v + a (t - t ) = (135 m/s) + (-5.0 m/s )(12 s - 9.0 s) =120 m/s That is, the cars speed as it passes the judges is 120 m/s. Assess: This is a very fast motion (~250 mph), but the acceleration is large and the long burn time of 9 s yields a high velocity. 2.79. Model: Use the particle model. Visualize: Solve: (a) Substituting into the constant-acceleration kinematic equation 2 2 1 1 2 1 1 2 1 1 1 2 1 2 1 ( ) 1 ( ) 100 m 10 0 m 2 3 100 10 3 x x v t t a t t x v t t x v = + - + - . = + . - . + . . . . - = + Let us now find 1 v and 1 x as follows: 2 1 0 0 1 0 2 2 2 1 0 0 1 0 0 1 0 ( ) 0 m/s (3.6 m/s ) 10 s 0 s 12 m/s 3 ( ) 1 ( ) 0 m 0 m 1 (3.6 m/s ) 10 s 0 s 20 m 2 2 3 v v a t t x x v t t a t t = + - = + . - . = . . . . = + - + - = + + . - . = . . . . The expression for 2 t can now be solved as 2 100 m 20 m 10 s 10 s 12 m/s 3 t - = + = (b) The top speed = 12 m/s which means 1 v =12 m/s. To find the acceleration so that the sprinter can run the 100-meter dash in 9.9 s, we use 1 0 0 1 0 01 1 0 2 2 2 1 0 0 1 0 0 1 0 0 1 0 1 ( ) 12 m/s 0 m/s 12 m/s ( ) 1 ( ) 0 m 0 m 1 1 2 2 2 v v a t t at t a x x v t t a t t at at = + - . = + . = = + - + - = + + = Since 1 2 2 1 1 2 1 2 1 2 1 x = x + v (t - t ) + a (t - t ) , we get 2 0 1 1 100 m 1 (12 m/s) (9.9 s ) 0 m 2 = a t + - t + Substituting the above equation for 1 t in this equation, ( ) 2 2 0 0 0 0 100 m 1 12 m/s 12 m/s 9.9 s 12 m/s 3.8 m/s 2 a a a a . . . . . . = . . . . + . - .. = . . . . . . (c) We see from parts (a) and (b) that the acceleration has to be increased from 3.6 m/s2 to 3.8 m/s2 for the sprint time to be reduced from 10 s to 9.9 s, that is, by 1%. This decrease of time by 1% corresponds to an increase of acceleration by 3.8 3.6 100% 5.6% 3.6 - = 2.80. Solve: (a) The acceleration is the time derivative of the velocity. x (1 bt ) bt x a dv d a e abe dt dt = = . - - . = - . . With a =11.81 m/s and b =0.6887 s-1, 8.134 0.6887t m/s2. xa = e- At the times t =0 s, 2 s, and 4 s, this has the values 8.134 m/s2, 2.052 m/s2, and 0.5175 m/s2. (b) Since x v dx dt = , the position x is the integral of the velocity. With bt x v dx a ae dt = = - - and the initial condition that 0 m i x = at 0 s i t = , x t t bt o o o . dx = . a dt - . ae- dt Thus t t bt o o x at a e b = + - = at a e bt a b b + - - This can be written a little more neatly as ( ) ( 0.6887 ) 1 17.15 0.6887 1 m bt t x a bt e b t e - - = + - = + - (c) By trial and error, t = 9.92 s yields x = 100.0 m. Assess: Lewiss actual time was 9.93 s. 2.81. Model: We will use the particle-model to represent the sprinter and the equations of kinematics. Visualize: Solve: Substituting into the constant-acceleration kinematic equations, 2 2 2 2 1 0 0 1 0 0 1 0 0 01 0 2 1 0 1 0 0 1 0 0 1 0 ( ) 1 ( ) 0 m 0 m 1 (4 s 0 s) 1 1 (4.0 s) 2 2 2 2 (8 s ) ( ) 0 m/s (4.0 s 0 s) (4.0 s) x x v t t a t t a at a x a v v a t t a v a = + - + - = + + - = = . = = + - = + - . = From these two results, we find that 1 1 x = (2 s)v . Now, 2 2 1 1 2 1 1 2 1 1 1 1 ( ) 1 ( ) 2 100 m (2 s) (10 s 4 s) 0 m 12.5 m/s x x v t t a t t v v v = + - + - . = + - + . = Assess: Using the conversion 2.24 mph = 1 m/s, 1 v =12.5 m/s = 28 mph. This speed as the sprinter reaches the finish line is physically reasonable. 2.82. Model: The balls are particles undergoing constant acceleration. Visualize: Solve: (a) The positions of each of the balls at 1 t is found from kinematics. ( ) ( ) ( ) 2 2 1 0 0 1 1 01 1 1 1 A A y A 2 2 y = y + v t - gt = v t - gt ( ) ( ) ( ) 2 2 1 0 0 1 1 1 1 1 2 2 B B y B y = y + v t - gt = h - gt In the particle model the balls have no physical extent, so they meet when ( ) ( ) 1 A 1 B y = y . This means 2 2 0 1 1 1 1 0 1 1 2 2 v t gt h gt t h v - = - . = Thus the collision height is 2 2 coll 1 2 0 1 2 2 gh y h gt hv = - = - . (b) We need the collision to occur while coll y = 0 . Thus 2 2 0 0 2 h gh v - = . 2 0 1 2 gh v = . 2 0 h 2v g = So 2 0 max h 2v g = . (c) Ball A is at its highest point when its velocity ( ) 1 0 y A v = . ( ) ( ) 1 0 1 0 1 0 y A y A v = v - gt . = v - gt 0 1 t v g . = In (a) we found that the collision occurs at 1 0 t h v = . Equating these, 2 0 0 0 h v h v v g g = . = . Assess: Interestingly, the height at which a collision occurs while Ball A is at its highest point is exactly half of max h . 2.83. Model: The space ships are represented as particles. Visualize: Solve: The difficulty with this problem is how to describe barely avoid. The Klingon ship is moving with constant speed, so its position-versus-time graph is a straight line from xK0 = 100 km. The Enterprise will be decelerating, so its graph is a parabola with decreasing slope. The Enterprise doesnt have to stop; it merely has to slow quickly enough to match the Klingon ship speed at the point where it has caught up with the Klingon ship. (You do the same thing in your car when you are coming up on a slower car; you decelerate to match its speed just as you come up on its rear bumper.) Thus the parabola of the Enterprise will be tangent to the straight line of the Klingon ship, showing that the two ships have the same speed (same slopes) when they are at the same position. Mathematically, we can say that at time t1 the two ships will have the same position (xE1 = xK1) and the same velocity E1 K1 (v = v ). Note that we are using the particle model, so the ships have zero length. At time t1, K1 K0 K0 1 2 E1 E0 1 1 1 2 x x v t x v t at = + = + K1 K0 E1 E0 1 v v v v at = = + Equating positions and velocities at t1: 2 K0 K0 1 E0 1 1 1 2 x + v t = v t + at K0 E0 1 v = v + at We have two simultaneous equations in the two unknowns a and t1. From the velocity equation, 1 K0 E0 t = (v - v )/a Substituting into the position equation gives 2 2 K0 E0 K0 E0 K0 E0 K0 K0 E0 ( ) ( ) 1 ( ) ( ) 2 2 x v v v v a v v v v a a a - . - . - = - - + . . = - . . 2 2 K0 E0 2 K0 ( ) (20,000 m/s 50,000 m/s) 4500 m/s 2 2(100,000 m) a v v x - - . = - = - = - Assess: The magnitude of the deceleration is 4500 m/s2, which is a rather extreme 460g. Fortunately, the Enterprise has other methods to keep the crew from being killed. 3.1. Visualize: Solve: (a) To find A + B .. .. , we place the tail of vector B .. on the tip of vector A .. and connect the tail of vector A .. with the tip of vector . B .. (b) Since A - B = A + (-B) .. .. .. .. , we place the tail of the vector (-B) .. on the tip of vector A .. and then connect the tail of vector A .. with the tip of vector (-B) .. . 3.2. Visualize: Solve: (a) To find A + B .. .. , we place the tail of vector B .. on the tip of vector A .. and then connect vector s A .. tail with vector s B .. tip. (b) To find A - B .. .. , we note that A - B = A + (-B) .. .. .. .. . We place the tail of vector -B .. on the tip of vector A .. and then connect vector s A .. tail with the tip of vector -B. .. 3.3. Visualize: Solve: Vector E .. points to the left and up, so the components x E and y E are negative and positive, respectively, according to the Tactics Box 3.1. (a) Ex = -Ecos. and sin . y E = E . (b) sin xE = -E f and cos . y E = E f Assess: Note that the role of sine and cosine are reversed because we are using a different angle. . and f are complementary angles. 3.4. Visualize: The position vector r.. whose magnitude r is 10 m has an x-component of 6 m. It makes an angle . with the +x-axis in the first quadrant. Solve: Using trigonometry, cos , xr = r . or 6 m = (10 m)cos. . This gives. = 53.1. Thus the y-component of the position vector r.. is sin (10 m)sin53.1 8 m. yr = r . = = Assess: The y-component is positive since the position vector is in the first quadrant. 3.5. Visualize: The figure shows the components vx and vy, and the angle .. Solve: We have, vy = -vsin 40, or -10 m/s = -vsin 40, or v =15.56 m/s. Thus the x-component is cos40 (15.56 m/s ) cos40 12 m/s. x v = v = = Assess: The x-component is positive since the position vector is in the fourth quadrant. 3.6. Visualize: We will follow rules in Tactics Box 3.1. Solve: (a) Vector r.. points to the right and down, so the components x r and y r are positive and negative, respectively: rx = r cos. = (100 m)cos45 = 70.7 m sin (100 m)sin 45 70.7 m yr = -r . = - = - (b) Vector v.. points to the right and up, so the components x v and y v are both positive: cos (300 m /s) cos20 282 m/s x v = v . = = sin (300 m/s)sin 20 103 m/s y v = v . = = (c) Vector a.. has the following components: cos (5.0 m/s2 )cos90 0 m/s2 xa = -a . = - = sin (5.0 m/s2 )sin90 5.0 m/s2 ya = -a . = - = - Assess: The components have same units as the vectors. Note the minus signs we have manually inserted according to Tactics Box 3.1. 3.7. Visualize: We will follow the rules given in Tactics Box 3.1. Solve: (a) (5 cm/s)sin90 5 cm/s x v = - = - (5 cm/s)cos90 0 cm/s y v= = (b) (10 m/s2 )sin 40 6.4 m/s2 x a = - = - (10 m/s2 )cos40 7.7 m/s2 y a = - = - (c) (50 N)sin36.9 30 N x F= = (50 N)cos36.9 40 N y F= = Assess: The components have the same units as the vectors. Note the minus signs we have manually inserted according to Tactics Box 3.1. 3.8. Visualize: The components of the vector C .. and , D .. and the angles . are shown. Solve: For C .. we have (3.15 m)cos15 3.04 m x C = - = - and (3.15 m)sin15 0.815 m. y C= = For D .. we have x D = 25.6sin30 =12.8 and 25.67cos30 22.2. y D = - = - Assess: The components of the vector C .. have the same units as C .. itself. x D and y D are unitless because D .. is without units. Note the minus signs we have manually inserted following rules of Tactics Box 3.1. 3.9. Visualize: Solve: The magnitude of the vector is ( )2 ( )2 (125 V/m)2 ( 250 V/m)2 280 V/m. x y E = E + E = + - = In the expression for , E .. the j - and i + means that E .. is in quadrant IV. The angle . is below the positive x-axis. We have: 1 1 1 | | 250 V/m tan tan tan 2 63.4 125 V/m y x E E . - - - . . = = . . = = . . Assess: Since | | | | y x E > E , the angle . made with the +x-axis is larger than 45. . = 45 for | | | | y x E = E . 3.10. Visualize: Solve: (a) Using the formulas for the magnitude and direction of a vector, we have: B = (-4)2 + (4)2 = 5.7 tan 1 4 tan 11 45 4 . = - = - = (b) r = (-2 cm)2 + (-1 cm)2 = 2.2 cm tan 1 1 tan 1 0.5 26.6 2 . = - = - = (c) v = (-10 m/s)2 + (-100 m/s)2 =100.5 m/s tan 1100 tan 110 84.3 10 . = - = - = (d) a = (10 m/s2 )2 + (20 m/s2 )2 = 22.4 m/s2 tan 1 10 tan 1 0.5 26.6 20 . = - = - = Assess: Note that . = 45 when | | | |, y x E = E where . is the angle made with the x-axis. On the other hand, . > 45 when | | | | . y x E > E 3.11. Visualize: Solve: (a) Using the formulas for the magnitude and direction of a vector, we have: A = (4)2 + (-6)2 = 7.21 tan 1 6 tan 11.5 56.3 4 . = - = - = (b) r = (50 m)2 + (80 m)2 = 94.3 m tan 1 tan 1 80 m 58.0 50 m y x r r . - - . . . . = . . = . . = . . . . (c) v = (-20 m/s)2 + (40 m/s)2 = 44.7 m/s tan 1 40 tan 1 2 63.4 20 . = - = - = (d) a = (2 m/s2 )2 + (-6 m/s2 )2 = 6.3 m/s2 tan 1 2 tan 1 0.33 18.4 6 . = - = - = Assess: Note that the angle . made with the x-axis is smaller than 45 whenever | | | | y x E < E , . = 45 for | | | |, y x E = E and . > 45 for | | | | y x E > E . In part (d), . is with the y-axis, where the opposite of this rule applies. 3.12. Visualize: We have C = A - B .. .. .. or C = A + (-B), .. .. .. where -B = (B, .. .. direction opposite ). B .. Look back at Tactics Box 1.2, which shows how to perform vector subtraction graphically. Solve: To obtain vector C .. from A .. and , B .. we place the tail of -B .. on the tip of , A .. and then use the tip-totail rule of graphical addition. 3.13. Visualize: The vectors A, B, .. .. and C = A + B .. .. .. are shown. Solve: (a) We have A = 5i + 2 j .. and B = -3i - 5 j. .. Thus, C = A + B .. .. .. = (5i + 2 j) + (-3i - 5 j) = 2i - 3 j. (b) Vectors , A .. , B .. and C .. are shown with their tails together. (c) Since 2 3 , x y C = i - j = C i + C j .. 2, x C = and 3. y C = - Therefore, the magnitude and direction of C .. are C = (2)2 + (-3)2 = 3.6 1 1 | | 3 tan tan 56 below the + -axis 2 y x C x C . = - = - .. .. = . . Assess: The vector C .. is to the right and down, thus implying a negative y-component and positive xcomponent, as obtained above. Also . > 45 since | | | | y x C > C . 3.14. Visualize: Solve: (a) We have A = 5i + 2 j, .. B = -3i - 5 j, .. and -B = +3i + 5 j. .. Thus, D = A + (-B) .. .. .. = 8i + 7 j. (b) Vectors , A .. B .. and D .. are shown in the above figure. (c) Since 8 7 , x y D = i + j = D i + D j .. 8 x D = and 7. y D = Therefore, the magnitude and direction of D .. are 2 2 1 1 7 (8) (7) 10.6 tan tan 41 8 y x D D D . - - . . . . = + = = . . = . . = . . . . Assess: Since | | | | y x D < D , the angle . is less than 45, as it should be. 3.15. Visualize: Solve: (a) We have A = 5i + 2 j .. and B = -3i - 5 j. .. This means 2A =10i + 4 j .. and 3B = -9i -15 j. .. Hence, E = 2A+ 3B .. .. .. =1i -11 j. (b) Vectors A .. , , B .. and E .. are shown in the above figure. (c) From the E .. vector, 1 x E = and 11 y E = - . Therefore, the magnitude and direction of E .. are E = (1)2 + (-11)2 =11.05 tan 1 tan 1 1 5.19 right of the -axis | | 11 x y E y E f - - . . . . = .. .. = . . = - . . . . Assess: Note that f is the angle made with the y-axis, and that is why tan 1( /| |) x y f = - E E rather than tan 1(| |/ ), y x - E E which would be the case if f were the angle made with the x-axis. 3.16. Visualize: Solve: (a) We have A = 5i + 2 j .. and B = -3i - 5 j. .. This means -4B = +12i + 20 j. .. Hence, F = A - 4B = .. .. .. 17i + 22 j = x y F i + F j with 17 x F = and 22. y F = (b) The vectors , A .. , B .. and F .. are shown in the above figure. (c) The magnitude and direction of F .. are 2 2 (17)2 (22)2 27.8 x y F = F + F = + = tan 1 tan 1 22 52.3 17 y x F F . - - . . . . = . . = . . = . . . . Assess: y x F > F implies . > 45, as is observed. 3.17. Solve: A different coordinate system can only mean a different orientation of the grid and a different origin of the grid. (a) False, because the size of a vector is fixed. (b) False, because the direction of a vector in space is independent of any coordinate system. (c) True, because the orientation of the vector relative to the axes can be different. 3.18. Visualize: Solve: In coordinate system I, A = -(4 m) j, .. so 0 m x A = and 4 m. y A = - The vector B .. makes an angle of 60 counterclockwise from vertical, which makes it have an angle of . = 30 with the x-axis. Since B .. points to the left and up, it has a negative x-component and a positive y-component. That is, (5.0 m)cos30 4.3 m x B = - = - and y B = +(5.0 m)sin30 = 2.5 m. Thus, B = -(4.3 m)i + (2.5 m) j. .. In coordinate system II, A .. points to the left and down, and makes an angle of 30 with the y-axis. Therefore, (4.0 m)sin30 2.0 m x A = - = - and (4.0)cos30 3.5 m. y A = - = - This implies A = -(2.0 m)i - (3.5 m) j. .. The vector B .. makes an angle of 30 with the +y-axis and is to the left and up. This means we have to manually insert a minus sign with the x-component. sin30 (5.0 m)sin30 2.5 m, xB = -B = - = - and y B = +Bcos30 = (5.0 m)cos30 = 4.3 m. Thus B = -(2.5 m)i + (4.3 m) j. .. 3.19. Visualize: Refer to Figure EX3.19. The velocity vector v.. points west and makes an angle of 30 with the x-axis. v.. points to the left and up, implying that x v is negative and y v is positive. Solve: We have vx = -vcos30 = -(100 m/s)cos30 = -86.6 m/s and sin30 (100 m/s)sin30 yv = +v = = 50.0 m/s. Assess: x v and y v have the same units as v... 3.20. Visualize: (a) Solve: (b) The components of the vectors A, B, .. .. and C .. are (3.0 m)cos 20 2.8 m and (3.0 m)sin 20 1.0 m x y A = = A = - = - ; 0 m x B = and 2 m; y B = (5.0 m)cos70 1.71 m and (5.0 m)sin70 4.7 m. x y C = - = - C = - = - This means the vectors can be written, A = (2.8 m)i - (1.0 m) j B = (2.0 m) j C = (-1.71 m)i - (4.7 m) j .. .. .. (c) We have D = A + B + C = (1.09 m)i - (3.7 m) j. .. .. .. .. This means (1.09 m)2 (3.7 m)2 3.9 m tan 1 3.9 tan 13.58 74 1.09 D = + = . = - = - = The direction of D .. is south of east, 74 below the positive x-axis. 3.21. Visualize: Solve: Using the method of tail-to-tip graphical addition, the diagram shows the resultant for D + E + F .. .. .. in (a), the resultant for D + 2E .. .. in (b), and the resultant for D - 2E + F .. .. .. in (c). 3.22. Solve: We have 2 3 , x y E = E i + E j = i + j .. which means 2 x E = and 3. y E = Also, x y F = F i + F j = .. 2i - 2 j, which means 2 x F = and 2. y F = - (a) The magnitude of E .. is given by 2 2 (2)2 (3)2 3.6 x y E = E + E = + = and the magnitude of F .. is given by 2 2 (2)2 (2)2 2.8. x y F = F + F = + = (b) Since E + F = 4i +1j, .. .. the magnitude of E + F .. .. is (4)2 + (1)2 = 4.1. (c) Since -E - 2F = -(2i + 3 j) - 2(2i - 2 j) .. .. = -6i +1 j, the magnitude of -E - 2F .. .. is (-6)2 + (1)2 = 6.1. 3.23. Solve: We have r.. = (5i + 4 j)t2 m. This means that r.. does not change the ratio of its components as t increases, that is, the direction of r.. is constant. The magnitude of r.. is given by r = (5t2 )2 + (4t2 )2m = (6.40t2 ) m. (a) The particles distance from the origin at t = 0 s, t = 2 s, and t = 5 s is 0 m, 25.6 m, and 160 m. (b) The particles velocity is 2 v dr (5i 4 j) dt m/s dt dt = = + .. .. = (5i + 4 j)2t m/s = (10i + 8 j)t m/s (c) The magnitude of the particles velocity is given by v = (10t)2 + (8t)2 =12.8t m/s. The particles speed at t = 0 s, t = 2 s, and t = 5 s is 0 m/s, 25.6 m/s , and 64.0 m/s . 3.24. Visualize: Solve: (a) Vector C .. is the sum of vectors A .. and , B .. which is obtained using the tip-to-tip rule of graphical addition. Its magnitude is measured to be 4.7 and its angle made with the +x-axis is measured to be 33. (b) Using the law of cosines, C2 = A2 + B2 - 2ABcosf , and the geometry of parallelograms, which shows that 180 ( ) 180 (60 20 ) 140 , B A f = - . -. = - - = we obtain C = (3)2 + (2)2 - 2(3)(2)cos(140) = 4.71 Using the law of sines: sin sin140 2 4.71 a = .a =15.8 Thus, 20 35.8 . C . =a + = (c) We have: cos 3cos20 2.82 sin 3sin 20 1.03 cos 2cos60 1.00 sin 2sin 60 1.73 x A y A x B y B A A A A B B B B . . . . = = = = = = = = = = = = This means: 3.82 x x x C = A + B = and 2.76. y y y C = A + B = The magnitude and direction of C are given by 2 2 (3.82)2 (2.76)2 4.71 tan 1 tan 1 2.76 35.8 3.82 y x y C x C C C C C . - - . . . . = + = + = = . . = . . = . . . . Assess: Using the method of vector components and their algebraic addition to find the resultant vector yields the same results as using the graphical addition of vectors. 3.25. Visualize: Refer to Figure P3.25 in your textbook. Solve: (a) We are given that A + B + C = -2i .. .. .. with A = 4i, .. and C = -2 j. .. This means A + C = 4i - 2 j. .. .. Thus, B = (A+ B + C) - (A+ C) .. .. .. .. .. .. = (-2i ) - (4i - 2 j) = -6i + 2 j. (b) We have x y B = B i + B j .. with 6 x B = - and 2. y B = Hence, B = (-6)2 + (2)2 = 6.3 tan 1 tan 1 2 18 | | 6 y x B B . = - = - = Since B .. has a negative x-component and a positive y-component, the angle . made by B .. is with the x-axis and it is above the x-axis. Assess: Since | | | |, y x B < B . < 45 as is obtained above. 3.26. Visualize: Solve: (a) 1 1 1 1 2 1 tan 45 tan 63.4 E F . . - . . . . . . - . . . . . . = = = = Thus 180 71.6 E F f = -. -. = (b) From the figure, E = 2 and F = 5. .. .. Using 2 2 2 2 cos ( 2)2 ( 5)2 2( 2)( 5)cos(180 71.6 ) 3.00. G E F EF G = + - f = + - - . = Furthermore, using sin sin(180 71.6 ) 45 5 2.975 a a - = . = Since 45 E . = , the angle made by the vector G .. with the +x-axis is ( ) 45 45 90 . G E . = a +. = + = (c) We have ( )2 ( )2 1 1 1.0, and 1.0 1.0, and 2.0 0.0, and 3.0 | | 3.0 0.0 3.0 3.0, and tan tan 90 | | 0.0 x y x y x y y x E E F F G G G G G . - - = + = + = - = + = = . = + = = = . . = . . . . That is, the vector G .. makes an angle of 90 with the x-axis. Assess: The graphical solution and the vector solution give the same answer within the given significance of figures. 3.27. Visualize: Refer to Figure P3.27. Solve: From the rules of trigonometry, we have 4cos40 3.1 x A = = and 4sin 40 2.6. y A = = Also, 2cos10 Bx = - = -1.97 and By = +2sin10 = 0.35. Since A + B + C = 0, .. .. .. .. C = -A - B = (-A) + (-B) = .. .. .. .. .. (-3.1i - 2.6 j) + (+1.97i - 0.35 j) = -1.1i-3.0 j. 3.28. Visualize: Solve: In the tilted coordinate system, the vectors A and B .. .. are expressed as: A = (2sin15 m)i + (2cos15 m) j and B = (4cos15 m)i - (4sin15 m) j. .. .. Therefore, D = 2A + B = (4 m)[(sin15 + cos15)i + (cos15 - sin15) j] = (4.9 m)i + (2.9 m) j. .. .. .. The magnitude of this vector is D = 5.7 m, and it makes an angle of . = tan-1(2.9 m/4.9 m) = 31 with the +x-axis. Assess: The resultant vector can be obtained graphically by using the rule of tail-to-tip addition. D u 3.29. Visualize: The magnitude of the unknown vector is 1 and its direction is along i + j . Let A = i + j .. as shown in the diagram. That is, A =1i +1j .. and the x- and y-components of A .. are both unity. Since tan 1( / ) 45 , y x . = - A A = the unknown vector must make an angle of 45 with the +x-axis and have unit magnitude. Solve: Let the unknown vector be x y B = B i + B j .. where cos45 1 and sin 45 1 2 2 x y B = B = B B = B = B We want the magnitude of B .. to be 1, so we have 2 2 2 2 2 1 1 1 1 1 1 2 2 x y B B B B B B B = + = . . . + . . = . = . = . . . . . . . . Hence, 1 2 x y B = B = Finally, x y B = B i + B j .. 1 1 2 2 = i + j 3.30. Model: Carlos will be represented as a particle and the particle model will be used for motion under constant acceleration kinetic equations. Visualize: Solve: Carlos runs at constant speed without changing direction. The total distance he travels is found from kinematics: ( r1 = r0 + v0.t = 0 m + 5 m/s)(600 s) = 3000 m This displacement is north of east, or . = 25 from the +x-axis. Thus the position 1 r.. becomes 1 r.. = (3000 m)(cos25i + sin 25 j) = 2.7 km i +1.27 km j That is, Carlos ends up 1.27 km north of his starting position. Assess: The choice of our coordinate system is such that the x-component of the displacement is along the east and the y-component is along the north. The displacement of 3.0 km is reasonable for Carlos to run in 10 minutes if he is an athlete. 3.31. Visualize: The coordinate system (x,y,z) is shown here. While +x denotes east and +y denotes north, the +z-direction is vertically up. The vectors morning S .. (shortened as m S .. ), afternoon S .. (shortened as a S .. ), and the total displacement vector Stotal = Sa + Sm .. .. .. are also shown. Solve: m S = (2000i + 3000 j + 200k) m, .. and a S = (-1500i + 2000 j - 300k) m. .. The total displacement is the sum of the individual displacements. (a) The sum of the z-components of the afternoon and morning displacements is Sza + Szm = -300 m + 200 m = -100 m, that is, 100 m lower. (b) total a m S = S + S .. .. .. = (500i + 5000 j -100k) m, that is, (500 m east) + (5000 m north) (100 m vertical). The magnitude of your total displacement is ( )2 ( )2 ( )2 total S = 500 + 5000 + -100 m = 5.03 km 3.32. Visualize: Only the minute hand is shown in the figure. Solve: (a) We have S8:00 = (2.0 cm) j .. and 8:20 S = (2.0 cm)cos30i - (2.0 cm)sin30 j. .. The displacement vector is 8:20 8:00 (2.0 cm)[cos30 (sin30 1) ] (2.0 cm)[0.87 1.50 ] (1.74 cm) (3.00 cm) r S S i j i j i j . = - = - + = - = - .. .. .. (b) We have 8:00 S = (2.0 cm) j .. and 9:00 S = (2.0 cm) j. .. The displacement vector is 9:00 8:00 .r = S - S .. .. .. = 0. Assess: The displacement vector in part (a) has a positive x-component and a negative y-component. The vector thus is to the right and points down, in quadrant IV. This is where the vector drawn from the tip of the 8:00 a.m. arm to the tip of the 8:20 a.m. arm will point. 3.33. Visualize: (a) Note that +x is along the east and +y is along the north. Solve: (b) We are given A = -(200 m) j, .. and can use trigonometry to obtain B = -(283 m)i - (283 m) j .. and C = (100 m)i + (173 m) j. .. We want A + B + C + D = 0. .. .. .. .. This means (200 m ) (283 m 283 m ) ( 100 m 173 m ) 183 m 310 m D A B C j i j i j i j = - - - = + + +- - = + .. .. .. .. The magnitude and direction of D .. are (183 m)2 (310 m)2 360 m and tan 1 tan 1 310 m 59.4 183 m y x D D D = + = . = - = - .. .. = . . This means D = (360 m, 59.4 .. north of east). (c) The measured length of the vector D .. on the graph (with a ruler) is approximately 1.75 times the measured length of vector A .. . Since A = 200 m, this gives D = 1.75 200 m = 350 m. Similarly, the angle . measured with the protractor is close to 60. These answers are in close agreement to part (b). 3.34. Visualize: (a) The figure shows Sparkys individual displacements and his net displacement. Solve: (b) Dnet = D1 + D2 + D3, .. .. .. .. where individual displacements are 1 2 3 (50cos45 50sin 45 ) m (35.4 35.4 ) m 70 m 20 m D i j i j D i D j = + = + = - = - .. .. .. Thus Sparkys displacement is net D = (-35i +15.4 j) m. .. (c) As a magnitude and angle, 2 2 net net net ( ) ( ) 38 m, x y D = D + D = 1 net net net ( ) tan 24 |( ) | y x D D . -. . = . . = . . Sparkys net displacement is 38 m in a direction 24 north of west. 3.35. Visualize: Solve: We are given A=5 m i .. and C = (-1 m)k. .. Using trigonometry, B = (3cos45 m)i - (3sin 45 m) j. .. The total displace-ment is r = A + B + C = (7.12 m)i - (2.12 m) j - (1 m)k. .. .. .. .. The magnitude of r.. is r = (7.12)2 + (2.12)2 + (1)2 m = 7.5 m. Assess: A displacement of 7.5 m is a reasonable displacement. 3.36. Visualize: Solve: We have v.. = vxi + vy j ||v i v j . = + = vcos. i + vsin. j. Thus, || v = vcos. = (100 m/s)cos30 = 86.6 m/s. Assess: For the small angle of 30, the obtained value of 86.6 m/s for the horizontal component is reasonable. 3.37. Visualize: Solve: (a) Since vx = vcos. , we have 2.5 m/s = (3.0 m/s)cos. cos 1 2.5 m/s 34 . 3.0 m/s . -. . . = . . = . . (b) The vertical component is sin y v = v . = (3.0 m/s) sin34 =1.7 m/s. 3.38. Visualize: The coordinate system used here is tilted with x-axis along the slope. Solve: The component of the velocity parallel to the x-axis is || v = -vcos70 = -vsin 20 = -10 m/s (0.34) = -3.4 m/s. This is the speed down the slope. The component of the velocity perpendicular v vsin 70 vcos20 10 m/s (0.94) 9.4 m/s. . = - = - = - = - This is the speed toward the ground. Assess: A speed of approximately 10 m/s implies a fall time of approximately 1 second under free fall. Note that g = 9.8 m/s2. This time is reasonable for a drop of approximately 5 m, or 16 feet. 3.39. Visualize: Solve: (a) The river is 100 m wide. If Mary rows due north at a constant speed of 2.0 m/s, it will take her 50 s to row across. But while shes doing so, the current sweeps her boat sideways a distance 1 m/s 50 s = 50 m. Marys net displacement is the vector sum of the displacement due to her rowing plus the displacement due to the rivers current. She lands 50 m east of the point that was directly across the river from her when she started. (b) Marys net displacement is shown on the figure. 3.40. Visualize: Establish a coordinate system with origin at the tree and with the x-axis pointing east. Let A .. be a displacement vector directly from the tree to the treasure. Vector A .. is A = (100i + 500 j) paces. .. This describes the displacement you would undergo by walking north 500 paces, then east 100 paces. Instead, you follow the road for 300 paces and undergo displacement B = (300sin 60i + 300cos60 j) paces = (260i +150 j) paces .. Solve: Now let C .. be the displacement vector from your position to the treasure. From the figure A = B + C. .. .. .. So the displacement you need to reach the treasure isC = A - B = (-160i + 350 j) paces. .. .. .. If . is the angle measured between C .. and the y-axis, tan 1 160 24.6 350 . = - .. .. = . . You should head 24.6 west of north. You need to walk distance 2 2 385 x y C = C + C = paces to get to the treasure. 3.41. Visualize: A 3% grade rises 3 m for every 100 m horizontal distance. The angle of the ground is thusa = tan-1(3/100) = tan-1(0.03) =1.72. Establish a tilted coordinate system with one axis parallel to the ground and the other axis perpendicular to the ground. Solve: From the figure, the magnitude of the component vector of v.. perpendicular to the ground is v vsina 15.0 m/s. .= = But this is only the size. We also have to note that the direction of v. .. is down, so the component is v. = -15.0 m/s. 3.42. Visualize: Solve: The resulting velocity is given by v.. = v..fly + v..wind , where wind v.. = 6 m/s i and fly v.. = -vsin. i - vcos. j. Substituting the known values we get v.. = -8 m/s sin. i - 8 m/s cos. j + 6 m/s i. We need to have 0. x v = This means 0 = -8 m/s sin. + 6 m/s, so sin 6 8 . = and . = 48.6. Thus the ducks should head 48.6 west of south. 3.43. Model: The car is treated as a particle in this problem. Visualize: Solve: (a) The tangential component is 2 2 || sin30 a = a = (2.0 m/s )(0.5) =1.0 m/s . (b) The perpendicular component is a acos30 (2.0 m/s2 )(0.866) 1.7 m/s2. .= = = Assess: Magnitudes of the tangential and perpendicular components of acceleration are reasonable. 3.44. Model: We will treat the knot in the rope as a particle in static equilibrium. Visualize: Solve: Expressing the vectors using unit vectors, we have F1 = 3.0i .. and 2 F = -5.0sin30i + 5.0cos30 j. .. Since 1 2 3 F +F + F =0, .. .. .. .. we can write 3 1 2 F = -F -F .. .. .. =-0.5i - 4.33 j . The magnitude of 3 F .. is given by 2 2 3 F = (-0.5) + (-4.33) = 4.4 units. The angle 3 F .. makes is . = tan-1(4.33/0.5) = 83 and is below the negative xaxis. Assess: The resultant vector has both components negative, and is therefore in quadrant III. Its magnitude and direction are reasonable. Note the minus sign that we have manually inserted with the force 2 F .. . 3.45. Visualize: Use a tilted coordinate system such that x-axis is down the slope. Solve: Expressing all three forces in terms of unit vectors, we have F1 = -(3.0 N)i, .. 2 F = +(6.0 N) j, .. and 3 F = .. (5.0 N)sin. i - (5.0 N)cos. j. (a) The component of net F .. parallel to the floor is net ( ) (3.0 N) 0 N (5.0 N)sin30 0.50 N, x F = - + + = - or 0.50 N up the slope. (b) The component of net F .. perpendicular to the floor is net ( ) 0 N (6.0 N) (5.0 N)cos30 1.67 N. y F = + - = (c) The magnitude of net F .. is net net net ( ) ( ) x y F = F + F = (-0.50 N)2 + (1.67 N)2 =1.74 N. The angle net F .. makes is 1 net 1 net ( ) 1.67 N tan tan 73 |( ) | 0.50 N y x F F f = - = - .. .. = . . .. net F .. is 73 above the floor on the left side of 2F . .. 00000 3-1 3.46. Visualize: Solve: Using trigonometry to calculate ., we get . = tan-1(100 cm/141 cm) = 35.3. Expressing the three forces in unit vectors, B F = -(3.0 N)i, .. C F = -(6.0 N) j, .. and D F = +(2.0 N)cos35.3i - .. (2.0 N)sin35.3 j = (1.63 N)i - (1.16 N) j. The total force is net B C D F = F + F + F = -1.37 Ni - 7.2 N j. .. .. .. .. The magnitude of net F .. is net F = (1.37 N)2 + (7.2 N)2 = 7.3 N. 1 net 1 net net |( ) | 7.2 N tan tan 79 |( ) | 1.37 N y x F F . = - = - .. .. = . . net F = (7.3 N,79 .. below -x in quadrant III). 4.1. Solve: (a) (b) A race car slows from an initial speed of 100 mph to 50 mph in order to negotiate a tight turn. After making the 90 turn the car accelerates back up to 100 mph in the same time it took to slow down. 4.2. Solve: (a) (b) A car drives up a hill, over the top, and down the other side at constant speed. 4.3. Solve: (a) (b) A ball rolls along a level table at 3 m/s. It rolls over the edge and falls 1 m to the floor. How far away from the edge of the table does it land? 4.4. Solve: (a) The figure shows a motion diagram of a pendulum as it swings from one side to the other. Its clear that the velocity at the lowest point is not zero. The velocity vector at this point is tangent to the circle. We can use the method of Tactics Box 1.3 to find the acceleration at the lowest point. The acceleration is not zero. Instead, you can see that the acceleration vector points toward the center of the circle. (b) The end of the arc is like the highest point of a ball tossed straight up. The velocity is zero for an instant as the vector changes from pointing outward to pointing inward. However, the acceleration is not zero at this point. The velocity is changing at the end point, and this requires an acceleration. The motion diagram shows that .v.., and thus a.., is tangent to the circle at the end of the arc. 4.5. Model: The boat is treated as a particle whose motion is governed by constant-acceleration kinematic equations in a plane. Visualize: Solve: Resolving the acceleration into its x and y components, we obtain a.. = (0.80 m/s2 )cos40i + (0.80 m/s2 )sin 40 j = (0.613 m/s2 )i + (0.514 m/s2 ) j From the velocity equation ( ) 1 0 1 0 v.. = v.. + a.. t - t , ( ) ( 2 ) ( 2 ) ( ) ( ) ( ) 1 v = 5.0 m/s i + .. 0.613 m/s i + 0.514 m/s j.. 6 s - 0 s = 8.68 m/s i + 3.09 m/s j .. The magnitude and direction of v.. are ( )2 ( )2 v = 8.68 m/s + 3.09 m/s = 9.21 m/s 1 1 1 1 tan tan 3.09 m/s 20 north of east 8.68 m/s y x v v . - - . . . . = . . = . . = . . . . Assess: An increase of speed from 5.0 m/s to 9.21 m/s is reasonable. 4.6. Solve: (a) At t = 0 s, x = 0 m and y =0 m, or r.. = (0i + 0j) m. At t = 4 s, x =0 m and y =0 m, or r.. = (0i+0j) m. In other words, the particle is at the origin at both t = 0 s and at t = 4 s. From the expressions for x and y, 3 2 4 ( 2) m/s 2 v dx i dy j t t i t j dt dt .. . . = + = .. - . + - . .. . . .. At t = 0 s, v.. = -2j m/s, v = 2 m/s. At t = 4 s, v.. = (8i + 2j) m/s, v = 8.3 m/s. (b) At 0 s, t = v.. is along j - , or 90 south of +x. At t = 4 s, tan 1 2 m/s 14 north of + 8 m/s . = - .. .. = x . . 4.7. Visualize: Refer to Figure EX4.7. Solve: From the figure, identify the following: x1 = 0 m 1 y = 0 m 2 x = 2000 m 2 y =1000 m 1 0 m/s x v = 1 200 m/s y v = 2 200 m/s x v = 2 100 m/s y v = - The components of the acceleration can be found by applying 2 2 2 1 v = v + 2a.s for the x and y directions. Thus ( ) ( ) ( ) 2 2 2 2 2 1 2 200 m/s 0 m/s 10.00 m/s 2 2 2000 m 0 m x x x a v v x - - = = = . - ( ) ( ) ( ) 2 2 2 100 m/s 200 m/s 15.00 m/s 2 1000 m 0 m y a - - = =- - So a.. = (10.00i -15.00 j ) m/s2. Assess: A time of 20 s is needed to change 1 0 m/s x v = to 2 200 m/s x v = at 10 m/s2. x a = This is the same time needed to change 2 1 2 to at 15 m/s . y y y v v a = - 4.8. Model: The puck is a particle and follows the constant-acceleration kinematic equations of motion. Visualize: Please refer to Figure EX4.8. Solve: (a) At t = 2 s, the graphs give 16 cm/s x v = and 30 cm/s. y v = The angle made by the vector v.. with the x-axis can thus be found as tan 1 tan 1 30 cm/s 62 16 cm/s y x v v . - - . . . . = . . = . . = . . . . above the x-axis (b) After t = 5 s, the puck has traveled a distance given by: 5 1 1 0 0 2 0 m area under - curve (40 cm/s)(5 s) 100 cm s x x x = x + . v dt = + v t = = 5 1 0 0 0 m area under - curve (30 cm/s)(5 s) 150 cm s y y y = y + . v dt = + v t = = 2 2 ( )2 ( )2 1 1 1 .r = x + y = 100 cm + 150 cm =180 cm 4.9. Model: Use the particle model for the puck. Visualize: Please refer to Figure EX4.9 Solve: (a) Since the x v vs t and y v vs t graphs are straight lines, the puck is undergoing constant acceleration along the x- and y- axes. The components of the pucks acceleration are ( 10 m/s 10 m/s) 2.0 m/s2 10 s 0 s x x x a dv v dt t . - - = = = =- . - (10 m/s 0 m/s) 1.0 m/s2 (10 s 0 s) y a - = = - The magnitude of the acceleration is 2 2 2.2 m/s2. x y a = a + a = (b) The puck is undergoing constant acceleration in both the x and y directions. Identify from the graphs 10 m/s, ix v = 0 m/s. iy v = Since the puck starts at the origin, 0 m, i i x = y = and set 0 s. i t = Using kinematics, 1 2 2 2 x = 0 m + (10 m/s)t + (-2.0 m/s )t 1 2 2 2 y = 0 m + 0 m/s + (1.0 m/s )t The distance from the origin at time t is r = x2 + y2 . The table below shows the values of x, y, and r at the times t = 0, 5, 10 s. x y R t = 0 s 0 m 0 m 0 m 5 s 25 m 12.5 m 28 m 10 s 0 m 50 m 50 m Assess: The puck turns around at t = 5 s in the x direction, and constantly accelerates in the y direction. Traveling 50 m from the starting point in 10 s is reasonable. 4.10. Model: Assume the particle model for the ball, and apply the constant-acceleration kinematic equations of motion in a plane. Visualize: Solve: (a) We know the velocity v..1 = (2.0i + 2.0j) m/s at t =1 s. The ball is at its highest point at t = 2 s, so 0 m/s. y v = The horizontal velocity is constant in projectile motion, so 2.0 m/s x v = at all times. Thus 2 v.. = 2.0i m/s at t = 2 s. We can see that the y-component of velocity changed by 2.0 m/s. y .v = - between t =1 s and t = 2 s. Because y a is constant, y v changes by 2.0 m/s in any 1-s interval. At t = 3 s, y v is 2.0 m/s less than its value of 0 at t = 2 s. At t = 0 s, y v must have been 2.0 m/s more than its value of 2.0 m/s at t =1 s. Consequently, at t = 0 s, 0 v.. = (2.0i + 4.0j) m/s At t =1 s, 0 v.. = (2.0i + 2.0j) m/s At t = 2 s, 0 v.. = (2.0i + 0.0j) m/s At t = 3 s, 0 v.. = (2.0i - 2.0j) m/s (b) Because y v is changing at the rate 2.0 m/s per s, the y-component of acceleration is 2.0 m/s2. y a = - But ya = -g for projectile motion, so the value of g on Exidor is g = 2.0 m/s2. (c) From part (a) the components of 0 v.. are 0 0 2.0 m/s and 4.0 m/s. x y v = v = This means 1 0 1 0 tan tan 4.0 m/s 63.4 2.0 m/s y x v v . - - . . . . = . . = . . = . . . . above +x Assess: The y-component of the velocity vector decreases from 2.0 m/s at t =1 s to 0 m/s at t = 2 s. This gives an acceleration of -2 m/s2. All the other values obtained above are also reasonable. 4.11. Model: The ball is treated as a particle and the effect of air resistance is ignored. Visualize: Solve: Using ( ) 1 ( )2 1 0 0x 1 0 2 x 1 0 x = x + v t - t + a t - t , 1 1 50 m = 0 m + (25 m/s)(t - 0 s) + 0 m.t = 2.0 s Now, using ( ) 1 ( )2 1 0 0 y 1 0 2 y 1 0 y = y + v t - t + a t - t , 1 2 2 1 2 y = 0 m+ 0 m + (-9.8 m/s )(2.0 s - 0 s) = -19.6 m Assess: The minus sign with 1 y indicates that the balls displacement is in the negative y direction or downward. A magnitude of 19.6 m for the height is reasonable. 4.12. Model: The bullet is treated as a particle and the effect of air resistance on the motion of the bullet is neglected. Visualize: Solve: (a) Using ( ) 1 ( )2 1 0 0 1 0 2 1 0 , y y y = y + v t - t + a t - t we obtain 2 1 2 2 2 1 1 (-2.010- m) = 0 m+ 0 m+ (-9.8 m/s )(t - 0 s) .t = 0.0639 s (b) Using ( ) 1 ( )2 1 0 0 1 0 2 1 0 , x x x = x + v t - t + a t - t 0 0 (50 m) 0 m (0.0639 s 0 s) 0 m 782 m/s x x = + v - + .v = Assess: The bullet falls 2 cm during a horizontal displacement of 50 m. This implies a large initial velocity, and a value of 782 m/s is understandable. 4.13. Model: We will use the particle model for the food package and the constant-acceleration kinematic equations of motion. Visualize: Solve: For the horizontal motion, 1 2 1 0 0 1 0 2 1 0 1 1 ( ) ( ) 0 m (150 m/s)( 0 s) 0 m (150 m/s) x x x = x + v t - t + a t - t = + t - + = t We will determine 1 t from the vertical y-motion as follows: 1 2 1 0 0 1 0 2 1 0 ( ) ( ) y y y = y + v t - t + a t - t 1 2 2 2 1 1 2 0 m 100 m 0 m ( 9.8 m/s ) 200 m 4.518 s 9.8 m/s . = + + - t .t = = From the above x-equation, the displacement is 1 x = (150 m/s)(4.518 s) = 678 m. Assess: The horizontal distance of 678 m covered by a freely falling object from a height of 100 m and with an initial horizontal velocity of 150 m/s ( 335 mph) is reasonable. 4.14. Model: Assume the particle model for the spyglass and use the projectile motion equations. Visualize: Solve: (a) The spyglass has an initial horizontal velocity equal to that of the ship. As the spyglass falls, it and the ship move forward together at the same velocity, so the spyglass lands at the bottom of the mast directly vertically below on the ship where the sailor dropped it. (b) The time the spyglass takes to fall can be found by considering the vertical motion: 1 2 1 2 2 1 0 0 1 0 2 1 0 2 1 1 ( ) ( ) 0 m 15 m 0 m ( 9.8 m/s )( 0 s) 1.749 s y y y = y + v t - t + a t - t . = + + - t - .t = Therefore, 1 x = 0 m/s + (4.0 m/s)(1.749 s) = 7.0 m. The spyglass lands 7.0 m to the right of the fisherman. 4.15. Model: The position vectors r.. and r..' in frames S and S' are related by the equation r = r' + R, .. .. .. where R .. is the position vector of the origin of frame S' as measured in frame S. S is Teds frame and S' is Stellas frame. Visualize: Solve: The relation between R .. and the velocity of Stella V .. is (100 m/s)cos45 (100 m/s)sin 45 5 s R =V = i - j .. .. (5.0 m) 100 100 (353.6 353.6 ) m 2 2 R i j i j . = . - . = - .. .. .. Because r = r' + R, .. .. .. r' = r - R = (200 m)i - ..(353.6 m)i - (353.6 m) j.. = -(154 m)i + (354 m) j .. .. .. The vector r..' determines the position of the exploding firecracker as seen by Stella. 4.16. Model: Assume motion along the x-direction. Let the earth frame be S and a frame attached to the water be S'. Frame S' moves relative to S with a velocity x V . Visualize: Solve: Let x v be the velocity of the boat in frame S, and x v' be the velocity of the boat in S'. Then for travel down the river, 30 km 10.0 km/h 3.0 hr vx = v'x +Vx = = For travel up the river, 30 km 6.0 km/h 5.0 hr x x -v' +V = -.. .. = - . . Adding these two equations yields 2.0 km/h. x V = That is, the velocity of the flowing river relative to the earth is 2.0 km/h. 4.17. Motion: Assume motion along the x-direction. Let the earth frame be S and a frame attached to the moving sidewalk be S . ' Frame S' moves relative to S with velocity . x V Solve: Let vx be your velocity in frame S and v'x be your velocity in S'. In the first case, when the moving sidewalk is broken, 0 m/s x V = and ( ) 1 0 W 50 s x x x v - = In the second case, when you stand on the moving sidewalk, 0 m/s. x v' = Therefore, using , x x x v = v' +V we get 1 0 S 75 s x x v V x x - = = In the third case, when you walk while riding, W. x x v' = v Using , x x x v = v' +V we get 1 0 1 0 1 0 50 s 75 s x x x x x x t - - - = + .t = 30 s Assess: A time smaller than 50 s was expected. 4.18. Model: Let the earth frame be S and a frame attached to the water be S . ' Frame S' moves relative to S with velocity V. We define the x-axis along the direction of east and the y-axis along the direction of north for both frames. The frames S and S' have their origins coincident at t = 0 s. Solve: (a) According to the Galilean transformation of position: r = r' + R .. .. .. = r' + tV. .. .. We need to find Marys position vector r.. from the earth frame S. The observer in frame S' will observe the boat move straight north and will find its position as r..' = (2.0t ) j m/s. We also know that V = (3.0)i m/s. .. Since r' =100 m = (2.0 m/s)t and t = 50 s, we have r.. = (2.0 m/s)(50 s) j + (50 s)(3.0 m/s)t i = (150 m)i + (100 m) j Thus she lands 150 m east of the point that was straight across the river from her when she started. (b) Note that r' .. is the displacement due to rowing, R .. is the displacement due to the rivers motion, and r.. is the net displacement. 4.19. Model: Let Susans frame be S and Shawns frame be S'. S' moves relative to S with velocity V. Both Susan and Shawn are observing the intersection point from their frames. Solve: The Galilean transformation of velocity is v = v' +V, .. .. .. where v.. is the velocity of the intersection point from Susans reference frame, v..' is the velocity of the intersection point from Shawns frame S', and V .. is the velocity of S' relative to S or Shawns velocity relative to Susan. Because v.. = -(60 mph) j and v..' = -(45 mph)i, we have V = v - v' .. .. ( ) ( ) = 45 mph i - 60 mph j. This means that Shawns speed relative to Susan is ( )2 ( )2 V = 45 mph + -60 mph = 75 mph 4.20. Solve: (a) From t = 0 s to t =1 s the particle does not rotate. From t =1 s to t = 3 s, the particle rotates clockwise from the angular position 0 rad to -2p rad. Therefore, .. = -2p rad in two seconds, or . = -p rad s. From t = 3 s to t = 4 s the particle rotates counterclockwise from the angular position -2p rad to +4p rad. Thus .. = 4p - (-2p ) = 6p rad and . = +6p rad s. (b) 4.21. Solve: Since . = (d. dt ) we have f i . =. + area under the . -versus-t graph between i t and f t From t = 0 s to t = 2 s, the area is 1 ( )( ) 2 20 rad/s 2 s = 20 rad. From t = 2 s to t = 4 s, the area is (20 rad/s)(2 s) = 40 rad. Thus, the area under the . -versus-t graph during the total time interval of 4 s is 60 rad or (60 rad) (1 rev/2p rad) = 9.55 revolutions. 4.22. Solve: Since . = (d. dt ) we have f i . =. + area under the . versus t graph between i t and f t Fromt = 0 s to s to t = 4 s, the area is 1 ( )( ) 2 20 rad/s 4 s = 40 rad. From t = 4 s to t = 8 s, the area is 1 ( )( ) 2 -10 rad/s 4 s = -20 rad. Thus, the area under the . versus t graph during the total time interval of 8 s is 20 rad or (20 rad) (1 rev/2p rad) = 3.2 revolutions. 4.23. Model: Treat the record on a turntable as a particle rotating at 45 rpm. Solve: (a) The angular velocity is 45 rpm 1min 2 rad 1.5 rad/s 60 s 1 rev p . = = p (b) The period is 2 rad 2 rad 1.33 s 1.5 rad/s T p p . p = = = 4.24. Model: The airplane is to be treated as a particle. Visualize: Solve: (a) The angle you turn through is 2 1 5000 miles 1.2500 rad 1.2500 rad 180 71.62 4000 miles rad s r . . p - = = = = = (b) The planes angular velocity is 2 1 5 2 1 1.2500 rad 0.13889 rad/h 0.13889 rad 1 h 3.858 10 rad/s t t 9 hr h 3600 s . . . - - = = = = = - Assess: An angular displacement of approximately one-fifth of a complete rotation is reasonable because the separation between Kampala and Singapore is approximately one-fifth of the earths circumference. 4.25. Solve: Let RE be the radius of the earth at the equator. This means RE + 300 m is the radius to the top of the tower. Letting T be the period of rotation, we have E E 2 top bottom 2 ( 300 m) 2 2 (300 m) 600 m 2.18 10 m/s 24 h 24(3600) s v v R R T T p p p p- + - = - = = = 4.26. Solve: The plane must fly as fast as the earths surface moves, but in the opposite direction. That is, the plane must fly from east to west. The speed is 2 rad (6.4 103 km) 1680 km 1680 km 1 mile 1040 mph 24 h h h 1.609 km v r p =. = .. .. = = = . . 4.27. Model: The rider is assumed to be a particle. Solve: Since ar = v2 / r, we have 2 (98 m/s2 )(12 m) 34 m/s r v = a r = .v = Assess: 34 m/s 70 mph is a large yet understandable speed. 4.28. Model: The earth is a particle orbiting around the sun. Solve: (a) The magnitude of the earths velocity is displacement divided by time: 11 2 2 (1.5 10 m) 3.0 104 m/s 365 days 24 h 3600 s 1 day 1 h v r T p p = = = (b) Since v = r., the angular acceleration is 4 7 11 3.0 10 m/s 2.0 10 rad/s 1.5 10 m v r . - = = = (c) The centripetal acceleration is 2 4 2 3 2 11 (3.0 10 m/s) 6.0 10 m/s 1.5 10 m r a v r - = = = Assess: A tangential velocity of 3.0104 m/s or 30 km/s is large, but needed for the earth to go through a displacement of 2p (1.51011 m) 9.4108 km in 1 year. 4.29. Solve: The pebbles angular velocity . = (3.0 rev/s)(2p rad/rev) =18.9 rad/s. The speed of the pebble as it moves around a circle of radius r = 30 cm = 0.30 m is v =.r = (18.9 rad/s)(0.30 m) = 5.7 m/s The radial acceleration is 2 ( )2 2 5.7 m/s 108 m/s 0.30 m r a v r = = = 4.30. Model: The crankshaft is a rotating rigid body. Solve: The crankshaft at t = 0 s has an angular velocity of 250 rad/s. It gradually slows down to 50 rad/s in 2 s, maintains a constant angular velocity for 2 s until t = 4 s, and then speeds up to 200 rad/s from t = 4 s to t = 7 s. The angular acceleration (a ) graph is based on the fact that a is the slope of the .-versus-t graph. 4.31. Model: The turntable is a rotating rigid body. Solve: The angular velocity is the area under the a -versus-t graph: 0 d (x)dt area under the graph dt . a = .. = .a =. + a The values of . at selected values of time (t) are: t (s) . (rad/s) 0 0 0.5 (5 + 3.75)(0.5)/2 = 2.18 1.0 (5 + 2.5)(1)/2 = 3.75 1.5 (5 +1.25)(1.5)/2 = 4.68 2.0 (5 + 0)(2)/2 = 5.0 2.5 5.0 3.0 5.0 4.32. Model: The wheel is a rotating rigid body. Solve: (a) The angular acceleration (a ) is the slope of the .-versus-t graph. (b) The car is at rest at t = 0 s. It gradually speeds up for 4 s and then slows down for 4 s. The car is at rest from t = 8 s to t =12 s, and then speeds up again for 4 s. 4.33. Model: The angular velocity and angular acceleration graphs correspond to a rotating rigid body. Solve: (a) The a-versus-t graph has a positive slope of 5 rad/s2 from t = 0 s to t = 2 s and a negative slope of -5 rad/s2 from t = 2 s to t = 4 s. (b) The angular velocity is the area under the a -versus-t graph: 0 d (x)dt area under graph. dt . a = .. = .a =. + a 4.34. Model: Model the car as a particle in nonuniform circular motion. Visualize: Note that halfway around the curve, the tangent is 45 south of east. The perpendicular component of the acceleration is 45 north of east. Solve: The radial and tangential components of the acceleration are cos25 (3.0 m s2 )cos25 2.7 m s2 r a = a = = sin 25 (3.0 m s2 )sin 25 1.27 m s2 t a = a = = 4.35. Model: Model the child on the merry-go-round as a particle in nonuniform circular motion. Visualize: Solve: (a) The speed of the child is v0 = r. = (2.5 m)(1.57 rad/s) = 3.9 m/s. (b) The merry-go-round slows from 1.57 rad/s to 0 in 20 s. Thus 0 2 1 0 1 1 0 (2.5 m)(1.57 rad/s) 0.197 m/s 20 s t t a t a r r t . . = =. + . = - = - = During these 20 s, the wheel turns through angle 2 2 2 1 0 01 1 0 (1.57 rad/s) (20 s) 0.197 m/s (20 s) 15.6 rad 2 2(2.5 m) t t a t r . =. +. + = + - = In terms of revolutions, 1 . = (15.6 rad)(1 rev/2p rad) = 2.49 rev. 4.36. Model: Model the particle on the crankshaft as being in nonuniform circular motion. Visualize: Solve: (a) The initial angular velocity is 0 2500 rpm . = (1 min/60 s) (2p rad/rev) = 261.8 rad/s. The crankshaft slows from 261.8 rad/s to 0 in 1.5 s. Thus 0 2 2 1 0 1 1 0 (0.015 m)(261.8 rad/s) 2.618 m/s 2.6 m/s 1.5 s t t a t a r r t . . = =. + . = - = - = = - (b) During these 1.5 s, the crankshaft turns through angle 2 2 2 1 0 01 1 0 (261.8 rad/s) (1.5 s) 2.618 m/s (1.5 s) 196 rad 2 2(0.015 m) t t a t r . =. +. + = + - = In terms of revolutions, 1 . = (196 rad)(1 rev/p rad) = 31.2 rev. 4.37. Model: The fan is in nonuniform circular motion. Visualize: Solve: Note 1800 rev/min min 30 rev/s. 60 s . . = . . . . Thus 30 rev/s 0 rev/s (4.0 s) =7.5 rev/s2. f i . =. +a.t . = +a .a This can be expressed as (7.5 rev/s) 2 rad 47 rad/s2. rev . p . . . = . . Assess: An increase in the angular velocity of a fan blade by 7.5 rev/s each second seems reasonable. 4.38. Model: The wheel is in nonuniform circular motion. Visualize: Solve: (a) Express i . in rad/s: ( ) i 50 rev/min min 2 rad 5.2 rad/s 60 s rev p . = .. .... ... . .. . After 10 s, ( 2 )( )2 f i f . =. +a.t .. = 5.2 rad/s + 0.50 rad/s 10 s = 55 rad/s. Converting to rpm, (55 rad/s) 60 s rev 53 rpm min 2p rad . .. .= . .. . . .. . (b) In 10 s, the wheel has turned a number of radians 1 2 1 2 2 f i 0 2 f 2 . =. +. .t + a.t .. = 0 rad/sec + (5.2 rad/s)(10 s) + (0.50 rad/s )(10 s) = 77 radians. Converting, 77 rad =12.3 revolutions. Assess: Making a bicycle wheel turn just over 12 revolutions in 10 s when it is initially turning almost one revolution per second to begin with seems attainable by a cyclist. 4.39. Model: We will assume that constant-acceleration kinematic equations in a plane apply. Visualize: Solve: (a) The particles position r..0 = (9.0 j) m implies that at 0 t the particles coordinates are 0 x = 0 m and 0 y = 9.0 m. The particles position 1 r.. = (20i) m at time 1 t implies that 1 x = 20 m and 1 y = 0 m. This is the position where the wire hoop is located. Let us find the time 1 t when the particle crosses the hoop at 1 x = 20 m. From the x v -versus-t curve and using the relation 1 0 x = x + area of the - x v t graph, we get 20 m = 0 m + area of the - graph x v t = area of the - x v t graph From Figure P4.39 we see that the area of the - x v t graph equals 20 m when 1 t = t = 3 s. (b) We can now look at the y-motion to find . y a Note that the slope of the - x v t graph (that is, ) y a is negative and constant, and we can determine y a by substituting into 1 2 3 0 0 3 0 2 3 0 ( ) ( ): y y y = y + v t - t + a t - t 1 2 2 2 0 m 9 m 0 m (3 s 0 s) 2 m/s y y = + + a - .a = - Therefore, 2 4 0 4 0 ( ) 0 m/s ( 2 m/s )(4 s 0 s) 8 m/s y y y v = v + a t - t = + - - = - (c) 4.40. Solve: From the expression for R, 2 Rmax = v0 /g. Therefore, 2 max 0 sin 2 sin 2 1 15 and 75 2 2 R R v g . = = . . = .. = Assess: The discussion of Figure 4.22 explains why launch angles . and (90 -. ) give the same range. 4.41. Model: Assume particle motion in a plane and constant-acceleration kinematics for the projectile. Visualize: Solve: (a) We know that v0 y = v0 sin. , , ya = -g and 1 0 y v = m/s. Using 2 1y v = 2 ( ) 0 1 0 2 , y y v + a y - y ( ) 2 2 2 2 2 2 0 0 0 m /s sin 2 sin 2 v v g h hg . = . + - . = (b) Using Equation 4.19 and the above expression for . = 30.0: ( ) ( ) 2 2 2 33.6 m/s sin 30.0 14.4 m 2 9.8 m/s h = = ( ) 2 0 2 0 x x v sin 2 g . - = ( ) ( ) ( ) 2 2 33.6 m/s sin 2 30.0 99.8 m 9.8 m/s = = For . = 45.0: ( ) ( ) 2 2 2 33.6 m/s sin 45.0 28.8 m 2 9.8 m/s h = = ( ) ( ) ( ) ( ) 2 2 0 2 33.6 m/s sin 2 45.0 115.2 m 9.8 m/s x x - = = For . = 60.0: ( ) ( ) 2 2 2 33.6 m/s sin 60.0 43.2 m 2 9.8 m/s h = = ( ) ( ) ( ) ( ) 2 2 0 2 33.6 m/s sin 2 60.0 99.8 m 2 9.8 m/s x x - = = Assess: The projectiles range, being proportional to sin(2. ), is maximum at a launch angle of 45, but the maximum height reached is proportional to sin2 (. ). These dependencies are seen in this problem. 4.42. Model: Assume the particle model for the projectile and motion in a plane. Visualize: Solve: (a) Using ( ) 1 ( )2 2 0 0 2 0 2 2 0 , y y y = y + v t - t + a t - t ( ) ( ) 1 ( 2 )( )2 2 2 y = 0 m+ 30 m/s sin60 7.5 s - 0 s + -9.8 m/s 7.5 s - 0 s = -80.8 m Thus the launch point is 81 m higher than where the projectile hits the ground. (b) Using 2 2 ( ) 1 0 1 0 2 , y y y v = v + a y - y 2 2 ( )2 ( 2 )( ) 1 1 0 m /s = 30sin 60 m/s + 2 -9.8 m/s y - 0 m . y = 34.4 m , or 1 y = 34 m (c) The x-component is ( ) 2 0 0cos60 30 m/s cos60 15 m/s. x x v = v = v = = The y-component is ( ) ( ) 2 0 2 0 0 2 1 sin 60 y y y v = v + a t - t = v - g t - t = (30 m/s)sin60 - (9.8 m/s2 )(7.5 s - 0 s) = -47.52 m/s ( )2 ( )2 .v = 15 m/s + -47.52 m/s = 50 m/s 1 2 1 2 tan tan 47.52 73 15 y x v v . = - = - = below +x Assess: An angle of 73 made with the ground, as the projectile hits the ground 81 m below its launch point, is reasonable in view of the fact that the projectile was launched at an angle of 60. 4.43. Model: Assume the particle model and motion under constant-acceleration kinematic equations in a plane. Visualize: Solve: (a) Using ( ) 1 ( )2 1 0 0 1 0 2 1 0 , y y y = y + v t - t + a t - t ( ) ( )( ) ( ) ( ) 1 2 2 0 1 2 1 2 2 1 1 1 0 m 1.80 m sin 40 0 s 9.8 m/s 0 s 1.80 m 7.713 m/s 4.9 m/s 0.206 s and 1.780 s v t t t t t = + - + - - = + - . =- The negative value of 1 t is unphysical for the current situation. Using 1 t =1.780 s and ( ) 1 0 0x 1 0 x = x + v t - t , we get ( )( ) 1 0 x = 0 + v cos40 m/s 1.780 s - 0 s = (12.0 m/s)cos40(1.78 s) =16.36 m (b) We can repeat the calculation for each angle. A general result for the flight time at angle . is ( 2 ) 1 t = 12sin. + 144sin . + 35.28 /9.8 s and the distance traveled is 1 1 x = (12.0)cos. t . We can put the results in a table. . 1 t 1 x 40.0 1.780 s 16.36 m 42.5 1.853 s 16.39 m 45.0 1.923 s 16.31 m 47.5 1.990 s 16.13 m Maximum distance is achieved at . 42.5. Assess: The well-known fact that maximum distance is achieved at 45 is true only when the projectile is launched and lands at the same height. That isnt true here. The extra 0.03 m = 3 cm obtained by increasing the angle from 40.0 to 42.5 could easily mean the difference between first and second place in a world-class meet. 4.44. Model: The golf ball is a particle following projectile motion. Visualize: (a) The distance traveled is x1 = v0xt1 = v0 cos. t1. The flight time is found from the y-equation, using the fact that the ball starts and ends at y = 0 : 1 2 1 0 1 0 0 1 2 1 0 2 1 1 1 y y 0 v sin t gt (v sin gt ) t t 2v sin g . - = = . - = . - . = Thus the distance traveled is 2 0 0 1 0 x v cos 2v sin 2v sin cos g g . . . = . = For . = 30, the distances are 2 2 0 1 earth 2 earth 2 2 2 0 0 0 1 moon 1 1 earth moon 6 earth earth ( ) 2 sin cos 2(25 m/s) sin30 cos30 55.2 m 9.80 m/s ( ) 2 sin cos 2 sin cos 6 2 sin cos 6( ) 331.2 m x v g x v v v x g g g . . . . . . . . = = = = = = = = The golf ball travels 331.2 m - 55.2 m = 276 m farther on the moon than on earth. (b) The flight times are 0 1 earth earth 0 0 1 moon 1 1 earth moon 6 earth ( ) 2 sin 2.55 s ( ) 2 sin 2 sin 6( ) 15.30 s t v g t v v t g g . . . = = = = = = The ball spends 15.30 s - 2.55 s =12.75 s longer in flight on the moon. 4.45. Model: The particle model for the ball and the constant-acceleration equations of motion are assumed. Visualize: Solve: (a) Using ( ) 1 ( )2 1 0 0 1 0 2 1 0 , y y y = y + v t - t + a t - t ( ) ( ) 1 ( 2 )( )2 2 h = 0 m+ 30 m/s sin60 4 s - 0 s + -9.8 m/s 4 s - 0 s = 25.5 m The height of the cliff is 26 m. (b) Using ( 2 ) 2 ( ) top top 0 2 , y y y v = v + a y - y ( ) ( )( ) ( ) ( ) ( ) 2 2 2 2 2 0 0 top top 2 sin 30 m/s sin60 0 m /s sin 2 34.4 m 2 2 9.8 m/s v v gy yg . . .. .. = + - . = = = The maximum height of the ball is 34 m. (c) The x and y components are ( ) ( ) ( 2 ) ( ) 1 0 1 0 0 1 sin 30 m/s sin 60 9.8 m/s 4.0 s y y y v = v + a t - t = v . - gt = - = -13.22 m/s ( ) 1 0 0 cos60 30 m/s cos60 15.0 m/s x y v = v = v = = 2 2 1 1 1 20.0 m/s x y .v = v + v = The impact speed is 20 m/s. Assess: Compared to a maximum height of 34.4 m, a height of 25.5 for the cliff is reasonable. 4.46. Model: The particle model for the ball and the constant-acceleration equations of motion in a plane are assumed. Visualize: Solve: The initial velocity is ( v0x = v0 cos5.0 = 20 m/s)cos5.0 =19.92 m/s ( ) 0 0sin5.0 20 m/s sin 5.0 1.743 m/s y v = v = = The time it takes for the ball to reach the net is ( ) 1 0 0x 1 0 x = x + v t - t ( )( ) 1 .7.0 m = 0 m+ 19.92 m/s t - 0 s .t = 0.351 s The vertical position at 1 t = 0.351 s is ( ) ( ) ( ) ( )( ) ( )( ) 1 2 1 0 0 1 0 2 1 0 1 2 2 2 2.0 m 1.743 m/s 0.351 s 0 s 9.8 m/s 0.351 s 0 s 2.01 m y y y = y + v t - t + a t - t = + - + - - = Thus the ball clears the net by 1.01 m. Assess: The vertical free fall of the ball, with zero initial velocity, in 0.351 s is 0.6 m. The ball will clear by approximately 0.4 m if the ball is thrown horizontally. The initial launch angle of 5 provides some initial vertical velocity and the ball clears by a larger distance. The above result is reasonable. 4.47. Model: The particle model for the ball and the constant-acceleration equations of motion in a plane are assumed. Visualize: Solve: (a) The time for the ball to fall is calculated as follows: ( ) 1 ( )2 1 0 0 y 1 0 2 y 1 0 y = y + v t - t + a t - t 1 ( 2 )( )2 2 1 1 .0 m = 4 m+ 0 m+ -9.8 m/s t - 0 s .t = 0.9035 s Using this result for the horizontal velocity: ( ) 1 0 0x 1 0 x = x + v t - t . ( ) 0 0 25 m 0 m v 0.9035 s 0 s 27.7 m/s x x= + - .v = The friends pitching speed is 28 m/s. (b) We have0 0sin , y v = v . where we will use the plus sign for up 5 and the minus sign for down 5. We can write ( ) ( )2 1 0 0 1 0 1 0 sin 2 y = y v . t - t - g t - t 2 0 1 1 0 m 4 m sin 2 . = v . t - g t Let us first find 1 t from ( ) 1 0 0 1 0 : x x = x + v t - t 0 1 25 m = 0 m+ v cos. t . 1 0 25 m cos t v . = Now substituting 1 t into the y-equation above yields 2 0 0 0 0 m 4 m sin 25 m 25 m cos 2 cos v g v v . . . . . . . = . . - . . . . . . ( ) ( ) 2 2 0 2 25 m 1 22.3 m/s and 44.2 m/s 2cos 4 m 25 m tan g v . . .. .. . = . . = .. .. The range of speeds is 22 m/s to 44 m/s, which is the same as 50 mph to 92 mph. Assess: These are reasonable speeds for baseball pitchers. 4.48. Model: We will use the particle model and the constant-acceleration kinematic equations in a plane. Visualize: Solve: The x- and y-equations of the ball are ( ) ( ) 1 ( ) ( )2 1B 0B 0B x 1B 0B 2 B x 1B 0B x = x + v t - t + a t - t ( ) 0B 1B .65 m = 0 m+ v cos30 t + 0 m ( ) ( ) 1 ( ) ( )2 1B 0B 0B y 1B 0B 2 B y 1B 0B y = y + v t - t + a t - t ( ) 1 ( ) 2 0B 1B 2 1B .0 m = 0 m+ v sin30 t + -g t From the y-equation, ( ) 1B 0B 2sin30 v = gt Substituting this into the x-equation yields 2 1B 1B
65 m cos30 2sin30 2.77 s g t t = . = For the runner: 1R 20 m 2.50 s 8.0 m/s t = = Thus, the throw is too late by 0.27 s. Assess: The times involved in running the bases are small, and a time of 2.5 s is reasonable. 4.49. Model: Use the particle model for the ball and the constant-acceleration kinematic equations. Visualize: Solve: (a) The distance from the ground to the peak of the house is 6.0 m. From the throw position this distance is 5.0 m. Using the kinematic equation 2 2 ( ) 1 0 1 0 2 , y y y v = v + a y - y 2 2 2 ( 2 )( ) 0 0 m /s 2 9.8 m/s 5.0 m 0 m y = v + - - . 0 9.899 m/s y v = The time for up and down motion is calculated as follows: ( ) 1 ( )2 2 0 0 y 2 0 2 y 2 0 y = y + v t - t + a t - t ( ) 1 ( 2 ) 2 2 2 2 .0 m = 0 m+ 9.899 m/s t - 9.8 m/s t 2 .t = 0 s and 2.02 s The zero solution is not of interest. Having found the time 2 t = 2.02 s, we can now find the horizontal velocity needed to cover a displacement of 18.0 m: ( ) 2 0 0x 2 0 x = x + v t - t ( ) 0 18.0 m 0 m 2.02 s 0 s x . = + v - 0 8.911 m/s x .v = ( )2 ( )2 0 .v = 8.911 m/s + 9.899 m/s = 13.3 m/s (b) The direction of 0 v.. is given by 1 0 1 0 tan tan 9.899 48 8.911 y x v v . = - = - = Assess: Since the maximum range corresponds to an angle of 45, the value of 48 corresponding to a range of 18 m and at a modest speed of 13.3 m/s is reasonable. 4.50. Model: We will assume a particle model for the sand, and use the constant-acceleration kinematic equations. Visualize: Solve: Using the equation ( ) 1 ( )2 1 0 0 1 0 2 1 0 , x x x = x + v t - t + a t - t ( )( ) 1 0 1 x = 0 m+ v cos15 t - 0 s + 0 m ( ) 1 = 60 m/s (cos15)t We can find 1 t from the y-equation, but note that 0 0sin15 y v = -v because the sand is launched at an angle below horizontal. ( ) 1 ( )2 1 0 0 y 1 0 2 y 1 0 y = y + v t - t + a t - t ( ) 1 2 0 1 2 1 .0 m = 3.0 m- v sin15 t - gt ( ) 1 ( 2 ) 2 1 2 1 = 3.0 m - 6.0 m/s (sin15)t - 9.8 m/s t . 2 1 1 4.9t +1.55t - 3.0 = 0 . 1 t = 0.6399 s and - 0.956 s (unphysical) Substituting this value of 1 t in the x-equation gives the distance ( ) ( ) 1 d = x = 6.0 m/s cos15 0.6399 s = 3.71 m 4.51. Model: We will assume a particle model for the cannonball, and apply the constant-acceleration kinematic equations. Visualize: Solve: (a) The cannonball that was accidentally dropped can be used to find the height of the wall: ( ) ( ) 1 ( ) ( )2 1A 0A 0A y 1A 0A 2 A y 1A 0A y = y + v t - t + a t - t 1 2 0A 2 1A .0 m = y + 0 m- gt 1 ( 2 )( )2 0A 2 . y = 9.8 m/s 1.5 s =11.03 m For the cannonball that was shot: 0S 0S ( ) cos30 (50 m/s)cos30 43.30 m/s x v = v = = 0S 0s ( ) sin30 (50 m/s)sin30 25.0 m/s y v = v = = We can now find the time it takes the cannonball to hit the ground: 1 2 2S 0S 0S 2S 0S 2 S 2S 0S ( )( ) ( )( ) y y y = y + v t - t + a t - t 2 2 2S 2S 0 m (11.03 m) (25.0 m/s) (9.8 m/s ) 2 . = + t - t 2 2 2S 2S 2S .(4.9 m/s )t - (25.0 m/s)t - (11.03 m) = 0.t = 5.51 s There is also an unphysical root 2S t = -0.41 s. Using this time 2St , we can now find the horizontal distance from the wall as follows: 2s 0s 0s 2s 0s ( ) ( ) 0 m (43.30 m/s)(5.51 s) 239 m x x = x + v t - t = + = The cannonball hits the ground 2.4102 m from the castle wall. (b) At the top of the trajectory 1S y (v ) = 0 m/s. Using 2 2 1S 0S 1S 0S ( ) ( ) 2 ( ), y y v = v - g y - y 2 2 2 2 1s 1 0 m /s (25.0 m/s) 2(9.8 m/s )( 11.03 m) 42.9 m s = - y - . y = The maximum height above the ground is 43 m. Assess: In view of the fact that the cannonball has a speed of approximately 110 mph, a distance of 239 m for the cannonball to hit the ground is reasonable. 4.52. Model: We will use the particle model and the constant-acceleration kinematic equations for the car. Visualize: Solve: (a) The initial velocity is 0 0cos v x = v . = (20 m/s)cos20 =18.79 m/s 0 0sin (20 m/s)sin 20 6.840 m/s y v = v . = = Using 1 2 1 0 0 1 0 2 1 0 ( ) ( ), y y y = y + v t - t + a t - t 1 2 2 2 1 2 1 1 1 0 m = 30 m+ (6.840 m/s)(t - 0 s) + (-9.8 m/s )(t - 0 s) .4.9t - 6.840t - 30 = 0 The positive root to this equation is 1 t = 3.269 s. The negative root is physically unreasonable in the present case. Using 1 2 1 0 0 1 0 2 1 0 ( ) ( ), x x x = x + v t - t + a t - t we get 1 x = 0 m+ (18.79 m/s)(3.269 s - 0 s) + 0 = 61.4 m The car lands 61 m from the base of the cliff. (b) The components of the final velocity are 1 0 18.79 m/s x x v = v = and 2 1 0 1 0 ( ) 6.840 m/s (9.8 m/s )(3.269 s 0 s) 25.2 m/s y y y v = v + a t - t = - - = - .v = (18.79 m/s) + (-25.2 m/s)2 = 31.4 m/s The cars impact speed is 31 m/s. Assess: A car traveling at 45 mph and being driven off a 30-m high cliff will land at a distance of approximately 200 feet (61.4 m). This distance is reasonable. 4.53. Model: Use the particle model for the cat and apply the constant-acceleration kinematic equations. Visualize: Solve: The relative velocity of the cat from the mouses reference frame is 0 (4.0cos30 -1.5) m/s =1.964 m/s = v Thus, 0 0 1.964 m/s x v = v = and 0 0sin30 (4.0 m/s)sin30 2.0 m/s y v = v = = . The time for the cat to land on the floor is found as follows: 1 2 1 2 2 1 0 0 1 0 2 1 0 1 2 1 ( ) ( ) 0 m 0 m (2.0 m/s) (9.8 m/s) y y y = y + v t - t + a t - t . = + t + - t 1 .t = 0 s (trivial solution) and 0.408 s The horizontal distance covered in time 1 t is 1 2 1 0 0 1 0 2 1 0 ( ) ( ) 0 m (1.964 m/s)(0.408 s) 0.802 m x x x = x + v t - t + a t - t = + = That is, the cat should leap when he is 80 cm behind the mouse. 4.54. Model: Assume motion along the x-direction. Let the earth frame be S and a frame attached to the staple gun be S'. Frame S' moves relative to S with velocity x V . Solve: The velocity x v of the parts to be stapled in frame S is +3.0 m/s. Also 1.0 m/s. x V = - Using , x x x v = v' +V we get 3.0 m/s ( 1.0 m/s) 4.0 m/s x x x v' = v -V = + - - = That is, the velocity of the parts in the frame of the staple gun is 4.0 m/s. In this frame 10 staples are fired per second. That is, 1 10 4.0 m/s distance between two staples distance Distance 0.4 m time between two staples s = = . = Assess: Note that the staple gun is in frame S' and we had to find the velocity of the moving parts in this frame to solve the problem. 4.55. Model: If a frame S' is in motion with velocity V .. relative to another frame S and has a displacement R .. relative to S, the positions and velocities (r.. and v..) in S are related to the positions and velocities (r..' and v..') in S' as r = r' + R .. .. and v = v' +V. .. .. .. In the present case, ship A is frame S and ship B is frame S . ' Both ships have a common origin at t = 0 s. The position and velocity measurements are made in S and S' relative to their origins. Solve: (a) The velocity vectors of the two ships are: A v.. = (20 mph)[cos30i - sin30 j] = (17.32 mph)i - (10.0 mph) j B v.. = (25 mph)[cos20i + sin 20 j] = (23.49 mph)i + (8.55 mph) j Since r.. = v...t, A A r.. = v.. (2 h) = (34.64 miles)i - (20.0 miles) j B B r.. = v.. (2 h) = (46.98 miles)i + (17.10 miles) j As A B r = r + R, .. .. .. A B R = r - r = (-12.34 miles)i - (37.10 miles) j. R = 39.1 miles .. .. .. The distance between the ships two hours after they depart is 39 miles. (b) Because A B v = v +V, .. .. .. A B V = v - v = -(6.17 mph)i - (18.55 mph) j.V =19.5 mph .. .. .. The speed of ship A as seen by ship B is 19.5 mph. Assess: The value of the speed is reasonable. 4.56. Model: Let the earth frame be S and a frame attached to the water be S'. Frame S' moves relative to S with velocity V. We define the x-axis along the direction of east and the y-axis along the direction of north for both frames. Solve: (a) The kayakers speed of 3.0 m/s is relative to the water. Since hes being swept toward the east, he needs to point at angle . west of north. In frame S', the water frame, his velocity is v..' = (3.0 m/s, . west of north) = (-3.0sin. m/s)i + (3.0cos. m/s)j We can find his velocity in earth frame S from the Galilean transformation v = v' +V .. .. .. , with V = (2.0 m/s)i. .. Thus v.. = ((-3.0sin. + 2.0) m/s)i + (3.0cos. m/s)j In order to go straight north in the earth frame, the kayaker needs 0. x v = This will be true if sin 2.0 sin 1 2.0 41.8 3.0 3.0 . = . . = - .. .. = . . Thus he must paddle in a direction 42 west of north. (b) His northward speed is 3.0 cos(41.8 ) m/s 2.236 m/s. y v= = The time to cross is 100 m 44.7 s 2.236 m/s t= = The kayaker takes 45 s to cross. 4.57. Model: Let Mikes frame be S and Nancys frame be S'. Frame S' moves relative to S with velocity V. x is the horizontal direction and y is the vertical direction for motion. The frames S and S' coincide at t = 0 s. Solve: (a) According to the Galilean transformation of velocity v = v' +V. .. .. .. Mike throws the ball with velocity v.. = (22 m/s)cos63i + (22 m/s)sin63 j, and V = (30 m/s)i. .. Thus in Nancys frame v' = v -V .. .. .. = (22cos63 - 30)i m/s + (22sin63) j m/s = (-20.0i +19.6j) m/s tan 1 tan 1 19.6 m/s 44.4 20.0 m/s y x v v . - - ' = = = ' The direction of the angle is 44.4 above the -x' axis (in the second quadrant). (b) In Nancys frame S' the equation 1 2 0 0 2 0 ( ) ( ) x x x' = x' + v' t' - t' + a' t' - t' becomes x' = -(20.0 m/s)t' and the equation 1 2 0 0 2 0 ( ) ( ) y y y' = y' + v' t' - t' + a' t' - t' becomes 1 2 2 2 y' = 0 m + (19.6 m/s)t' - (9.8 m/s )t' = (19.6 m/s)t' - (4.9 m/s2 )t'2 4.58. Model: Let the earth frame be S and a frame attached to the sailboat be S'. Frame S' moves relative to S with velocity V = 8.0i mph. .. Solve: (a) From Equation 6.24, v = v' +V, .. .. .. where v..' = (12.0)cos45i mph + (12.0)sin 45 j mph is the velocity of the wind in S', or the apparent wind velocity. Thus, v.. = (12.0)cos45i mph + (12.0)sin 45 j mph + (8.0)i mph = (8.0 +12.0cos45)i mph + (12.0sin45) j mph = (16.48i + 8.48j) mph tan 1 8.48 mph 27.2 16.48 mph . = - = The wind speed is v = 18.5 mph and the direction is from 27.2 south of west. (b) In this case, v..' = -(12.0 mph)cos45i - (12.0 mph)sin 45 j and V = (8.0 mph)i. .. So, v = ..-(12.0 mph)cos45 + 8.0 mph.. i - (12.0 mph)sin 45 j .. = -0.485i - 8.485 j tan 1 8.485 mph 86.7 0.485 mph . = - = The speed is 8.5 mph and the direction is from 86.7 north of east or from 3.3 east of north. 4.59. Model: Let the ground frame be S and the car frame be S'. S' moves relative to S with a velocity V along the x-direction. Solve: The Galilean transformation of velocity is v = v' +V .. .. .. where v.. and v..' are the velocities of the raindrops in frames S and S'. While driving north, V = (25 m/s)i .. and R R v = -v cos. j - v sin. i. Thus, v' = v -V .. .. .. ( ) R R= -v sin. - 25 m/s i - v cos. j Since the observer in the car finds the raindrops making an angle of 38 with the vertical, we have R R sin 25 m/s tan38 cos v v . . + = While driving south, V = -(25 m/s)i, .. and R R v.. = -v cos. j - v sin. i. Thus, ( ) R Rv..' = -v sin. + 25 m/s i - v cos. j Since the observer in the car finds the raindrops falling vertically straight, we have R R sin 25 m/s tan0 0 cos v v . . - + = = R .v sin. = 25 m/s Substituting this value of sin R v . into the expression obtained for driving north yields: R 25 m/s 25 m/s tan38 v cos. + = R cos 50 m/s 64.0 m/s tan38 .v . = = Therefore, we have for the velocity of the raindrops: ( )2 ( )2 ( )2 ( )2 R R sin cos 25 m/s 64.0 m/s v v . . + = + . ( )2 2R v = 4721 m/s . R v = 68.7 m/s R R tan sin 25 m/s 21.3 cos 64 m/s v v . . . . = = . = The raindrops fall at 69 m/s while making an angle of 21 with the vertical. 4.60. Model: Let the ground be frame S and the wind be frame S'. S' moves relative to S. The ground frame S has the x-axis along the direction of east and the y-axis along the direction of north. Visualize: Solve: (a) The Galilean velocity transformation is v = v' +V , .. .. .. where V = (50 mph)cos30i + (50 mph)sin30 j .. v..' = (200 mph)cos. i - (200 mph)sin. j Thus, v = ..(50cos30 + 200cos. )i + (50sin30 - 200sin. ) j.. mph. .. Because v.. should have no j-component, 50sin30 - 200sin. = 0.. = 7.18 (b) The pilot must head 7.18 south of east. Substituting this value of . in the v.. equation gives v.. = (242)i mph, along the direction of east. At a speed of 242 mph, the trip takes 600 mi 2.48 h 242 mph t= = 4.61. Model: We will use the particle model for the test tube which is in nonuniform circular motion. Solve: (a) The radial acceleration is ( ) 2 2 0.1 m 4000 rev 1 min 2 rad 1.75 104m/s2 min 60 s 1 rev r a r p = . = .. .. = . . (b) An object falling 1 meter has a speed calculated as follows: 2 2 ( ) ( 2 )( ) 1 0 1 0 1 2 0 m 2 9.8 m/s 1.0 m 4.43 m/s y v = v + a y - y = + - - .v = When this object is stopped in 110-3 s upon hitting the floor, ( ) ( 3 ) 3 2 2 1 2 1 0 m s 4.43 m s 1 10 s 4.4 10 m/s y y y v = v + a t - t . = - + a - .a = This result is one-fourth of the above radial acceleration. Assess: The radial acceleration of the centrifuge is large, but it is also true that falling objects are subjected to large accelerations when they are stopped by hard surfaces. 4.62. Model: We will use the particle model for the astronaut undergoing nonuniform circular motion. Solve: (a) The initial conditions are .0 = 0 rad/s, 0 . = 0 rad, 0 t = 0 s, and r = 6.0 m. After 30 s, 1 1 rev 1 rev 2 rad 4.83 rad/s 1.3 s 1.3 s rev p . = = = Using these values at 1 t = 30 s, ( )( ) ( ) 1 0 1 0 1 0 t t . =. + a r t - t = + a r t ( )( ) 2 t 6.0 m 4.83 rad/s 1 0.97 m/s 30 s .a = .. .. = . . (b) The radial acceleration is ( )( ) ( ) 2 2 1 2 6.0 m 4.83 rad/s 14.3 9.8 m/s r a = r. = g = g Assess: The above acceleration is typical of what astronauts experience during liftoff. 4.63. Model: Model the car as a particle in nonuniform circular motion. Visualize: Note that the tangential acceleration stays the same at 1.0 m/s2. As the tangential velocity increases, the radial acceleration increases as well. After a time t1, as the car goes through an angle 1 0. -. , the total acceleration will increase to 2.0 m/s2. Our objective is to find this angle. Solve: Using ( ) 1 0 t 1 0 , v = v + a t - t we get ( 2 )( ) ( 2 ) 1 1 1 v = 0 m/s + 1.0 m/s t - 0 s = 1.0 m/s t ( ) ( ) 2 2 2 2 2 1 1 1 4 2 1.0 m/s m/s 120 m 120 r rv t t a r . = = = ( ) ( ) 2 2 2 2 2 2 2 1 4 total 1 2.0 m/s 1.0 m/s m/s 14.4 s 120 t r a a a t t . . . = = + = + . . . = . . We can now determine the angle 1 . using ( ) 1 ( )2 1 0 0 1 0 2 1 0 t t t a t t r . =. +. - + .. .. - . . ( ) ( ) ( ) 2 2 1 1.0 m/s 0 rad 0 rad 14.4 s 0.864 rad 49.5 2 120 m = + + = = The car will have traveled through an angle of 50. 4.64. Model: The earth is a rigid, rotating, and spherical body. Visualize: Solve: At a latitude of . degrees, the radius is r = Re cos . with 6 e R = 6400 km = 6.40010 m. (a) In Miami . = 26, and we have r = (6.400106 m)(cos 26) = 5.752106 m . The angular velocity of the earth is 2 2 7.272 10 5 rad/s T 24 3600 s p p . = = = - Thus, 6 5 student v = r. = (5.75210 m)(7.27210- rad/s) = 418 m/s. (b) In Fairbanks . = 65, so r = (6.400106 m)cos 65 = 2.705106 m and 6 5 student v = r. = (2.70510 m)(7.27210- rad/s) = 197 m/s. 4.65. Model: The satellite is a particle in uniform circular motion. Visualize: Solve: (a) The satellite makes one complete revolution in 24 h about the center of the earth. The radius of the motion of the satellite is r = 6.37106 m+ 3.58107 m = 4.22107 m The speed of the satellite is ( ) ( ) 3 distance traveled 2 3.07 10 m/s. time taken 24 h v r p = = = (b) The acceleration of the satellite is centripetal, with magnitude ( )3 2 2 2 7 3.07 10 m/s 0.223 m/s 4.22 10 m r a v r = = = Assess: The small centripetal acceleration makes sense when realized it is for an object traveling in a circle with radius 26,400 miles. 4.66. Model: The magnetic computer disk is a rigid rotating body. Visualize: Solve: Using the rotational kinematic equation.f =.i +a.t, we get 2 1 . = 0 rad + (600 rad/s )(0.5 s - 0 s) = 300 rad/s 2 2 . = (300 rad/s) + (0 rad/s )(1.0 s - 0.5 s) = 300 rad/s The speed of the painted dot 2 2 v = r. = (0.04 m)(300 rad/s) =12 m/s. The number of revolutions during the time interval 0 2 t to t is 2 2 2 1 0 0 1 0 0 1 0 2 2 1 1 2 1 1 2 1 ( ) 1 ( ) 0 rad 0 rad 1 (600 rad/s )(0.5 s 0 s) 75 rad 2 2 ( ) 1 ( ) 2 75 rad (300 rad/s)(1.0 s 0.5 s) 0 rad 225 rad (225 rad) 1 rev 35.8 rev 2 rad t t t t t t t t . . . a . . . a p = + - + - = + + - = = + - + - = + - + = = . . = . . . . 4.67. Model: The drill is a rigid rotating body. Visualize: The figure shows the drills motion from the top. Solve: (a) The kinematic equation .f =.i +a (tf - ti ) becomes, after using i . =2400 rpm=(2400)(2p )/60 = 251.3 rad/s, f i t - t = 2.5 s - 0 s = 2.5 s, and f w = 0 rad/s, 0 rad = 251.3 rad/s +a (2.5 s).a = -100 rad/s2 (b) Applying the kinematic equation for angular position yields: 2 f i i f i f i 2 2 2 ( ) 1 ( ) 2 0 rad (251.3 rad/s)(2.5 s 0 s) 1 ( 100 rad/s )(2.5 s 0 s) 2 3.2 10 rad 50 rev . =. +. t - t + a t - t = + - + - - = = 4.68. Model: The turbine is a rigid rotating body. Solve: The known values are .i = 3600 rpm = (3600)(2p )/60 =120p rad/s, i t = 0 s, f t =10 min = 600 s, f . = 0 rad/s, and i . = 0 rad Using the rotational kinematic equation f i f i . =. +a (t - t ), we get 0 rad = (120p rad/s) +a (600 s - 0 s). Thus, a = -0.628 rad/s2. Now, 2 f i i f i f i 2 2 4 ( ) 1 ( ) 2 0 rad (120 rad/s)(600 s 0 s) 1 ( 0.628 rad/s )(600 s 0 s) 2 113,100 rad 1.80 10 rev . . . t t a t t p = + - + - = + - + - - = = Assess: 18,000 revolutions during 10 minutes when the starting angular velocity is 3600 rpm is reasonable. 4.69. Visualize: Please refer to Figure P4.69. Solve: Since f i . =. + (area under a vs t curve), at t = 3 s, the angular velocity is ( )( ) ( ) 2 f 60 rpm 4.0 rad/s 2 s 1 s 60 rpm 4 rad/s 20 rev 60 s 2 rad 1 min 60 rpm 38 rpm = 98 rpm . p = + - = + . . . . . . = + 4.69. Visualize: Please refer to Figure P4.69. Solve: Since .f =.i + (area under a vs t curve), at t = 3 s, the angular velocity is ( )( ) ( ) 2 f 60 rpm 4.0 rad/s 2 s 1 s 60 rpm 4 rad/s 20 rev 60 s 2 rad 1 min 60 rpm 38 rpm = 98 rpm . p = + - = + . . . . . . = + 4.70. Model: The cars wheels are in nonuniform circular motion with constant angular acceleration. Solve: The problem must be solved in two steps: first, the angular acceleration a is found for the case of stopping in one revolution, then a is applied to the case of stopping with twice the initial angular velocity .i . To find a for the case of stopping in 1.0 revolutions, set .. =1.0 rev = 2p rad in 2 2 f i . =. + 2a.. . 2 i 0 m =. + 2a (2p rad). 2 i 4 rad . a p . . = -. . . . Now use the same relationship, with that value of a, but this time .. is unknown and the initial angular velocity has been doubled. ( ) 2 2 2 i 2 i i i 0 m 2 2 0 m 4 8 rad 4.0 rev 4 rad 2 rad . . . . . . . p p p . - . = + . .. . = - . .. = = . . Assess: Doubling the initial velocity quadruples the number of turns the wheels make before locking. This is within reason. 4.71. Model: The bicycle wheel undergoes nonuniform circular motion with constant angular acceleration. Visualize: Solve: First find the angular acceleration a , then use it to find . f . Using kinematics, 2 f i . =. +a.t .0 rpm =100 rpm+a (1.0 min).a = -100 (rpm)/min = -100 rev/min The minus sign indicates the wheel is slowing down. The total number of revolutions the wheel makes while stopping is 1 2 2 f 2 . = 0 rev + (100 rpm)(1.0 min) + (-100 rev/min )(1.0 min) = 50 rev Assess: A total of 50 revolutions in 60 s is on average less than one revolution per second, which is quite reasonable. 4.72. Model: Treat the rock as a particle in nonuniform circular motion with constant angular acceleration. Visualize: Solve: (a) The angular acceleration a is 2 1.00 m/s 1.667 rad/s2 0.600 m t a r a - = = =- The minus sign indicates the angular acceleration is clockwise, as shown in the figure above, opposite to the angular velocity. The initial angular velocity is i 3.00 m/s 5.00 rad/s. 0.600 m . = = Using angular kinematics, the angular velocity at some time t after braking starts is ( ) ( ) f i . t =. + 0a.t = 5.00 rad/s -1.667 rad/s t - 0 s At t =1.5 s, f . = 2.50 rad/s whilea = -1.667 rad/s. (b) The total acceleration of the rock is 2 2. t r a = a + a The value of 1.00 m/s2 t a = is given. The radial acceleration at a certain time t is 2 t . r a v r = Since , tv =.r we can use the expression for f . (t) to find ( ) t v t by multiplying through by r = 0.600 m: ( ) ( ) ( ) ( ) ( 2 ) f 5.00 rad/s 1.667 rad/s 3.00 m/s 1.00m/s t . t r = r - rt.v t = - t 2 3.00 m/s 1.00 m/s t t v - . = The value of t v for a = g is needed. Setting a = g and substituting 2 t , r a v r = ( ) ( ) ( ) 2 2 4 2 2 2 2 2 2 2 4 2 2 24 9.80 m/s 1.00 m/s 95.0 m /s 0.600 m 95.0 m /s 2.42 m/s t t t t t t g a v g a v r r v v . . = + . . . - = = - = . . . = . = Finally, ( ) 2 3.00 m/s 2.42 m/s 0.58 s. 1.00 m/s t - = = Assess: The time of 0.58 s occurs early during the braking. This is reasonable since an acceleration equal to that supplied by gravity is fairly strong. 4.73. Model: The string is wrapped around the spool in such a way that it does not pile up on itself, and unwinds without slipping. Visualize: Solve: Since the string unwinds without slipping, the angular distance the spool turns as the string is pulled 1.0 m is 2 1.0 m 33 radians. 3.0 10 m x r . - . . = = = The angular acceleration of the spool due to the pull on the string is 2 2 2 1.5 m/s 50 rad/s 3.0 10 m t a r a - = = = The angular velocity of the spool after pulling the string is found with kinematics. 2 2 2 2 2 ( 2 )( ) f i f 2 f 2 0 rad /s 2 50 rad/s 33 rad 57 rad/s . . a . . . = + . . = + . = Converting to revolutions per minute, (57 rad/s) rev 60 s 5.5 102 rpm 2p rad min . .. . = . .. . . .. . Assess: The angular speed of 57 rad/s 9 rev/s is reasonable for a medium-sized spool. 4.74. Solve: (a) A golfer hits an iron shot with a new club as she approaches the green. She is pretty sure, based on past experience, that she hit the ball with a speed of 50 m/s, but she is not sure at what angle the golf ball took flight. She observed that the ball traveled 100 m before hitting the ground. What angle did she hit the ball? (b) From the second equation, 2 2 (4.9 m/s )t1 - (50sin. m/s) t1 = 0 1 1 2 0 s and (50 m/s)sin 4.9 m/s t t . . = = Using the above value for 1 t in the first equation yields: 2 2 2 100 m (50cos )(50sin ) m /s 4.9 m/s . . = 2cos sin sin 2 9.8 0.392 25 . . . = . = = .2. = 23.1.. =11.5 Assess: Although the original speed is reasonably high (50 m/s =112 mph), the ball travels a distance of only 100 m, implying either a small launch angle around 10 or an angle closer to 80. The calculated angle of 11.5 is thus pretty reasonable. 4.75. Solve: (a) A submarine moving east at 3.0 m/s sees an enemy ship 100 m north of its path. The submarines torpedo tube happens to be stuck in a position pointing 45 west of north. The tube fires a torpedo with a speed of 6.0 m/s relative to the submarine. How far east or west of the ship should the sub be when it fires? (b) Relative to the water, the torpedo will have velocity components 6.0cos45 m/s 3.0 m/s 4.24 m/s 3 m/s 1.24 m/s x v = - + = - + = - 6.0cos45 m/s 4.2 m/s y v = + = + The time to travel north to the ship is 100 m = (4.2 m/s) t1 1 .t = 24 s Thus, x = (1.24 m/s)(24 s) = -30 m. That is, the ship should be 30 m west of the submarine. 4.76. Solve: (a) A 1000 kg race car enters a 50 m radius curve and accelerates around the curve for 10.0 s. The forward force provided by the cars wheels is 1500 N. After 10.0 s the car has moved 125 m around the track. Find the initial and final angular velocities. (b) From Newtons second law, 1500 N (1000 kg) 1.5 m/s2 t t t t F = ma . = a .a = 125 m 2.5 rad 50 m s r . . . = = = ( ) ( )( ) 2 2 2 f i i i 2.5 rad 0 rad 10 s 1.5 m/s 10 s 0.10 rad/s 2 250 m t i t a t r . =. +. + . = +. + .. = ( ) 2 f i 0.1 rad/s 1.5 m/s 10 s 0.40 rad/s 50 m t a t r . =. + = + = 4.77. Solve: You decide to test fly your model airplane off of a 125 m tall building. The models engine starts fine and gets the airplane moving at 4.0 m/s but quits just as it gets to the edge of the building. The model proceeds to fall like a rock. How far from the edge of the building will it crash into the ground? (Assume g =10 m/s2 for easier calculation.) Visualize: Using the equation 1 ( )2 1 0 0 2 0 1 , y y y = y +. t + a t - t we get 2 2 1 1 y = -(5 m/s ) t = -125 m 2 1 2 125m 25 s 5 s 5m/s .t = = = The distance ( )( ) 1 x = 4 m/s 5 s = 20 m. 4.78. Model: The ions are particles that move in a plane. They have vertical acceleration while between the acceleration plates, and they move with constant velocity from the plates to the tumor. The flight time will be so small, because of the large speeds, that well ignore any deflection due to gravity. Visualize: Solve: Theres never a horizontal acceleration, so the horizontal motion is constant velocity motion at 5.0 106 m/s. x v = The times to pass between the 5.0-cm-long acceleration plates and from the plates to the tumor are 8 1 0 1 6 7 2 1 6 0.050 m 1.00 10 s 5.0 10 m/s 1.50 m 3.00 10 s 5.0 10 m/s t t t t t - - - = = = - = = Upon leaving the acceleration plates, the ion has been deflected sideways to position 1 y and has velocity 1 . y v These are 1 2 1 2 1 0 0 1 2 1 2 1 1 0 1 1 y y y y y y y y y v t a t a t v v a t a t = + + = = + = In traveling from the plates to the tumor, with no vertical acceleration, the ion reaches position 1 2 (1 2 ) 2 1 1 2 1 2 1 1 2 1 2 1 1 2 1 ( ) ( )( ) ( ) y y y y y = y + v t - t = a t + a t t - t = t + t t - t a We know 2 y = 2.0 cm = 0.020 m, so we can solve for the acceleration y a that the ion had while between the plates: 2 12 2 1 2 1 8 2 8 7 2 1 1 2 1 2 0.020 m 6.6 10 m/s ( ) (1.00 10 s) (1.00 10 s)(3.00 10 s) y a y t t t t - - - = = = + - + Assess: This acceleration is roughly 1012 times larger than the acceleration due to gravity. This justifies our assumption that the acceleration due to gravity can be neglected. 4.79. Model: We will use the particle model for the balls motion under constant-acceleration kinematic equations. Note that the balls motion on the smooth, flat board is ay = -g sin 20 = -3.352 m/s2. Visualize: Solve: The balls initial velocity is ( ) 0 0cos 3.0 m/s cos x v = v . = . ( ) 0 0sin 3.0 m/s sin y v = v . = . Using ( ) 1 ( )2 1 0 0 1 0 2 1 0 , x x x = x + v t - t + a t - t ( ) ( ) 1 2.5 m = 0 m + 3.0 m/s cos. t - 0 s + 0 m ( ) 1 ( ) 2.5 m 0.833 s 3.0 m/s cos cos t . . . = = Using ( ) 1 ( )2 1 0 0 y 1 0 2 y 1 0 y = y + v t - t + a t - t and the above equation for 1t , ( ) ( )( )2 1 2 2 2 0.833 s 0.833 s 0 m 0 m 3.0 m/s sin 3.352 m/s cos cos . . . = + . . - . . . . ( ) 2 2.5 m sin 1.164 cos cos . . . . = . 2.5sin. cos. =1.164 .2. = 68.6.. = 34.3 4.80. Model: Use the particle model for the arrow and the constant-acceleration kinematic equations. Visualize: Solve: Using ( ) 1 0 1 0 , v y = v y + ay t - t we get 1 1 1 1 0 m/s y y v = - gt .v = -gt Also using ( ) 1 ( )2 1 0 0 1 0 2 1 0 , x x x = x + v t - t + a t - t 0 1 60 m 0 m 0 m x = + v t + .0 1 1 60 m x x v v t = = Since 1 1/ tan3 0.0524, y x v v = - = - using the components of 0 v gives ( ) 1 1 0.0524 60 m/ gt t - = - ( )( ) 1 ( 2 ) 0.0524 60 m 0.566 s 9.8 m/s .t = = Having found 1t , we can go back to the x-equation to obtain 0 60 m/0.566 s 106 m/s. x v = = Assess: In view of the fact that the arrow took only 0.566 s to cover a horizontal distance of 60 m, a speed of 106 m/s or 237 mph for the arrow is understandable. 4.81. Model: Use the particle model for the arrow and the constant-acceleration kinematic equations. We will assume that the archer shoots from 1.75 m above the slope (about 5' 9''). Visualize: Solve: For the y-motion: ( ) 1 ( )2 1 0 0 y 1 0 2 y 1 0 y = y + v t - t + a t - t ( ) 1 2 1 0 1 2 1 . y =1.75 m + v sin 20 t - gt ( ) 1 2 1 1 2 1 . y =1.75 m + 50 m/s sin 20t - gt For the x-motion: ( ) 1 ( )2 1 0 0 x 1 0 2 x 1 0 x = x + v t - t + a t - t ( ) 0 1 = 0 m+ v cos20 t + 0 m ( ) 1 = 50 m/s (cos20)t Because 1 1 y x = -tan15 = -0.268, ( ) ( ) 1 2 1 2 1 1 1 1.75 m 50 m/s (sin 20 ) 0.268 6.12 s 50 m/s (cos20 ) t gt t t + - = - . = and -0.058 s (unphysical) Using 1 t = 6.12 s in the x- and y-equations above, we get 1 y = -77.0 m and 1 x = 287 m. This means the distance down the slope is 2 2 ( )2 ( )2 1 1 x + y = 287 m + -77.0 m = 297 m. Assess: With an initial speed of 112 mph (50 m/s) for the arrow, which is shot from a 15 slope at an angle of 20 above the horizontal, a horizontal distance of 287 m and a vertical distance of 77.0 m are reasonable numbers. 4.82. Model: Treat the ball as a particle and apply the constant-acceleration equations of kinematics. Visualize: Solve: After the first bounce, the ball leaves the surface at 40 relative to the vertical or 50 relative to the horizontal. We first calculate the time 1 t between the second bounce and the first bounce as follows: ( ) 1 ( )2 1 0 0x 1 0 2 x 1 0 x = x + v t - t + a t - t ( ) 0 1 .3.0 m = 0 m+ v cos50 t + 0 m 1 0 3.0 m cos50 t v . = In this time, the ball undergoes a vertical displacement of 1 0 y - y = -(3.0 m) tan 20 = -1.092 m. Substituting these values in the equation for the vertical displacement yields: ( ) 1 ( )2 1 0 0 y 1 0 2 y 1 0 y = y + v t - t + a t - t ( ) 1 2 0 1 2 1 -1.092 m = 0 m + v sin50 t - gt ( ) ( ) 2 1 2 0 2 0 0 sin50 3.0 m 9.8 m/s 3.0 m cos50 cos50 v v v . . . . = . . - . . . . . . 3 2 2 0 0 1.092 m 3.575 m 106.73 m /s v 4.78 m/s, v - . - - = . = or 0 v = 4.8 m/s Assess: A speed of 4.8 m/s or 10.7 mph on the first bounce is reasonable. 4.83. Model: Treat the skateboarder as a particle. Visualize: This is a two-part problem. Use an s-axis parallel to the slope for the first part, regular xycoordinates for the second. The skateboarders final velocity at the top of the ramp is her initial velocity as she becomes airborne. Solve: Without friction, the skateboarders acceleration on the ramp is 2 a0 = -g sin30 = -4.90 m/s . The length of the ramp is 1 s = (1.0 m) / sin30 = 2.0 m. We can use kinematics to find her speed at the top of the ramp: 2 2 2 1 0 0 1 0 0 01 2 2 1 2 ( ) 2 (7.0 m/s) 2( 4.90 m/s ) (2.0 m) 5.4 m/s v v a s s v as v = + - = + . = + - = This is the skateboarders initial speed into the air, giving her velocity components 1 1 cos30 4.7 m/s x v = v = and 1 1 cos30 2.7 m/s. y v = v = We can use the y-equation of projectile motion to find her time in the air: 1 2 2 2 2 1 1 2 2 1 2 2 2 0 m 1.0 m (2.7 m/s) (4.90 m/s ) y y y = = y + v t + a t = + t - t This quadratic equation has roots 2 t = -0.253 s (unphysical) and 2 t = 0.805 s. The x-equation of motion is thus 2 1 1 2 2 0 m (4.7 m/s) 3.8 m x x = x + v t = + t = She touches down 3.8 m from the end of the ramp. 4.84. Model: Use the particle model for the motorcycle daredevil and apply the kinematic equations of motion. Visualize: Solve: We need to find the coordinates of the landing ramp (x1, y1). We have ( ) ( ) 1 0 0 1 0 1 0 m 40 m/s x x = x + v t - t = + t ( ) 1 ( )2 1 0 0 y 1 0 2 y 1 0 y = y + v t - t + a t - t 1 ( 2 ) 2 2 1 = 0 m+ 0 m+ -9.8 m/s t ( 2 ) 2 1 = - 4.9 m/s t This means we must find 1t . Since 1 v.. makes an angle of 20 below the horizontal we can find 1y v as follows: 1 1 tan 20 y x v v = ( ) 1 1tan 20 40 m/s tan 20 14.56 m/s y x .v = -v = - = - We now use this value of 1y v , 9.8 m/s2 y a = - , and 0y v = 0 m/s in the following equation to obtain 1t : ( ) 1y 0 y y 1 0 v = v + a t - t . ( 2 ) 1 1 -14.56 m/s = 0 m/s - 9.8 m/s t .t =1.486 s Now, we are able to obtain 1 x and 1 y using the above x- and y-equations: ( )( ) 1 x = 40 m/s 1.486 s = 59.4 m ( 2 )( )2 1 y = - 4.9 m/s 1.486 s = -10.82 m That is, the landing ramp should be placed 10.8 m lower and 59 m away from the edge of the horizontal platform. 4.85. Model: The train and projectile are treated in the particle model. The height of the cannon above the tracks is ignored. Visualize: Solve: In the ground reference frame, the projectile is launched with velocity components ( 0 )P 0 train cos x v = v . + v ( ) 0 P 0 sin y v = v . While the projectile is in free fall, f i . y y v = v - g.t The time for the projectile to rise to the highest point is f (with 0) y v = iy . v t g . = So the time to vertically rise and fall is ( ) 0 P 0 1 2 2 2 sin y v v t t g g = . = = . During this time the projectile travels a horizontal distance ( ) ( ) 2 0 0train 1 P 0 P 1 2 sin cos 2 sin x x v t v v v g g = = . . + . During the same time, the train travels a horizontal distance ( ) ( ) 2 2 0 train 0 2 1 T 0 T 1 1 2 1 2 sin 2 sin 2 g x x v t at v v v a g = + = . + . The range R is the difference between the two horizontal distances: ( ) ( ) 2 0 2 1 P 1 T R x x 2v sin cos a sin g g . . . . . = - = . - . . . Note that the range is independent of train v the trains steady motion. This makes sense, since the train and projectile share that motion when the projectile is launched. Maximizing the range R requires dR 0. d. = Thus (ignoring the constant 2 0 2v g ) dR cos2 sin2 a (2sin cos ) cos2 a sin 2 0 d g g . . . . . . . = - - = - = Solving for . , sin 2 tan 2 1 tan 1 cos2 2 g g a a . . . . = = . = - .. .. . . Note that a > 0 (train speeding up) gives. < 45.. and a < 0 (train slowing down) gives . > 45.. since tan 1 g a - . . . . . . will be in the 2nd quadrant. Assess: As a check, see what the angle . is for the limiting case in which the train does not accelerate: 0 1 tan 1 ( ) 1 90 45 2 2 a = .. = - 8 = = This is the expected answer. 4.86. Model: Let the earth be frame S and the river be frame S'. Assume the river flows toward the east, which is the x and x'-axis. Visualize: Solve: In frame S , ' the rivers frame, the child is at rest. The boat can go directly to the child at angle . = tan-1(200/1500) = 7.595. The boats speed is 8.0 m/s, so the components of the boats velocity in S' are (8.0 m/s)cos7.595 7.93 m/s x v' = - = - (8.0 m/s)sin7.595 1.06 m/s y v'= = The river flows with velocity V 2.0i m/s . = relative to the earth. In the earths frame, which is also the frame of the riverbank and the boat dock, the boats velocity is 5.93 m/s x x x v = v' =V = - and 1.06 m/s y y y v = v' +V = Thus the boats angle with respect to the riverbank is . = tan-1(5.93/1.06) =10.1. Assess: The boat, like the child, is being swept downstream. This moves the boats angle away from the shore. 4.87. Model: The airplanes are modeled as particles, and are undergoing relative motion according to the Galilean transformations of position and velocity. We designate Uris plane as frame S' and the earth as frame S. Frame S' moves relative to frame S with velocity V. .. Visualize: Solve: According to the Galilean transformation of velocity v = v' +V, .. .. .. where v.. is the velocity of Vals plane relative to the earth, v..' is the velocity of Vals plane relative to Uris plane, and V .. is the velocity of Uris plane relative to the earth. We have v.. = -(500 mph)cos30i + (500 mph)sin30 j V = -(500 mph)cos20i - (500 mph)sin 20 j .. v' = v -V .. .. .. ( ) ( ) = 36.8 mph i + 421 mph j tan 1 421 85 36.8 . = - .. .. = . . The fuselage of Vals plane points 30 north of west. Val sees her plane moving in a direction 85 north of east. Thus the angle between the fuselage and the direction of motion is f =180 - 30 - 85 = 65 4.87. Model: The airplanes are modeled as particles, and are undergoing relative motion according to the Galilean transformations of position and velocity. We designate Uris plane as frame S' and the earth as frame S. Frame S' moves relative to frame S with velocity V. .. Visualize: Solve: According to the Galilean transformation of velocity v = v' +V, .. .. .. where v.. is the velocity of Vals plane relative to the earth, v..' is the velocity of Vals plane relative to Uris plane, and V .. is the velocity of Uris plane relative to the earth. We have v.. = -(500 mph)cos30i + (500 mph)sin30 j V = -(500 mph)cos20i - (500 mph)sin 20 j .. v' = v -V .. .. .. ( ) ( ) = 36.8 mph i + 421 mph j tan 1 421 85 36.8 . = - .. .. = . . The fuselage of Vals plane points 30 north of west. Val sees her plane moving in a direction 85 north of east. Thus the angle between the fuselage and the direction of motion is 4-1 4.88. Model: The ball is a particle launched into projectile motion by the wheel. Visualize: Solve: The initial velocity of the projectile is the tangential velocity at the point of release, and the direction is tangential to the wheel. The strategy is to find the initial velocity vi =.f r, then use f . to find the required angular acceleration a. Since the release point is 11 12 .. = of a complete circle, 1 (2 rad) 30 12 . = p = remains. In the coordinate system of the figure, i x = -20sin30cm =10.0 cm i y = 20cos30cm =17.3 cm f x =100 cm f y = 0 ya = -g i icos x v = v . i isin y v = v . 0 x a = The launch angle . is identified by calculating the angles in the right triangle highlighted in the figure and realizing the initial velocity is tangential to the circle. Thus the launch angle is also .. We will use Equations 4.17 for the position of a projectile as a function of time, and use the fact that the time to travel the horizontal distance to the cup is the same for the vertical motion. First, find the flight time required for the ball to hit the cup from the horizontal motion: ( ) i i i 100 cm 10.0 cm cos30 110 cm 127 cm cos30 v t t v v = - + . = = Substitute this time in the equations for the vertical motion: 2 i i i 0 cm 17.3 cm sin30 127 cm 1 127 cm 2 v g v v . . . . = + . . - . . . . . . Solving for the required initial velocity, ( ) ( ) 127 cm . 2 80.8 cm g v = The centimeter units cancel, and i v = 2.78 m/s. Now it is time to consider the angular motion. The required final angular velocity i f 2.78 m/s 13.9 rad/s 0.20 m v r . = = = Using 2 2 f i . =. + 2a.. , ( ) ( ) 2 2 2 f i 2 13.9 rad/s 16.8 rad/s 2 2 11 2 rad 12 . . a . p - = = = . . . . . . . Assess: An angular acceleration of 16.8 rad/s2 2.7 rev/s2 seems reasonable for a spring-loaded wheel. Kinematics in Two Dimensions 4-2 5.1. Visualize: Assess: Note that the climber does not touch the sides of the crevasse so there are no forces from the crevasse walls. 5.2. Visualize: 5.3. Visualize: 5.4. Model: Assume friction is negligible compared to other forces. Visualize: 5.5. Visualize: Assess: The bow and archer are no longer touching the arrow, so do not apply any forces after the arrow is released. 5.6. Model: An objects acceleration is linearly proportional to the net force. Solve: (a) One rubber band produces a force F, two rubber bands produce a force 2F, and so on. Because F . a and two rubber bands (force 2F) produce an acceleration of 1.2 m/s2, four rubber bands will produce an acceleration of 2.4 m/s2. (b) Now, we have two rubber bands (force 2F) pulling two glued objects (mass 2m). Using F = ma, 2F = (2m)a.a = F/m = 0.6 m/s2 5.7. Solve: Let the object have mass m and each rubber band exert a force F. For two rubber bands to accelerate the object with acceleration a, we must have a 2F . m = We will need N rubber bands to give acceleration 3a to a mass 1 . 2 m Find N: 1 2 3a NF 3 2F 2NF N 3. m m m = . . . = . = . . . . Three rubber bands are required. 5.8. Visualize: Please refer to Figure EX5.8. Solve: Mass is defined to be 1 slope of the acceleration-versus-force graph m = A larger slope implies a smaller mass. We know m2 = 0.20 kg, and we can find the other masses relative to 2 m by comparing their slopes. Thus 1 2 1 2 1/slope 1 slope 2 1 2 0.40 1/slope 2 slope 1 5 2 5 0.40 0.40 0.20 kg 0.08 kg m m m m = = = = = . = = = Similarly, 3 2 3 2 1/slope 3 slope 2 1 5 2.50 1/slope 2 slope 3 2 5 2 2.50 2.50 0.20 kg 0.50 kg m m m m = = = = = . = = = Assess: From the initial analysis of the slopes we had expected 3 2 m > m and 1 2m < m . This is consistent with our numerical answers. 5.9. Visualize: Please refer to Figure EX5.9. Solve: Mass is defined to be 1 slope of the acceleration-versus-force graph m = Thus ( ) ( ) 1 1 1 1 1 1 2 1 5 3 slope of line 1 9 3 55 slope of line 2 3 25 5 3 a m N m a N - - - - . . . . = = . . = = . . . . . . The ratio of masses is 1 2 9 25 m m = Assess: More rubber bands produce a smaller acceleration on object 2, so it should be more massive. 5.10 Solve: Use proportional reasoning. Given that distance traveled is proportional to the square of the time, d . t2 , so 2 d t should be constant. We have ( )2 ( )2 2.0 furlongs 2.0 s 4.0 s = x Thus the distance traveled by the object in 4.0 s is x = 8.0 furlongs. Assess: A longer time should result in a longer distance traveled. 5.11 Solve: Use proportional reasoning. Let T = period of the pendulum, L = length of pendulum. We are told T . L, so T L should be constant. We have 3.0 s 2.0 m 3.0 m = x Solving, the period of the 3.0 m long pendulum is x = 3.7 s. Assess: Increasing the length increases the period, as expected. 5.12. Force is not necessary for motion. Constant velocity motion occurs in the absence of forces, that is, when the net force on an object is zero. Thus, it is incorrect to say that force causes motion. Instead, force causes acceleration. That is, force causes a change in the motion of an object, and acceleration is the kinematic quantity that measures a change of motion. Newtons second law quantifies this idea by stating that the net force net F .. on an object of mass m causes the object to undergo an acceleration: a Fnet m = .. .. The acceleration vector and the net force vector must point in the same direction. 5.13. Visualize: Solve: (a) Newtons second law is F = ma. When F = 2 N, we have 2 N = (0.5 kg)a, hence a = 4 m/s2. (b) When F =1 N, we have 1 N = (0.5 kg)a, hence a = 2 m/s2. After repeating this procedure at various points, the above graph is obtained. 5.14 Solve: Newtons second law tells us that F = ma. Compute F for each case: (a) F = (0.200 kg)(5 m/s2 ) =1N. (b) F = (0.200 kg)(10 m/s2 ) = 2N. Assess: To double the acceleration we must double the force, as expected. 5.15. Visualize: Please refer to Figure EX5.15. Solve: Newtons second law is F = ma. We can read a force and an acceleration from the graph, and hence find the mass. Choosing the force F =1 N gives us a = 4 m/s2. Newtons second law yields m = 0.25 kg. 5.16. Solve: (a) This problem calls for an estimate so we are looking for an approximate answer. Table 5.3 gives us no information on laptops, but does give the weight of a one-pound object. Place a pound weight in one hand and the laptop on the other. The sensation on your hand is the weight of the object. The sensation from the laptop is about five times the sensation from the pound weight. So we conclude the weight of the laptop is about five times the weight of the one-pound object or about 25 N. (b) According to Table 5.3, the propulsion force on a car is 5000 N. A bicycle (including the rider) is about 100 kg. This is about one-tenth of the mass of a car, which is about 1000 kg for a compact model. The acceleration of a bicycle is somewhat less than that of a car, lets guess about one-fifth. We can write Newtons second law as follows: (bicycle) 1 (mass of car) 1 (acceleration of car) 5000 N 100 N 10 5 50 F = = = So we would roughly estimate the propulsion force of a bicycle to be 100 N. 5.17. Solve: (a) This problem calls for an estimate so we are looking for an approximate answer. Table 5.3 gives us no information on pencils, but does give us the weight of the U.S. quarter. Put the quarter on one hand and a pencil on the other hand. The sensation on your hand is the weight of the object. The sensation from the quarter is about the same as the sensation from the pencil. So they both have about the same weight. We can estimate the weight of the pencil to be 0.05 N. (b) According to Table 5.3, the propulsion force on a car is 5000 N. The mass of a sprinter is about 100 kg. This is about one-tenth of the mass of a car, which is about 1000 kg for a compact model. The acceleration of a sprinter is somewhat less than that of a car, lets guess about one-fifth. We can write Newtons second law as follows: (sprinter) 1 (mass of car) 1 (acceleration of car) 5000 N 100 N 10 5 50 F = = = So, we would roughly estimate the propulsion force of a sprinter to be 100 N. Assess: This is the same estimated number as we obtained in Exercise 5.16. This is reasonable since in both the cases the propulsion force comes from a human and it probably does not matter how the human is providing that force. 5.18. Visualize: Solve: The object will be in equilibrium if 3 F .. has the same magnitude as F1 + F2 .. .. but is in the opposite direction so that the sum of all the three forces is zero. 5.19. Visualize: Solve: The object will be in equilibrium if 3 F .. has the same magnitude as F1 + F2 .. .. but is in the opposite direction so that the sum of all three forces is zero. 5.20. Visualize: Solve: The object will be in equilibrium if 3 F .. has the same magnitude as F1 + F2 .. .. but is in the opposite direction so that the sum of all the three forces is zero. 5.21. Visualize: Solve: The free-body diagram shows two equal and opposite forces such that the net force is zero. The force directed down is labeled as a gravitational force, and the force directed up is labeled as a tension. With zero net force the acceleration is zero. So, a possible description is: An object hangs from a rope and is at rest. Or, An object hanging from a rope is moving up or down with a constant speed. 5.22. Visualize: Solve: The free-body diagram shows three forces with a net force (and therefore net acceleration) upward. There is a force labeled G F .. directed down, a force thrust F .. directed up, and a force D .. directed down. So a possible description is: A rocket accelerates upward. 5.23. Visualize: Solve: The free-body diagram shows three forces. There is a gravitational force FG, .. which is down. There is a normal force labeled n.., which is up. The forces G F .. and n .. are shown with vectors of the same length so they are equal in magnitude and the net vertical force is zero. So we have an object on the ground which is not moving vertically. There is also a force k f .. to the left. This must be a frictional force and we need to decide whether it is static or kinetic friction. The frictional force is the only horizontal force so the net horizontal force must be k f . .. This means there is a net force to the left producing an acceleration to the left. This all implies motion and therefore the frictional force is kinetic. A possible description is: A baseball player is sliding into second base. 5.24. Visualize: Assess: There is a gravitational force FG. .. You are touching the park bench, so it exerts a contact force n.. on you. 5.25. Visualize: Assess: The problem says that there is no friction and it tells you nothing about any drag; so we do not include either of these forces. The only remaining forces are the weight and the normal force. 5.26. Visualize: Assess: Since the velocity is constant, the acceleration is zero, and the net force is zero. 5.27. Visualize: Assess: The problem uses the word sliding. Any real situation involves friction with the surface. Since we are not told to neglect it, we show that force. 5.28. Visualize: Figure (a) shows velocity as downward, so the object is moving down. The length of the vector increases with each step showing that the speed is increasing (like a dropped ball). Thus, the acceleration is directed down. Since F = ma .. .. the force is in the same direction as the acceleration and must be directed down. Figure (b), however, shows the velocity as upward, so the object is moving upward. But the length of the vector decreases with each step showing that the speed is decreasing (like a ball thrown up). Thus, the acceleration is also directed down. As in part (a) the net force must be directed down. 5.29. Visualize: The velocity vector in figure (a) is shown downward and to the left. So movement is downward and to the left. The velocity vectors get successively longer, which means the speed is increasing. Therefore the acceleration is downward and to the left. By Newtons second law F = ma, .. .. the net force must be in the same direction as the acceleration. Thus, the net force is downward and to the left. The velocity vector in (b) is shown to be upward and to the right. So movement is upward and to the right. The velocity vector gets successively shorter, which means the speed is decreasing. Therefore the acceleration is downward and to the left. From Newtons second law, the net force must be in the direction of the acceleration and so it is directed downward and to the left. 5.30. Visualize: Solve: According to Newtons second law F = ma, the force at any time is found simply by multiplying the value of the acceleration by the mass of the object. 5.31. Visualize: Solve: According to Newtons second law F = ma, the force at any time is found simply by multiplying the value of the acceleration by the mass of the object. 5.32. Visualize: Solve: According to Newtons second law F = ma, the acceleration at any time is found simply by dividing the value of the force by the mass of the object. 5.33. Visualize: Solve: According to Newtons second law F = ma, the acceleration at any time is found simply by dividing the value of the force by the mass of the object. 5.34. Model: Use the particle model for the object. Solve: (a) We are told that for an unknown force (call it F0 ) acting on an unknown mass 0 (call it m ) the acceleration of the mass is 10 m/s2. According to Newtons second law, 2 0 0F = m (10 m/s ). The force then becomes 1 2 0F . Newtons second law gives ( 2 ) 0 0 0 1 1 10 m s 2 2 F = m a = ..m .. This means a is 5 m/s2. (b) The force is F0 and the mass is now 1 2 0m . Newtons second law gives ( 2 ) 0 0 0 1 10 m s 2 F = m a = m This means a = 20 m/s2. (c) A similar procedure gives a =10 m/s2. (d) A similar procedure gives a = 2.5 m/s2. 5.35. Model: Use the particle model for the object. Solve: (a) We are told that for an unknown force (call it F0 ) acting on an unknown mass 0 (call it m ) the acceleration of the mass is 8 m/s2. According to Newtons second law, 2 0 0F = m (8 m/s ). The force then becomes 0 2F . Newtons second law gives ( 2 ) 0 0 0 2F = m a = 2..m 8 m s .. This means a is 16 m/s2. (b) The force is 0 F and the mass is now 0 2m . Newtons second law gives ( 2 ) 0 0 0 F = 2m a = m 8 m s This means a = 4 m/s2. (c) A similar procedure gives a = 8 m/s2. (d) A similar procedure gives a = 32 m/s2. 5.36. Visualize: Solve: (d) There are a normal force and a gravitational force which are equal and opposite, so this is an object on a horizontal surface. The description could be: A tow truck pulls a stuck car out of the mud. 5.37. Visualize: Solve: (d) There is a normal force and a gravitational force which are equal and opposite, so this is an object on a horizontal surface, or at least balanced in the vertical direction. The description of this free-body diagram could be: A jet plane is flying at constant speed. 5.38. Visualize: Solve: (d) This is an object on a surface because FG = n. It must be moving to the left because the kinetic friction is to the right. The description of the free-body diagram could be: A compressed spring is shooting a plastic block to the left. 5.39. Visualize: Solve: (d) There is only a single force of weight. We are unable to tell the direction of motion. The description could be: Galileo has dropped a ball from the Leaning Tower of Pisa. 5.40. Visualize: Solve: (d) There is an object on an inclined surface. The net force is down the plane so the acceleration is down the plane. The net force includes both the frictional force and the component of the gravitational force. The direction of the force of kinetic friction implies that the object is moving upward. The description could be: A car is skidding up an embankment. 5.41. Visualize: Solve: (d) There is an object on an inclined surface with a tension force down the surface. There is a small frictional force up the surface implying that the object is sliding down the slope. A description could be: A sled is being pulled down a slope with a rope that is parallel to the slope. 5.42. Visualize: Solve: (d) There is a thrust at an angle to the horizontal and a gravitational force. There is no normal force so the object is not on a surface. The description could be: A rocket is fired at an angle to the horizontal and there is no drag force. 5.43. Visualize: Tension is the only contact force. The downward acceleration implies that FG > T. 5.44. Visualize: 5.45. Visualize: The normal force is perpendicular to the ground. The thrust force is parallel to the ground and in the direction of acceleration. The drag force is opposite to the direction of motion. 5.46. Visualize: The normal force is perpendicular to the hill. The frictional force is parallel to the hill. 5.47. Visualize: The normal force is perpendicular to the hill. The kinetic frictional force is parallel to the hill and directed upward opposite to the direction of motion. The wind force is given as horizontal. Since the skier stays on the slope (that is, there is no acceleration away from the slope) the net force must be parallel to the slope. 5.48. Visualize: As the rock slides there is kinetic friction between it and the rough concrete sidewalk. Since the rock stays on the level surface, the net force must be along that surface, and is equal to the kinetic friction. 5.49. Visualize: The drag force due to air is opposite the motion. 5.50. Visualize: The ball rests on the floor of the barrel because the gravitational force is equal to the normal force. There is a force of the spring to the right which causes an acceleration. 5.51. Visualize: There are no contact forces on the rock. The gravitational force is the only force acting on the rock. 5.52. Visualize: The gymnast experiences the long range force of gravity. There is also a contact force from the trampoline that is the normal force of the trampoline on the gymnast. The gymnast is moving downward and the trampoline is decreasing her speed, so the acceleration is upward and there is a net force upward. Thus the normal force must be larger than the gravitational force. The actual behavior of the normal force over time will be complicated as it involves the stretching of the trampoline and therefore tensions. 5.53. Visualize: You can see from the motion diagram that the box accelerates to the right along with the truck. According to Newtons second law, F = ma, .. .. there must be a force to the right acting on the box. This is friction, but not kinetic friction. The box is not sliding against the truck. Instead, it is static friction, the force that prevents slipping. Were it not for static friction, the box would slip off the back of the truck. Static friction acts in the direction needed to prevent slipping. In this case, friction must act in the forward (toward the right) direction. 5.54. Visualize: You can see from the motion diagram that the bag accelerates to the left along with the car as the car slows down. According to Newtons second law, F = ma, .. .. there must be a force to the left acting on the bag. This is friction, but not kinetic friction. The bag is not sliding across the seat. Instead, it is static friction, the force that prevents slipping. Were it not for static friction, the bag would slide off the seat as the car stops. Static friction acts in the direction needed to prevent slipping. In this case, friction must act in the backward (toward the left) direction. 5.55. Visualize: (a) (b) (c) (d) The ball accelerates downward until the instant when it makes contact with the ground. Once it makes contact, it begins to compress and to slow down. The compression takes a short but nonzero distance, as shown in the motion diagram. The point of maximum compression is the turning point, where the ball has an instantaneous speed of v = 0 m/s and reverses direction. The ball then expands and speeds up until it loses contact with the ground. The motion diagram shows that the acceleration vector a.. points upward the entire time that the ball is in contact with the ground. An upward acceleration implies that there is a net upward force net F .. on the ball. The only two forces on the ball are the gravitational force downward and the normal force of the ground upward. To have a net force upward requires Gn > F . So the ball bounces because the normal force of the ground exceeds the gravitational force, causing a net upward force during the entire time that the ball is in contact with the ground. This net upward force slows the ball, turns it, and accelerates it upward until it loses contact with the ground. Once contact with the ground is lost, the normal force vanishes and the ball is simply in free fall. 5-1 5.56. Visualize: (a) You are sitting on a bench driving along to the right. Both you and the bench are moving with a constant speed. There is a force on you due to gravity, which is directed down. There is a contact force between you and the bench, which is directed up. Since you are not accelerating up or down the net vertical force on you is zero, which means the two vertical forces are equal in magnitude. The statement of the problem gives no indication of any other contact forces. Specifically, we are told that the bench is very slippery. We can take this to mean there is no frictional force. So our force diagram includes only the normal force up, the gravitational force down, and no horizontal force. (b) The above considerations lead to the free-body diagram that is shown. (c) The car (and therefore the bench) slows down. Does this create any new force on you? No. The forces remain the same. This means the pictorial representation and the free-body diagram are unchanged. (d) The car slows down because of some new contact force on the car (maybe the brakes lock the wheels and the road exerts a force on the tires). But there is no new contact force on you. So the force diagram for you remains unchanged. There are no horizontal forces on you. You do not slow down and you continue at an unchanged velocity until something in the picture changes for you (for example, you fall off the bench or hit the windshield). (e) The net force on you has remained zero because the net vertical force is zero and there are no horizontal forces at all. According to Newtons first law if the net force on you is zero, then you continue to move in a straight line with a constant velocity. That is what happens to you when the car slows down. You continue to move forward with a constant velocity. The statement that you are thrown forward is misleading and incorrect. To be thrown there would need to be a net force on you and there is none. It might be correct to say that the car has been thrown backward leaving you to continue onward (until you part company with the bench). (f) We are now asked to consider what happens if the bench is NOT slippery. That implies there is a frictional force between the bench and you. This force is certainly horizontal (parallel to the surface of the bench). Is the frictional force directed forward (in the direction of motion) or backward? The car is slowing down and you are staying on the bench. That means you are slowing down with the bench. Your velocity to the right is decreasing (you are moving right and slowing down) so you are accelerating to the left. By Newtons second law that means the force producing the acceleration must be to the left. That force is the force of static friction and it is shown on the free-body diagram below. Of course, when the car accelerates (increases in speed to the right) and you accelerate with it, then your acceleration is to the right and the frictional force must be to the right. Force and Motion 5-2 6.1. Model: We can assume that the ring is a single massless particle in static equilibrium. Visualize: Solve: Written in component form, Newtons first law is ( net )x x 1x 2x 3x 0 N F = SF = T + T + T = ( ) net 1 2 3 0 N y y y y y F = SF = T + T + T = Evaluating the components of the force vectors from the free-body diagram: 1 1 2 0 N x x T = -T T = 3 3cos30 x T = T 1 0 N y T = 2 y 2 T = T 3 3sin30 y T = -T Using Newtons first law: 1 3-T + T cos30 = 0 N 2 3T -T sin30 = 0 N Rearranging: ( )( ) 1 3T = T cos30 = 100 N 0.8666 = 86.7 N ( )( ) 2 3T = T sin30 = 100 N 0.5 = 50.0 N Assess: Since 3 T .. acts closer to the x-axis than to the y-axis, it makes sense that 1 2T > T . 6.2. Model: We can assume that the ring is a particle. Visualize: This is a static equilibrium problem. We will ignore the weight of the ring, because it is very light, so the only three forces are the tension forces shown in the free-body diagram. Note that the diagram defines the angle . . Solve: Because the ring is in equilibrium it must obey net F = 0 N. .. This is a vector equation, so it has both xand y-components: ( ) net 3 2 cos 0 N x F = T . -T = 3 2 .T cos. = T ( ) net 1 3 3 1 sin 0 N sin y F = T -T . = .T . = T We have two equations in the two unknowns 3 T and . . Divide the y-equation by the x-equation: 3 1 1 ( ) 3 2 sin tan 80 N 1.6 tan 1.6 58 cos 50 N T T T T . . . . = = = = . = - = Now we can use the x-equation to find 2 3 50 N 94 N cos cos58 T T . = = = The tension in the third rope is 94 N directed 58 below the horizontal. 6.3. Model: We assume the speaker is a particle in static equilibrium under the influence of three forces: gravity and the tensions in the two cables. Visualize: Solve: From the lengths of the cables and the distance below the ceiling we can calculate . as follows: sin 2 m 0.677 sin 1 0.667 41.8 3 m . = = .. = - = Newtons first law for this situation is ( ) net 1 2 1 2 0 N cos cos 0 N x x x x F = SF = T + T = .-T . + T . = ( ) net 1 2 1 2 0 N sin sin 0 N y y y y y F = SF = T + T + w = .T . + T . - w = The x-component equation means 1 2T = T . From the y-component equation: 1 2T sin. = w ( )( 2 ) 1 20 kg 9.8 m/s 196 N 147 N 2sin 2sin 2sin 41.8 1.333 T w mg . . . = = = = = Assess: Its to be expected that the two tensions are equal, since the speaker is suspended symmetrically from the two cables. That the two tensions add to considerably more than the weight of the speaker reflects the relatively large angle of suspension. 6.4. Model: We can assume that the coach and his sled are a particle being towed at a constant velocity by the two ropes, with friction providing the force that resists the pullers. Visualize: Solve: Since the sled is not accelerating, it is in dynamic equilibrium and Newtons first law applies: ( net )x x 1x 2x kx 0 N F = SF = T + T + f = ( ) net 1 2 k 0 N y y y y y F = SF = T + T + f = From the free-body diagram: 1 2 k cos 1 cos 1 0 N 2 2 T .. . .. + T .. . .. - f = . . . . 1 2 sin 1 sin 1 0 N 0 N 2 2 T .. . .. -T .. . .. + = . . . . From the second of these equations 1 2T = T . Then from the first: 1 2T cos10 =1000 N 1 1000 N 1000 N 508 N 2cos10 1.970 .T = = = Assess: The two tensions are equal, as expected, since the two players are pulling at the same angle. The two add up to only slightly more than 1000 N, which makes sense because the angle at which the two players are pulling is small. 6.5. Visualize: Please refer to the Figure EX6.5. Solve: Applying Newtons second law to the diagram on the left, ( net ) 4 N 2 N 2 1.0 m/s 2 kg x x F a m - = = = ( ) net 3 N 3 N 0 m/s2 2 kg y y F a m - = = = For the diagram on the right: ( ) net 4 N 2 N 1.0 m/s2 2 kg x x F a m - = = = ( ) net 3 N 1 N 2 N 0 m/s2 2 kg y y F a m - - = = = 6.6. Visualize: Please refer to Figure EX6.6. Solve: For the diagram on the left, three of the vectors lie along the axes of the tilted coordinate system. Notice that the angle between the 3 N force and the y-axis is the same 20 by which the coordinates are tilted. Applying Newtons second law, ( ) ( ) net 2 5 N 1 N 3sin 20 N 1.49 m/s 2 kg x x F a m - - = = = ( net ) ( ) 2 2.82 N 3cos20 N 0 m/s 2 kg y y F a m - = = = For the diagram on the right, the 2-newton force in the first quadrant makes an angle of 15 with the positive xaxis. The other 2-newton force makes an angle of 15 with the negative y-axis. The accelerations are ( ) ( ) ( ) ( ) ( ) ( ) net 2 net 2 2cos15 N 2sin15 N 3 N 0.28 m/s 2 kg 1.414 N 2sin15 N 2cos15 N 0 m/s 2 kg x x y y F a m F a m + - = = =- + - = = = 6.7. Visualize: Please refer to Figure EX6.7. Solve: (a) Apply Newtons second law in both the x and y directions. ( net )x (5.0 N)cos37 2.0 N (5.0 kg) x F = - = a 0.40 m/s2 x .a = ( ) ( ) ( ) net 2.0 N 5.0 N sin37 5.0 N 5.0 kg y y F = + - = a 0.0 m/s2 y .a = (b) The angle that the 5.0 N force makes with the y-axis is 37. Apply Newtons second law for both the x and y direction. ( ) ( ) ( ) net 3.0 N 5.0 N sin37 2.0 N 5.0 kg x x F = + - = a 0.80 m/s2 x .a = ( ) ( ) ( ) net 4.0 N 5.0 N cos37 5.0 kg y y F = - = a 0.0 m/s2 y .a = Assess: The orientation of the coordinate axes is chosen for convenience, and does not always need to conform to the horizontal and vertical. 6.8. Visualize: Please refer to Figure EX6.8. Solve: We can use the constant slopes of the three segments of the graph to calculate the three accelerations. For t between 0 s and 3 s, 12 m/s 0 s 4 m/s2 3 s x x a v t . - = = = . For t between 3 s and 6 s, 0 m/s, x .v = so 0 m/s2. x a = For t between 6 s and 8 s, 0 m/s 12 m/s 6 m/s2 2 s x x a v t . - = = =- . From Newtons second law, at t =1 s we have 2 net (2.0 kg)(4 m/s ) 8 N x F = ma = = At 4 s, 0 m/s2 , x t = a = so net F = 0 N. At t = 7 s, 2 net (2.0 kg)( 6.0 m/s ) 12 N x F = ma = - = - Assess: The magnitudes of the forces look reasonable, given the small mass of the object. The positive and negative signs are appropriate for an object first speeding up, then slowing down. 6.9. Visualize: Please refer to Figure EX6.9. Positive forces result in the object gaining speed and negative forces result in the object slowing down. The final segment of zero force is a period of constant speed. Solve: We have the mass and net force for all the three segments. This means we can use Newtons second law to calculate the accelerations. The acceleration from t = 0 s to t = 3 s is 4 N 2 m/s2 2.0 kg x x a F m = = = The acceleration from t = 3 s to t = 5 s is 2 N 1 m/s2 2.0 kg x x a F m - = = =- The acceleration from t = 5 s to 8 s is 0 m/s2. x a = In particular, (at 6 s) 0 m/s2. xa t= = We can now use one-dimensional kinematics to calculate v at t = 6 s as follows: ( ) ( ) ( )( ) ( )( ) 0 1 1 0 2 2 0 0 2 m/s2 3 s 1 m/s2 2 s 6 m/s 2 m/s 4 m/s v = v + a t - t + a t - t = + + - = - = Assess: The positive final velocity makes sense, given the greater magnitude and longer duration of the positive 1F . .. A velocity of 4 m/s also seems reasonable, given the magnitudes and directions of the forces and the mass involved. 6.10. Model: We assume that the box is a particle being pulled in a straight line. Since the ice is frictionless, the tension in the rope is the only horizontal force. Visualize: Solve: (a) Since the box is at rest, 0 m/s2 , x a = and the net force on the box must be zero. Therefore, according to Newtons first law, the tension in the rope must be zero. (b) For this situation again, 0 m/s2 , x a = so net F = T = 0 N. (c) Here, the velocity of the box is irrelevant, since only a change in velocity requires a nonzero net force. Since 5.0 m/s2 , x a = ( )( 2 ) net 50 kg 5.0 m/s 250 N x F = T = ma = = Assess: For parts (a) and (b), the zero acceleration immediately implies that the rope is exerting no horizontal force on the box. For part (c), the 250 N force (the equivalent of about half the weight of a small person) seems reasonable to accelerate a box of this mass at 5.0 m/s2. 6.11. Model: We assume that the box is a point particle that is acted on only by the tension in the rope and the pull of gravity. Both the forces act along the same vertical line. Visualize: Solve: (a) Since the box is at rest, 0 m/s2 y a = and the net force on it must be zero: ( )( 2 ) net G G F = T - F = 0 N.T = F = mg = 50 kg 9.8 m/s = 490 N (b) Since the box is rising at a constant speed, again 0 m/s2 , y a = net F = 0 N, and G T = F = 490 N. (c) The velocity of the box is irrelevant, since only a change in velocity requires a nonzero net force. Since 5.0 m/s2 , y a = ( )( 2 ) net G 50 kg 5.0 m/s 250 N 250 N 250 N 490 N 740 N y F T F ma T w = - = = = . = + = + = (d) The situation is the same as in part (c), except that the rising box is slowing down. Thus 5.0 m/s2 y a = - and we have instead ( )( 2 ) net G G 50 kg 5.0 m/s 250 N 250 N 250 N 490 N 240 N y F T F ma T F = - = = - = - . =- + = - + = Assess: For parts (a) and (b) the zero accelerations immediately imply that the gravitational force on the box must be exactly balanced by the upward tension in the rope. For part (c) the tension not only has to support the gravitational force on the box but must also accelerate it upward, hence, T must be greater than GF . When the box accelerates downward, the rope need not support the entire gravitational force, hence, T is less than GF . 6.12. Model: We assume the rocket is a particle moving in a vertical straight line under the influence of only two forces: gravity and its own thrust. Visualize: Solve: (a) Using Newtons second law and reading the forces from the free-body diagram, thrust F - FG = ma. Fthrust = ma + mgEarth ( )( ) = 0.200 kg 10 m/s2 + 9.80 m/s2 = 3.96 N (b) Likewise, the thrust on the moon is (0.200 kg)(10 m/s2 +1.62 m/s2 ) = 2.32 N. Assess: The thrust required is smaller on the moon, as it should be, given the moons weaker gravitational pull. The magnitude of a few newtons seems reasonable for a small model rocket. 6.13. Model: The astronaut is treated as a particle. Solve: The mass of the astronaut is earth 2 earth 800 N 81.6 kg 9.80 m/s m w g = = = Therefore, the weight of the astronaut on Mars is ( )( 2 ) Mars Mars w = mg = 81.6 kg 3.76 m/s = 307 N Assess: The smaller acceleration of gravity on Mars reveals that objects are less strongly attracted to Mars than to the earth, so the smaller weight on Mars makes sense. 6.14. Model: Use the particle model for the woman. Solve: (a) The womans weight on the earth is ( )( 2 ) earth earth w = mg = 55 kg 9.80 m/s = 540 N (b) Since mass is a measure of the amount of matter, the womans mass is the same on the moon as on the earth. Her weight on the moon is ( )( 2 ) moon moon w = mg = 55 kg 1.62 m/s = 89 N Assess: The smaller acceleration due to gravity on the moon reveals that objects are less strongly attracted to the moon than to the earth. Thus the womans smaller weight on the moon makes sense. 6.15. Model: We assume that the passenger is a particle subject to two vertical forces: the downward pull of gravity and the upward push of the elevator floor. We can use one-dimensional kinematics and Equation 6.10. Visualize: Solve: (a) The weight is 1 y 1 0 (60 kg)(9.80 m/s2 ) 590 N a w mg mg mg g g . . . . = . + . = . + . = = = . . . . (b) The elevator speeds up from 0 0 m/s y v = to its cruising speed at 10 m/s. y v = We need its acceleration before we can find the apparent weight: 10 m/s 0 m/s 2.5 m/s2 4.0 s y a v t . - = = = . The passengers weight is ( ) 2 2 1 590 N 1 2.5 m/s (590 N)(1.26) 740 N 9.80 m/s y a w mg g . . . . = . + . = . + . = = . . . . (c) The passenger is no longer accelerating since the elevator has reached its cruising speed. Thus, w = mg = 590 N as in part (a). Assess: The passengers weight is the gravitational force on the passenger in parts (a) and (c), since there is no acceleration. In part (b), the elevator must not only support the gravitational force but must also accelerate him upward, so its reasonable that the floor will have to push up harder on him, increasing his weight. 6.16. Model: We assume that the passenger is a particle acted on by only two vertical forces: the downward pull of gravity and the upward force of the elevator floor. Visualize: Please refer to Figure EX6.16. The graph has three segments corresponding to different conditions: (1) increasing velocity, meaning an upward acceleration; (2) a period of constant upward velocity; and (3) decreasing velocity, indicating a period of deceleration (negative acceleration). Solve: Given the assumptions of our model, we can calculate the acceleration for each segment of the graph and then apply Equation 6.10. The acceleration for the first segment is 1 0 2 1 0 8 m/s 0 m/s 4 m/s 2 s 0 s y a v v t t - - = = = - - ( )( ) 2 2 2 1 1 4 m/s 75 kg 9.80 m/s 1 4 1035 N 9.80 m/s 9.80 y a w mg mg g . . . . . . . = . + . = . + . = . + . = . . . . . . For the second segment, 0 m/s2 y a = and the weight is ( )( ) 2 w mg 1 0 m/s mg 75 kg 9.80 m/s2 740 N g . . = . + . = = = . . For the third segment, ( )( )( ) 3 2 2 3 2 2 2 2 0 m/s 8 m/s 2 m/s 10 s 6 s 1 2 m/s 75 kg 9.80 m/s 1 0.2 590 N 9.80 m/s y a v v t t w mg - - = = =- - - . - . . = . + . = - = . . Assess: As expected, the weight is greater than the gravitational force on the passenger when the elevator is accelerating upward and lower than normal when the acceleration is downward. When there is no acceleration the weight is the gravitational force. In all three cases the magnitudes are reasonable, given the mass of the passenger and the accelerations of the elevator. 6.17. Model: We assume that the safe is a particle moving only in the x-direction. Since it is sliding during the entire problem, we can use the model of kinetic friction. Visualize: Solve: The safe is in equilibrium, since its not accelerating. Thus we can apply Newtons first law in the vertical and horizontal directions: ( net )x x B C k 0 N k B C 350 N 385 N 735 N F = SF = F + F - f = . f = F + F = + = ( ) ( )( 2 ) 3 net G G 0 N 300 kg 9.80 m/s 2.94 10 N y y F = SF = n - F = .n = F = mg = = Then, for kinetic friction: k k k k 3 735 N 0.250 2.94 10 N f n f n = . = = = Assess: The value of k = 0.250 is hard to evaluate without knowing the material the floor is made of, but it seems reasonable. 6.18. Model: We assume that the mule is a particle acted on by two opposing forces in a single line: the farmers pull and friction. The mule will be subject to static friction until (and if!) it begins to move; after that it will be subject to kinetic friction. Visualize: Solve: Since the mule does not accelerate in the vertical direction, the free-body diagram shows that n = FG = mg. The maximum friction force is ( )( )( 2 ) smax s f = mg = 0.8 120 kg 9.80 m/s = 940 N The maximum static friction force is greater than the farmers maximum pull of 800 N; thus, the farmer will not be able to budge the mule. Assess: The farmer should have known better. 6.19. Model: We will represent the crate as a particle. Visualize: Solve: (a) When the belt runs at constant speed, the crate has an acceleration a = 0 m/s2 .. .. and is in dynamic equilibrium. Thus net F = 0. .. .. It is tempting to think that the belt exerts a friction force on the crate. But if it did, there would be a net force because there are no other possible horizontal forces to balance a friction force. Because there is no net force, there cannot be a friction force. The only forces are the upward normal force and the gravitational force on the crate. (A friction force would have been needed to get the crate moving initially, but no horizontal force is needed to keep it moving once it is moving with the same constant speed as the belt.) (b) If the belt accelerates gently, the crate speeds up without slipping on the belt. Because it is accelerating, the crate must have a net horizontal force. So now there is a friction force, and the force points in the direction of the crates motion. Is it static friction or kinetic friction? Although the crate is moving, there is no motion of the crate relative to the belt. Thus, it is a static friction force that accelerates the crate so that it moves without slipping on the belt. (c) The static friction force has a maximum possible value s max s ( f ) = n. The maximum possible acceleration of the crate is ( ) s max s max f n a m m = = If the belt accelerates more rapidly than this, the crate will not be able to keep up and will slip. It is clear from the free-body diagram that G n = F = mg. Thus, 2 2 max s a = g = (0.5)(9.80 m/s ) = 4.9 m/s 6.20. Model: We assume that the truck is a particle in equilibrium, and use the model of static friction. Visualize: Solve: The truck is not accelerating, so it is in equilibrium, and we can apply Newtons first law. The normal force has no component in the x-direction, so we can ignore it here. For the other two forces: ( ) ( ) ( ) ( )( 2 )( ) net s G s G 0 N sin 4000 kg 9.80 m/s sin15 10,145 N x x x x F = SF = f - F = . f = F = mg . = = Assess: The trucks weight (mg) is roughly 40,000 N. A friction force that is 25% of the trucks weight seems reasonable. 6.21. Model: The car is a particle subject to Newtons laws and kinematics. Visualize: Solve: Kinetic friction provides a horizontal acceleration which stops the car. From the figure, applying Newtons first and second laws gives x k x SF = - f = ma G G 0 y SF = n - F = .n = F = mg Combining these two equations with k k f = n yields ( )( 2 ) k 0.50 9.80 m/s 4.9 m/s xa = - g = - = - Kinematics can be used to determine the initial velocity. 2 2 2 f i i 2 2x v = v + a.x.v = - a .x Thus ( 2 )( ) 2 i v = -2 -4.9 m/s 65 m- 0 m = 25 m/s Assess: The initial speed of 25 m/s2 56 mph is a reasonable speed to have initially for a vehicle to leave 65- meter-long skid marks. 6.22. Model: We assume that the plane is a particle accelerating in a straight line under the influence of two forces: the thrust of its engines and the rolling friction of the wheels on the runway. We can use one-dimensional kinematics. Visualize: Solve: We can use the definition of acceleration to find a, and then apply Newtons second law. We obtain: 82 m/s 0 m/s 2.34 m/s2 35 s a v t . - = = = . ( ) net x thrust r thrust r F = SF = F - f = ma. F = f + ma For rubber rolling on concrete, 0.02 r = (Table 6.1), and since the runway is horizontal, G n = F = mg. Thus: ( ) ( )( )( ) thrust r G r r 75,000 kg 0.02 9.8 m/s2 2.34 m/s2 190,000 N F = F + ma = mg + ma = m g + a = .. + .. = Assess: Its hard to evaluate such an enormous thrust, but comparison with the planes mass suggests that 190,000 N is enough to produce the required acceleration. 6.23. Model: We treat the train as a particle subject to rolling friction but not to drag (because of its slow speed and large mass). We can use the one-dimensional kinematic equations. Visualize: Solve: The locomotive is not accelerating in the vertical direction, so the free-body diagram shows us that n = FG = mg. Thus, ( )( )( 2 ) r r f = mg = 0.002 50,000 kg 9.80 m/s = 980 N From Newtons second law for the decelerating locomotive, r 2 980 N 0.01960 m/s 50,000 kg x a f m - - = = =- Since were looking for the distance the train rolls, but we dont have the time: ( ) ( ) ( ) ( ) 2 2 2 2 2 2 1 0 3 1 0 2 0 m/s 10 m/s 2 2.5510 m 2 2 0.01960 m/s x x v v a x x v v a - - - = . .. = = = - Assess: The locomotives enormous inertia (mass) and the small coefficient of rolling friction make this long stopping distance seem reasonable. 6.24. Model: We can treat the sliding player as a particle experiencing kinetic friction. Visualize: Solve: We can assume a mass of 80 kg (which corresponds to a gravitational force of about 175 pounds). We have no value for k of cloth sliding on loose dirt. It is probably greater than k for wood on wood, due to the roughness of both surfaces. Lets guess 0.40. From the free-body diagram, n = FG = mg, since theres no vertical acceleration. Thus, ( )( )( 2 ) k k f = mg = 0.40 80 kg 9.80 m/s = 314 N Assess: A frictional force of 314 N would produce an acceleration of net 2 314 N 4.0 m/s 80 kg a F m - = = =- A player running initially at 8 m/s would thus be brought to a stop in about 2 seconds, which seems somewhat too long. Our estimate of k is probably a bit low but 8 m/s is a large speed as well. 6.25. Model: We assume that the skydiver is shaped like a box and is a particle. Visualize: The skydiver falls straight down toward the earths surface, that is, the direction of fall is vertical. Since the skydiver falls feet first, the surface perpendicular to the drag has the cross-sectional area A = 20 cm 40 cm. The physical conditions needed to use Equation 6.16 for the drag force are satisfied. The terminal speed corresponds to the situation when the net force acting on the skydiver becomes zero. Solve: The expression for the magnitude of the drag with v in m/s is 1 2 0.25(0.20 0.40) 2 N 0.020 2 N 4 D Av = v = v The gravitational force on the skydiver is ( )( 2 ) G F = mg = 75 kg 9.8 m/s = 735 N. The mathematical form of the condition defining dynamical equilibrium for the skydiver and the terminal speed is net G F = F + D = 0 N .. .. .. 2 term term 0.02 N 735 N 0 N 735 192 m/s 0.02 . v - = .v = Assess: The result of the above simplified physical modeling approach and subsequent calculation, even if approximate, shows that the terminal velocity is very high. This result implies that the skydiver will be very badly hurt at landing if the parachute does not open in time. 6.26. Model: We will represent the tennis ball as a particle. Visualize: The tennis ball falls straight down toward the earths surface. The ball is subject to a net force that is the resultant of the gravitational and drag force vectors acting vertically, in the downward and upward directions, respectively. Once the net force acting on the ball becomes zero, the terminal velocity is reached and remains constant for the rest of the motion. Solve: The mathematical equation defining the dynamical equilibrium situation for the falling ball is net G 0 N F F D = + = .. .. .. .. Since only the vertical direction matters, one can write: net G 0 N 0 N ySF = . F = D - F = When this condition is satisfied, the speed of the ball becomes the constant terminal speed termv = v . The magnitudes of the gravitational and drag forces acting on the ball are: ( ) ( ) ( ) ( )( ) ( ) 2 G 2 2 2 2 2 term term 9.80 m/s 1 0.25 0.25 0.0325 m 26 m/s 0.56 N 4 F mg m D Av p R v p = = = = = The condition for dynamic equilibrium becomes: ( 2 ) 2 9.80 m/s 0.56 N 0 N 0.56 N 57 g 9.80 m/s m - = .m = = Assess: The value of the mass of the tennis ball obtained above seems reasonable. 6.27. Visualize: We used the force-versus-time graph to draw the acceleration-versus-time graph. The peak acceleration was calculated as follows: max 2 max 10 N 2 m/s 5 kg a F m = = = Solve: The acceleration is not constant, so we cannot use constant acceleration kinematics. Instead, we use the more general result that 0 v(t) = v + area under the acceleration curve from 0 s to t The object starts from rest, so 0 v = 0 m/s. The area under the acceleration curve between 0 s and 6 s is 1 2 (4 s) (2 m/s2 ) = 4.0 m/s.Weve used the fact that the area between 4 s and 6 s is zero. Thus, at t = 6 s, 4.0 m/s. x v = 6.28. Visualize: The acceleration is ax = Fx m, so the acceleration-versus-time graph has exactly the same shape as the forceversus- time graph. The maximum acceleration is ( ) ( ) 2 max max a = F m = 6 N 2 kg = 3 m/s . Solve: The acceleration is not constant, so we cannot use constant-acceleration kinematics. Instead, we use the more general result that 0 v(t) = v + area under the acceleration curve from 0 s to t The object starts from rest, so 0 v = 0 m/s. The area under the acceleration curve between 0 s and 4 s is a rectangle (3 m/s2 2 s = 6 m/s) plus a triangle (1 2 ) 2 3 m/s 2 s = 3 m/s . Thus 9 m/s x v = at t = 4 s. 6.29. Model: You can model the beam as a particle in static equilibrium. Visualize: Solve: Using Newtons first law, the equilibrium equations in vector and component form are: ( ) ( ) net 1 2 G net 1 2 G net 1 2 G 0 N 0 N 0 N x x x x y y y y F T T F F T T F F T T F = + + = = + + = = + + = .. .. .. .. .. Using the free-body diagram yields: 1 1 2 2 1 1 2 2 G -T sin. + T sin. = 0 N T cos. + T cos. - F = 0 N The mathematical model is reduced to a simple algebraic system of two equations with two unknowns, 1 T and 2T . Substituting 1 2 . = 20, . = 30, and G F = mg = 9800 N, the simultaneous equations become 1 2 1 2 -T sin 20 + T sin30 = 0 N T cos20 + T cos30 = 9800 N You can solve this system of equations by simple substitution. The result is 1 2 T = 6397 N and T = 4376 N. Assess: The above approach and result seem reasonable. Intuition indicates there is more tension in the left rope than in the right rope. 6.30. Model: The plastic ball is represented as a particle in static equilibrium. Visualize: Solve: The electric force, like the weight, is a long-range force. So the ball experiences the contact force of the strings tension plus two long-range forces. The equilibrium condition is ( ) ( ) ( ) ( ) net elec elec net G sin 0 N cos 0 N x x x y y y F T F T F F T F T mg . . = + = - = = + = - = We can solve the y-equation to get (0.001 kg)(9.8 m/s2 ) 0.0104 N cos cos20 T mg . = = = Substituting this value into the x-equation, ( 2 ) elec F = T sin. = 1.0410- N sin 20 = 0.0036 N (b) The tension in the string is 0.0104 N. 6.31. Model: The piano is in static equilibrium and is to be treated as a particle. Visualize: Solve: (a) Based on the free-body diagram, Newtons second law is ( ) ( ) net 1 2 2 2 1 1 net 1 2 3 G 3 1 1 2 2 0 N cos cos 0 N sin sin x x x y y y y y F T T T T F T T T F TT T mg . . . . = = + = - = = + + + = - - - Notice how the force components all appear in the second law with plus signs because we are adding forces. The negative signs appear only when we evaluate the various components. These are two simultaneous equations in the two unknowns 2 T and 3T . From the x-equation we find ( ) 1 1 2 2 cos 500 N cos15 533 N cos cos25 T T . . = = = (b) Now we can use the y-equation to find 3 3 1 1 2 2 T = T sin. + T sin. + mg = 5.2510 N 6.32. Model: We will represent Henry as a particle. His motion is governed by constant-acceleration kinematic equations. Visualize: Please refer to the Figure EX6.32. Solve: (a) Henry undergoes an acceleration from 0 s to 2.0 s, constant velocity motion from 2.0 s to 10.0 s, and another acceleration as the elevator brakes from 10.0 s to 12.0 s. The weight is the same as the gravitational force during constant velocity motion, so Henrys weight w = FG = mg is 750 N. His weight is less than the gravitational force on him during the initial acceleration, so the acceleration is in a downward direction (negative a). Thus, the elevators initial motion is down. (b) Because the gravitational force on Henry is 750 N, his mass is Gm = F /g = 76.5 kg. (c) The apparent weight (see Equation 6.10) during vertical motion is given by G w mg 1 a a g w 1 g F . . . . = . + .. = . - . . . . . During the interval 0 s = t = 2 s, the elevators acceleration is 600 N 1 1.96 m/s2 750 N a = g .. - .. = - . . At t = 2 s, Henrys position is ( )2 ( )2 1 0 0 0 0 0 1 1 3.92 m 2 2 y = y + v .t + a .t = a .t = - and his velocity is 1 0 0 0 v = v + a.t = a.t = -3.92 m/s During the interval 2 s = t =10 s, a = 0 m/s2. This means Henry travels with a constant velocity 1 v = -3.92 m/s. At t =10 s he is at position 2 1 1 1 y = y + v .t = -35.3 m and he has a velocity 2 1 v = v = -3.92 m/s. During the interval 10 s = t =12.0 s, the elevators acceleration is 900 N 1 1.96 m/s2 750 N a = g .. - .. = + . . The upward acceleration vector slows the elevator and Henry feels heavier than normal. At t =12.0 s Henry is at position 2 3 2 2 2 2 ( ) 1 ( ) 39.2 m 2 y = y + v .t + a .t = - Thus Henry has traveled distance 39.2 m. 6.33. Model: Well assume Zach is a particle moving under the effect of two forces acting in a single vertical line: gravity and the supporting force of the elevator. Visualize: Solve: (a) Before the elevator starts braking, Zach is not accelerating. His weight (see Equation 6.10) is ( )( ) 2 w mg 1 a mg 1 0 m/s mg 80 kg 9.80 m/s2 784 N g g . . . . = . + . = . + . = = = . . . . Zachs weight is 7.8102 N. (b) Using the definition of acceleration, ( ) 1 0 2 1 0 0 10 m/s 3.33 m/s 3.0 s a v v v t t t . - - - = = = = . - ( )( ) 2 2 2 1 80 kg 9.80 m/s 1 3.33 m/s (784 N)(1 0.340) 1050 N 9.80 m/s w mg a g . . . . . = . + . = . + . = + = . . . . Now Zachs weight is 1.05103 N. Assess: While the elevator is braking, it not only must support the gravitational force on Zach but must also push upward on him to decelerate him, so his weight is greater than the gravitational force. 6.34. Model: We can assume your body is a particle moving in a straight line under the influence of two forces: gravity and the support force of the scale. Visualize: Solve: The weight (see Equation 6.10) of an object moving in an elevator is w mg 1 a a w 1 g g mg . . . . = . + .. = . - . . . . . When accelerating upward, the acceleration is 170 lb 1 (9.80 m/s2 ) 1.3 m/s2 150 lb a = .. - .. = . . When braking, the acceleration is 120 lb 1 (9.80 m/s2 ) 2.0 m/s2 150 lb a = .. - .. = - . . Assess: A 10-20% change in apparent weight seems reasonable for a fast elevator, as the ones in the Empire State Building must be. Also note that we did not have to convert the units of the weights from pounds to newtons because the weights appear as a ratio. 6.35. Model: We can assume the foot is a single particle in equilibrium under the combined effects of gravity, the tensions in the upper and lower sections of the traction rope, and the opposing traction force of the leg itself. We can also treat the hanging mass as a particle in equilibrium. Since the pulleys are frictionless, the tension is the same everywhere in the rope. Because all pulleys are in equilibrium, their net force is zero. So they do not contribute to T. Visualize: Solve: (a) From the free-body diagram for the mass, the tension in the rope is ( )( 2 ) G T = F = mg = 6 kg 9.80 m/s = 58.8 N (b) Using Newtons first law for the vertical direction on the pulley attached to the foot, ( ) ( ) net G foot sin sin15 0 N y y F = SF = T . -T - F = ( ) G foot foot sin15 sin sin15 T F m g T T . + . = = + (4 kg)(9.80 m/s2 ) 0.259 0.259 0.667 0.926 58.8 N = + = + = .. = sin-1 0.926 = 67.8 (c) Using Newtons first law for the horizontal direction, ( ) net traction cos cos15 0 N x x F = SF = T . + T - F = ( ) traction . F = T cos. + T cos15 = T cos67.8 + cos15 = (58.8 N)(0.3778 + 0.9659) = (58.8 N)(1.344) = 79.0 N Assess: Since the tension in the upper segment of the rope must support the foot and counteract the downward pull of the lower segment of the rope, it makes sense that its angle is larger (a more direct upward pull). The magnitude of the traction force, roughly one-tenth of the gravitational force on a human body, seems reasonable. 6.36. Model: We can assume the person is a particle moving in a straight line under the influence of the combined decelerating forces of the air bag and seat belt or, in the absence of restraints, the dashboard or windshield. Visualize: Solve: (a) In order to use Newtons second law for the passenger, well need the acceleration. Since we dont have the stopping time: 2 2 ( ) 1 0 1 0 v = v + 2a x - x ( ) ( ) ( ) 2 2 2 2 2 1 0 2 1 0 0 m /s 15 m/s 112.5 m/s 2 21 m0 m a v v x x - - . = = = - - - ( )( 2 ) net . F = F = ma = 60 kg -112.5 m/s = -6750 N The net force is 6750 N to the left. (b) Using the same approach as in part (a), ( ) ( ) ( ) ( ) 2 2 2 2 2 1 0 1 0 0 m /s 15 m/s 60 kg 1,350,000 N 2 2 0.005 m F ma m v v x x - - = = = =- - The net force is 1,350,000 N to the left. (c) The passengers weight is mg = (60 kg)(9.8 m/s2 ) = 588 N. The force in part (a) is 11.5 times the passengers weight. The force in part (b) is 2300 times the passengers weight. Assess: An acceleration of 11.5g is well within the capability of the human body to withstand. A force of 2300 times the passengers weight, on the other hand, would surely be catastrophic. 6.37. Model: The ball is represented as a particle that obeys constant-acceleration kinematic equations. Visualize: Solve: This is a two-part problem. During part 1 the ball accelerates upward in the tube. During part 2 the ball undergoes free fall (a = -g). The initial velocity for part 2 is the final velocity of part 1, as the ball emerges from the tube. The free-body diagram for part 1 shows two forces: the air pressure force and the gravitational force. We need only the y-component of Newtons second law: ( ) net air G air 2 2 2 N 9.80 m/s 30.2 m/s 0.05 kg y y F F F F a a g m m m - = = = = - = - = We can use kinematics to find the velocity 1 v as the ball leaves the tube: 2 2 ( ) ( 2 )( ) 1 0 1 0 1 1 v = v + 2a y - y . v = 2ay = 2 30.2 m/s 1 m = 7.77 m/s For part 2, free-fall kinematics 2 2 ( ) 2 1 2 1 v = v - 2g y - y gives 2 1 2 1 3.1 m 2 y y v g - = = 6.38. Model: We will represent the bullet as a particle. Visualize: Solve: (a) We have enough information to use kinematics to find the acceleration of the bullet as it stops. Then we can relate the acceleration to the force with Newtons second law. (Note that the barrel length is not relevant to the problem.) The kinematic equation is ( ) ( ) 2 2 2 2 0 5 2 1 0 400 m/s 2 6.6710 m/s 2 2 0.12 m v v a x a v x = + . . = - = - = - . Notice that a is negative, in agreement with the vector a.. in the motion diagram. Turning to forces, the wood exerts two forces on the bullet. First, an upward normal force that keeps the bullet from falling through the wood. Second, a retarding frictional force k f .. that stops the bullet. The only horizontal force is k f , .. which points to the left and thus has a negative x-component. The x-component of Newtons second law is ( ) ( )( 5 2 ) net k k 0.01 kg 6.67 10 m/s 6670 N x F = - f = ma. f = -ma = - - = Notice how the signs worked together to give a positive value of the magnitude of the force. (b) The time to stop is found from 1 0 v = v + a.t as follows: t v0 6.00 10 4 s 600 s a . = - = - = (c) Using the above kinematic equation, we can find the velocity as a function of t. For example at t = 60 s, 400 m/s ( 6.667 105 m/s2 )(60 10 6 s) 360 m/s x v = + - - = 6.39. Model: Represent the rocket as a particle that follows Newtons second law. Visualize: Solve: (a) The y-component of Newtons second law is ( ) 5 net thrust 2 2 3.0 10 N 9.80 m/s 5.2 m/s 20,000 kg y y F F mg a a m m - = = = = - = (b) At 5000 m the acceleration has increased because the rocket mass has decreased. Solving the equation of part (a) for m gives 5 thrust 4 5000 m 2 2 5000 m 3.0 10 N 1.9 10 kg 6.0 m/s 9.80 m/s m F a g = = = + + The mass of fuel burned is m 3 fuel initial 5000 m = m - m =1.010 kg. 6.40. Model: The steel block will be represented by a particle. Steel-on-steel has a static coefficient of friction s = 0.80 and a kinetic coefficient of friction k = 0.60. Visualize: Solve: (a) While the block is at rest, Newtons second law is ( ) ( ) net s s net G G 0 N x y F = T - f = .T = f F = n - F .n = F = mg The static friction force has a maximum value ( )s max s f = mg. The string tension that will cause the block to slip and start to move is ( )( )( 2 ) s T = mg = 0.80 2.0 kg 9.80 m/s =15.7 N Any tension less than this will not be sufficient to cause the block to move, so this is the minimum tension for motion. (b) As the block is moving with a tension of 20 N in the string, we can find its acceleration from the xcomponent of Newtons second law as follows: ( ) k net x k x x F T f ma a T f m - = - = . = The kinetic friction force k k f = mg. The acceleration of the block is ( )( )( 2 ) 2 20 N 0.60 2 kg 9.8 m/s 4.12 m/s 2 kg x a - = = Using kinematics, the blocks speed after moving 1.0 m will be 2 2 2 ( 2 )( ) 1 1 v = 0 m /s + 2 4.12 m/s 1.0 m .v = 2.9 m/s (c) The only difference in this case is the coefficient of kinetic friction whose value is 0.050 instead of 0.60. The acceleration of the block is ( )( )( 2 ) 2 20 N 0.050 2.0 kg 9.80 m/s 9.51 m/s 2.0 kg x a - = = The blocks speed after moving 1.0 m will be 2 2 2 ( 2 )( ) 1 1 v = 0 m /s + 2 9.51 m/s 1.0 m .v = 4.4 m/s 6.41. Model: We assume that Sam is a particle moving in a straight horizontal line under the influence of two forces: the thrust of his jet skis and the resisting force of friction on the skis. We can use one-dimensional kinematics. Visualize: Solve: (a) The friction force of the snow can be found from the free-body diagram and Newtons first law, since theres no acceleration in the vertical direction: ( )( 2 ) G n = F = mg = 75 kg 9.80 m/s = 735 N ( )( ) k k . f = n = 0.10 735 N = 73.5 N Then, from Newtons second law: ( ) thrust k 2 net thrust k 0 0 200 N 73.5 N 1.687 m/s x 75 kg F F f ma a F f m - - = - = . = = = From kinematics: ( 2 )( ) 1 0 01 v = v + a t = 0 m/s + 1.687 m/s 10 s =16.9 m/s (b) During the acceleration, Sam travels to 2 ( 2 )( )2 1 0 01 01 1 11.687 m/s 10 s 84 m 2 2 x = x + v t + a t = = After the skis run out of fuel, Sams acceleration can again be found from Newtons second law: ( ) net 2 net k 1 73.5 N 73.5 N 0.98 m/s x 75 kg F F f am - = - = - . = = = - Since we dont know how much time it takes Sam to stop: 2 2 ( ) 2 1 1 2 1 v = v + 2a x - x ( ) ( ) 2 2 2 2 2 2 1 2 1 2 1 0 m /s 16.9 m/s 145 m 2 2 0.98 m/s x x v v a - - . - = = = - The total distance traveled is ( ) 2 1 1 x - x + x =145 m+ 84 m = 229 m. Assess: A top speed of 16.9 m/s (roughly 40 mph) seems quite reasonable for this acceleration, and a coasting distance of nearly 150 m also seems possible, starting from a high speed, given that were neglecting air resistance. 6.42. Model: We assume Sam is a particle moving in a straight line down the slope under the influence of gravity, the thrust of his jet skis, and the resisting force of friction on the snow. Visualize: Solve: From the height of the slope and its angle, we can calculate its length: 1 0 1 0 sin 50 m 288 m sin sin10 h x x h x x . . = . - = = = - Since Sam is not accelerating in the y-direction, we can use Newtons first law to calculate the normal force: ( ) net G cos 0 N y y F = SF = n - F . = G .n = F cos. = mg cos. = (75 kg)(9.80 m/s2 )(cos10) = 724 N One-dimensional kinematics gives us Sams acceleration: 2 2 ( ) 1 0 0 2 x v = v + a x - x ( ) ( ) ( ) 2 2 2 2 2 1 0 2 1 2 40 m/s 0 m /s 2.78 m/s 2 2 288 m x a v v x x - - . = = = - Then, from Newtons second law and the equation k k f = n : ( ) net G thrust k thrust k sin sin x x x F F F F f ma mg F ma n . . = S = + - = + - . = (75 kg)(9.80 m/s2 )(sin10 ) 200 N (75 kg)(2.78 m/s2 ) 0.165 724 N + - = = Assess: This coefficient seems a bit high for skis on snow, but not impossible. 6.43. Model: We assume the suitcase is a particle accelerating horizontally under the influence of friction only. Visualize: Solve: Because the conveyor belt is already moving, friction drags your suitcase to the right. It will accelerate until it matches the speed of the belt. We need to know the horizontal acceleration. Since theres no acceleration in the vertical direction, we can apply Newtons first law to find the normal force: ( )( 2 ) G n = F = mg = 10 kg 9.80 m/s = 98.0 N The suitcase is accelerating, so we use k to find the friction force ( )( ) k k f = mg = 0.3 98.0 N = 29.4 N We can find the horizontal acceleration from Newtons second law: ( ) net x x k F = SF = f = ma k 2 29.4 N 2.94 m/s 10 kg a f m . = = = From one of the kinematic equations: 2 2 ( ) 1 0 1 0 v = v + 2a x - x ( ) ( ) ( ) 2 2 2 2 1 0 1 0 2 2.0 m/s 0 m/s 0.68 m 2 2 2.94 m/s x x v v a - - . - = = = The suitcase travels 0.68 m before catching up with the belt and riding smoothly. Assess: If we imagine throwing a suitcase at a speed of 2.0 m/s onto a motionless surface, 0.68 m seems a reasonable distance for it to slide before stopping. 6.44. Model: The box of shingles is a particle subject to Newtons laws and kinematics. Visualize: Solve: Newtons laws can be used in the coordinate system in which the direction of motion of the box of shingles defines the +x-axis. The angle that G F .... makes with the y-axis is 25. ( ) G k sin25 x SF = F - f = ma ( ) G G cos25 0 cos25 y SF = n - F = .n = F We have used the observation that the shingles do not leap off the roof, so the acceleration in the y-direction is zero. Combining these equations with k k f = n and GF = mg yields k mg sin 25 - mg cos25 = ma ( ) 2 k .a = sin25 - cos25 g = -0.743 m/s where the minus sign indicates the acceleration is directed up the incline. The required initial speed to have the box come to rest after 5.0 m is found from kinematics. 2 2 2 ( 2 )( ) f i i i v = v + 2a.x.v = -2 -0.743 m/s 5.0 m .v = 2.7 m/s Assess: To give the shingles an initial speed of 2.7 m/s requires a strong, determined push, but is not beyond reasonable. 6.45. Model: We will model the box as a particle, and use the models of kinetic and static friction. Visualize: The pushing force is along the +x-axis, but the force of friction acts along the x-axis. A component of the gravitational force on the box acts along the x-axis as well. The box will move up if the pushing force is at least equal to the sum of the friction force and the component of the gravitational force in the x-direction. Solve: Lets determine how much pushing force you would need to keep the box moving up the ramp at steady speed. Newtons second law for the box in dynamic equilibrium is ( ) net G k push ( )x x x x ( )x ( )x 0 N sin k push 0 N F = SF = n + F + f + F = - mg . - f + F = ( ) net G k push ( ) ( ) ( ) cos 0 N 0 N 0 N y y y y y y F = SF = n + F + f + F = n - mg . + + = The x-component equation and the model of kinetic friction yield: push k k F = mg sin. + f = mg sin. + n Let us obtain n from the y-component equation as n = mg cos. , and substitute it in the above equation to get push k k F = mg sin. + mg cos. = mg(sin. + cos. ) = (100 kg)(9.80 m/s2 )(sin 20 + 0.60 cos20) = 888 N The force is less than your maximum pushing force of 1000 N. That is, once in motion, the box could be kept moving up the ramp. However, if you stop on the ramp and want to start the box from rest, the model of static friction applies. The analysis is the same except that the coefficient of static friction is used and we use the maximum value of the force of static friction. Therefore, we have 2 push s F = mg(sin. + cos. ) = (100 kg)(9.80 m/s )(sin 20 + 0.90 cos20) =1160 N Since you can push with a force of only 1000 N, you cant get the box started. The big static friction force and the weight are too much to overcome. 6.46. Model: We will represent the wood block as a particle, and use the model of kinetic friction and kinematics. Assume w sin . > fs , so it does not hang up at the top. Visualize: The block ends where it starts, so 2 0 x = x = 0 m. We expect 2 v to be negative, because the block will be moving in the x-direction, so well want to take 2 v as the final speed. Because of friction, we expect to find 2 0v < v . Solve: (a) The friction force is opposite to v, . so k f .. points down the slope during the first half of the motion and up the slope during the second half. G F .. and n.. are the only other forces. Newtons second law for the upward motion is ( ) ( ) net G k k 0 2 net G sin sin 0 m/s cos cos x x y y F F f mg f a a m m m F n F n mg a m m m . . . . - - - - = = = = - - = = = = The friction model is k k f = n. First solve the y-equation to give n = mg cos. . Use this in the friction model to get k k f = mg cos. . Now substitute this result for k f into the x-equation: k ( ) ( 2 )( ) 2 0 k a mg sin mg cos g sin cos 9.8 m/s sin30 0.20cos30 6.60 m/s m . . . . - - = =- + =- + =- Kinematics now gives ( ) ( ) ( ) 2 2 2 2 2 2 2 1 0 1 0 0 1 0 1 2 0 0 m /s 10 m/s 2 7.6 m 2 2 6.60 m/s v v a x x x v v a - - = + - . = = = - The blocks height is then 1 h = x sin. = (7.6 m)sin30 = 3.8 m. (b) For the return trip, k f .. points up the slope, so the x-component of the second law is ( ) net G k k 1 sin sin x x F F f mg f a a m m m - . + - . + = = = = Note the sign change. The y-equation and the friction model are unchanged, so we have ( ) 2 1 k a = -g sin. - cos. = -3.20 m/s The kinematics for the return trip are 2 2 ( ) ( 2 )( ) 2 1 1 2 1 2 11 v = v + 2a x - x .v = -2a x = 2 -3.20 m/s -7.6 m = -7.0 m/s Notice that we used the negative square root because 2 v is a velocity with the vector pointing in the x-direction. The final speed is 2 v = 7.0 m/s. 6.47. Model: We will model the sled and friend as a particle, and use the model of kinetic friction because the sled is in motion. Visualize: The net force on the sled is zero (note the constant speed of the sled). That means the component of the pulling force along the +x-direction is equal to the magnitude of the kinetic force of friction in the x-direction. Also note that net ( ) 0 N, y F = since the sled is not moving along the y-axis. Solve: Newtons second law is ( ) net G k pull k ( )x x x x ( )x ( )x 0 N 0 N pull cos 0 N F = SF = n + F + f + F = + - f + F . = ( ) net G k pull pull ( ) ( ) ( ) 0 N sin 0 N y y y y y y F = SF = n + F + f + F = n - mg + + F . = The x-component equation using the kinetic friction model k k f = n reduces to k pull n = F cos. The y-component equation gives pull n = mg - F sin. We see that the normal force is smaller than the gravitational force because pull F has a component in a direction opposite to the direction of the gravitational force. In other words, pull F is partly lifting the sled. From the xcomponent equation, k can now be obtained as pull k pull cos sin F mg F . . = = - ( )( ) ( )( 2 ) ( )( ) 75 N cos30 0.12 60 kg 9.80 m/s 75 N sin30 = - Assess: A quick glance at the various k values in Table 6.1 suggests that a value of 0.12 for k is reasonable. 6.48. Model: As long as the static force of friction between the box and the sled is sufficiently large for the box not to slip, the acceleration a of the box is the same as the acceleration of the sled. We will therefore model the box as a particle and use the model of static friction. Visualize: The force on the box that is responsible for its acceleration is the force of static friction provided by the sled because the box would slide backward without this force. Solve: Newtons second law for the box is (( ) ) net G box s s max max ( ) ( ) 0 N 0 N x x x x x x F = SF = n + F + f = + + f = ma = ma (( ) ) net G box s ( ) ( ) 0 N 0 N y y y y y y F = SF = n + F + f = n - mg + = ma = The above equations can be combined along with the friction model to get s max s s max s a f n mg g m m m = = = = To find max T , we write Newtons second law for the box-sled system, with total mass m + M, as (( ) ) net G sled+box s max max max ( ) ( ) ( ) 0 N 0 N 0 N ( ) x x x x x x F = SF = N + F + f + T = + + + T = m+ M a (( ) ) net G sled+box s max ( ) ( ) ( ) ( ) 0 N 0 N 0 N y y y y y y F = SF = N + F + f + T = N - m+ M g + + = Referring to Table 6.1 for the coefficient of friction, the x-component equation for the box-sled system yields 2 max max s T = (m + M)a = (m + M) g = (0.50)(9.80 m/s )(5.0 kg +10 kg) = 74 N That is, the largest tension force for which the box does not slip is 74 N. Assess: If the box were glued to the sled, there would be no resistance to motion because the ice is frictionless, and the tension force could be as large as one wants. On the other hand, if the friction between the box and the sled were zero, one could not pull at all without causing the box to slide. 6.49. Model: We will model the steel cabinet as a particle. It touches the trucks steel bed, so only the steel bed can exert contact forces on the cabinet. As long as the cabinet does not slide, the acceleration a of the cabinet is equal to the acceleration of the truck. Visualize: Solve: The shortest stopping distance is the distance for which the static friction force has its maximum value ( fs )max . Newtons second law for the box and the model of static friction are ( ) ( ) ( ) net G s s max s max ( ) ( ) 0 N 0 N x x x x x x F = SF = n + F + f = + - f = ma = ma.- f = ma ( ) net G s ( ) ( ) 0 N 0 N y y y y y F = SF = n + F + f = n - mg + = .n = mg ( )s max s s - f = - n = - mg = ma These three equations can be combined together to get s a = - g. Because constant-acceleration kinematics gives 2 2 1 0 1 0 v = v + 2a(x - x ) and s = 0.80 (Table 6.1), we find ( ) ( ) ( ) ( )( )( ) 2 2 2 2 1 0 0 1 0 2 s 15 m/s 14.3 m 2 2 2 0.80 9.80 m/s x x v v v a g - - - = = = = - Assess: The truck was moving at a speed of 15 m/s or at approximately 34 mph. A stopping distance without making the contents slide of about 14.3 m or approximately 47 feet looks reasonable. 6.50. Model: The antiques (mass = m) in the back of your pickup (mass = M) will be treated as a particle. The antiques touch the trucks steel bed, so only the steel bed can exert contact forces on the antiques. The pickup-antiques system will also be treated as a particle, and the contact force on this particle will be due to the road. Visualize: Solve: (a) We will find the smallest coefficient of friction that allows the truck to stop in 55 m, then compare that to the known coefficients for rubber on concrete. For the pickup-antiques system, with mass m + M, Newtons second law is (( ) ) net G PA ( ) ( ) 0 N 0 N ( ) ( ) x x x x x x F = SF = N + F + f = + - f = m+ M a = m+ M a (( ) ) net G PA ( ) ( ) N ( ) 0 N 0 N y y y y y F = SF = N + F + f = - m + M g + = The model of static friction is f = N, where is the coefficient of friction between the tires and the road. These equations can be combined to yield a = - g. Since constant-acceleration kinematics gives 2 2 1 0 1 0 v = v + 2a(x + x ), we find ( ) ( ) ( ) ( )( )( ) 2 2 2 2 1 0 0 min 2 1 0 1 0 25 m/s 0.58 2 2 2 9.8 m/s 55 m a v v v x x g x x - = . = = = - - The truck cannot stop if is smaller than this. But both the static and kinetic coefficients of friction, 1.00 and 0.80 respectively (see Table 6.1), are larger. So the truck can stop. (b) The analysis of the pickup-antiques system applies to the antiques, and it gives the same value of 0.58 for min . This value is smaller than the given coefficient of static friction s ( = 0.60) between the antiques and the truck bed. Therefore, the antiques will not slide as the truck is stopped over a distance of 55 m. Assess: The analysis of parts (a) and (b) are the same because mass cancels out of the calculations. According to the California Highway Patrol Web site, the stopping distance (with zero reaction time) for a passenger vehicle traveling at 25 m/s or 82 ft/s is approximately 43 m. This is smaller than the 55 m over which you are asked to stop the truck. 6.51. Model: The box will be treated as a particle. Because the box slides down a vertical wood wall, we will also use the model of kinetic friction. Visualize: Solve: The normal force due to the wall, which is perpendicular to the wall, is here to the right. The box slides down the wall at constant speed, so a = 0 .. .. and the box is in dynamic equilibrium. Thus, Fnet = 0. .... .. Newtons second law for this equilibrium situation is net push ( ) 0 N cos45 x F = = n - F net k push G k push ( ) 0 N sin45 sin45 y F = = f + F - F = f + F - mg The friction force is k k f = n. Using the x-equation to get an expression for n, we see that k k push f = F cos45. Substituting this into the y-equation and using Table 6.1 to find k = 0.20 gives, k push push F cos45 + F sin 45 - mg = 0 N ( )( 2 ) push k 2.0 kg 9.80 m/s 23 N cos45 sin 45 0.20cos45 sin 45 F mg . = = = + + 6.52. Model: Use the particle model for the block and the model of static friction. Visualize: Solve: The block is initially at rest, so initially the friction force is static friction. If the 12 N push is too strong, the box will begin to move up the wall. If it is too weak, the box will begin to slide down the wall. And if the pushing force is within the proper range, the box will remain stuck in place. First, lets evaluate the sum of all the forces except friction: SFx = n - Fpush cos30 = 0 N.n = Fpush cos30 2 push G push sin30 sin30 (12 N)sin30 (1 kg)(9.8 m/s ) 3.8 N y SF = F - F = F - mg = - = - In the first equation we utilized the fact that any motion is parallel to the wall, so 0 m/s2. x a = These three forces add up to -3.8 j N. This means the static friction force will be able to prevent the box from moving if s f = +3.8 j N. Using the x-equation and the friction model we get ( )s max s s push f = n = F cos30 = 5.2 N where we used s = 0.5 for wood on wood. The static friction force s f .. needed to keep the box from moving is less than ( ) s max f . Thus the box will stay at rest. 6.53. Model: We will model the skier along with the wooden skis as a particle of mass m. The snow exerts a contact force and the wind exerts a drag force on the skier. We will therefore use the models of kinetic friction and drag. Visualize: We choose a coordinate system such that the skiers motion is along the +x-direction. While the forces of kinetic friction k f .. and drag D .. act along the x-direction opposing the motion of the skier, the gravitational force on the skier has a component in the +x-direction. At the terminal speed, the net force on the skier is zero as the forces along the +x-direction cancel out the forces along the x-direction. Solve: Newtons second law and the models of kinetic friction and drag are ( ) 2 net G k k ( ) ( ) ( ) 0 N sin 1 0 N 4 x x x x x x x F = SF = n + F + f + D = + mg . - f Av = ma = ( ) net G k ( ) ( ) ( ) cos 0 N 0 N = 0 N y y y y y y F = SF = n + F + f + D = n - mg . + + k k f = n These three equations can be combined together as follows: 2 k k k (1/4)Av = mg sin. - f = mg sin. - n = mg sin. - mg cos. 1 2 k term 1 4 v mg sin cos A . . - . . . =. . . . Using k = 0.06 and A =1.8 m 0.40 m = 0.72 m2 , we find ( )( ) ( )( ) 1 2 2 term 1 3 2 4 80 kg 9.8 m/s sin 40 0.06cos40 51 m/s kg m 0.72 m v . . - .. = . . .. = . . .. . . .. Assess: A terminal speed of 51 m/s corresponds to a speed of 100 mph. This speed is reasonable but high due to the steep slope angle of 40 and a small coefficient of friction. 6.54. Model: The ball is a particle experiencing a drag force and traveling at twice its terminal velocity. Visualize: Solve: (a) An object falling at greater than its terminal velocity will slow down to its terminal velocity. Thus the drag force is greater than the force of gravity, as shown in the free-body diagrams. When the ball is shot straight up, ( ) ( ) 2 ( )2 G term 1 1 2 4 5 y 4 4 F ma F D mg Av mg A v mg A mg mg A = = - + = -. + . = - - = - - . . = - . . . . . . . . S Thus a = -5 g, where the minus sign indicates the downward direction. We have used Equations 6.16 for the drag force and 6.19 for the terminal velocity. (b) When the ball is shot straight down, ( ) ( )2 G term 12 3 y 4 SF = ma = D - F = A v - mg = mg Thus a = 3 g, this time directed upward. (c) The ball will slow down to its terminal velocity, slowing quickly at first, and more slowly as it gets closer to the terminal velocity because the drag force decreases as the ball slows. 6.55. Model: We will model the sculpture as a particle of mass m. The ropes that support the sculpture will be assumed to have zero mass. Visualize: Solve: Newtons first law in component form is net 1 2 G 1 2 ( ) sin30 F x = SFx = T x + T x + F x = -T + T sin 60 + 0 N = 0 N net 1 2 G 1 2 G ( ) cos30 cos60 0 N y y y y y F = SF = T + T + F = -T + T - F = Using the x-component equation to obtain an expression for 1 T and substituting into the y-component equation yields: ( )( ) G 2 500 lbs = 250 lbs sin60 cos30 2 cos60 sin30 T = F = + Substituting this value of 2 T back into the x-component equation, 1 2 sin 60 sin30 T T = 250 lbs sin 60 433 lbs sin30 = = We will now find a rope size for a tension force of 433 lbs, that is, the diameter of a rope with a safety rating of 433 lbs. Since the cross-sectional area of the rope is 1 2 4 p d , we have ( ) ( ) 1 2 2 4 433 lbs 0.371 inch 4000 lbs/inch d p . . = . . = .. ..
Any diameter larger than 0.371 inch will ensure a safety rating of at least 433 lbs. The rope size corresponding to a diameter of 3/8 of an inch will therefore be appropriate. Assess: If only a single rope were used to hang the sculpture, the rope would have to support a gravitational force of 500 lbs. The diameter of the rope for a safety rating of 500 lbs is 0.399 inches, and the rope size jumps from a diameter of 3/8 to 4/8 of an inch. Also note that the gravitational force on the sculpture is distributed in the two ropes. It is the sum of the y-components of the tensions in the ropes that will equal the gravitational force on the sculpture. 6.56. Model: We will model the container as a particle of mass m. The steel cable of the crane will be assumed to have zero mass. Visualize: Solve: As long as the container is stationary or it is moving with a constant speed (zero acceleration), the net force on the container is zero. In these cases, the tension in the cable is equal to the gravitational force on the container: T = mg = 44,000 N The cable should safely lift the load. More tension is required to accelerate the load. Newtons second law is ( ) net G ( ) ( ) y y y y y F = SF = F + T = -mg + T = ma The cranes maximum acceleration is 2 max a =1.0 m/s . So the maximum cable tension is max max T = mg + ma = 48,600 N This is less than the cables rating, so the cable must have been defective. 6.57. Model: We will model the skier as a particle, and use the model of kinetic friction. Visualize: Solve: Your best strategy, if its possible, is to travel at a very slow constant speed (a = 0 so Fnet = 0). .. .. .. .. Alternatively, you want the smallest positive . x a A negative x a would cause you to slow and stop. Lets find the value of k that gives net F = 0. .. .. Newtons second law for the skier and the model of kinetic friction are ( ) net G k k ( ) ( ) ( ) 0 sin cos 0 N x x x x x x F = SF = n + F + f + D = + mg . - f - D . = ( ) net G k ( ) ( ) ( ) cos 0 N sin 0 N y y y y y y F = SF = n + F + f + D = n - mg . + - D . = k k f = n The x- and y-component equations are k f = +mg sin. - Dcos. n = mg cos. + Dsin. From the model of kinetic friction, ( ) ( ) ( ) ( ) 2 k k 2 sin cos 75 kg 9.8 m/s sin15 50 N cos15 0.196 cos sin 75 kg 9.8 m/s cos15 50 N sin15 f mg D n mg D . . . . - - = = = = + + Yellow wax with k = 0.20 applied to skis will make the skis stick and hence cause the skier to stop. The skiers next choice is to use the green wax with k = 0.15. 6.58. Model: The ball hanging from the ceiling of the truck by a string is represented as a particle. Visualize: Solve: (a) You cannot tell from within the truck. Newtons first law says that there is no distinction between at rest and constant velocity. In both cases, the net force acting on the ball is zero and the ball hangs straight down. (b) Now you can tell. If the truck is accelerating, then the ball is tilted back at an angle. (c) The ball moves with the truck, so its acceleration is 5 m/s2 in the forward direction. (d) The free-body diagram shows that the horizontal component of T .. provides a net force in the forward direction. This is the net force that causes the ball to accelerate in the forward direction along with the truck. (e) Newtons second law for the ball is ( ) ( ) ( ) net sin10 0 m2/s2 net G cos10 x x y y y x y F T T F T F T mg a a m m m m m m + - = = = = = = = We can solve the second equation for the magnitude of the tension: (1.0 kg)(9.8 m/s2 ) 9.95 N cos10 cos10 T = mg = = Then the first equation gives the acceleration of the ball and truck: ( ) 2 sin10 9.95 N sin10 =1.73 m/s x a T m m = = The trucks velocity cannot be determined. 6.59. Model: You will be treated as a particle. You will experience your weight force and the force (P) of the scale pulling up. Visualize: Solve: The weight of an object is the magnitude of the contact force supporting it. Here, the contact force is , P .. so the weight is w = P. Newtons second law is ( ) net y y G y y F = SF = P + F = ma net G . F = P - F = ma w P mg ma mg 1 a g . . . = = + = . + . . . 6.60. Solve: Using x , x a dv dt = we express Newtons second law as a differential equation, which we then use to solve for . x v x x x x F m dv dv F dt ct dt dt m m = . = = Integrating from the initial to final conditions for each variable of integration, 0 2 0 0 2 x x v t x x x v dv c t dt v v ct m m . = . . - = Thus 2 0 2 x x v v ct m = + 6.61. Model: The astronaut is a particle oscillating on a spring. Solve: (a) The position versus time function x(t) can be used to find the velocity versus time function v(t) dx . dt = We have v(t) d {(0.30 m)sin(( rad/s)t )} (0.30 m/s)cos(( rad/s)t ) dt = p = p p This can then be used to find the acceleration a(t) dv . dt = a(t) dv (0.30 2 m/s2 )sin(( rad/s)t ) dt = = - p p Newtons second law yields a general expression for the force on the astronaut. ( 2 2 ) (( ) ) net F (t) = ma(t) = -(75 kg) 0.30p m/s sin p rad/s t Evaluating this at t =1.0 s gives ( ) net F 1.0 s = 0 N, since sin(p ) = 0. (b) Evaluating at t =1.5 s, 2 2 net 22.5 N sin 3 2.2 10 N 2 F p = - p .. .. = . . Assess: The force of 220 N is only one third of the astronauts weight on earth, so is easy for her to withstand. 6.62. Solve: (a) The terminal velocity for a falling object is reached when the downward gravitational force is balanced by the upward drag force. FG = D term term mg = bv = 6p.Rv term 6 v mg p. R . = (b) The mass of the spherical sand grain of density p = 2400 kg/m3 is 4 3 . 3 m = . .. p R .. . . Thus ( )( )( )3 2 4 2 2 term 3 2 2 2 2400 kg/m 9.80 m/s 5.0 10 m 1.3 m/s 9 9 1.0 10 Ns m v gR . . - - = = = . . . . . . The time required for the sand grain to fall 50 m at this speed is 50 m 38 s. 1.3 m/s t= = Assess: The speed of 1.3 m/s for a sand grain falling through water seems about right. 6.63. Solve: (a) A 1.0 kg block is pulled across a level surface by a string, starting from rest. The string has a tension of 20 N, and the blocks coefficient of kinetic friction is 0.50. How long does it take the block to move 1.0 m? (b) Newtons second law for the block is ( net ) k k 2 ( net ) G x 0 m/s y x y F T f T n F n F n mg a a a m m m m m m - - - - = = = = = = = = where we have incorporated the friction model into the first equation. The second equation gives n = mg. Substituting this into the first equation gives k 2 20 N 4.9 N 15.1 m/s 1.0 kg a T mg m - - = = = Constant acceleration kinematics gives ( )2 ( )2 1 ( ) 1 0 0 2 1 1 2 2 1.0 m 0.36 s 2 2 15.1 m/s x x v t a t a t t x a = + . + . = . .. = = = 6.64. Solve: (a) A 15,000 N truck starts from rest and moves down a 15 hill with the engine providing a 12,000 N force in the direction of the motion. Assume the frictional force between the truck and the road is very small. If the hill is 50 m long, what will be the speed of the truck at the bottom of the hill? (b) Newtons second law is SFy = ny + FGy + fy + Ey = may = 0 x x Gx x x x SF = n + F + f + E = ma G .0 N + F sin. + 0 N +12,000 N = ma ( ) ( ) 2 2 sin 12,000 N 15,000 N sin15 12,000 N 10.4 m/s 15,000 N/9.8 m/s a mg m . + + . = = = where we have calculated the mass of the truck from the gravitational force on it. Using the constant-acceleration kinematic equation 2 2 0 2 , x v - v = ax 2 2 2(10.4 m/s2 )(50 m) 32 m/s x x x v = a x = .v = 6.65. Solve: (a) A 1.0 kg box is pushed along an assembly line by a mechanical arm. The arm exerts a 20 N force at a downward angle of 30. The box has a coefficient of kinetic friction on the assembly line of 0.25. What is the speed of the box after traveling 50 cm, starting from rest? (b) Newtons second law for the box is ( ) ( ) net push k push k 2 net G push push cos30 cos30 sin30 sin30 0 m/s x x y y F F f F n a a m m m F n F F n mg F a m m m - - = = = = - - - - = = = = The second equation is solved to give an expression for n. Substituting into the first equation: push k ( push ) ( ) 2 cos30 sin30 20 N cos30 4.9 N 12.4 m/s 1.0 kg F F mg a m - + - = = = Using kinematics, 2 2 ( 2 )( ) 1 0 1 v = v + 2a.x = 2a.x.v = 2a.x = 2 12.4 m/s 0.50 m = 3.5 m/s 6.66. Solve: (a) A driver traveling at 40 m/s in her 1500 kg auto slams on the brakes and skids to rest. How far does the auto slide before coming to rest? (b) (c) Newtons second law is y y ( G )y y 0 N x 0.80 x SF = n + F = n - mg = ma = SF = - n = ma The y-component equation gives n = mg = (1500 kg)(9.8 m/s2 ). Substituting this into the x-component equation yields (1500 kg) 0.80(1500 kg)(9.8 m/s2 ) x a = - ( 0.80)(9.8 m/s2 ) 7.8 m/s2 x .a = - = - Using the constant-acceleration kinematic equation 2 2 1 0 v = v + 2a.x, we find ( ) ( ) 2 2 0 2 40 m/s 102 m 2 2 7.8 m/s x v a . = - = - = - 6.67. Solve: (a) A 20.0 kg wooden crate is being pulled up a 20 wooden incline by a rope that is connected to an electric motor. The crates acceleration is measured to be 2.0 m/s2. The coefficient of kinetic friction between the crate and the incline is 0.20. Find the tension T in the rope. (b) (c) Newtons second law for this problem in the component form is 2 2 net ( ) 0.20 (20 kg)(9.80 m/s )sin 20 (20 kg)(2.0 m/s ) x x F = SF = T - n - = 2 net ( ) (20 kg)(9.80 m/s )cos20 0 N y y F = SF = n - = Solving the y-component equation, n =184.18 N. Substituting this value for n in the x-component equation yields T =144 N. 6.68. Solve: (a) You wish to pull a 20 kg wooden crate across a wood floor (k = 0.20) by pulling on a rope attached to the crate. Your pull is 100 N at an angle of 30 above the horizontal. What will be the acceleration of the crate? (b) (c) Newtons equations and the model of kinetic friction are ( ) ( ) ( ) ( ) ( ) G k k G k k 0 N 100 N cos30 0 N 100 N cos30 100 N sin30 0 N 0 N x x x x x x y y y y y y F n P F f f f ma F n P F f n mg ma f n S = + + + = + + - = - = S = + + + = + - - = = = From the y-component equation, n =150 N. From the x-component equation and using the model of kinetic friction with k = 0.20, (100 N)cos30 (0.20)(150 N) (20 kg) 2.8 m/s2 x x - = a .a = 6.69. Model: Take the coin to be a particle held against the palm of your hand (which is like a vertical wall) by friction. The friction needs to be vertical and equal to the weight of the coin to hold the coin in place. Visualize: The hand pushes on the coin giving a normal force to the coin and causing a friction force. Solve: (a) We need a force (push) to create a normal force. Since the force accelerates the coin, a minimum acceleration min a is needed. (b) Newtons second law and the model of friction are SFy = ( fmin )y - (FG )y = fmin - mg = 0 N x min min SF = n = ma min min f = n Since you do not want the coin to slip down the hand you need s . Combining the above three equations yields min f = mg . s min n = mg s min . ma = mg . 2 2 min s 9.8 m/s 12.3 m/s 0.80 a g = = = 6.70. Model: We will model the shuttle as a particle and assume the elastic cord to be massless. We will also use the model of kinetic friction for the motion of the shuttle along the square steel rail. Visualize: Solve: The upward tension component Ty = T sin 45 =14.1 N is larger than the gravitational force on the shuttle. Consequently, the elastic cord pulls the shuttle up against the rail and the rails normal force pushes downward. Newtons second law in component form is ( ) net k x G k ( ) ( ) ( ) cos45 0 N 0 N x x x x x x x F = SF = T + f + n + F = T - f + + = ma = ma ( ) net k y G ( ) ( ) ( ) sin45 0 N 0 N y y y y y y F = SF = T + f + n + F = T + - n - mg = ma = The model of kinetic friction is k k f = n. We use the y-component equation to get an expression for n and hence k f . Substituting into the x-component equation and using the value of k in Table 6.1 gives us ( ) ( ) ( ) ( ) ( )( ) k 2 2 cos45 sin 45 20 N cos45 0.60 20 N sin 45 0.800 kg 9.80 m/s 13.0 m/s 0.800 kg x T T mg a m - - = - ..+ - .. = = Assess: The x-component of the tension force is 14.1 N. On the other hand, the net force on the shuttle in the xdirection is (0.800 kg)(13.0 m/s2 ) 10.4 N. x ma= = This value for ma is reasonable since a part of the 14.1 N tension force is used up to overcome the force of kinetic friction. 6.71. Model: Assume the ball is a particle on a slope, and that the slope increases as the x-displacement increases. Assume that there is no friction and that the ball is being accelerated to the right so that it remains at rest on the slope. Visualize: Although the ball is on a slope, it is accelerating to the right. Thus well use a coordinate system with horizontal and vertical axes. Solve: Newtons second law is G sin cos SFx = n . = max SFy = n . - F = may = 0 N Combining the two equations, we get G sin tan tan cosx x ma F . mg . a g . . = = . = The curve is described by y = x2. Its slope a position x is tan., which is also the derivative of the curve. Hence, tan 2 (2 ) x dy x a xg dx = . = . = (b) The acceleration at x = 0.20 m is (2)(0.20)(9.8 m/s2 ) 3.9 m/s2. x a= = 6.72. Model: We will represent the widget as a particle. Visualize: Please refer to Figure CP6.72. Solve: (a) There are only two forces on the widget: the normal force of the table and the gravitational force. (b) Newtons second law along the y-axis is ( net )y y ( G )y y y y ( y ) F = n + F = n - mg = ma .n = m a + g We know what the -vs- ya tgraph looks like. To get the -vs- yn tgraph, we need to add y g to the graph, which amounts to shifting the whole graph up by 9.8 m/s2 , and multiplying by m = 5 kg. (c) The normal force is negative for t > 0.75 s. Physically, this means that the normal force is pointed in the downward direction. In other words, the table is pulling down on the widget rather than pushing up on the widget. It can do this because the widget is glued to the table rather than simply sitting on the table. (d) The weight is a maximum at t = 0 s, when the upward acceleration is maximum. (e) The weight is zero at t = 0.75 s when 9.8 m/s2 . ya = - = -g (f) If not glued down, the widget will fly off the table at t = 0.75 s, the instant at which its weight becomes zero. The table is accelerating in the negative direction so quickly after t = 0.75 s that the widget can stay on only if the table pulls downward on it. That is the significance of the negative value for . y n If the widget is not glued down, the largest downward acceleration it can achieve is the free-fall acceleration. 6.73. Visualize: Solve: (a) The horizontal velocity as a function of time is determined by the horizontal net force. Newtons second law as the x-direction gives ( net )x x cos cos x F = ma = -D . = -bv . = -bv Note that D ...... points opposite to , v .. so the angle . with the x-axis is the same for both vectors, and the x components of both vectors have the same cos. term. As the particle changes direction as it falls, the evolution of the horizontal motion depends only on the horizontal component of the velocity. Thus x x m dv bv dt = - Separating and integrating, ( ) 0 0 vx t t x v x dv b dt v m . = - . ( ) 0 ln x v t b t v m . . . . . = - . . Solving, ( ) 6 0 0 bt Rt m m x v t ve ve p. - - = = (b) The time to reach ( ) 0 1 2 v t = v is found by solving for the time when 6 0 0 1 2 Rt v v e m p. - = Hence ln(2) 6 m t p. R = With . =1.010-3 Ns/m2 , R = 2.010-2 m, and m = 0.033 kg, we get t = 61 s. Assess: The magnitude of the acceleration is 6 (1.1 10 2 s-1 ) . x x a R v v m p. = = - This is a small fraction of the velocity, so a time of about one minute to slow to half the initial speed is reasonable. 6.74. Visualize: Solve: (a) Using the chain rule, x x x . x x a dv dv dx v dv dt dx dt dx = = . .. . = . .. . . .. . (b) The horizontal motion is determined by using Newtons second law in the horizontal direction. Using the free-body diagram at a later time t, ( ) net cos cos x x x F = ma = -D . = -bv . = -bv Note that since D ...... points opposite to , v .. the angle . with the x-axis is the same for both vectors, and the xcomponents of both vectors have the same cos. term. Thus x x x x ma mv dv bv dx = =- Separating and integrating, ( ) ( ) 0 0 vx x x t x v x dv b dx m . = - . ( ) ( ( ) ) x 0 0 v x v b x t x m . - =- - Solving with 0 x = 0, ( ) 0 0 6 x v x v b x v R x m m p. = - = - (c) The marble stops after traveling a distance d when ( ) 0. x v d = Hence 0 v 6 R d m p. = 0 6 d mv p.R . = Using 0 v =10 cm/s, R = 0.50 cm, m =1.010-3 kg, and using . =1.010-3 Ns/m, ( )( ) ( )( ) 3 3 2 3 1.0 10 kg 0.10 m/s 1.1 m 6 1.0 10 Ns/m 5.0 10 m d p - - - = = Assess: The equation for d indicates that a marble with a faster initial velocity travels a further distance. 6.75. Model: We will model the object as a particle, and use the model of drag. Visualize: Solve: (a) We cannot use the constant-acceleration kinematic equations since the drag force causes the acceleration to change with time. Instead, we must use ax = dvx/dt and integrate to find . x v Newtons second law for the object is net k ( ) ( ) ( ) x x x x x x F = SF = n + w + f + D 0 N 0 N 0 N 1 2 4 x x x Av ma m dv dt = + + - = = This can be written 2 4 x x dv A dt v m = We can integrate this from the start 0 ( at 0) x v t= to the end ( at ): xv t 0 2 0 0 1 1 4 4 x x v x t v x x x dv A dt A t v m v v m . = . .- + = Solving for x v gives 0 0 1 4 x x x v v Av t m = + (b) Using 2 0 (1.6 m)(1.4 m) 2.24 m , 20 m/s, x A = = v = and m =1500 kg, we get ( ) 20 m/s 2.24 20 1 4 1500 x v t = + 20 1 0.007467t = + 1 20 m/s1 0.007467 x t v . .. . . = . .. - . . .. . where t is in seconds. We can now obtain the time t for v =10 m/s: 1 20 m/s 1 133.93 20 1 134 s 0.007467 10 m/s 10 t . .. . . . = . .. - . = . - . = . .. . . . When 5 m/s, x v = then t = 402 s. (c) If the only force acting on the object was kinetic friction with, say, k = 0.05, that force would be (0.05)(1500 kg) (9.8 m/s2 ) = 735 N. The drag force at an average speed of 10 m/s is 1 ( )( )2 4 D = 2.24 10 N = 56 N. We conclude that it is not reasonable to neglect the kinetic friction force. 7.1. Visualize: Solve: (a) The weight lifter is holding the barbell in dynamic equilibrium as he stands up, so the net force on the barbell and on the weight lifter must be zero. The barbells have an upward contact force from the weight lifter and the gravitational force downward. The weight lifter has a downward contact force from the barbells and an upward one from the surface. Gravity also acts on the weight lifter. (b) The system is the weight lifter and barbell, as indicated in the figure. (c) 7.2. Visualize: Solve: (a) Both the bowling ball and the soccer ball have a normal force from the surface and gravitational force on them. The interaction forces between the two are equal and opposite. (b) The system consists of the soccer ball and bowling ball, as indicated in the figure. (c) Assess: Even though the soccer ball bounces back more than the bowling ball, the forces that each exerts on the other are part of an action/reaction pair, and therefore have equal magnitudes. Each balls acceleration due to the forces on it is determined by Newtons second law, a = Fnet /m, which depends on the mass. Since the masses of the balls are different, their accelerations are different. 7.3. Visualize: Solve: (a) Both the mountain climber and bag of supplies have a normal force from the surface on them, as well as a gravitational force vertically downward. The rope has gravity acting on it, along with pulls on each end from the mountain climber and supply bag. Both the mountain climber and supply bag also experience a frictional force with the surface of the mountain. In the case of the motionless mountain climber it is static friction, but the sliding supply bag experiences kinetic friction. (b) The system consists of the mountain climber, rope, and bag of supplies, as indicated in the figure. (c) Assess: Since the motion is along the surface, it is convenient to choose the x-coordinate axis along the surface. The free-body diagram of the rope shows pulls that are slightly off the x-axis since the rope is not massless. 7.4. Visualize: Solve: (a) The car and rabbit both experience a normal force and friction from the floor and a gravitational force from the Earth. The push each exerts on the other is a Newtons third law force pair. (b) The system consists of the car and stuffed rabbit, as indicated in the figure. (c) 7.5. Visualize: Please refer to Figure EX7.5. Solve: (a) Gravity acts on both blocks, and where Block A is in contact with the floor there is a normal force and friction. The string tension is the same on both blocks since the rope and pulley are massless, and the pulley is frictionless. There are two third law pairs of forces at the surface where the two blocks meet. Block B pushes against Block A with a normal force, while Block A has a reaction force that pushes back against Block B. There is also friction between the two blocks at the surface. (b) A string that will not stretch constrains the two blocks to accelerate at the same rate but in opposite directions. Block A accelerates down the incline with the same acceleration that Block B has up the incline. The system consists of the two blocks, as indicated in the figure. (c) Assess: The inclined coordinate systems allow the acceleration a to be purely along the x-axis. This is convenient since the one component of a is zero, simplifying the mathematical expression of Newtons second law. 7.6. Visualize: Please refer to Figure EX7.6. Solve: (a) For each block, there is a gravitational force with the earth, a normal force and kinetic friction with the surface, and a tension force due to the rope. (b) The tension in the massless ropes over the frictionless pulley is the same on both blocks. Block A accelerates down the incline with the same acceleration that Block B has up the incline. The system consists of the two blocks, as indicated in the figure. (c) Assess: The inclined coordinate systems allow the acceleration a to be purely along the x-axis. This is convenient since then one component of a is zero, simplifying the mathematical expression of Newtons second law. 7.7. Model: We will model the astronaut and the chair as particles. The astronaut and the chair will be denoted by A and C, respectively, and they are separate systems. The launch pad is a part of the environment. Visualize: Solve: (a) Newtons second law for the astronaut is ( on A ) C on A ( G )A A A 0 N y S F = n - F = m a = ( ) C on A G A A .n = F = m g By Newtons third law, the astronauts force on the chair is ( )( 2 ) 2 A on C Con A A n = n = m g = 80 kg 9.8 m/s = 7.810 N (b) Newtons second law for the astronaut is: ( ) ( ) on A y C on A G A A A S F = n - F = m a ( ) ( ) C on A G A A A A A .n = F + m a = m g + a By Newtons third law, the astronauts force on the chair is ( ) ( )( 2 2 ) 3 A on C Con A A A n = n = m g + a = 80 kg 9.8 m/s +10 m/s =1.610 N Assess: This is a reasonable value because the astronauts acceleration is more than g. 7.8. (a) Visualize: The upper magnet is labeled U, the lower magnet L. Each magnet exerts a long-range magnetic force on the other. Each magnet and the table exert a contact force (normal force) on each other. In addition, the table experiences a normal force due to the surface. (b) Solve: Each object is in static equilibrium with (Fnet ) = 0. Start with the lower magnet. Because U on L G L F = 3(F ) = 6.0 N, equilibrium requires T on L n = 4.0 N. For the upper magnet, L on U U on L F = F = 6.0 N because these are an action/ reaction pair. Equilibrium for the upper magnet requires T on U n = 8.0 N. For the table, action/reaction pairs are L on T n = T on L n = 4.0 N and U on T T on U n = n = 8.0. The tables gravitational force is G T (F ) = 20 N, leaving S on T n = 24 N in order for the table to be in equilibrium. Summarizing, Upper magnet Table Lower magnet G U (F ) = 2.0 N G T (F ) = 20 N G L (F ) = 2.0 N T on U n = 8.0 N U on T n = 8.0 N T on L n = 4.0 N L on U F = 6.0 N L on T n = 4.0 N U on L F = 6.0 N S on T n = 24 N Assess: The result S on T n = 24 N makes sense. The combined gravitational force on the table and two magnets is 24 N. Because the table is in equilibrium, the upward normal force of the surface has to exactly balance the total gravitational force on the table and magnets. 7.9. Model: The car and the truck will be modeled as particles and denoted by the symbols C and T, respectively. The surface of the ground will be denoted by the symbol S. Visualize: Solve: (a) The x-component of Newtons second law for the car is ( on C )x S on C T on C C C S F = F - F = m a The x-component of Newtons second law for the truck is ( ) on T x C on T T T S F = F = m a Using C T T on C C on T a = a = a and F = F , we get ( ) Con S Con T C F F 1 a m . . - . . = . . ( ) C on T T F 1 a m . . . . = . . Combining these two equations, ( ) ( ) Con S Con T Con T C T F F 1 F 1 m m . . . . - . . = . . . . . . ( ) C on T Con S C T C F 1 1 F 1 m m m . . . . . . + . = . . . . . . ( ) T C on T C on S C T F F m m m . . . = . . . + . (4500 N) 2000 kg 3000 N 1000 kg 2000 kg . . = . . = . + . (b) Due to Newtons third law, T on C F = 3000 N. 7.10. Model: The blocks are to be modeled as particles and denoted as 1, 2, and 3. The surface is frictionless and along with the earth it is a part of the environment. The three blocks are our three systems of interest. Visualize: The force applied on block 1 is FA on 1 =12 N. The acceleration for all the blocks is the same and is denoted by a. Solve: (a) Newtons second law for the three blocks along the x-direction is ( ) on 1 x A on 1 2 on 1 1 S F = F - F = m a ( ) on 2 x 1on 2 3 on 2 2 S F = F - F = m a ( ) on 3 x 2 on 3 3 S F = F = m a Adding these three equations and using Newtons third law 2 on 1 1 on 2 3 on 2 2 on 3 (F = F and F = F ), we get ( ) A on1 1 2 3 F = m + m + m a .(12 N) = (1 kg + 2 kg + 3 kg)a .a = 2 m/s2 Using this value of a, the force equation on block 3 gives ( )( 2 ) 2 on 3 3 F = m a = 3 kg 2 m/s = 6 N (b) Substituting into the force equation on block 1, ( )( 2 ) 2 on1 12 N - F = 1 kg 2 m/s 2 on1 . F =10 N Assess: Because all three blocks are pushed forward by a force of 12 N, the value of 10 N for the force that the 2 kg block exerts on the 1 kg block is reasonable. 7.11. Model: The block (B) and the steel cable (C), the two objects of interest to us, are treated like particles. The motion of these objects is governed by the constant-acceleration kinematic equations. Visualize: Solve: Using 2 2 ( ) 1 0 1 0 2 , x x x v = v + a x - x (4.0 m/s)2 0 m2/s2 2 (2.0 m) x = + a . 4.0 m/s2 x a = From the free-body diagram on the block: ( ) on B x C on B B x S F = F = m a ( )( 2 ) C on B . F = 20 kg 4.0 m/s = 80 N Also, according to Newtons third law B on C C on B F = F = 80 N. Newtons second law on the cable is: ( ) on C x ext B on C C x S F = F - F = m a ( 2 ) C .100 N - 80 N = m 4.0 m/s C .m = 5 kg 7.12. Model: The man (M) and the block (B) are interacting with each other through a rope. We will assume the pulley to be frictionless. This assumption implies that the tension in the rope is the same on both sides of the pulley. The system is the man and the block. Visualize: Solve: Clearly the entire system remains in equilibrium since mB > mM. The block would move downward but it is already on the ground. From the free-body diagrams, we can write down Newtons second law in the vertical direction as ( ) ( ) on M R on M G M 0 N y S F = T - F = ( ) ( )( 2 ) R on M G M .T = F = 60 kg 9.8 m/s = 588 N Since the tension is the same on both sides, Bon R Mon R T = T = T = 588 N. 7.13. Model: Together the carp (C) and the trout (T) make up the system that will be represented through the particle model. The fishing rod line (R) is assumed to be massless. Visualize: Solve: Jimmys pull 2 T is larger than the total weight of the fish, so they accelerate upward. They are tied together, so each fish has the same acceleration a. Newtons second law along the y-direction for the carp and the trout is ( on C )y 2 1 ( G )C C S F = T -T - F = m a ( ) ( ) on T y 1 G T T S F = T - F = m a Adding these two equations gives ( ) ( ) ( ) ( )( 2 ) ( )( 2 ) 2 G C G T 2 C T 60 N 1.5 kg 9.8 m/s 3 kg 9.8 m/s 3.533 m/s 1.5 kg 3.0 kg T F F a m m - - - - = = = + + Substituting this value of acceleration back into the force equation for the trout, we find that ( ) ( )( 2 2 ) 1 T T = m a + g = 3 kg 3.533 m/s + 9.8 m/s = 40 N ( )( 2 ) G T T (F ) = m g = 3 kg 9.8 m/s = 29.4 N ( ) ( )( 2 ) G C C F = m g = 1.5 kg 9.8 m/s =14.7 N Thus, ( ) ( ) 2 1 G T G C T > T > F > F . 7.14. Model: The block of ice (I) is a particle and so is the rope (R) because it is not massless. We must therefore consider both the block of ice and the rope as objects in the system. Visualize: Solve: The force ext F .. acts only on the rope. Since the rope and the ice block move together, they have the same acceleration. Also because the rope has mass, ext F on the front end of the rope is not the same as I on R F that acts on the rear end of the rope. Newtons second law along the x-axis for the ice block and the rope is ( ) ( )( 2 ) on I R on I I 10 kg 2.0 m/s 20 N x S F = F = m a = = ( ) on R x ext I on R R S F = F - F = m a ext R on I R .F - F = m a ( )( 2 ) ext R on I R . F = F + m a = 20 N + 0.500 kg 2.0 m/s = 21 N 7.15. Model: The hanging block and the rail car are objects in the systems. Visualize: Solve: The mass of the rope is very small in comparison to the 2000-kg block, so we will assume a massless rope. In that case, the forces 1 T .. and 1 T' .. act as if they are an action/reaction pair. The hanging block is in static equilibrium, with Fnet = 0 N, .. so 1 block T'= m g =19,600 N. The rail car with the pulley is also in static equilibrium: 2 3 1 T + T -T = 0 N Notice how the tension force in the cable pulls both the top and bottom of the pulley to the right. Now, 1 1 T = T'=19,600 N by Newtons third law. Also, the cable tension is 2 3 T = T = T. Thus, 1 2 1 T = T'= 9800 N. 7.16. Visualize: Solve: The rope is treated as two 1.0-kg interacting objects. At the midpoint of the rope, the rope has a tension TB on T = TT on B = T. Apply Newtons first law to the bottom half of the rope to find T. ( ) ( ) G 0 net y B F = = T - F (1.0 kg)(9.80 m/s2 ) 9.8 N B .T = m g = = Assess: 9.8 N is half the gravitational force on the whole rope. This is reasonable since the top half is holding up the bottom half of the rope against gravity. 7.17. Model: The two hanging blocks, which can be modeled as particles, together with the two knots where rope 1 meets with rope 2 and rope 2 meets with rope 3 form a system. All the four objects in the system are in static equilibrium. The ropes are assumed to be massless. Visualize: Solve: (a) We will consider both the two hanging blocks and the two knots. The blocks are in static equilibrium with Fnet = 0 N. .. Note that there are three action/reaction pairs. For Block 1 and Block 2, net F = 0 N .. and we have ( ) 4 G 1 1 T' = F = m g ( ) 5 G 2 2 T' = F = m g Then, by Newtons third law: 4 4 1 T = T' = m g 5 5 2 T = T' = m g The knots are also in equilibrium. Newtons law applied to the left knot is ( ) net 2 1 1 cos 0 N x F = T -T . = ( ) net 1 1 4 1 1 1 sin sin 0 N y F = T . -T = T . - m g = The y-equation gives 1 1 1 T = m g sin. . Substitute this into the x-equation to find 1 1 1 2 1 1 cos sin tan T m g m g . . . = = Newtons law applied to the right knot is ( ) net 3 3 2 cos 0 N x F = T . -T' = ( ) net 3 3 5 3 3 2 sin sin 0 N y F = T . -T = T . - m g = These can be combined just like the equations for the left knot to give 2 3 2 2 3 3 cos sin tan T m g m g . . . '= = But the forces 2 T .. and 2 T' .. are an action/reaction pair, so 2 2T = T'. Therefore, 1 2 2 3 1 1 3 1 tan tan tan tan m g m g m m . . . . = . = 1 ( ) 3 .. = tan- 2tan 20 = 36 We can now use the y-equation for the right knot to find 3 2 3 T = m g sin. = 67 N. 7.18. Visualize: Please refer to Figure P7.18. Solve: Since the ropes are massless we can treat the tension force they transmit as a Newtons third law force pair on the blocks. The connection shown in figure P7.18 has the same effect as a frictionless pulley on these massless ropes. The blocks are in equilibrium as the mass of A is increased until block B slides, which occurs when the static friction on B is at its maximum value. Applying Newtons first law to the vertical forces on block B gives nB = (FG )B = mBg. The static friction force on B is thus ( )s B s B s B f = n = m g. Applying Newtons first law to the horizontal forces on B gives ( )s B A on Bf = T , and the same analysis of the vertical forces on A gives ( ) B on A G A A T = F = m g. Since A on B B on A T = T , we have ( )s B A f = m g, so s B A m g = m g ( )( ) A s B .m = m = 0.60 20 kg =12.0 kg 7.19. Model: The astronaut and the satellite, the two objects in our system, will be treated as particles. Visualize: Solve: The astronaut and the satellite accelerate in opposite directions for 0.50 s. The force on the satellite and the force on the astronaut are an action/reaction pair, so both are 100 N. Newtons second law for the satellite along the x-direction is ( on S )x A on S S S S F = F = m a ( ) A on S 2 S S 100 N 0.156 m/s 640 kg F a m - . = = =- Newtons second law for the astronaut along the x-direction is ( ) on A x S on A A A S F = F = m a S on A A on S 2 A A A 100 N 1.25 m/s 80 kg F F a m m . = = = = Let us first calculate the positions and velocities of the astronaut and the satellite at 1 t = 0.50 s under the accelerations A a and S a : ( ) 1 ( )2 1A 0A 0A 1 0 2 A 1 0 x = x + v t - t + a t - t 1 ( 2 )( )2 2 = 0 m+ 0 m+ 1.25 m/s 0.50 s - 0 s = 0.156 m ( ) 1 ( )2 1S 0S 0S 1 0 2 S 1 0 x = x + v t - t + a t - t 1 ( 2 )( )2 2 = 0 m+ 0 m + -0.156 m/s 0.50 s - 0 s = -0.020 m ( ) 1A 0A A 1 0 v = v + a t - t = 0 m/s + (1.25 m/s2 )(0.50 s - 0 s) = 0.625 m/s ( ) 1S 0S S 1 0 v = v + a t - t = 0 m/s + (-0.156 m/s2 )(0.5 s - 0 s) = -0.078 m/s With 1A x and 1S x as initial positions, 1A v and 1S v as initial velocities, and zero accelerations, we can now obtain the new positions at ( ) 2 1 t - t = 59.5 s : ( ) 2A 1A 1A 2 1 x = x + v t - t = 0.156 m+ (0.625 m/s)(59.5 s) = 37.34 m ( ) 2S 1S 1S 2 1 x = x + v t - t = -0.02 m+ (-0.078 m/s)(59.5 s) = -4.66 m Thus the astronaut and the satellite are ( ) ( ) 2A 2S x - x = 37.34 m - -4.66 m = 42 m apart. 7.20. Model: The block (B) and the steel cable (C), the two objects in the system, are considered particles, and their motion is determined by the constant-acceleration kinematic equations. Visualize: Solve: Using ( ) 1 0 1 0 , v x = v x + ax t - t 4.0 m/s 0 m/s (2.0 s 0 s) x = + a - 2.0 m/s2 x .a = Newtons second law along the x-direction for the block is ( ) on B x C on B B x S F = F = m a = (20 kg)(2.0 m/s2 ) = 40 N ext F acts on the right end of the cable and B on C F acts on the left end. According to Newtons third law, B on C C on B F = F = 40 N. The difference in tension between the two ends of the cable is thus ext B on C F - F =100 N - 40 N = 60 N 7.21. Model: The block (B) and the steel cable (C), the two objects in the system, are treated as particles, and their motion is determined by constant-acceleration kinematic equations. We ignore the hanging shape of the cable. Visualize: Solve: Using ( ) 1 ( )2 1 0 0 1 0 2 1 0 , x x x = x + v t - t + a t - t 1 ( )2 2 4.0 m 0 m 0 m 2.0 s 0 s x = + + a - 2.0 m/s2 x .a = Newtons second law along the x-direction for the block is ( ) ( )( 2 ) on B C on B B 20 kg 2.0 m/s 40 N x x S F = F = m a = = Therefore, the change in tension (T) in the cable from one end to the other is ext B on C F - F =100 N - 40 N = 60 N. The tension in the cable as a function of x is (60 N) ( ) 40 N 1 m T x = + x = (40 + 60x) N with x in meters. x = 0 m is where the cable attaches to the box and x =1.0 m is at the right end of the cable. 7.22. Visualize: Consider a segment of the rope of length y, starting from the bottom of the rope. The weight of this segment of rope is a downward force. It is balanced by the tension force at height y. Solve: The mass m of this segment of rope is the same fraction of the total mass M = 2.2 kg as length y is a fraction of the total length L = 3.0 m. That is, m/M = y/L, from which we can write the mass of the rope segment m M y L = This segment of rope is in static equilibrium, so the tension force pulling up on it is 2 G (2.2 kg)(9.8 m /s ) 7.19 N 3.0 m T F mg Mg y y y L = = = = = where y is in m. The tension increases linearly from 0 N at the bottom ( y = 0 m) to 21.6 N at the top ( y = 3 m). This is shown in the graph. 7.23. Model: Sled A, sled B, and the dog (D) are treated like particles in the model of kinetic friction. Visualize: Solve: The acceleration constraint is ( A )x ( B )x x . a = a = a Newtons second law on sled A is ( ) ( ) on A A G A 0 N y S F = n - F = .. ( ) A G A A .n = F = m g ( ) on A x 1on A A A x S F = T - f = m a .. Using A k A , f = n the x-equation yields 1 on A k A A x T - n = m a 150 N (0.1)(100 kg)(9.8 m/s2 ) (100 kg) x . - = a 0.52 m/s2 x .a = On sled B: ( ) ( ) on B B G B 0 N y S F = n - F = .. ( ) B G B B .n = F = m g ( ) on B x 2 1on B B B x S F = T -T - f = m a .. 1 on B T and 1 on A T act as if they are an action/reaction pair, so 1 on B T =150 N. Using ( )( ) B k B f = n = 0.10 80 kg (9.8 m/s2 ) = 78.4 N, we get ( )( 2 ) 2 T -150 N - 78.4 N = 80 kg 0.52 m/s 2 .T = 270 N Thus the tension 2 2 T = 2.710 N. 7.24. Model: The coffee mug (M) is the only object in the system, and it will be treated as a particle. The model of friction and the constant-acceleration kinematic equations will also be used. Visualize: Solve: The mug and the car have the same velocity. If the mug does not slip, their accelerations will also be the same. Using 2 2 ( ) 1 0 1 0 2 , x x x v = v + a x - x we get 0 m2/s2 (20 m/s)2 2 (50 m) x = + a 4.0 m/s2 x .a = - The static force needed to stop the mug is ( ) ( )( 2 ) net s 0.5 kg 4.0 m/s 2.0 N x x F = - f = ma = - = - s . f = 2.0 N The maximum force of static friction is s max s s G s ( f ) = n = F = mg = (0.50)(0.50 kg)(9.8 m/s2 ) = 2.45 N Since s max s max ( f ) < ( f ) , the mug does not slide. 7.25. Visualize: The car and the ground are denoted by C and S, respectively. Solve: (a) The car has an internal source of energyfuelthat allows it to turn the wheels and exert the force FC on G. .. As the drive wheels turn they push backward against the ground. This is a static friction force C on S F .. because the wheels dont slip against the ground. By Newtons third law, the ground exerts a reaction force Son CF . .. This reaction force is opposite in direction to C on SF , .. hence, is in the forward direction. This is the force that accelerates the car. Houses do not have an internal source of energy that allows them to push sideways against the ground. They also arent on wheels, which let the car slide across the ground with minimal friction. (b) The car presses down against the ground at both the drive wheels (assumed to be the front wheels F, although that is not critical) and the nondrive wheels. For this car, two-thirds of the gravitational force rests on the front wheels. Physically, force Son C F .. is a static friction force. The maximum acceleration of the car on the ground (or concrete surface) occurs when the static friction reaches its maximum possible value. ( ) ( ) S on C s max s F s G F F f n F = = = ( )( )( )( ) 2 23 = 1.00 1500 kg 9.8 m/s = 9800 N Son C 2 max 9800 N 6.533 m/s 1500 kg F a m . = = = 7.26. Model: The starship and the shuttlecraft will be denoted as M and m, respectively, and both will be treated as particles. We will also use the constant-acceleration kinematic equations. Visualize: Solve: (a) The tractor beam is some kind of long-range force FM on m. .. Regardless of what kind of force it is, by Newtons third law there must be a reaction force m on M F .. on the starship. As a result, both the shuttlecraft and the starship move toward each other (rather than the starship remaining at rest as it pulls the shuttlecraft in). However, the very different masses of the two crafts means that the distances they each move will also be very different. The pictorial representation shows that they meet at time 1 t when M1 m1x = x . Theres only one force on each craft, so Newtons second law is very simple. Furthermore, because the forces are an action/reaction pair, 4 M on m m on M tractor beam F = F = F = 4.010 N The accelerations of the two craft are 4 m onM 2 6 4.0 10 N 0.020 m/s 2.0 10 kg M F a M = = = 4 Mon m 2 4 4.0 10 N 2.0 m/s 2.0 10 kg m F a m - = = =- .. Acceleration m a is negative because the force and acceleration vectors point in the negative x-direction. The accelerations are very different even though the forces are the same. Now we have a constant-acceleration problem in kinematics. At a later time 1 t the positions of the crafts are ( ) 1 ( )2 1 2 M1 M0 M0 1 0 2 M 1 0 2 M 1 x = x + v t - t + a t - t = a t ( ) 1 ( )2 1 2 m1 m0 m0 1 0 2 m 1 0 m0 2 m 1 x = x + v t - t + a t - t = x + a t The craft meet when M1 m1x = x , so 1 2 1 2 2 M 1 m0 2 m 1 a t = x + a t ( ) m0 m0 1 2 M m M m 2 2 2 10,000 m 99.5 s 2.02 m/s t x x a a a a . = = = = - + Knowing 1t , we can now find the starships position as it meets the shuttlecraft: 1 2 M1 2 M 1 x = a t = 99 m The starship moves 99 m as it pulls in the shuttlecraft from 10 km away. 7.27. Model: The rock (R) and Bob (B) are the two objects in our system, and will be treated as particles. We will also use the constant-acceleration kinematic equations. Visualize: Solve: (a) Bob exerts a forward force B on R F .. on the rock to accelerate it forward. The rocks acceleration is calculated as follows: ( ) ( ) 2 2 2 2 1R 2 1R 0R 0R R 30 m/s 2 450 m/s 2 21.0 m v v a x a v x = + . . = = = . The force is calculated from Newtons second law: ( )( 2 ) Bon R R R F = m a = .500 kg 450 m/s =225 N Bob exerts a force of 2.3102 N on the rock. (b) Because Bob pushes on the rock, the rock pushes back on Bob with a force R on BF . .. Forces R on B F .. and B on R F .. are an action/reaction pair, so R on B B on R F = F = 225 N. The force causes Bob to accelerate backward with an acceleration equal to ( ) net on B R on B 2 B B B 225 N 3.0 m/s 75 kg x F F a m m = =- =- =- This is a rather large acceleration, but it lasts only until Bob releases the rock. We can determine the time interval by returning to the kinematics of the rock: 1R 1R 0R R R R v v a t a t t v 0.0667 s a = + . = . . . = = At the end of this interval, Bobs velocity is 1B 0B B B v = v + a .t = a .t = -0.20 m/s Thus his recoil speed is 0.20 m/s.
7.28. Model: The boy (B) and the crate (C) are the two objects in our system, and they will be treated in the particle model. We will also use the static and kinetic friction models. Visualize: Solve: The fact that the boys feet occasionally slip means that the maximum force of static friction must exist between the boys feet and the sidewalk. That is, fsB = sBnB. Also kC kC Cf = n . Newtons second law for the crate is ( ) ( ) on C C G C C C 0 N y S F = n - F = .n = m g ( ) on C B on C kC 0 N x S F = F - f = B on C kC kC C kC C . F = f = n = m g Newtons second law for the boy is ( ) ( ) on B B G B B B 0 N y S F = n - F = .n = m g ( ) on B sB C on B 0 N x S F = f - F = C on B sB sB B sB B . F = f = n = m g C on B F .. and B on C F .. are an action/reaction pair, so C on B B on C sB B kC C F = F . m g = m g ( )( ) ( ) sB B 2 C kC 0.8 50 kg 2.0 10 kg 0.2 m m . = = = 7.29. Model: Assume package A and package B are particles. Use the model of kinetic friction and the constant-acceleration kinematic equations. Visualize: Solve: Package B has a smaller coefficient of friction. It will try to overtake package A and push against it. Package A will push back on B. The acceleration constraint is (aA )x = (aB )x = a. Newtons second law for each package is ( ) ( ) ( ) on A B on A G A kA A Bon A A kA A A sin sin cos x F F F f ma F mg mg ma . . . = + - = . + - = S ( ) ( ) on B A on B kB G B B sin x S F = -F - f + F . = m a ( ) A on B kB B B B .-F - m g cos. + m g sin. = m a where we have used A A B B n = m cos. g and n = m cos. g. Adding the two force equations, and using A on B B on A F = F because they are an action/reaction pair, we get ( )( ) kA A kB B A B cos sin m m g a g m m . . + = - + =1.82 m/s2 Finally, using ( ) 1 ( )2 1 0 0x 1 0 2 1 0 x = x + v t - t + a t - t , 1 ( 2 )( )2 2 1 1 2.0 m = 0 m+ 0 m+ 1.82 m/s t - 0 s .t =1.48 s 7.30. Model: The two blocks form a system of interacting objects. Visualize: Please refer to Figure P7.30. Solve: It is possible that the left-hand block (Block L) is accelerating down the slope faster than the right-hand block (Block R), causing the string to be slack (zero tension). If that were the case, we would get a zero or negative answer for the tension in the string. Newtons first law applied to the y-direction on Block L yields ( L ) L ( G )L L L S = 0 = - cos20. = cos20 y F n F n mg Therefore ( ) ( ) ( )( )( 2 ) k L k L L f = m g cos20 = 0.20 1.0 kg 9.80 m/s cos20 =1.84 N A similar analysis of the vertical forces on Block R gives ( ) k R f =1.84 N as well. Using Newtons second law in the x-direction for Block L, ( ) ( ) ( ) L L R on L k L G L S = = - + sin 20 x F ma T f F L R on L L .m a = T -1.84 N + m g sin 20. For Block R, ( ) ( ) R R G R L on R sin 20 1.84 N x SF = m a = F - -T R R L on R .m a = m g sin 20 -1.84 N -T These are two equations in the two unknowns a and L on R R on L T = T = T. Solving them, we obtain a = 2.12 m/s2 and T = 0.61 N. Assess: The tension in the string is positive, and is about 1/3 of the kinetic friction force on each of the blocks, which is reasonable. 7.31. Model: The two ropes and the two blocks (A and B) will be treated as particles. Visualize: Solve: (a) The two blocks and two ropes form a combined system of total mass M = 2.5 kg. This combined system is accelerating upward at a = 3.0 m/s2 under the influence of a force F and the gravitational force -Mg j. Newtons second law applied to the combined system is ( ) ( ) net 32 N y F = F - Mg = Ma. F = M a + g = (b) The ropes are not massless. We must consider both the blocks and the ropes as systems. The force F acts only on block A because it does not contact the other objects. We can proceed to apply the y-component of Newtons second law to each system, starting at the top. Each has an acceleration a = 3.0 m/s2. For block A: ( ) net on A y A 1 on A A F = F - m g -T = m a ( ) 1on A A .T = F - m a + g =19.2 N (c) Applying Newtons second law to rope 1: ( ) net on 1 y A on 1 1 B on 1 1 F = T - m g -T = m a A on 1 T .. and 1 on A T .. are an action/reaction pair. But, because the rope has mass, the two tension forces A on 1 T .. and B on 1 T .. are not the same. The tension at the lower end of rope 1, where it connects to B, is ( ) Bon1 A on1 1 T = T - m a + g =16.0 N (d) We can continue to repeat this procedure, noting from Newtons third law that 1on B Bon1 T = T and 2on B Bon 2 T = T Newtons second law applied to block B is ( ) net on B y 1on B B 2 on B B F = T - m g -T = m a ( ) 2on B 1on B B .T = T - m a + g = 3.2 N 7.32. Model: The two blocks (1 and 2) are the systems of interest and will be treated as particles. The ropes are assumed to be massless, and the model of kinetic friction will be used. Visualize: Solve: (a) The separate free-body diagrams for the two blocks show that there are two action/reaction pairs. Notice how block 1 both pushes down on block 2 (force 1 n..' ) and exerts a retarding friction force 2 top f .. on the top surface of block 2. Block 1 is in static equilibrium 2 1 (a = 0 m/s ) but block 2 is accelerating. Newtons second law for block 1 is ( ) net on 1 1 rope rope 1 0 N x F = f -T = .T = f ( ) net on 1 1 1 1 1 0 N y F = n - m g = .n = m g Although block 1 is stationary, there is a kinetic force of friction because there is motion between block 1 and block 2. The friction model means 1 1 1 .k k f = n = m g Substitute this result into the x-equation to get the tension in the rope: rope 1 k 1 T = f = m g = 3.92 N (b) Newtons second law for block 2 is ( ) net on 2 pull 2 top 2 bot 2 2 x x F T f f a a m m - - = = = ( ) 2 net on 2 2 1 2 2 2 0 m/s y y F n n m g a m m - ' - = = = Forces 1 n.. and 1 n..' are an action/reaction pair, so 1 1 1 . n' = n = m g Substituting into the y-equation gives ( ) 2 1 2 . n = m + m g This is not surprising because the combined weight of both objects presses down on the surface. The kinetic friction on the bottom surface of block 2 is then ( ) 2 bot k 2 k 1 2 f = n = m + m g The forces 1 f .. and 2 top f .. are an action/reaction pair, so 2 bot 1 k 1 f = f = m g. Inserting these friction results into the x-equation gives ( net on 2 ) pull k 1 k ( 1 2 ) 2 2 2 x 2.16 m/s F T mg m mg a m m - - + = = = 7.33. Model: The 3-kg and 4-kg blocks are to be treated as particles. The models of kinetic and static friction and the constant-acceleration kinematic equations will be used. Visualize: Solve: Minimum time will be achieved when static friction is at its maximum possible value. Newtons second law for the 4-kg block is S(Fon 4 )y = n3 on 4 - (FG )4 = 0 N ( ) ( )( 2 ) 3 on 4 G 4 4 .n = F = m g = 4.0 kg 9.8 m/s = 39.2 N ( )( ) s4 s max s 3 on 4 . f = ( f ) = n = 0.60 39.2 N = 23.52 N Newtons second law for the 3-kg block is ( ) ( ) on 3 3 4 on 3 G 3 0 N y S F = n - n - F = . ( ) ( )( 2 ) 3 4on3 G3 n = n + F = 39.2 N + 3.0 kg 9.8 m/s = 68.6 N Friction forces f and s4 f are an action/reaction pair. Thus ( ) on 3 x s3 k3 3 3 S F = f - f = m a s4 k 3 3 3 . f - n = m a . ( )( ) ( ) 3 23.52 N - 0.20 68.6 N = 3.0 kg a 2 3 .a = 3.267 m/s Since block 3 does not slip, this is also the acceleration of block 4. The time is calculated as follows: ( ) 1 ( )2 1 0 0x 1 0 2 1 0 x - x + v t - t + a t - t 1 ( 2 )( )2 2 1 1 .5.0 m = 0 m+ 0 m+ 3.267 m/s t - 0 s .t =1.75 s 7.34. Model: Blocks 1 and 2 make up the system of interest and will be treated as particles. Assume a massless rope and frictionless pulley. Visualize: Solve: The blocks accelerate with the same magnitude but in opposite directions. Thus the acceleration constraint is a2 = a = -a1, where a will have a positive value. There are two real action/reaction pairs. The two tension forces will act as if they are action/reaction pairs because we are assuming a massless rope and a frictionless pulley. Make sure you understand why the friction forces point in the directions shown in the freebody diagrams, especially force 1 f ' .. exerted on block 2 by block 1. We have quite a few pieces of information to include. First, Newtons second law for blocks 1 and 2: ( ) net on 1 x 1 1 k 1 1 1 1 1 F = f -T = n -T = m a = -m a .. ( ) net on 1 1 1 1 1 0 N y F = n - m g = .n = m g ( ) net on 2 x pull 1 2 2 pull 1 k 2 2 2 2 2 F = T - f '- f -T = T - f '- n -T = m a = m a ( ) net on 2 2 1 2 2 1 2 0 N y F = n - n' - m g = .n = n' + m g Weve already used the kinetic friction model in both x-equations. Next, Newtons third law: 1 1 1 n' = n = m g 1 1 k 1 k 1 f '= f = n = m g 1 2 T = T = T Knowing 1n', we can now use the y-equation of block 2 to find 2n . Substitute all these pieces into the two xequations, and we end up with two equations in two unknowns: k 1 1 m g -T = -m a ( ) pull k 1 k 1 2 2 T -T - m g - m + m g = m a Subtract the first equation from the second to get ( ) ( ) pull k 1 2 1 2 T - 3m + m g = m + m a ( ) pull k 1 2 2 1 2 3 1.77 m/s T m mg a m m - + . = = + 7.35. Model: The sled (S) and the box (B) will be treated in the particle model, and the model of friction will be used. Visualize: In the sleds free-body diagram S n is the normal (contact) force on the sled due to the snow. Similarly kS f is the force of kinetic friction on the sled due to snow. Solve: Newtons second law on the box in the y-direction is ( ) ( )( 2 ) S on B G B S on B n - F cos20 = 0 N.n = 10 kg 9.8 m/s cos20 = 92.09 N The static friction force S on B f .. accelerates the box. The maximum acceleration occurs when static friction reaches its maximum possible value. ( )( ) s max S SonB ( f ) = n = 0.50 92.09 N = 46.05 N Newtons second law along the x-direction thus gives the maximum acceleration ( ) S on B G B B f - F sin 20 = m a .46.05 N - (10 kg)(9.8 m/s2 )sin 20 = (10 kg)a .a =1.25 m/s2 Newtons second law for the sled along the y-direction is ( ) S BonS GS n - n - F cos20 = 0 N S BonS S .n = n + m g cos20 = (92.09 N) + (20 kg)(9.8 m/s2 )cos20 = 276.27 N Therefore, the force of friction on the sled by the snow is kS k S f = ( )n = (0.06)(276.27 N) =16.58 N Newtons second law along the x-direction is pull S kS B on S S T - w sin 20 - f - f = m a The friction force Bon S Son B f = f because these are an action/reaction pair. Were using the maximum acceleration, so the maximum tension is ( )( 2 ) ( )( 2 ) max T - 20 kg 9.8 m/s sin 20 -16.58 N - 46.05 N = 20 kg 1.25 m/s max .T =155 N 7.36. Model: The masses m and M are to be treated in the particle model. We will also assume a massless rope and frictionless pulley, and use the constant-acceleration kinematic equations for m and M. Visualize: Solve: Using ( ) 1 ( )2 1 0 0 1 0 2 M 1 0 , y y = y + v t - t + a t - t ( ) 1 ( )2 2 M -1 m = 0 m+ 0 m+ a 6.0 s - 0 s 2 M .a = -0.0556 m/s Newtons second law for m and M is: ( ) ( ) on m y R on m G m m S F = T - F = ma ( ) ( ) on M y R on M G M M S F = T - F = Ma The acceleration constraint is . m M a = -a Also, the tensions are an action/reaction pair, thus R on m R on M T = T . With these, the second law equations are R on M M R on M M T Mg Ma T mg ma - = - =- Subtracting the second from the first gives M M -Mg + mg = Ma + ma M M m M g a g a . + . . = . . . - . (100 kg) 9.8 0.556 99 kg 9.8 0.556 . - . = . . = . + . Assess: Note that 2 m M a = -a = 0.0556 m/s . For such a small acceleration, a mass of 99 kg for m compared to M =100 kg is understandable. 7.37. Model: Use the particle model for the block of mass M and the two massless pulleys. Additionally, the rope is massless and the pulleys are frictionless. The block is kept in place by an applied force F. .. Visualize: Solve: Since there is no friction on the pulleys, 2 3 T = T and 2 5T = T . Newtons second law for mass M is ( )( 2 ) 1 G 1 T - F = 0 N.T = Mg = 10.2 kg 9.8 m/s =100 N Newtons second law for the small pulley is 1 2 3 1 2 3 5 0 N 50 N 2 T + T -T = .T = T = T = = T = F Newtons second law for the large pulley is 4 2 3 5 4 2 3 5 T -T -T -T = 0 N.T = T + T + T =150 N 7.38. Model: Assume the particle model for m1, 2 m , and 3m , and the model of kinetic friction. Assume the ropes to be massless, and the pulleys to be frictionless and massless. Visualize: Solve: Newtons second law for 1 m is 1 G1 11 T - (F ) = m a . Newtons second law for 2 m is ( ) ( ) ( )( ) 2 2 on 2 G 2 2 0 N 2.0 kg 9.8 m/s 19.6 N m y S F = n - F = .n = = on 2 2 k2 2 2 2 k 2 1 2 ( ) (2.0 kg) m x S F = T - f -T = m a .T - n -T = a Newtons second law for 3 m is ( ) ( ) on m3 y 2 G 3 3 3 S F = T - F = m a Since 1 2 3 m , m , and m move together, 1 2 3 a = a = -a = a. The equations for the three masses thus become ( ) ( ) 1 G 1 1 T - F = m a = 1.0 kg a ( ) 2 k 2 1 2 T - n -T = m a = 2.0 kg a ( ) ( ) 2 G 3 3 T - F = -m a = - 3.0 kg a Subtracting the third equation from the sum of the first two equations yields: ( ) ( ) ( ) G 1 k 2 G 3 - F - n + F = 6.0 kg a .-(1.0 kg)(9.8 m/s2 ) - (0.30)(19.6 N) + (3.0 kg)(9.8 m/s2 ) = (6.0 kg)a .a = 2.3 m/s2 7.39. Model: Assume the particle model for the two blocks, and the model of kinetic and static friction. Visualize: Solve: (a) If the mass m is too small, the hanging 2.0 kg mass will pull it up the slope. We want to find the smallest mass that will stick because of friction. The smallest mass will be the one for which the force of static friction is at its maximum possible value: ( s )max s . s f = f = n As long as the mass m is stuck, both blocks are at rest with net F = 0 N. .. Newtons second law for the hanging mass M is ( ) net M M 0 N 19.6 N y F = T -Mg = .T = Mg = For the smaller mass m, ( ) net m s sin 0 N x F = T - f - mg . = ( ) net cos cos y F = n - mg . .n = mg . For a massless string and frictionless pulley, forces m T .. and M T .. act as if they are an action/reaction pair. Thus m MT = T . Mass m is a minimum when ( )s max s s f = n = mg cos. . Substituting these expressions into the xequation, M s T - mg cos. - mg sin. = 0 N ( ) M s 1.83 kg cos sin m T . . g . = = + (b) Because k S < the 1.83 kg block will begin to slide up the ramp, and the 2.0 kg mass will begin to fall, if the block is nudged ever so slightly. Now the net force and the acceleration are not zero. Notice how, in the pictorial representation, we chose different coordinate systems for the two masses. This gives block M an acceleration with only a y-component and block m an acceleration with only an x-component. The magnitudes of the accelerations are the same because the blocks are tied together. But block M has a negative acceleration component y a (vector a.. points down) whereas block m has a positive . x a Thus the acceleration constraint is m M ( ) ( ) , x y a = - a = a where a will have a positive value. Newtons second law for block M is ( ) ( ) net y M y F = T -Mg = M a = -Ma For block m we have ( ) ( ) net k k m sin cos sin x x F = T - f - mg . = T - mg . - mg . = m a = ma In writing these equations, we used Newtons third law to write m M T = T = T. Also, we noticed that the yequation and the friction model for block m dont change, except for s becoming k , so we already know k f from part (a). Notice that the tension in the string is not the gravitational force Mg. We have two equations in the two unknowns T and a: T -Mg = -Ma ( ) k T - cos. + sin. mg = ma Subtracting the second equation from the first to eliminate T, ( ) k -Mg + cos. + sin. mg = -Ma - ma = -(M + m)a ( ) k 2 cos sin 1.32 m/s M m a g M m - . + . . = = + 7.40. Model: Assume the particle model for the two blocks. Visualize: Solve: (a) The slope is frictionless, so the blocks stay in place only if held. Once m is released, the blocks will move one way or the other. As long as m is held, the blocks are in static equilibrium with Fnet = 0 N. .. Newtons second law for the hanging block M is ( ) net on M M M 0 N 19.6 N y F = T -Mg = .T = Mg = By Newtons third law, M m T = T = T =19.6 N is the tension in the string. (b) The free-body diagram shows box m after it is released. Whether it moves up or down the slope depends on whether the acceleration a is positive or negative. The acceleration constraint is m M ( ) ( ) . x y a = a = - a Newtons second law for each system gives ( ) ( ) net on m m sin35 x x F = T - mg = m a = ma ( ) ( ) net on M y M y F = T -Mg = M a = -Ma We have two equations in two unknowns. Subtract the second from the first to eliminate T: -mg sin35 + Mg = (m + M )a a M msin35 g 0.481 m/s2 M m - . = = - + Since a < 0 m/s2 , the box accelerates down the slope. (c) It is now straightforward to compute T = Mg -Ma = 21 N. Notice how the tension is larger than when the blocks were motionless. 7.41. Model: Assume the particle model for the book (B) and the coffee cup (C), the models of kinetic and static friction, and the constant-acceleration kinematic equations. Visualize: Solve: (a) Using 2 2 ( ) 1 0 1 0 2 , x x v = v + a x - x 2 2 ( )2 ( ) 1 0 m /s = 3.0 m/s + 2a x 2 2 1 .ax = -4.5 m /s To find 1x , we must first find a. Newtons second law for the book and the coffee cup is ( ) ( ) ( )( 2 ) on B B G B B cos20 0 N 1.0 kg 9.8 m/s cos20 9.21 N y S F = n - F = .n = = ( ) ( ) on B k G B B B sin 20 x S F = -T - f - F = m a ( ) ( ) on C y G C C C S F = T - F = m a The last two equations can be rewritten, using C B a = a = a, as k B B B -T - n - m g sin 20 = m a C C T - m g = m a Adding the two equations, ( ) ( ) ( ) C B C B k a m + m = -g m + m sin 20 - 9.21 N .(1.5 kg)a = -(9.8 m/s2 )..0.500 kg + (1.0 kg)sin 20.. - (0.20)(9.21 N) .a = -6.73 m/s2 Using this value for a, we can now find 1 x as follows: 2 2 2 2 1 2 4.5 m /s 4.5 m /s 0.67 m 6.73 m/s x a - - = = = - (b) The maximum static friction force is ( )( ) s max s B ( f ) = n = 0.50 9.21 N = 4.60 N. Well see if the force s f needed to keep the book in place is larger or smaller than s max ( f ) . When the cup is at rest, the string tension is C T = m g. Newtons first law for the book is ( ) on B s B s C B sin 20 sin 20 0 x S F = f -T - w = f - m g - m g = . fs = (MC + MB sin 20) g = 8.25 N Because s smax f > ( f ) , the book slides back down. 7.42. Model: Use the particle model for the cable car and the counterweight. Assume a massless cable. Visualize: Solve: (a) Notice the separate coordinate systems for the cable car (object 1) and the counterweight (object 2). Forces 1 T .. and 2 T .. act as if they are an action/reaction pair. The braking force B F .. works with the cable tension 1 T .. to allow the cable car to descend at a constant speed. Constant speed means dynamic equilibrium, so Fnet = 0 .. N for both systems. Newtons second law for the cable car is ( ) net on 1 1 B 1 1 sin 0 N x F = T + F - m g . = ( ) net on 1 1 1 1 cos 0 N y F = n - m g . = Newtons second law for the counterweight is ( ) net on 2 2 2 2 sin 0 N x F = m g . -T = ( ) net on 2 2 2 2 cos 0 N y F = n - m g . = From the x-equation for the counterweight, 2 2 2 T = m g sin. . By Newtons third law, 1 2T = T . Thus the x-equation for the cable car becomes B 1 1 1 1 1 2 2 F = m g sin. -T = m g sin. - m g sin. = 3770 N (b) If the brakes fail, then B F = 0 N. The car will accelerate down the hill on one side while the counterweight accelerates up the hill on the other side. Both will have negative accelerations because of the direction of the acceleration vectors. The constraint is 1 2 , x x a = a = a where a will have a negative value. Using 1 2 T = T = T, the two x-equations are ( ) net on 1 1 1 1 1 1 sin x x F = T - m g . = m a = m a ( ) net on 2 2 2 2 2 2 sin x x F = m g . -T = m a = m a Note that the y-equations arent needed in this problem. Add the two equations to eliminate T: ( ) 1 1 2 1 2 -m g sin. + m g sin. = m + m a 1 1 2 2 2 1 2 a m sin m sin g 0.991 m/s m m . - . . = - = - + Now we have a problem in kinematics. The speed at the bottom is calculated as follows: 2 2 ( ) 1 0 1 0 1 v = v + 2a x - x = 2ax ( 2 )( ) 1 1 .v = 2ax = 2 -0.991 m/s -400 m = 28.2 m/s Assess: A speed of approximately 60 mph as the cable car travels a distance of 2000 m along a frictionless slope of 30 is reasonable. 7.43. Model: Assume the cable mass is negligible compared to the car mass and that the pulley is frictionless. Use the particle model for the two cars. Visualize: Please refer to Figure P7.43. Solve: (a) The cars are moving at constant speed, so they are in dynamic equilibrium. Consider the descending car D. We can find the rolling friction force on car D, and then find the cable tension by applying Newtons first law. In the y-direction for car D, (Fnet )y = 0 = nD - (FG )D cos35 D D .n = m g cos35 So the rolling friction force on car D is ( )R D R D R D f = n = m g cos35 Applying Newtons first law to car D in the x-direction, ( ) ( ) ( ) A on D R D G D sin35 0 net x F = T + f - F = Thus ( )( )( ) A on D D R D 2 3 sin35 cos35 1500 kg 9.80 m/s sin35 0.020cos35 8.2 10 N T = m g - m g = - = (b) Similarly, we find that for car A, ( )R A R A f = m g cos35. In the x-direction for car A, ( ) ( ) ( ) net motor D on A R A G A sin35 0 x F = T +T - f - F = ( ) motor A R A D R D .T = m g sin35 + m g cos35 - m g sin35 - m g cos35 Here, we have used A on D D on A T = T . If we also use A D m = m , then 2 motor R A T = 2 m g cos35 = 4.810 N. Assess: Careful examination of the free-body diagrams for cars D and A yields the observation that motor R A T = 2(F ) in order for the cars to be in dynamic equilibrium. It is a tribute to the design that the motor must only provide such a small force compared to the tension in the cable connecting the two cars. 7.44. Model: The painter and the chair are treated as a single object and represented as a particle. We assume that the rope is massless and that the pulley is massless and frictionless. Visualize: Solve: If the painter pulls down on the rope with force F, Newtons third law requires the rope to pull up on the painter with force F. This is just the tension in the rope. With our model of the rope and pulley, the same tension force F also pulls up on the painters chair. Newtons second law for (painter + chair) is ( ) G P C 2F - F = m + m a ( 1 ) ( ) ( ) 1 ( )( ) 2 P C P C 2 P C . F = .. m + m a + m + m g.. = m + m a + g ( 1 )( )( 2 2 ) 2 2 = 70 kg +10 kg 0.20 m/s + 9.8 m/s = 4.010 N Assess: A force of 400 N, which is approximately one-half the total gravitational force, is reasonable since the upward acceleration is small. 7.45. Model: Use the particle model for the tightrope walker and the rope. The rope is assumed to be massless, so the tension in the rope is uniform. Visualize: Solve: Newtons second law for the tightrope walker is ( ) ( )( 2 2 ) 3 R onW G R onW F - F = ma. F = m a + g = 70 kg 8.0 m/s + 9.8 m/s =1.2510 N Now, Newtons second law for the rope gives ( ) on R Won R sin sin 0 N y S F = T . + T . - F = 3 Won R R on W 1.25 10 N 3.6 103 N 2sin10 2sin10 2sin10 F F T . = = = = We used Won R R onW F = F because they are an action/reaction pair. 7.46. Model: Use the particle model for the wedge and the block. Visualize: The block will not slip relative to the wedge if they both have the same horizontal acceleration a. Note: n1 on 2 = n2 on 1. Solve: The y-component of Newtons second law for block 2 m is ( ) ( ) 2 on 2 1 on 2 G 2 1 on 2 cos 0 N y cos F n . F n m g . S = - = . = Combining this equation with the x-component of Newtons second law yields: ( ) 1 on 2 on 2 1 on 2 2 2 sin sin tan x F n m a a n g m . S = . = . = = . Now, Newtons second law for the wedge is ( ) on 1 2 on 1 1 sin x S F = F - n . = m a 1 2 on 1 1 2 1 2 1 2 . F = m a + n sin. = m a + m a = (m + m )a = (m + m )g tan. 7.47. Model: Treat the basketball player (P) as a particle, and use the constant-acceleration kinematic equations. Visualize: Solve: (a) While in the process of jumping, the basketball player is pressing down on the floor as he straightens his legs. He exerts a force P on F F on the floor. The player experiences a gravitational force ( ) G P F .. as well as a normal force by the floor n..F on P. The only force that the floor experiences is the one exerted by the player P on FF . .. (b) The player standing at rest exerts a force P on F F .. on the floor. The normal force F on P n.. is the reaction force to P on FF . .. But F on P P on Fn = F , so net F = 0 N. .. When the basketball player accelerates upward by straightening his legs, his speed has to increase from zero to 1y v with which he leaves the floor. Thus, according to Newtons second law, there must be a net upward force on him during this time. This can be true only if ( ) F on P G P n > F . In other words, the player presses on the floor with a force P on F F larger than the gravitational force on him, which is equal to his weight. The reaction force F on P n.. then exceeds his weight and accelerates him upward until his feet leave the floor. (c) The height of 80 cm = 0.80 m is sufficient to determine the speed 1y v with which he leaves the floor. Once his feet are off the floor, he is simply in free fall, with 1 a = -g. From kinematics, 2 2 ( ) 2 1 1 2 1 2 y y v = v + a y - y 2 2 2 ( )( ) 1 0 m /s 2 0.80 m y. = v + -g ( ) 1 2 0.80 m 3.96 m/s y.v = g = The basketball player reached 1 4.0 m/s y v = by accelerating from rest through a distance of 0.60 m. (d) Assuming 0 a to be constant during the jump, we find 2 2 ( ) 2 2 ( ) 1 0 0 1 0 0 1 2 0 m/s2 0 m y y v = v + a y - y = + a y - ( ) ( ) 2 2 1 2 0 1 3.96 m/s 13.1 m/s 2 2 0.60 m y v a y . = = = (e) The scale reads the value of F on P n , the force exerted by the scale on the player. Before jumping, ( ) ( ) ( )( 2 ) F on P G P F on P G P n - F = 0 N.n = F = mg = 100 kg 9.8 m/s = 980 N While accelerating up, nF on P - mg = ma0 .nF on P = ma0 + mg mg 1 a0 g . . = . + . . . (980 N) 1 13.1 2290 N 9.8 = . + . = . . . . After leaving the scale, F on P n = 0 N because there is no contact with the scale. 7.48. A 1.0 kg wood block is placed on top of a 2.0 kg wood block. A horizontal rope pulls the 2.0 kg block across a frictionless floor with a force of 21.0 N. Does the 1.0 kg block on top slide? Visualize: Solve: The 1.0 block is accelerated by static friction. It moves smoothly with the lower block if fs < ( fs )max . It slides if the force that would be needed to keep it in place exceeds s max ( f ) . Begin by assuming the blocks move together with common acceleration a. Newtons second is on 1 s 1 on 2 pull s 2 Top block: ( ) Bottom block: ( ) x x F f ma F T f ma = = = - = S S Adding these two equations gives pull 1 2 T = (m + m )a, or a = (21.0 N)/(1.0 kg + 2.0 kg) = 7.0 m/s2. The static friction force needed to accelerate the top block at 7.0 m/s2 is 2 s 1 f m a = (1.0 kg)(7.0 m/s ) = 7.0 N To find the maximum possible static friction force s max s 1 ( f ) = n , the y-equation of Newtons second law for the top block shows that 1 1 n = m g. Thus 2 s max s 1 ( f ) = m g = (0.50)(1.0 kg)(9.80 m/s ) = 4.9 N Because 7.0 N > 4.9 N, static friction is not sufficient to accelerate the top block. It slides. 7.49. A 1.0 kg wood block is placed behind a 2.0 kg wood block on a horizontal table. The coefficients of kinetic friction with the table are 0.3 for the 1.0 kg block and 0.5 for the 2.0 kg block. The 1.0 kg block is pushed forward, against the 2.0 block, and released with a speed of 2.0 m/s. How far do the blocks travel before stopping? Visualize: Solve: The 2.0 kg block in front has a larger coefficient of friction. Thus the 1.0 kg block pushes against the rear of the 2.0 kg block and, in reaction, the 2.0 kg block pushes backward against the 1.0 kg block. Theres no vertical acceleration, so n1 = m1g and 2 2 n = m g, leading to 1 1 1 f = m g and 2 2 2 . f = m g Newtons second law along the x-axis is on 1 2 on 1 1 2 on 1 1 1 1 on 2 1 on 2 2 2 on 1 2 2 2 1 kg block: ( ) 2 kg block: ( ) x x F F f F mg ma F F f F mg ma = - - = - - = = - = - = S S where we used 1 2 a = a = a. Also, 1 on 2 2 on 1 F = F because they are an action/reaction pair. Adding these two equations gives 1 1 2 2 1 2 1 1 2 2 2 2 1 2 ( ) ( ) (0.3)(1.0 kg) + (0.5)(2.0 kg) 9.80 m/s 4.25 m/s 1.0 kg + 2.0 kg m m g m m a a m m g m m - + = + + = - = - = - + We can now use constant-acceleration kinematics to find 2 2 2 2 0 1 0 1 0 1 2 0 2 ( ) (2.0 m/s) 0.47 m 2 2(4.25 m/s) x x x v v v a x x xa = = + - . = - = - = - 7.50. Model: Treat the ball of clay and the block as particles. Visualize: Solve: (a) Forces C on B F .. and B on C F .. are an action/reaction pair, so FB on C = FC on B. Note that B C a . a because the clay is decelerating while the block is accelerating. Newtons second law in the x-direction is on C B on C C C on B C on B B on C B B Clay: ( ) Block: ( ) x x F F ma F F F ma = - = = = = S S Equating the two expressions for B on C F gives B C B C a m a m = - Turning to kinematics, the velocity of each after .t is C 1 C 0 C B 1 B 0 B B ( ) ( ) ( ) ( ) v v a t v v a t a t = + . = + . = . But C 1 B 1 (v ) = (v ) because the clay and the block are moving together after .t has elapsed. Equating these two expressions gives C 0 C B (v ) + a .t = a .t, from which we find C 0 C B a a (v ) t = - . We can now equate the two expressions for Ca : B C 0 C 0 2 B B B C BC ( ) ( ) / (10 m/s)(0.01 s) 100 m/s 1 / 1 (900 g)(100 g) m a a v a v t m t mm . - = - . = = = . + + Then 2 C B a = -9a = -900 m/s . With the acceleration now known, we can use either kinematic equation to find 2 C 1 B 1 (v ) = (v ) = (100 m/s )(0.010 s) =1.0 m/s (b) 2 C on B B B F = m a = (0.90 kg)(100 m/s ) = 90 N. (c) 2 B on C C C F = m a = (0.10 kg)(900 m/s ) = 90 N. Assess: The two forces are of equal magnitude, as expected from Newtons third law. 7.51. Model: Use the particle model for the two blocks. Assume a massless rope, and massless, frictionless pulleys. Visualize: Note that for every meter block 1 moves forward, one meter is provided to block 2. So each rope on 2 m has to be lengthened by one-half meter. Thus the acceleration constraint is 1 2 2 1a = - a . Solve: Newtons second law for block 1 is 1 1T = m a . Newtons second law for block 2 is G 2 2 2 2T - (F ) = m a . Combining these two equations gives ( ) ( 1 ) 1 1 2 2 2 1 2 m a - m g = m - a [ ] 1 1 2 2 .a 4m + m = 2m g 2 1 1 2 2 4 a m g m m . = + where we have used 1 2 2 1a = - a . Assess: If 1 m = 0 kg, then 2 a = -g. This is what is expected for a freely falling object. 7.52. Model: Use the particle model for the two blocks. Assume a massless rope and massless, frictionless pulleys. Visualize: For every one meter that the 1.0-kg block goes down, each rope on the 2.0-kg block will be shortened by onehalf meter. Thus the acceleration constraint is a1 = -2a2. Solve: Newtons second law for the two blocks is 2 2 G 1 1 1 2T = m a T - (F ) = m a Since 1 2 a = -2a , the above equations become 2 2 1 1 2 2T = m a T - m g = m (-2a ) 2 ( ) 2 1 2 1 2 2 .m a + m a = m g ( )( ) ( ) 2 1 2 2 2 1 2 2 1.0 kg 9.8 m/s 3.3 m/s 4 2.0 kg 4.0 kg a m g m m . = = = + + Assess: If 1 m = 0 kg, then 2 2 a = 0 m/s , which is expected. 7.53. Model: The hamster of mass m and the wedge with mass M will be treated as objects 1 and 2, respectively. They will be treated as particles. Visualize: The scale is denoted by the letter s. Solve: (a) The reading of the scale is the magnitude of the force 2 n.. that the scale exerts upward. There are two action/reaction pairs. Initially the hamster of mass m is stuck in place and is in static equilibrium with Fnet = 0 .. N. Because of the shape of the blocks, it is not clear whether the scale has to exert a horizontal friction force s on 2 f .. to prevent horizontal motion. Weve included one just in case. Newtons second law for the hamster is ( ) net on 1 2 on 1 2 on 1 sin 0 N sin x F = mg . - f = . f = mg . ( ) net on 1 1 cos 0 N y F = n - mg . = 1 .n = mg cos. For the wedge, we see from Newtons third law that 1 1 n' = n = mg cos. and that 2 on1 1on 2 f = f = mg sin. . Using these equations, Newtons second law for the wedge is ( ) net on 2 1on 2 s on 2 1 cos sin x F = f . + f - n' . s on 2 = mg sin. cos. + f - mg cos. sin. = 0 N s on 2 . f = 0 N ( ) net on 2 2 1 1on 2 cos sin y F = n - n' . - f . -Mg 2 2 2 = n - mg cos . - mg sin . -Mg = 0 N ( 2 2 ) ( ) 2 .n = mg cos . + sin . + Mg = M + m g = (0.800 kg + 0.200 kg)(9.8 m/s2 ) = 9.80 N First we find that s on 2 f = 0 N, so no horizontal static friction is needed to prevent motion. More interesting, the scale reading is (M + m) g which is the total gravitational force resting on the scale. This is the expected result. (b) Now suppose that the hamster is accelerating down the wedge. The total mass is still M + m, but is the reading still (M + m)g ? The frictional forces between the systems 1 and 2 have now vanished, and system 1 now has an acceleration. However, the acceleration is along the hamsters x-axis, so 2 1 0 m/s . y a = The hamsters yequation is still ( ) net on 1 1 1 cos 0 N cos y F = n - mg . = .n = mg . We still have 1 1 n' = n = mg cos. , so the y-equation for block 2 2 2 (with 0 m/s ) y a = is ( ) 2 net on 2 2 1 2 cos cos 0 N y F = n - n' . -Mg = n - mg . -Mg = 2 ( 2 ) 2 .n = mg cos . + Mg = M + mcos . g = 8.99 N Assess: The scale reads less than it did when the hamster was at rest. This makes sense if you consider the limit . .90, in which case cos. .0. If the face of the wedge is vertical, then the hamster is simply in free fall and can have no effect on the scale (at least until impact!). So for . = 90 we expect the scale to record Mg only, and that is indeed what the expression for 2 n gives. 7.54. Model: The hanging masses m1, m2 , and m3 are modeled as particles. Pulleys A and B are massless and frictionless. The strings are massless. Visualize: Solve: (a) The length of the string over pulley B is constant. Therefore, ( ) ( ) B 3 B A B A B 3 B y - y + y - y = L . y = 2y - y - L The length of the string over pulley A is constant. Thus, ( ) ( ) A 2 A 1 A A 1 2 y - y + y - y = L = 2y - y - y ( ) B 3 B 1 2 A .2 2y - y - L - y - y = L 3 2 1 .2y + y + y = constant This constraint implies that 2 dy3 dy2 dy1 0 m/s dt dt dt + + = 3 2 1 2 y y y = v + v + v Also by differentiation, 2 3 2 1 2 0 m/s. y y y a + a + a = (b) Newtons second law for the masses 3 2 1 m , m , m , and pulley A is B 3 3 3 y A 2 2 2 y A 1 1 1y T - m g = m a T - m g = m a T - m g = m a B A T - 2T = 0 N The pulley equation is zero because the pulley is massless. These four equations plus the acceleration constraint are five equations for the five unknowns (two tensions and three accelerations). To solve for TA, multiply the 3 m equation by 2, substitute B A 2T = 4T , then divide each of the mass equations by the mass. This gives the three equations A 3 3 A 2 2 A 1 1 4 / 2 2 / / y y y T m g a T m g a T m g a - = - = - = If these three equations are added, the right side adds to zero because of the acceleration constraint. Thus 3 2 2 A A 3 2 2 ( 4/ 1/ 1/ ) 4 0 4 ( 4/ 1/ 1/ ) m m m T g T g m m m + + - = . = + + (c) Using numerical values, we find A T =18.97 N. Then 2 1 A 1 2 2 A 2 2 3 A 3 / 2.2 m/s / 2.9 m/s 2 / 0.32 m/s y y y a T m g a T m g a Tm g = - =- = - = = - =- (d) 3 1 2 , m = m + m so it looks at first like 3 m should hang in equilibrium. For it to do so, tension B T would need to equal 3 m g. However, B T is not 1 2 (m + m )g because masses 1 m and 2 m are accelerating rather than hanging at rest. Consequently, tension B T is not able to balance the weight of 3m . 8.1. Model: The model rocket and the target will be treated as particles. The kinematics equations in two dimensions apply. Visualize: Solve: For the rocket, Newtons second law along the y-direction is ( ) ( ) ( )( ) net R R 2 2 R R 1 115 N 0.8 kg 9.8 m/s 8.95 m/s 0.8 kg y F F mg ma a F mg m = - = . = - = .. - .. = Using the kinematic equation ( ) 1 ( )2 1R 0R 0R 1R 0R 2 R 1R 0R ( ) , y y = y + v t - t + a t - t 1 ( 2 )( )2 2 1R 30 m = 0 m+ 0 m+ 8.95 m/s t - 0 s 1R .t = 2.589 s For the target 1T 1R (noting t = t ), ( ) 1 ( )2 1T 0T 0T 1T 0T 2 T 1T 0T ( )x x = x + v t - t + a t - t = 0 m+ (15 m/s)(2.589 s - 0 s) + 0 m = 39 m You should launch when the target is 39 m away. Assess: The rocket is to be fired when the target is at 0Tx . For a net acceleration of approximately 9 m/s2 in the vertical direction and a time of 2.6 s to cover a vertical distance of 30 m, a horizontal distance of 39 m is reasonable. 8.2. Model: The model rocket will be treated as a particle. Kinematic equations in two dimensions apply. Air resistance is neglected. Visualize: The horizontal velocity of the rocket is equal to the speed of the car, which is 3.0 m/s. Solve: For the rocket, Newtons second law along the y-direction is: (Fnet )y = FR - mg = maR ( ) ( )( ) 1 8.0 N 0.5 kg 9.8 m/s2 0.5 kg y .a = .. - .. = 6.2 m/s2 Thus using ( ) 1 ( )2 1 0 0 1 0 2 1 0 ( ) , y y y = y + v t - t + a t - t ( ) 1 ( 2 )( )2 2 1R 20 m = 0 m+ 0 m+ 6.2 m/s t - 0 s ( ) ( 2 ) 2 1 . 20 m = 3.1 m/s t 1 .t = 2.54 s Since 1 t is also the time for the rocket to move horizontally up to the hoop, ( ) 1 ( )2 1 0 0 1 0 2 1 0 ( )x x x = x + v t - t + a t - t = 0 m+ (3.0 m/s)(2.54 s - 0 s) + 0 m = 7.6 m Assess: In view of the rockets horizontal speed of 3.0 m/s and its vertical thrust of 8.0 N, the above-obtained value for the horizontal distance is reasonable. 8.3. Model: The asteroid and the giant rocket will be treated as particles undergoing motion according to the constant-acceleration equations of kinematics. Visualize: Solve: (a) The time it will take the asteroid to reach the earth is 6 displacement 4.0 10 km 2.0 105 s 56 h velocity 20 km/s = = = (b) The angle of a line that just misses the earth is 1 1 6 0 0 tan R tan R tan 6400 km 0.092 y y 4.0 10 km . . - - . . . . = . = . . = . . = . . . . (c) When the rocket is fired, the horizontal acceleration of the asteroid is 9 2 10 5.0 10 N 0.125 m/s 4.0 10 kg x a = = (Note that the mass of the rocket is much smaller than the mass of the asteroid and can therefore be ignored completely.) The velocity of the asteroid after the rocket has been fired for 300 s is ( ) ( 2 )( ) 0 0 0 m/s 0.125 m/s 300 s 0 s 37.5 m/s x x x v = v + a t - t = + - = After 300 s, the vertical velocity is 2 104 m/s y v = and the horizontal velocity is 37.5 m/s. x v = The deflection due to this horizontal velocity is 1 4 tan tan 37.5 m/s 0.107 2 10 m/s x y v v . . -. . = . = . . = . . That is, the earth is saved. 8.4. Model: We are using the particle model for the car in uniform circular motion on a flat circular track. There must be friction between the tires and the road for the car to move in a circle. Visualize: Solve: The centripetal acceleration is 2 ( )2 2 25 m/s 6.25 m/s 100 m r a v r = = = The acceleration points to the center of the circle, so the net force is ( )( ) ( ) 1500 kg 6.25 m/s2 , toward center 9380 N, toward center r F = ma = = .. .. This force is provided by static friction s 9.4 kN r f = F = 8.5. Model: We will use the particle model for the car which is in uniform circular motion. Visualize: Solve: The centripetal acceleration of the car is 2 ( )2 2 15 m/s 4.5 m/s 50 m r a v r = = = The acceleration is due to the force of static friction. The force of friction is ( )( 2 ) s 1500 kg 4.5m s 6750 N r f = ma = = = 6.8 kN. Assess: The model of static friction is ( )s max s s f = n = mg mg 15,000 N since s 1 for a dry road surface. We see that ( ) s smax f < f , which is reasonable. 8.6. Model: Treat the block as a particle attached to a massless string that is swinging in a circle on a frictionless table. Visualize: Solve: (a) The angular velocity and speed are 75 rev 2 rad 471.2 rad/min min 1 rev p . = = (0.50 m)(471.2 rad min) 1min 3.93 m/s 60 s tv = r. = = The tangential velocity is 3.9 m/s. (b) The radial component of Newtons second law is 2 r F T mv r S = = Thus ( )( )2 3.93 m/s 0.20 kg 6.2 N 0.50 m T= = 8.7. Solve: Newtons second law is Fr = mar = mr. 2. Substituting into this equation yields: ( )( ) 8 31 11 8.2 10 N 9.1 10 kg 5.3 10 m r F mr . - - - = = 4.37 1016 rad/s 4.37 1016 rad 1 rev 6.6 1015 rev/s s 2p rad = = = Assess: This is a very high number of revolutions per second. 8.8. Model: The vehicle is to be treated as a particle in uniform circular motion. Visualize: On a banked road, the normal force on a vehicle has a horizontal component that provides the necessary centripetal acceleration. The vertical component of the normal force balances the gravitational force. Solve: From the physical representation of the forces in the r-z plane, Newtons second law can be written 2 sin r F n mv r S = . = cos 0 cos z SF = n . - mg = .n . = mg Dividing the two equations and making the conversion 90 km h = 25 m/s yields: ( ) ( ) 2 2 2 25 m/s tan 0.128 7.3 9.8 m/s 500 m v rg . = = = .. = Assess: Such a banking angle for a speed of approximately 55 mph is clearly reasonable and within our experience as well. 8.9. Model: The motion of the moon around the earth will be treated through the particle model. The circular motion is uniform. Visualize: Solve: The tension in the cable provides the centripetal acceleration. Newtons second law is 2 2 moon 2 r F T mr mrT p . . . = = = . . . . S ( )( ) 2 7.36 1022 kg 3.84 108 m 2 1 day 1 h 2.01 1020 N 27.3 days 24 h 3600 s . p . = . . = . . Assess: This is a tremendous tension, but clearly understandable in view of the moons large mass and the large radius of circular motion around the earth. 8.10. Model: Model the ball as a particle in uniform circular motion. Rolling friction is ignored. Visualize: Solve: The track exerts both an upward normal force and an inward normal force. From Newtons second law, 2 Fnet = n2 = mr. ( )( ) 2 0.030 kg 0.20 m 60 rev 2 rad 1min 0.24 N min 1 rev 60 s . p . = . . = . . 8.11. Model: The satellite is considered to be a particle in uniform circular motion around the moon. Visualize: Solve: The radius of the moon is 1.738106 m and the satellites distance from the center of the moon is the same quantity. The angular velocity of the satellite is 2 2 rad 1min 9.52 10 4 rad/s T 110 min 60 s p p . = = = - and the centripetal acceleration is ( )( )2 1.738 106 m 9.52 10 4 rad/s 2 1.58 m/s2 r a = r. = - = The acceleration of a body in orbit is the local g experienced by that body. 8.12. Model: The earth is considered to be a particle in uniform circular motion around the sun. Solve: The earth orbits the sun in 365 days and is 1.51011 m from the sun. The angular velocity and centripetal acceleration are 2 rad 1 day 1h 2.0 10 7 rad/s 365 days 24 h 3600 s p . = = - ( )( )2 1.5 1011 m 2.0 10 7 rad/s 2 6.0 10 3 m/s2 r a = g = r. = - = - Assess: The smallness of this acceleration due to gravity is essentially due to the large earth-sun distance. 8.13. Model: Use the particle model for the car which is undergoing circular motion. Visualize: Solve: The car is in circular motion with the center of the circle below the car. Newtons second law at the top of the hill is ( ) 2 r G r r r F F n mg n ma mv r S = - = - = = v2 r g n m . = . - . . . . . Maximum speed is reached when n = 0 and the car is beginning to lose contact with the road. ( )( 2 ) max v = rg = 50 m 9.8 m/s = 22 m/s Assess: A speed of 22 m/s is equivalent to 49 mph, which seems like a reasonable value. 8.14. Model: The passengers are particles in circular motion. Visualize: Solve: The center of the circle of motion of the passengers is directly above them. There must be a net force pointing up that provides the needed centripetal acceleration. The normal force on the passengers is their weight. Ordinarily their weight is FG , so if their weight increases by 50%, G n =1.5 F . Newtons second law at the bottom of the dip is ( ) 2 G G 1.5 1 0.5 r F n F F mg mv r S = - = - = = .v = 0.5gr = 0.5(9.8 m/s2 )(30 m) =12.1 m/s Assess: A speed of 12.1 m/s is 27 mph, which seems very reasonable. 8.15. Model: Model the roller coaster car as a particle at the top of a circular loop-the-loop undergoing uniform circular motion. Visualize: Notice that the r-axis points downward, toward the center of the circle. Solve: The critical speed occurs when n.. goes to zero and G F .. provides all the centripetal force pulling the car in the vertical circle. At the critical speed 2 mg = mvc r , therefore c v = rg. Since the cars speed is twice the critical speed, c 2 tv = v and the centripetal force is 2 ( 2 ) ( ) c G 4 4 4 r mv m v m rg F n F mg r r r S = + = = = = Thus the normal force is n = 3 mg. Consequently, G n F =3. 8.16. Model: Model the roller coaster car as a particle undergoing uniform circular motion along a loop. Visualize: Notice that the r-axis points downward, toward the center of the circle. Solve: In this problem the normal force is equal to the gravitational force: n = FG = mg. We have 2 r G F n F mv r S = + = = mg + mg .v = 2rg = 2(20 m)(9.8 m/s2 ) =19.8 m/s 8.17. Model: Model the bucket of water as a particle in uniform circular motion. Visualize: Solve: Let us say the distance from the bucket handle to the top of the water in the bucket is 35 cm. This makes the shoulder to water distance 65 cm+ 35 cm =1.00 m. The minimum angular velocity for swinging a bucket of water in a vertical circle without spilling any water corresponds to the case when the speed of the bucket is critical. In this case, n = 0 N when the bucket is in the top position of the circular motion. We get 2 c 2 G c 2 c 0 N / 9.8 m/s 3.13 rad/s 3.13 rad s 1 rev 60 s 30 rpm 1.00 m 2 rad 1 min r F n F mg mv mr r g r . . p = + = + = = . = = = = = S 8.18. Model: Use the particle model for the car, which is undergoing nonuniform circular motion. Visualize: Solve: The car is in circular motion with radius 100 m. 2 r = d = We require 2 2 2 1.5 m/s2 1.5 m/s 1.5 m/s 0.122 s 1 100 m ra r r =. = .. = = = - The definition of the angular velocity can be used to determine the time .t using the angular acceleration 2 1.5 m/s 1.5 10 2 s 2. 100 m t a r a = = = - - i . =. +a.t 1 1 i 2 0.122 s 0 s 8.2 s 0.015 s . . a - - - - - ..t = = = 8.19. Model: The train is a particle undergoing nonuniform circular motion. Visualize: Solve: (a) Newtons second law in the vertical direction is (Fnet )y = n - FG = 0 from which n = mg. The rolling friction is R R R . f = n = mg This force provides the tangential acceleration R t R a f g m = - = - The angular acceleration is ( )( 2 ) R 2 0.10 9.8 m/s 1.96 rad/s 0.50 m ta g r r a - - = = = =- (b) The initial angular velocity is 30 rev 1 min 2 rad 3.14 rad/s. min 60 sec rev p . .. .. .= . .. .. . . .. .. . The time to come to a stop due to the rolling friction is f i 2 0 3.14 rad/s 1.60 s 1.96 rad/s t . . a - - . = = = - Assess: The original angular speed of p rad/s means the train goes around the track one time every 2 seconds, so a stopping time of less than 2 s is reasonable. 8.20. Model: The object is treated as a particle in the model of kinetic friction with its motion governed by constant-acceleration kinematics. Visualize: Solve: The velocity 1x v as the object sails off the edge is related to the initial velocity 0x v by 2 2 ( ) 1 0 1 0 2 . x x x v = v + a x - x Using Newtons second law to determine x a while sliding gives x k x SF = - f = ma 0 N y .SF = n - mg = .n = mg Using this result and the model of kinetic friction ( ) k k f = n , the x-component equation can be written as k . x - mg = ma This implies ( )( 2 ) 2 k 0.50 9.8 m/s 4.9 m/s xa = - g = - = - Kinematic equations for the objects free fall can be used to determine 1 : x v ( ) 1 ( )( )2 2 1 1y 2 1 2 2 1 y = y + v t - t + -g t - t ( )2 2 1 0 m 1.0 m 0 m 2 . = + - g t - t ( ) 2 1 . t - t = 0.4518 s ( ) 2 1 1x 2 1 x = x + v t - t ( ) 1 2.30 m 2.0 m 0.4518 s x = = + v 1 0.664 m/s x .v = Having determined 1x v and , x a we can go back to the velocity equation 2 2 ( ) 1 0 1 0 2 : x x x v = v + a x - x ( )2 2 ( 2 )( ) 0 0 0.664 m/s 2 4.9 m/s 2.0 m 4.5 m/s x x = v + - .v = Assess: 0 4.5 m/s x v = is about 10 mph and is a reasonable speed. 8.21. Model: The rocket and puck together make a particle moving on frictionless ice. The thrust of the rocket motor is assumed to be constant. Visualize: Solve The kinematics equations can be used to examine the motion for each coordinate axis independently as a function of time t, then combined to eliminate t. In the x-direction, xf = xi + v0xt = 0 m+ (2.0 m/s)t So . 2.0 m/s f x t = The acceleration of the rocket and puck in the y-direction is ( ) 2 rocket puck 8.0 N 13.3 m/s 0.600 kg net y F y a m m = = = + In the ydirection, ( ) ( ) ( ) 2 22 f i 0 2 2 1 0 m 0 m/s 1 13.3 m/s 2 2 6.67 m/s y y y y v t at t t t = + + = + + = Substituting t from the equation for the x-direction, ( ) 2 2 2 f f 6.67 m/s 1.67 2.0 m/s y = . x . = x . . . . A graph of this equation is shown in the figure. Assess: When the puck has traveled 9 m in the x-direction it has traveled 15 m in the y-direction. 8.22. Model: Treat Sam as a particle. Visualize: This is a two-part problem. Use an s-axis parallel to the slope for the first part, regular xycoordinates for the second. Sams final velocity at the top of the slope is his initial velocity as he becomes airborne. Solve: Sams acceleration up the slope is given by Newtons second law: net 0 2 2 0 ( ) sin10 sin10 200 N (9.8 m/s ) sin10 0.965 m/s 75 kg s F F mg ma a F g m = - = = - = - = The length of the slope is 1 s = (50 m)/ sin10 = 288 m. His velocity at the top of the slope is 2 2 2 1 0 0 1 0 01 1 v = v + 2a (s - s ) = 2a s .v = 2(0.965 m/s ) (288 m) = 23.6 m/s This is Sams initial speed into the air, giving him velocity components 1 1cos10 23.2 m/s x v = v = and 1 1sin10 410 m/s. y v = v = This is not projectile motion because Sam experiences both the force of gravity and the thrust of his skis. Newtons second law for Sams acceleration is net 2 1 2 net 2 1 ( ) (200 N)cos10 2.63 m/s 75 kg ( ) (200 N)sin10 (75 kg)(9.80 m/s ) 9.34 m/s 75 kg x x y y a F m F a m = = = - = = =- The y-equation of motion allows us to find out how long it takes Sam to reach the ground: 1 2 2 2 2 1 1 2 2 1 2 2 2 0 m 50 m (4.10 m/s) (4.67 m/s ) y y y = = y + v t + a t = + t - t This quadratic equation has roots 2 t = -2.86 s (unphysical) and 2 t = 3.74 s. The x-equation of motionthis time with an accelerationis 1 2 1 2 2 2 1 1 2 2 1 2 2 2 2 0 m (23.2 m/s) (2.63 m/s ) 105 m x x x = x + v t + a t = + t - t = Sam lands 105 m from the base of the cliff. 8.23. Model: Treat the motorcycle and rider as a particle. Visualize: This is a two-part problem. Use an s-axis parallel to the slope for the first part, regular xycoordinates for the second. The motorcycles final velocity at the top of the ramp is its initial velocity as it becomes airborne. Solve: The motorcycles acceleration on the ramp is given by Newtons second law: net r r r 0 2 2 0 r ( ) sin 20 sin 20 cos20 sin 20 ( cos20 sin 20 ) (9.8 m/s )((0.02)cos20 sin 20 ) 3.536 m/s s F f mg nmg mg mg ma a g = - - = - - = - - = = - + = - + = - The length of the ramp is s1 = (2.0 m)/ sin 20 = 5.85 m. We can use kinematics to find its speed at the top of the ramp: 2 2 2 1 0 0 1 0 0 01 2 2 1 2 ( ) 2 (11.0 m/s) 2( 3.536 m/s ) (5.85 m) 8.92 m/s v v a s s v as v = + - = + . = + - = This is the motorcycles initial speed into the air, with velocity components 1 1cos20 8.38 m/s x v = v = and 1 1sin 20 y v = v = 3.05 m/s. We can use the y-equation of projectile motion to find the time in the air: 1 2 2 2 2 1 1 2 2 1 2 2 2 0 m 2.0 m (3.05 m/s) (4.90 m/s ) y y y = = y + v t + a t = + t - t This quadratic equation has roots 2 t = -0.399 s (unphysical) and 2 t =1.021 s. The x-equation of motion is thus 2 1 1 2 2 0 m (8.38 m/s) 8.56 m x x = x + v t = + t = 8.56 m <10.0 m, so it looks like crocodile food. 8.24. Model: Use the particle model and the constant-acceleration equations of kinematics for the rocket. Solve: (a) The acceleration of the rocket in the launch direction is obtained from Newtons second law F = ma: .. .. 140,700 N = (5000 kg)a . a = 28.14 m/s2 Therefore, cos44.7 20.0 m/s2 x a = a = and sin 44.7 19.8 m/s2. y a = a = The net acceleration in the y-direction is thus ( ) net y y a = a - g = (19.8 - 9.8) m/s2 =10.0 m/s2 With this acceleration, we can write the equations for the x- and y-motions of the rocket. ( ) 1 ( ) ( )2 0 0 y 0 2 net y 0 y = y + v t - t + a t - t 1 ( 2 ) 2 2 = 0 m+ 0 m + 10.0 m/s t = (5.00 m/s2 )t2 ( ) 1 ( ) ( )2 0 0x 0 2 net x 0 x = x + v t - t + a t - t 1 ( 2 ) 2 ( 2 ) 2 2 = 0 m+ 0 m + 20.0 m/s t = 10.0 m/s t From these two equations, ( ) ( ) 2 2 2 2 10.0 m/s 2 5.00 m/s x t y t = = The equation that describes the rockets trajectory is 1 2 y = x. (b) It is a straight line with a slope of 1 2 . (c) In general, ( ) ( ) y 0 y net y 1 0 v = v + a t - t ( 2 ) 1 = 0 + 10.0 m/s t ( ) ( ) x 0x net x 1 0 v = v + a t - t ( 2 ) 1 = 0 + 20.0 m/s t ( ) ( ) ( ) 2 2 2 2 2 2 2 1 1 1 v = 10.0 m/s t + 20.0 m/s t = 22.36 m/s t The time required to reach the speed of sound is calculated as follows: ( 2 ) 1 330 m/s = 22.36 m/s t 1 .t =14.76 s We can now obtain the elevation of the rocket. From the y-equation, ( 2 ) 2 1 y = 5.00 m/s t = (5.00 m/s2 )(14.76 s)2 =1090 m 8.25. Model: The hockey puck will be treated as a particle whose motion is determined by constantacceleration kinematic equations. We break this problem in two parts, the first pertaining to motion on the table and the second to free fall. Visualize: Solve: Newtons second law is: Fx = max 2.0 N 2.0 m/s2 1.0 kg x x a F m . = = = The kinematic equation 2 2 ( ) 1 0 1 0 2 x x x v = v + a x - x yields: 2 2 2 ( 2 )( ) 1 0 m /s 2 2.0 m/s 4.0 m x v = + 1 4.0 m/s x .v = Let us now find the time of free fall ( ) 2 1 t - t : ( ) 1 ( )2 2 1 1y 2 1 2 1y 2 1 y = y + v t - t + a t - t 1 ( 2 )( )2 2 2 1 .0 m = 2.0 m+ 0 m+ -9.8 m/s t - t ( ) 2 1 . t - t = 0.639 s Having obtained 1x v and ( ) 2 1 t - t , we can now find 2 1 (x - x ) as follows: ( ) 1 ( )2 2 1 1x 2 1 2 x 2 1 x = x + v t - t + a t - t ( )( ) 1 ( 2 )( )2 2 1 2 . x - x = 4.0 m/s 0.639 s + 2.0 m/s 0.639 s = 3.0 m Assess: For a modest horizontal thrust of 2.0 N, a landing distance of 3.0 m is reasonable. 8.26. Model: The model rocket is treated as a particle and its motion is determined by constant-acceleration kinematic equations. Visualize: Solve: As the rocket is accidentally bumped 0 0.5 m/s x v = and 0 0 m/s. y v = On the other hand, when the engine is fired 20 N 40 m/s2 0.500 kg x x x x F F ma a m = . = = = (a) Using ( ) 1 ( )2 1 0 0 1 0 2 1 0 , y y y = y + v t - t + a t - t 1 ( 2 ) 2 2 1 1 0 m = 40 m+ 0 m+ -9.8 m/s t .t = 2.857 s The distance from the base of the wall is ( ) 1 ( )2 1 0 0x 1 0 2 x 1 0 x = x + v t - t + a t - t ( )( ) 1 ( 2 )( )2 2 = 0 m+ 0.5 m/s 2.857s + 40 m/s 2.857 s =165 m (b) The x- and y-equations are ( ) 1 ( )2 0 0 y 0 2 y 0 y = y + v t - t + a t - t = 40 - 4.9t2 ( ) 1 ( )2 0 0x 0 2 x 0 x = x + v t - t + a t - t = 0.5t + 20t2 Except for a brief interval near t = 0, 20t2 ..0.5t. Thus x 20t2 , or t2 = x / 20. Substituting this into the yequation gives y = 40 - 0.245x This is the equation of a straight line, so the rocket follows a linear trajectory to the ground. 8.27. Model: Assume the particle model for the satellite in circular motion. Visualize: To be in a geosynchronous orbit means rotating at the same rate as the earth, which is 24 hours for one complete rotation. Because the altitude of the satellite is 3.58107 m, r = 3.58107 m, 7 6 7 e r = 3.5810 m + 6.3710 m = 4.2210 m. Solve: (a) The period (T ) of the satellite is 24.0 hours. (b) The acceleration due to gravity is ( ) 2 2 2 2 4.22 107m 2 1 hr 0.223 m/s2 24.0 hr 3600 s r g a r r T p p = = . = .. .. = .. .. = . . . . (c) There is no normal force on a satellite, so the weight is zero. It is in free fall. 8.28. Model: Treat the man as a particle. The man at the equator undergoes uniform circular motion as the earth rotates. Visualize: Solve: The scale reads the mans weight FG = n, the force of the scale pushing up against his feet. At the north pole, where the man is in static equilibrium, P G n = F = mg = 735 N At the equator, there must be a net force toward the center of the earth to keep the man moving in a circle. The raxis points toward the center, so 2 2 2 r G E E P SF = F - n = m. r.n = mg - m. r = n - m. r The equator scale reads less than the north pole scale by the amount m. 2r. The mans angular velocity is that of the equator, or 2 2 rad 7.27 10 5 rad/s T 24 hours (3600 s/1 h) p p . = = = - Thus the north pole scale reads more than the equator scale by .w = (75 kg)(7.2710-5 rad/s)2 (6.37106 m) = 2.5 N Assess: The man at the equator appears to have lost .m = .w/g 0.25 kg, or the equivalent of 1 2 lb. 8.29. Model: Use the particle model for the (cart + child) system which is in uniform circular motion. Visualize: Solve: Newtons second law along r and z directions can be written: cos20 sin 20 r r SF = T - n = ma sin 20 cos20 0 z SF = T - n - mg = The carts centripetal acceleration is ( ) 2 2 2.0cos20 m 14 rev 1 min 2 rad 4.04 m/s2 min 60 s 1 rev r a r p = . = .. .. = . . The above force equations can be rewritten as 0.94T - 0.342n = (25 kg)(4.04 m/s2 ) =101 N 0.342T + 0.94n = (25 kg)(9.8 m/s2 ) = 245 N Solving these two equations yields T =179 N for the tension in the rope. Assess: In view of the child + cart weight of 245 N, a tension of 179 N is reasonable. 8.30. Model: Model the ball as a particle which is in a vertical circular motion. Visualize: Solve: At the bottom of the circle, ( ) ( )( ) ( ) ( ) 2 2 2 G 0.500 kg 15 N 0.500 kg 9.8 m/s 5.5 m/s 1.5 m r mv v F T F v r S = - = . - = . = 8.31. Model: We will use the particle model for the car, which is undergoing uniform circular motion on a banked highway, and the model of static friction. Visualize: Note that we need to use the coefficient of static friction s , which is 1.0 for rubber on concrete. Solve: Newtons second law for the car is 2 s cos sin r F f n mv r S = . + . = s G cos sin 0 z SF = n . - f . - F = N Maximum speed is when the static friction force reaches its maximum value ( )s max s f = n. Then ( ) 2 s n cos15 sin15 mv r + = ( ) s n cos15 - sin15 = mg Dividing these two equations and simplifying, we get 2 s s s s tan15 tan15 1 tan15 1 tan15 v v gr gr + + = . = - - ( )( )( ) ( ) 2 1.0 0.268 9.80 m/s 70 m 34 m/s 1 0.268 + = = - Assess: The above value of 34 m/s 70 mph is reasonable. 8.32. Model: Use the particle model for the rock, which is undergoing uniform circular motion. Visualize: Solve: Newtons second law is 2 cos10 r F T mv r S = = sin10 0 z SF = T - mg = N where the radius of the circular motion is r = (1.0 m)cos10 = 0.985 m. Dividing these two equations, we get ( 2 )( ) 2 9.8 m/s 0.985 m tan10 7.40 m/s tan10 tan10 gr v gr v = . = = = 7.40 m/s 7.51 rad/s 7.51rad 1 rev 60 s 72 rpm 0.985 m s 2 rad 1 min v r . p . = = = = = 8.33. Model: Use the particle model and static friction model for the coin, which is undergoing circular motion. Visualize: Solve: The force of static friction is s s s . f = n = mg This force is equivalent to the maximum centripetal force that can be applied without sliding. That is, ( ) ( )( ) 2 2 2 s s max max 0.80 9.8 m/s 7.23 rad/s 0.15 m tmg mv m r g r r = = . .. = = = 7.23 rad 1 rev 60 s 69 rpm s 2p rad 1 min = = So, the coin will stay still on the turntable. Assess: A rotational speed of approximately 1 rev per second for the coin to stay stationary seems reasonable. 8.34. Model: Use the particle model for the car, which is in uniform circular motion. Visualize: Solve: Newtons second law is 2 sin 20 r r F T ma mv r S = = = G cos20 0 z SF = T - F = N These equations can be written as 2 T sin 20 mv r = T cos20 = mg Dividing these two equations gives tan 20 = v2 rg .v = rg tan 20 = (4.55 m)(9.8 m/s2 )tan 20 = 4.03 m/s 8.35. Model: Use the particle model for the ball in circular motion. Visualize: Solve: (a) The mass moves in a horizontal circle of radius r = 20 cm. The acceleration a.. and the net force vector point to the center of the circle, not along the string. The only two forces are the string tension T, .. which does point along the string, and the gravitational force GF . .. These are shown in the free-body diagram. Newtons second law for circular motion is G cos cos 0 N z SF = T . - F = T . - mg = 2 sin r r F T ma mv r S = . = = From the z-equation, (0.500 kg)(9.8 m/s2 ) 5.00 N cos cos11.54 T mg . = = = (b) We can find the tangential speed from the r-equation: v rT sin 0.63 m/s m . = = The angular speed is 0.63 ms 3.15 rad/s 30 rpm 0.20 m v r . = = = = 8.36. Model: Consider the passenger to be a particle and use the model of static friction. Visualize: Solve: The passengers stick to the wall if the static friction force is sufficient to support the gravitational force on them: fs = FG The minimum angular velocity occurs when static friction reaches its maximum possible value ( )s max s f = n. Although clothing has a range of coefficients of friction, it is the clothing with the smallest coefficient s ( = 0.60) that will slip first, so this is the case we need to examine. Assuming that the person is stuck to the wall, Newtons second law is 2 s s 0 r z SF = n = m. r SF = f - w = . f = mg The minimum frequency occurs when ( ) 2 s s max s s min f = f = n = mr. Using this expression for s f in the z-equation gives 2 s s min 2 min s 9.80 m/s 2.56 rad/s 2.56 rad/s 1 rev 60 s 24 rpm 0.60(2.5 m) 2 rad 1 min f mr mg g r . . p = = . = = = = = Assess: Note the velocity does not depend on the mass of the individual. Therefore, the minimum mass sign is not necessary. 8.37. Model: Use the particle model for the marble in uniform circular motion. Visualize: Solve: The marble will roll in a horizontal circle if the static friction force is sufficient to support the gravitational on it: fs = FG If ( ) s max mg > f then static friction is not sufficient and the marble will slip down the side as it rolls around the circumference. The r-equation of Newtons second law is 2 2 (0.010 kg)(0.060 m) 150 rpm 2 rad 1 min 0.148 N 1 rev 60 s r F n mr p = = . = .. .. = . . S Thus the maximum possible static friction is ( )s max s f = n = (0.80)(0.148 N) = 0.118 N. The friction force needed to support a 10 g marble is s f = mg = 0.098 N. We see that ( ) s smax f < f , therefore friction is sufficient and the marble spins in a horizontal circle. Assess: In reality, rolling friction will cause the marble to gradually slow down until ( ) s max f < mg. At that point, it will begin to slip down the inside wall. 8.38. Model: Use the particle model for the car and the model of kinetic friction. Visualize: Solve: We will apply Newtons second law to all three cars. Car A: x x ( k )x ( G )x 0 N k 0 N x SF = n + f + F = - f + = ma ( ) k 0 N 0 N y y y y SF = n + f + y = n + - mg = The y-component equation means n = mg. Since k k f = n, we have k k f = mg. From the x-component equation, k k 2 kg 9.8 m/s x a f mg m m - - = = = - = - Car B: Car B is in circular motion with the center of the circle above the car. ( ) ( ) 2 k G 0 N r r r r r F n f F n mg ma mv r S = + + = + - = = ( ) ( ) k G k 0 N 0 N t t t t t SF = n + f + F = - f + = +ma From the r-equation 2 2 k k k n mg mv f n m g v r r . . = + . = = . + . . . Substituting back into the t-equation, 2 ( )2 k k 2 2 k 25 m/s 9.8 m/s 12.9 m/s 200 m t a f m g v m m r . . . . = - = - . + . = - . + . = - . . . . . . Car C: Car C is in circular motion with the center of the circle below the car. ( ) ( ) 2 k G 0 N r r r r r F n f F n mg ma mv r S = + + = - + + = = ( ) ( ) k G k 0 N 0 N t t t t t SF = n + f + F = - f + = ma From the r-equation n = m(g - v2 r ). Substituting this into the t-equation yields k k ( 2 ) 2 k 6.7 m/s t a f n g v r m m - - = = =- - =- 8.39. Model: Model the ball as a particle that is moving in a vertical circle. Visualize: Solve: (a) The balls gravitational force ( )( 2 ) G F = mg = 0.500 kg 9.8 m/s = 4.9 N. (b) Newtons second law at the top is 2 r 1 G r v F T F ma mr S = + = = ( ) ( )2 2 2 1 4.0 m/s 0.500 kg 9.8 m/s 1.02 m T m v g r . . . . . = . - . = . - . . . .. .. = 2.9 N (c) Newtons second law at the bottom is 2 r 2 G F T F mv r S = - = ( ) ( )2 2 2 2 7.5 m/s 0.500 kg 9.8 m/s 32 N 1.02 m T m g v r . . . . . = . + . = . + . = . . .. .. 8.40. Model: Use the particle model for yourself while in uniform circular motion. Visualize: Solve: (a) The speed and acceleration are 2 2 (15 m) 3.77 m/s 25 s v r T p p = = = 2 ( )2 2 3.77 m/s 0.95 m/s 15 m r a v r = = = So the speed is 3.8 m/s and the centripetal acceleration is 0.95 m/s2. (b) The weight w = m, the normal force. On the ground, your weight is the same as the gravitational force GF . Newtons second law at the top is 2 r G r F F n ma mv r S = - = = ( ) ( ) 2 2 2 2 3.77 m/s 9.8 m/s 8.85 m/s 15 m n w m g v m m r . . . . . = = . - . = . - . = . . . . . . 2 2 G 8.85 m s 0.90 9.8 m s w F . = = (c) Newtons second law at the bottom is 2 r G r F n F ma mv r S = - = = ( ) ( ) 2 2 2 2 3.77 m/s 9.8 m/s 10.75 m/s 15 m n w m g v m m r . . . . . = = . + . = . + . = . . . . . . 2 2 G 10.75 m s 1.10 9.8 m s w F . = = 8.41. Model: Model a passenger as a particle rotating in a vertical circle. Visualize: Solve: (a) Newtons second law at the top is 2 r T G r F n F ma mv r S = + = = 2 T n mg mv r . + = The speed is 2 2 (8.0 m) 11.17 m/s 4.5 s v r T p p = = = ( ) ( )2 2 2 T 11.17 m/s 55 kg 9.8 m/s 319 N 8.0 m n m v g r . . . . . = . - . = . - . = . . .. .. That is, the ring pushes on the passenger with a force of 3.2102 N at the top of the ride. Newtons second law at the bottom: ( ) ( ) 2 2 2 B G B 2 2 11.17 m/s 55 kg 9.8 m/s 1397 N 8.0 m r r F n F ma mv n mv mg m v g r r r . . = - = = . = + = . + . . . . . = . + . = .. .. S Thus the force with which the ring pushes on the rider when she is at the bottom of the ring is 1.4 kN. (b) To just stay on at the top, T n = 0 N in the r-equation at the top in part (a). Thus, 2 2 2 max mg mv mr mr 2 r T p . . . = = = . . . . max 2 2 2 8.0 m 5.7 s 9.8 m/s T r g . = p = p = 8.42. Model: Model the chair and the rider as a particle in uniform circular motion. Visualize: Solve: Newtons second law along the r-axis is r r ( G )r r SF = T + F = ma .T sin. + 0 N = mr. 2 Since r = Lsin. , this equation becomes ( )( ) 2 2 150 kg 9.0 m 2 rad 3330 N 4.0 s T mL p = . = .. .. = . . Thus, the 3000 N chain is not strong enough for the ride. 8.43. Model: Model the ball as a particle in motion in a vertical circle. Visualize: Solve: If the ball moves in a complete circle, then there is a tension force T .. when the ball is at the top of the circle. The tension force adds to the gravitational force to cause the centripetal acceleration. The forces are along the r-axis, and the center of the circle is below the ball. Newtons second law at the top is ( ) 2 net r G mv F T F T mgr = + = + = top rT v rgm . = + The tension T cant become negative, so T = 0 N gives the minimum speed min v at which the ball moves in a circle. If the speed is less than min v , then the string will go slack and the ball will fall out of the circle before it reaches the top. Thus, ( ) ( ) 2 min min min 9.8 m/s 3.13 rad / s 30 rpm 1.0 m v rg g v rg r r r = .. = = = = = = 8.44. Model: The ball is a particle on a massless rope in circular motion about the point where the rope is attached to the ceiling. Visualize: Solve: Newtons second law in the radial direction is ( ) 2 r G F T F T mg mv r S = - = - = Solving for the tension in the rope and evaluating, ( ) ( )2 2 2 5.5 m/s 10.2 kg 9.8 m/s 168 N 4.5 m T m g v r . . . . = . + . = . + . = . . . . . . Assess: The tension in the rope is greater than the gravitational force on the ball in order to keep the ball moving in a circle. 8.45. Model: Model the person as a particle in uniform circular motion. Visualize: Solve: The only force acting on the passengers is the normal force of the wall. Newtons second law along the r-axis is: SFr = n = mr. 2 To create normal gravity, the normal force by the inside surface of the space station equals mg. Therefore, 2 2 2 2 2 500 m 45.0 s 9.8 m/s mg mr g T r T r g p = . .. = = . = p = p = Assess: This is a fast rotation. The tangential speed is 2 2 (500 m) 70 m/s 140 mph 45 s v r T p p = = = 8.46. Model: Masses 1 m and 2 m are considered particles. The string is assumed to be massless. Visualize: Solve: The tension in the string causes the centripetal acceleration of the circular motion. If the hole is smooth, it acts like a pulley. Thus tension forces 1 T .. and 2 T .. act as if they were an action/reaction pair. Mass 1 m is in circular motion of radius r, so Newtons second law for 1 m is 2 1 r 1 F T m v r S = = Mass 2 m is at rest, so the y-equation of Newtons second law is 2 2 2 2 0 N y SF = T - m g = .T = m g Newtons third law tells us that 1 2T = T . Equating the two expressions for these quantities: 2 1 2 2 1 mv m g v m rg r m = . = 8.47. Model: Model the ball as a particle swinging in a vertical circle, then as a projectile. Visualize: Solve: Initially, the ball is moving in a circle. Once the string is cut, it becomes a projectile. The final circularmotion velocity is the initial velocity for the projectile. The free-body diagram for circular motion is shown at the bottom of the circle. SinceT > FG , there is a net force toward the center of the circle that causes the centripetal acceleration. The r-equation of Newtons second law is ( ) 2 net r G mv F T F T mgr = - = - = ( ) ( )( 2 ) bottom 0.60 m 5.0 N 0.10 kg 9.8 m/s 4.91 m/s 0.100 kg v r T mg m . = - = .. - .. = As a projectile the ball starts at 0 y =1.4 m with 0 v.. = 4.91i m/s. The equation for the y-motion is 1 ( )2 1 2 1 0 0 2 0 2 1 0 m y y = = y + v .t - g .t = y - gt This is easily solved to find that the ball hits the ground at time 0 1 t 2y 0.535 s g = = During this time interval it travels a horizontal distance ( )( ) 1 0 0 1 4.91 m/s 0.535 s 2.63 m x x = x + v t = = So the ball hits the floor 2.6 m to the right of the point where the string was cut. 8.48. Model: Use the particle model for a ball in motion in a vertical circle and then as a projectile. Visualize: Solve: For the circular motion, Newtons second law along the r-direction is 2 G t r F T F mv r S = + = Since the string goes slack as the particle makes it over the top, T = 0 N. That is, ( )( ) 2 2 G t 9.8 m/s 0.5 m 2.21 m/s t F mg mv v gr r = = . = = = The ball begins projectile motion as the string is released. The time it takes for the ball to hit the floor can be found as follows: ( ) 1 ( )2 1 ( 2 )( )2 1 0 0 1 0 2 1 0 2 1 1 0 m 2.0 m 0 m 9.8 m/s 0 s 0.639 s y y y = y + v t - t + a t - t . = + + - t - .t = The place where the ball hits the ground is ( ) 1 0 0x 1 0 x = x + v t - t = 0 m+ (+2.21 m/s)(0.639 s - 0 s) = +1.41 m The ball hits the ground 1.41 m to the right of the point beneath the center of the circle. 8.49. Model: Model the ball as a particle undergoing circular motion in a vertical circle. Visualize: Solve: Initially, the ball is moving in circular motion. Once the string breaks, it becomes a projectile. The final circular-motion velocity is the initial velocity for the projectile, which we can find by using the kinematic equation 2 2 ( ) 2 2 ( )2 ( 2 )( ) 1 0 1 0 0 0 2 0 m s 2 9.8 m/s 4.0 m 0 m 8.85 m/s y v = v + a y - y . = v + - - .v = This is the speed of the ball as the string broke. The tension in the string at that instant can be found by using the r-component of the net force on the ball: r SF = 2 0 y v T m r . . = .. ... . . ( )( )2 8.85 m/s 0.100 kg 13.1 N 0.60 m T= = 8.50. Model: Model the car as a particle on a circular track. Visualize: Solve: (a) Newtons second law along the t-axis is SFt = Ft = mat ( ) 1000 N 1500 kg 2 3 m/s2 t t . = a .a = With this tangential acceleration, the cars tangential velocity after 10 s will be ( ) ( 2 )( ) 1 0 1 0 0 m s 2 3 m/s 10 s 0 s 20 3 m/s t t t v = v + a t - t = + - = The radial acceleration at this instant is 2 ( )2 1 2 20 3 m/s 16 m/s 25 m 9 t r a v r = = = The cars acceleration at 10 s has magnitude ( ) ( ) 2 2 2 2 2 2 2 1 2 3 m/s 16 9 m/s 1.90 m/s t r a = a + a = + = tan 1 tan 1 2 3 21 16 9 t r a a . - -. . = = . . = . . where the angle is measured from the r-axis. (b) The car will begin to slide out of the circle when the static friction reaches its maximum possible value ( )s max s f = n. That is, ( ) 2 2 s max s s t r F f n mg mv r S = = = = ( )( 2 ) 2 25 m 9.8 m/s 15.7 m/s t .v = rg = = In the above equation, n = mg follows from Newtons second law along the z-axis. The time when the car begins to slide can now be obtained as follows: ( ) 2t 0t t 2 0 v = v + a t - t . ( 2 )( ) 2 15.7 m/s = 0 m s + 2 3 m/s t - 0 2 .t = 24 s 8.51. Model: Model the steel block as a particle and use the model of kinetic friction. Visualize: Solve: (a) The components of thrust (F) .. along the r-, t-, and z-directions are sin 20 (3.5 N)sin 20 1.20 N r F = F = = cos20 (3.5 N)cos20 3.29 N t F = F = = z F = 0 N Newtons second law is ( ) 2 net r r F = T + F = mr. ( ) net t t k t F = F - f = ma ( ) net 0 N z F = n - mg = The z-component equation means n = mg. The force of friction is ( )( )( 2 ) k k k f = n = mg = 0.60 0.500 kg 9.8 m/s = 2.94 N Substituting into the t-component of Newtons second law (3.29 N) (2.94 N) (0.500 kg) 0.70 m/s2 t t - = a .a = Having found , t a we can now find the tangential velocity after 10 revolutions = 20p rad as follows: 2 1 1 1 1 1 0 1 1 2 18.95 s 2 6.63 rad/s t t t a t t r r a a t r . . . . = . . . = = . . . . = + . . = . . . . The blocks angular velocity after 10 s is 6.6 rad/s. (b) Substituting 1 . into the r-component of Newtons second law yields: 2 1 r 1 T + F = mr. ( ) ( )( )( )2 1 1 .T + 1.20 N = 0.500 kg 2.0 m 6.63 rad/s .T = 44 N 8.52. Model: Assume the particle model for a ball in vertical circular motion. Visualize: Solve: (a) Newtons second law in the r- and t-directions is ( ) 2 net cos t r r F T mg ma mv r = + . = = ( ) net sin t t F = -mg . = ma Substituting into the r-component, ( ) ( )( ) ( )( ) 2 20 N 2.0 kg 9.8 m/s2 cos30 2.0 kg 0.80 m t + = v 3.85 m/s t .v = The tangential velocity is 3.8 m/s. (b) Substituting into the t-component, (9.8 m/s2 )sin30 t - = a 4.9 m/s2 t .a = - The radial acceleration is 2 ( )2 2 3.85 m/s 18.5 m/s 0.80 m t r a v r = = = Thus, the magnitude of the acceleration is ( ) ( ) 2 2 18.5 m/s2 2 4.9 m/s2 2 19.1 m/s2 r t a = a + a = + - = The angle of the acceleration vector from the r-axis is tan 1 tan 1 4.9 14.8 18.5 t r a a f = - = - = The angle is below the r-axis. 8.53. Solve: (a) You are spinning a lead fishing weight in a horizontal 1.0 m diameter circle on the ice of a pond when the string breaks. You know that the test weight (breaking force) of the line is 60 N and that the lead weight has a mass of 0.30 kg. What was the weights angular velocity in rad/s and in rpm? (b) ( )( ) 2 60 N 20 rad/s rev 60 s 191 rpm 0.3 kg 0.5 m 2 rad min . . p = . = = 8.54. Solve: (a) At what speed does a 1500 kg car going over a hill with a radius of 200 m have a weight of 11,760 N? (b) The weight is the normal force. 1500 kg 2 2940 N 19.8 m/s 200 m = v .v = 8.55. Model: Assume the particle model and apply the constant-acceleration kinematic equations. Visualize: Solve: (a) Newtons second law for the projectile is ( net )x wind x F = -F = ma .. x a F m - . = where wind F is shortened to F. For the y-motion: ( ) 1 ( )2 1 0 0 y 1 0 2 y 1 0 y = y + v t - t + a t - t ( ) 1 2 0 1 2 1 .0 m = 0 m + v sin. t - gt 1 .t = 0 s and 0 1 t 2v sin g . = Using the above expression for 1 t and defining the range as R we get from the x motion: ( ) 1 ( )2 1 0 0x 1 0 2 x 1 0 x = x + v t - t + a t - t 1 2 1 0 0x 1 2 1 x x R v t F t m . - = = + .- . . . . . ( ) 2 0 0 0 cos 2 sin 2 sin 2 v v F v g m g . . . . . . . = . . - . . . . . . 2 2 0 0 2 2 2v cos sin 2v F sin g mg = . . - . We will now maximize R as a function of . by setting the derivative equal to 0: ( ) 2 2 0 2 2 0 2 dR 2v cos sin 2Fv 2sin cos 0 d g mg . . . . . = - - = 2 2 2 0 2 2 0 cos sin cos2 2 sin 2 2 Fv g mg v . . . . . .. . . - = = . .. . . .. . tan 2 mg F . . = Thus the angle for maximum range is 1 1 ( ) 2 . = tan- mg /F (b) We have (0.50 kg)(9.8 m/s2 ) 8.167 0.60 N mg F = = 1 1 ( ) 2 .. = tan- 8.167 = 41.51 The maximum range without air resistance is 2 2 0 0 R 2v sin 45 cos45 v g g ' = = Therefore, we can write the equation for the range R as R 2R sin 41.51 cos41.51 2F R sin2 41.51 mg = ' - ' = R'(0.9926 - 0.1076) = 0.885R' . R 0.8850 R = ' . R R 1 0.8850 0.115 R ' - = - = ' Thus R is reduced from R' by 11.5%. Assess: The condition for maximum range (tan 2. = mg /F ) means 2. .90 as F .0. That is, . = 45 when F = 0, as is to be expected. 8.56. Visualize: Solve: From Chapter 6 the drag on a projectile is 1 2 , direction opposite to motion , 4 D = .. Av .. . . .. where A is the cross- sectional area. Using the free-body diagram above, apply Newtons second law to each direction. In the x-direction, ( ) ( ) 1 2 net 4 1 2 net G 4 cos cos sin sin x x y y F D Av a m n m F D F Av a g m m m . . . . = =- =- - = =- =- - Since 2 2, x y v = v + v cos , x v = v . and sin , y v = v . we can rewrite these as ( ) ( ) 2 2 2 2 cos 4 m 4 m sin 4 m 4 m x x y x y x y y A v v Av v v a A v v Av v v a gg . . + = - = - + = - - = - - 8.57. Model: Use the particle model and apply the constant-acceleration kinematic equations of motion. Visualize: Solve: (a) The thrust of the rocket burn must be such that x v decreases from 2.0 km/s to 0 during the same interval that y v increases from 0 to 1.0 km/s. The forward distance in which to accomplish this is x1 = (500 km)cos30 = 433 km. We can find the burn time by combining two kinematic equations: 1 0 1 1 0 1 2 1 1 1 0 1 2 1 0 1 2 0 1 2 0 1 1 1 0 0 0 () 2 2(433 km) 433 s 2.0 km/s x x x x x x x x x x x v v a t a t v x v t a t v t v t v t t x v = = + . =- = + + = + - = . = = = The required acceleration is 0 1 / (2000 m/s)/(433 s) 4.62 m/s. x y a = -v t = - = - During this interval, y v increases from 0 to 1.0 km/s, thus 2 1 1 0 1000 m/s 2.31 m/s 433 s y y y v = + a t .a = = The thrust force is given by Newtons second law: 2 thrust F = ma = (20,000 kg)(4.62i + 2.31j m/s ) = (-92,400i + 46,200 j) N .. .. The magnitude of the thrust force is 103,300 N and it is at angle . = tan-1(46,200/92,400) = 26.6 below the +xaxis (i.e., to the right and down). In order to point the thrusters in this direction, you must rotate the rocket counterclockwise 180 -. =153.4. To make it home with one rocket burn, you need to rotate your rocket to 153.4 and fire with a thrust of 103,300 N for 433 s. (b) The x- and y-positions during the burn are 1 2 2 2 1 2 2 2 (2000 m/s) (4.62 m/s ) (2.31 m/s ) x t t y t = - = The y-position at t1 = 433 s is 216,500 m = 216.5 km, 33.5 km away from the entrance. With the rocket off, youre now coasting straight toward the entrance at 1.0 km/s and will cover this distance in 33.5 s. Thus you pass through the entrance at 433 s + 33.5 s = 466.5 s. The following table calculates your position at intervals of 50 s and, for accuracy, at 1 t = 433 s when you end the burn. The trajectory is a parabola that intersects the x = 433 km line at y = 216.5 km, then a vertical line into the entrance. 8.58. Model: Use the particle model for the ball, which is in uniform circular motion. Visualize: Solve: From Newtons second law along r and z directions, 2 cos r F n mv r S = . = sin 0 sin z SF = n . - mg = .n . = mg Dividing the two force equations gives 2 tan gr v . = From the geometry of the cone, tan. = r y. Thus 2 r gr v gy y v = . = 8.59. Model: Model the block as a particle and use the model of kinetic friction. Visualize: Solve: The only radial force is tension, so we can use Newtons second law to find the angular velocity max . at which the tube breaks: 2 max max 50 N 9.12 rad/s (0.50 kg)(1.2 m) r F T m r T mr S = = . .. = = = The compressed air and friction exert tangential forces, and the second law along the tangential direction is k k k 2 2 k 4.0 N (0.60)(9.80 m/s ) 2.12 m/s 0.50 kg t t t t t t t F F f F n F mg ma a F g m = - = - = - = = - = - = S The time needed to accelerate to 9.12 rad/s is given by max 1 max 1 1 2 0 (1.2)(9.12 rad/s) 5.16 s 2.12 m/s t t a t t r r a . . =. = + .. .. . = = = . . During this interval, the block turns through angle 2 1 2 1 2 1 0 0 1 2 1 2 0 2.12 m/s (5.16 s) 23.52 rad 1 rev 3.7 rev 1.2 m 2 rad t t a t r . . . . p . . . . . = - = + . . = + . . = = . . . . 8.60. Model: Assume the particle model for a sphere in circular motion at constant speed. Visualize: Solve: (a) Newtons second law along the r and z axes is: 2 1 2 sin30 sin 60 t r F T T mv r S = + = 1 2 G cos30 cos60 0 N z SF = T + T - F = Since we want 1 2 T = T = T, these two equations become ( ) 2 sin30 sin 60 t T mv r + = T (cos30 + cos60) = mg Since sin30 + sin60 = cos30 + cos60, 2 t t mg mv v rg r = . = The triangle with sides 1L , 2 L , and 1.0 m is isosceles, so 2 L =1.0 m and 2 r = L cos30. Thus ( ) ( )( 2 ) 2 L cos 30g = 1.0 m cos 30g = 0.866 m 9.8 m/s = 2.9 m/s (b) The tension is (2.0 kg)(9.8 m/s2 ) 14.3 N cos30 cos60 0.866 0.5 T = mg = = + + 8.61. Model: Use the particle model for a sphere revolving in a horizontal circle. Visualize: Solve: Newtons second law in the r- and z-directions is ( ) 2 1 2 cos30 cos30 t r F T T mv r S = + = ( ) 1 2 G sin30 sin30 0 N z S F = T -T - F = Using r = (1.0 m)cos30 = 0.886 m, these equations become ( )( ) ( )( ) 2 2 1 2 0.300 kg 7.5 m/s 22.5 N cos30 0.866 m 0.866 t T T mv r + = = = ( )( ) ( ) 2 1 2 0.300 kg 9.8 m/s 5.88 N sin30 0.5 T -T = mg = = Solving for 1 T and 2 T yields 1 T =14.2 N and 2 T = 8.3 N. 8.62. Model: Use the particle model for the ball. Visualize: Solve: (a) Newtons second law along the r- and z-directions is cos 2 r SF = n . = mr. G sin 0 N z SF = n . - F = Using GF = mg and dividing these equations yields: 2 tan g R y r r . . - = = where you can see from the figure that tan. = (R - y) r. Thus g . R y . = - (b) . will be minimum when (R - y) is maximum or when y = 0 m. Then min . = g / R. (c) Substituting into the above expression, 9.8 m/s2 rad 60 s 1 rev 9.9 95 rpm 0.20 m 0.10 m s 1 min 2 rad g R y . p = = = = - - 8.63. Model: Use the particle model for the airplane. Visualize: Solve: In level flight, the lift force L .. balances the gravitational force. When turning, the plane banks so that the radial component of the lift force can create a centripetal acceleration. Newtons second law along the r- and z-directions is 2 sin t r F L mv r S = . = cos 0 N z SF = L . - mg = These can be written: 2 sin mv rL . = cos mg L . = Dividing the two equations gives: 2 2 2 2 400miles 1 hr 1610 m tan hour 3600 s 1 mile 18.5 km tan 9.8 m s tan10 v r v gr g . . . . .. .. = . = = = .. .. The diameter of the airplanes path around the airport is 218.5 km = 37 km. 8-1 8.64. Model: Use the particle model for a small volume of water on the surface. Visualize: Solve: Consider a particle of water of mass m at point C on the surface. Newtons second law along the r- and zdirections is ( ) 2 2 net cos cos r F n mr mr n . = . = . . . = ( ) net sin 0 N sin z F n mg mg n = . - = . . = Dividing both equations gives tan. = g r. 2 . For a parabola z = ar2. This means dz 2ar dr = = slope of the curve at tan tan(90 ) 1 tan 1 tan 2 C ar f . . . = = - = . = Equating the two equations for tan. , we get 2 2 1 2 2 g a ar r g . . = . = Thus the surface is described by the equation 2 2 2 z r g . = which is the equation of a parabola. 9.1. Model: Model the car and the baseball as particles. Solve: (a) The momentum p = mv = (1500 kg)(10 m/s) =1.5104 kg m/s. (b) The momentum p = mv = (0.2 kg)(40 m/s) = 8.0 kg m/s. 9.2. Model: Model the bicycle and its rider as a particle. Also model the car as a particle. Solve: From the definition of momentum, pcar = pbicycle car car car bicycle bicycle bicycle car bicycle m v m v v m v m . = . = 1500 kg (5.0 m/s) 75 m/s 100 kg . . = . . = . . Assess: This is a very high speed (168 mph). This problem shows the importance of mass in comparing two momenta. 9.3. Visualize: Please refer to Figure EX9.3. Solve: The impulse x J is defined in Equation 9.6 as ( ) f i t x t x J = . F t dt = area under the ( ) x F t curve between i t and f t 1 (4 ms)(1000 N) (6 4 ms)(1000 N) 4 Ns 2 x J= + - = 9.4. Model: The particle is subjected to an impulsive force. Visualize: Please refer to Figure EX9.4. Solve: Using Equation 9.6, the impulse is the area under the curve. From 0 s to 2 ms the impulse is 1 ( )( 3 ) 2 . Fdt = -500 N 210- s = -0.5 N s From 2 ms to 8 ms the impulse is 1 ( )( ) 2 . Fdt = +2000 N 8 ms - 2 ms = +6.0 N s From 8 ms to 10 ms the impulse is 1 ( )( ) 2 . Fdt = -500 N 10 ms - 8 ms = -0.5 N s Thus, from 0 s to 10 ms the impulse is (-0.5 + 6.0 - 0.5) N s = 5.0 N s. 9.5. Visualize: Please refer to Figure EX9.5. Solve: The impulse is defined in Equation 9.6 as ( ) f i t x t x J = . F t dt = area under the ( ) x F t curve between i t and f t 1 ( )( ) 3 2 max max .6.0 N s = F 8 ms . F =1.510 N 9.6. Model: Model the object as a particle and the interaction as a collision. Visualize: Please refer to Figure EX9.6. Solve: The object is initially moving to the right (positive momentum) and ends up moving to the left (negative momentum). Using the impulse-momentum theorem pfx = pix + Jx , 2 kg m/s 6 kg m/s x - = + + J 8 kg m/s 8 N s x . J = - = - Since avg , x J = F .t we have avg F .t = -8 N s 2 avg 8 N s 8 10 N 10 ms F - . = =- The force is F = (8102 N, left). .. 9.7. Model: Model the object as a particle and the interaction with the force as a collision. Visualize: Please refer to Figure EX9.7. Solve: Using the equations fx ix x p = p + J and f i ( ) t x x t J = . F t dt = area under force curve ( ) ( )( ) f 2.0 kg 2.0 kg 1.0 m/s x v= +(area under the force curve) ( ) ( )( ) f 1.0 m/s 1 1.0 s 2.0 N 2.0 m/s 2.0 kg x .v = + = Becaue fx v is positive, the object moves to the right at 2.0 m/s. Assess: For an object with positive velocity, a positive impulse increases the objects speed. The opposite is true for an object with negative velocity. 9.8. Model: Model the object as a particle and the interaction with the force as a collision. Visualize: Please refer to Figure EX9.8. Solve: Using the equations fx ix x p = p + J and f i ( ) t x x t J = . F t dt = area under force curve ( ) ( )( ) f 2.0 kg 2.0 kg 1.0 m/s x v= +(area under the force curve) ( ) ( )( ) f 1.0 m/s 1 2.0 N 0.50 s 0.50 m/s 2.0 kg x v . . . = + . . - = . . Assess: For an object with positive velocity, a negative impulse slows the object. The opposite is true for an object with negative velocity. 9.9. Model: Use the particle model for the sled, the model of kinetic friction, and the impulse-momentum theorem. Visualize: Note that the force of kinetic friction k f imparts a negative impulse to the sled. Solve: Using .px = Jx , we have f f i i f i k k ( ) t t x x x t t p - p = . F t dt = - f .dt = - f .t fx ix k k .mv - mv = - n.t = - mg.t We have used the model of kinetic friction k k f = n, where k is the coefficient of kinetic friction and n is the normal (contact) force by the surface. The force of kinetic friction is independent of time and was therefore taken out of the impulse integral. Thus, ( ) i f k 1 x x t v v g . = - ( )( )( ) 2 1 8.0 m/s 5.0 m/s 1.22 s 0.25 9.8 m/s = - = 9.10. Model: Use the particle model for the falling object and the impulse-momentum theorem. Visualize: Note that the object is acted on by the gravitational force, whose magnitude is mg. Solve: Using the impulse-momentum theorem, f i f i ( ) t y y y y t p - p = J = . F t dt fy iy .mv -mv = -mg.t iy fy v v t g - . . = ( ) 2 5.5 m/s 10.4 m/s 0.50 s 9.8 m/s - - - = = Assess: Since yF = -mg is independent of time, we have taken it out of the impulse integral. 9.11. Model: Model the tennis ball as a particle, and its interaction with the wall as a collision. Visualize: The force increases to max F during the first two ms, stays at max F for two ms, and then decreases to zero during the last two ms. The graph shows that x F is positive, so the force acts to the right. Solve: Using the impulse-momentum theorem pfx = pix + Jx , ( )( ) ( )( ) ( ) 6 ms 0 0.06 kg 32 m/s 0.06 kg 32 m/s x = - + . F t dt The impulse is ( ) ( ) ( ) ( ) ( ) ( )( ) ( )( ) 6 ms max max max max 0 2 max area under force curve 1 0.002 s 0.002 s 1 0.002 s 0.004 s 2 2 0.06 kg 32 m/s 0.06 kg 32 m/s 9.6 10 N 0.004 s x F t dx F F F F F = = + + = + . = = . 9.12. Model: Model the ball as a particle, and its interaction with the wall as a collision in the impulse approximation. Visualize: Please refer to Figure EX9.12. Solve: Using the equations fx ix x p = p + J and f i ( ) t x x t J = . F t dt = area under force curve ( ) ( )( ) f 0.250 kg 0.250 kg 10 m/s x v = - + (500 N)(8.0 ms) ( ) f 10 m/s 4.0 N 6 m/s 0.250 kg x v . . . = - + . . = . . Assess: The balls final velocity is positive, indicating it has turned around. 9.13. Model: Model the glider cart as a particle, and its interaction with the spring as a collision. Visualize: Solve: Using the impulse-momentum theorem pfx - pix = . Fdt, ( )( ) ( )( ) 1 ( )( ) 2 0.6 kg 3 m/s - 0.6 kg -3 m/s = area under force curve = 36 N .t ..t = 0.20 s 9.14. Model: Choose car + gravel to be the system. Ignore friction in the impulse approximation. Visualize: Solve: There are no external forces on the car + gravel system, so the horizontal momentum is conserved. This means f i. x x p = p Hence, ( ) ( )( ) ( )( ) f 10,000 kg 4000 kg 10,000 kg 2.0 m/s 4000 kg 0.0 m/s x + v = + f 1.43 m/s x .v = 9.15. Model: Choose car + rainwater to be the system. Visualize: There are no external horizontal forces on the car + water system, so the horizontal momentum is conserved. Solve: Conservation of momentum is f i. x x p = p Hence, ( )( ) ( )( ) ( )( ) car water car water m + m 20 m/s = m 22 m/s + m 0 m/s ( )( ) ( )( ) water . 5000 kg + m 20 m/s = 5000 kg 22 m/s 2 water .m = 5.010 kg 9.16. Model: Choose skydiver + glider to be the system in the impulse approximation. Visualize: Note that there are no external forces along the x-direction (ignoring friction in the impulse approximation), implying conservation of momentum along the x-direction. Solve: The momentum conservation equation fx ix p = p is ( )( ) ( )( ) ( )( ) G D 680 kg 60kg 60 kg 680 kg 30 m/s x x - v + v = Immediately after release, the skydivers horizontal velocity is still ( ) D 30 m/s. x v = Thus ( )( ) ( )( ) ( )( ) G 620 kg 60 kg 30 m/s 680 kg 30 m/s x v + = ( ) G 30 m/s x . v = Assess: The skydivers motion in the vertical direction has no influence on the gliders horizontal motion. 9.17. Model: We will define our system to be bird + bug. This is the case of an inelastic collision because the bird and bug move together after the collision. Horizontal momentum is conserved because there are no external forces acting on the system during the collision in the impulse approximation. Visualize: Solve: The conservation of momentum equation fx ix p = p is ( ) ( ) ( ) 1 2 fx 1 ix 1 2 ix 2 m + m v = m v + m v ( ) ( )( ) ( )( ) f 300 g 10 g 300 g 6.0 m/s 10 g 30 m/s x . + v = + - f 4.8 m/s x .v = Assess: We left masses in grams, rather than convert to kilograms, because the mass units cancel out from both sides of the equation. Note that i 2 ( ) x v is negative. 9.18. Model: The two cars are not an isolated system because of external frictional forces. But during the collision friction is not going to be significant. Within the impulse approximation, the momentum of the Cadillac + Volkswagen system will be conserved in the collision. Visualize: Solve: The momentum conservation equation fx ix p = p is ( ) ( ) ( ) C VW fx C ix C VW ix VW m + m v = m v + m v . ( )( ) ( )( ) i VW 0 kg mph 2000 kg 1.0 mph 1000 kg x = + v ( ) i VW 2.0 mph x . v = - You need a speed of 2.0 mph. 9.19. Model: Because of external friction and drag forces, the car and the blob of sticky clay are not exactly an isolated system. But during the collision, friction and drag are not going to be significant. The momentum of the system will be conserved in the collision, within the impulse approximation. Visualize: Solve: The conservation of momentum equation fx ix p = p is ( )( ) ( ) ( ) C B f x B ix B C ix C m + m v = m v + m v ( )( ) ( )( ) 2 i B i B 0 kg m/s 10 kg 1500 kg 2.0 m/s ( ) 3.0 10 m/s x x . = v + - . v = Assess: This speed of the blob is around 600 mph, which is very large. However, we must point out that a very large speed is expected in order to stop a car with only 10 kg of clay. 9.20. Model: We will define our system to be archer + arrow. The force of the archer (A) on the arrow (a) is equal to the force of the arrow on the archer. These are internal forces within the system. The archer is standing on frictionless ice, and the normal force by ice on the system balances the weight force. Thus ext F = 0 .. .. on the system, and momentum is conserved. Visualize: The initial momentum ix p of the system is zero, because the archer and the arrow are at rest. The final moment fx p must also be zero. Solve: We have A A 0 kg m/s. a a M v + m v = Therefore, ( )( ) A A 0.100 kg 100 m/s 0.20 m/s 50 kg a a v m v m - - = = =- The archers recoil speed is 0.20 m/s. Assess: It is the total final momentum that is zero, although the individual momenta are nonzero. Since the arrow has forward momentum, the archer will have backward momentum. 9.21. Model: We will define our system to be Bob + rock. Bobs (B) force on the rock (R) is equal to the rocks force on Bob. These are internal forces within the system. Bob is standing on frictionless ice, and the normal force by ice on the system balances the weight. ext F = 0 .. .. on the system, and thus momentum is conserved. Visualize: The initial momentum ix p of the system is zero because Bob and the rock are at rest. Thus f 0 x p = kg m/s. Solve: We have B B R R m v + m v = 0 kg m/s. Hence, R ( ) B R B 0.500 kg 30 m/s 0.20 m/s 75 kg v m v m . . = - = -. . = - . . Bobs recoil speed is 0.20 m/s. Assess: Since the rock has forward momentum, Bobs momentum is backward. This makes the total momentum zero. 9.22. Model: We will define our system to be Dan + skateboard, and their interaction as an explosion. While friction is present between the skateboard and the ground, it is negligible in the impulse approximation. Visualize: The system has nonzero initial momentum i . x p As Dan (D) jumps backward off the gliding skateboard (S), the skateboard will move forward in such a way that the final total momentum of the system fx p is equal to i . x p Solve: We have ( ) ( ) ( ) S f S D f D S D i . x x x m v + m v = m + m v Hence, ( )( ) ( )( ) ( )( ) f D 5.0 kg 8.0 m/s 50 kg 5.0 kg 50 kg 4.0 m/s x + v = + ( ) f D 3.6 m/s x . v = 9.23. Model: We assume that the momentum is conserved in the collision. Visualize: Please refer to Figure EX9.23. Solve: The conservation of momentum equation yields ( fx )1 ( fx )2 ( ix )1 ( ix )2 p + p = p + p ( ) f 1 0 kg m/s 2 kg m/s 4 kg m/s x . p + = - ( ) f 1 2 kg m/s x . p = - ( ) ( ) ( ) ( ) fy 1 fy 2 iy 1 iy 2 p + p = p + p ( ) f 1 1 kg m/s 2 kg m/s 1 kg m/s y . p - = + ( ) f 1 4 kg m/s y . p = Thus, the final momentum of particle 1 is (-2i + 4 j) kg m/s. 9.24. Model: This problem deals with the conservation of momentum in two dimensions in an inelastic collision. Visualize: Solve: The conservation of momentum equation p..before = p..after is ( ) ( ) ( ) 1 ix 1 2 ix 2 1 2 fx m v + m v = m + m v ( ) ( ) ( ) 1 iy 1 2 iy 2 1 2 fy m v + m v = m + m v Substituting in the given values, ( )( ) ( ) f .02 kg 3.0 m/s + 0 kg m/s = .02 kg + .03 kg v cos. ( )( ) ( ) f 0 kg m/s + .03 kg 2.0 m/s = .02 kg + .03 kg v sin. f .v cos. =1.2 m/s f v sin. =1.2 m/s ( )2 ( )2 f .v = 1.2 m/s + 1.2 m/s =1.7 m/s tan 1 y tan 1 (1) 45 x v v . = - = - = The ball of clay moves 45 north of east at 1.7 m/s. 9.25. Model: Model the ball as a particle. We will also use constant-acceleration kinematic equations. Solve: (a) The momentum just after throwing is ( )( ) 0 0 0 cos30 cos30 0.050 k p x = p = mv = g 25 m/s cos30 =1.083 kg m/s ( )( ) 0 0 0 sin30 sin30 0.050 kg 25 m/s sin30 0.625 kg m/s y p = p = mv = = The momentum just after throwing is (1.08, 0.63) kg m/s. The momentum at the top is 1 1 0 1.08 kg m/s x x x p = mv = mv = 1 1 0 kg m/s y y p = mv = Just before hitting the ground, the momentum is 2 2 0 1.08 kg m/s x x x p = mv = mv = 2 y 2 y p = mv 0 0.63 kg m/s y = -mv = - (b) x p is constant because no forces act on the ball in the x-direction. Mathematically, x 0 N x x F dp p dt = = . = constant (c) The change in the y-component of the momentum during the balls flight is 0.625 kg m/s 0.625 kg m/s 1.250 kg m/s y .p = - - = - From kinematics, the time to reach the top is obtained as follows: ( ) 1y 0 y y 1 0 v = v + a t - t ( ) 0 1 2 12.5 m/s 1.276 s 9.8 m/s y y v t a - - . = = = - The time of flight is thus ( ) 1 .t = 2t = 2 1.276 s = 2.552 s. Multiplying this time by mg, the y-component of the weight, yields 1.25 kg m/s. This follows from the impulse-momentum theorem: y.p = -mg.t 9.26. Model: Model the rocket as a particle, and use the impulse-momentum theorem. The only force acting on the rocket is due to its own thrust. Visualize: Please refer to Figure P9.26. Solve: (a) The impulse is Jx = . Fx dt = area of the ( ) x F t graph between t = 0 s and t = 30 s 1 ( )( ) 4 2 = 1000 N 30 s =1.510 N s (b) From the impulse-momentum theorem, 4 f i 1.5 10 N s. x x p = p + That is, the momentum or velocity increases as long as x J increases. When x J increases no more, the speed will be a maximum. This happens at t = 30 s. At this time, 4 ( ) ( )( ) 4 f i f 1.5 10 N s 425 kg 425 kg 75 m/s 1.5 10 N s x x x mv = mv + . v = + f 110 m/s x .v = 9.27. Solve: Using the equation ( ) ( ) ( ( ) ( )) 2.0 s 0 2.0 s 0 area under the force curve 0 m/s 1 10 N sin 2 0.250 kg 4.0 s 40 N/m 4.0 s cos 2 2 4.0 s 80 m/s cos cos 0 25 m/s fx ix v v m t dt t p p p p p = + = + . . . . . . . .. . . . = . ..- . . . . .. . . . . . = -. . - . . . . = . Assess: The force is applied for half the period of 4.0 s. During that time, sin 2 4.0 s . p t . . . . . is positive, so an object initially at rest acquires a positive velocity. 9.28. Model: Let the system be ball + racket. During the collision of the ball and racket, momentum is conserved because all external interactions are insignificantly small. Visualize: Solve: (a) The conservation of momentum equation fx ix p = p is ( ) ( ) ( ) ( ) R fx R B fx B R ix R B ix B m v + m v = m v + m v ( )( ) ( )( ) ( )( ) ( )( ) f R 1.000 kg 0.060 kg 40 m/s 1.000 kg 10 m/s 0.060 kg 20 m/s x v + = + - ( ) f R 6.4 m/s x . v = (b) The impulse on the ball is calculated from ( ) ( ) fx B ix B x p = p + J as follows: (0.060 kg)(40 m/s) (0.060 kg)( 20 m/s) x = - + J avg 3.6 N s x . J = = . Fdt = F .t . 2 avg 3.6 Ns 3.6 10 N 10 ms F = = Let us now compare this force with the gravitational force on the ball ( ) ( )( 2 ) G B B F = m g = 0.060 kg 9.8 m/s = 0.588 N. Thus, avg G B F = 612(F ) . Assess: This is a significant force and is reasonable because the impulse due to this force changes the direction as well as the speed of the ball from approximately 45 mph to 90 mph. 9.29. Model: Model the ball as a particle that is subjected to an impulse when it is in contact with the floor. We will also use constant-acceleration kinematic equations. Ignore any forces other than the interaction between the floor and the ball during the collision in the impulse approximation. Visualize: Solve: To find the balls velocity just before and after it hits the floor: ( ) ( )( ) ( ) ( )( ) 2 2 2 2 2 1 0 1 0 1 2 2 2 2 2 2 3 2 3 2 2 2 2 0 m /s 2 9.8 m/s 0 2.0 m 6.261 m/s 2 0 m /s 2 9.8 m/s 1.5 m 0 m 5.422 m/s y y y y y y y y y v v a y y v v v a y y v v = + - = + - - . =- = + - . = + - - . = The force exerted by the floor on the ball can be found from the impulse-momentum theorem: 2 1 1mv y = mv y + . Fdt = mv y + area under the force curve ( )( ) ( )( ) 1 ( 3 ) 2 max . 0.200 kg 5.422 m/s = - 0.200 kg 6.261 m/s + F 5.010- s 2 max . F = 9.310 N Assess: A maximum force of 9.3102 N exerted by the floor is reasonable. 9.30. Model: Model the rubber ball as a particle that is subjected to an impulsive force when it comes in contact with the floor. We will also use constant-acceleration kinematic equations and the impulse-momentum theorem. Visualize: Solve: (a) To find the magnitude and direction of the impulse that the floor exerts on the ball, we use the impulse-momentum theorem: fy iy y p = p + J ( ) y 2 y 1y 2 y 1y . J = p - p = m v - v Let us now find 1y v and 2 y v by using kinematics. For the falling and rebounding ball, ( ) ( )( ) ( ) ( )( ) 2 2 2 2 2 1 0 1 0 1 2 2 2 2 2 2 3 2 3 2 2 2 2 0 m /s 2 9.8 m/s 0 m 1.8 m 5.940 m/s 2 0 m /s 2 9.8 m/s 1.2 m 0 m 4.850 m/s y y y y y y y y y v v a y y v v v a y y v v = + - = + - - . =- = + - . = + - - . = Going back to the impulse-momentum equation, we find (0.040 kg) 4.850 m/s ( 5.940 m/s) 0.432 N s y J = .. - - .. = The impulse is 0.43 N s upward. (b) As the ball compresses, the force of contact increases and the ball slows to 0 m/s. y v = Then in decompression the ball is accelerated upward. To a good approximation, the force due to the floor as a function of time is shown in the figure. (c) For a rubber ball, .t is likely in the range 5 to 10 ms. For 10 ms, avg avg 0.432 N s 40 N y J = F .t = . F The force is 80 N if .t = 5 ms. Altogether, 40 to 80 N is a reasonable estimate. 9.31. Model: Model the cart as a particle rolling down a frictionless ramp. The cart is subjected to an impulsive force when it comes in contact with a rubber block at the bottom of the ramp. We will use the impulsemomentum theorem and the constant-acceleration kinematic equations. Visualize: Solve: From the free-body diagram on the cart, Newtons second law before the collision is ( )x G sin x S F = F . = ma sin sin30 2 x a mg g g m . . = = = Using this acceleration, we can find the carts speed just before its contact with the rubber block: 2 2 ( ) 1 0 1 0 2 x x x v = v + a x - x 2 2 1 2 = 0 m /s + 2 ( g)(1.0 m - 0 m) 1 3.13 m/s x .v = Now we can use the impulse-momentum theorem to obtain the velocity just after the collision: 2 1 1 a rea under the force graph x x x x mv = mv + . F dt = mv + ( ) ( )( ) 1 ( )( 3 ) 2 2 0.500 kg 0.500 kg 3.13 m/s 200 N 26.7 10 s x v - . = - 2 2.21 m/s x .v = - Note that the given force graph is positive, but in this coordinate system the impulse of the force is to the left up the slope. That is the reason to put a minus sign while evaluating the x . F dt integral. We can once again use a kinematic equation to find how far the cart will roll back up the ramp: 2 2 ( ) 3 2 3 2 2 x x x v = v + a x - x ( )2 ( )2 ( 1 )( ) 2 3 2 . 0 m/s = -2.21 m/s + 2 - g x - x ( ) 3 2 . x - x = 0.50 m 9.32. Solve: Using Newtons second law for the xdirection, x . x F dp dt = Therefore, (6 2 kg m/s) 12 N x F d t t dt = = Assess: The x-component of the net force on an object is equal to the time rate of change of the x-component of the objects momentum. 9.33. Visualize: Solve: Using Newtons second law for the y-direction and the chain rule, ( ) ( ) ( ) ( )( ) ( )( ) net 2 2 0.50 kg/s 120 m/s 48 kg 18 m/s 8.0 10 N y y y y y dp d dm dv F mv vm dt dt dt dt . . = = = + . . . . = - + = Assess: Since the rocket is losing mass, dm 0. dt < The time derivative of the velocity is the acceleration. 9.34. Model: Model the train cars as particles. Since the train cars stick together, we are dealing with perfectly inelastic collisions. Momentum is conserved in the collisions of this problem in the impulse approximation, in which we ignore external forces during the time of the collision. Visualize: Solve: In the collision between the three-car train and the single car: mv1x + (3m)v2x = 4mv3x 1 2 3 3 4 x x x .v + v = v ( ) ( ) 3 4.0 m/s 3 2.0 m/s 4 x v . + =3 2.5 m/s x .v = In the collision between the four-car train and the stationary car: ( ) ( ) 3 4 5 4 5 x x x m v + mv = m v 3 5 4 0 m/s 5 x x . v + = v 3 ( )( ) 5 4 0.8 2.5 m/s 2.0 m/s 5 x x .v = v = = 9.35. Model: The dart and cork are particles in free fall until they have a perfectly inelastic head-on collision. Ignore air friction. Visualize: Solve: The positions of the dart and cork just before the collision are ( ) ( ) ( ) ( ) ( ) 2 1 D 0 D 0 D 1 1 2 1 C 0 C 1 1 2 1 2 y y v t gt y y gt = + - = - In the particle model, the dart and cork have no physical size, so ( ) ( ) 1 D 1 C y = y . Hence ( ) ( 2 ) 2 ( 2 ) 2 1 1 1 1 0 m 9.0 m/s 1 9.8 m/s 3 m 1 9.8 m/s 1 s 2 2 3 + t - t = - t .t = At this time, the velocities of the dart and cork are ( ) ( ) ( ) ( ) ( ) ( ) 2 1 D 0 D 1 2 1 C 0 C 1 9.0 m/s 9.8 m/s 1 s 5.73 m/s 3 v 9.8 m/s 1 s 3.27 m/s 3 v v gt v gt = - = - . . = . . . . = - = - . . = - . . . . These are the initial velocities to use with momentum conservation in a perfectly inelastic collision. ( ) ( ) ( ) f i D C D 1 D C 1 C y y p p m m v m v m v = + = + Thus ( )( ) ( )( ) ( ) 0.030 kg 5.73 m/s 0.020 kg 3.27 m/s 2.13 m/s, up 0.030 kg 0.20 kg v + - = = + Assess: The heavier, faster upward-going dart has more momentum than the falling cork, so the total momentum is upwards. 9.36. Model: Model the earth (E) and the asteroid (A) as particles. Earth + asteroid is our system. Since the two stick together during the collision, this is a case of a perfectly inelastic collision. Momentum is conserved in the collision since no significant external force acts on the system. Visualize: Solve: (a) The conservation of momentum equation fx ix p = p is ( ) ( ) ( ) A ix A E ix E A E fx m v + m v = m + m v ( 13 )( 4 ) ( 13 24 ) f 1.0 10 kg 4.0 10 m s 0 kg m/s 1.0 10 kg 5.98 10 kg x . + = + v 8 f 6.7 10 m/s x .v = - (b) The speed of the earth going around the sun is ( 11 ) 4 E 7 2 2 1.50 10 m 3.0 10 m/s 3.15 10 s v r T p p = = = Hence, 12 10 f E 2 10 2 10 %. x v v = - = - Assess: The earths recoil speed is insignificant compared to its orbital speed because of its large mass. 9.37. Model: Model the skaters as particles. The two skaters, one traveling north (N) and the other traveling west (W), are a system. Since the two skaters hold together after the collision, this is a case of a perfectly inelastic collision in two dimensions. Momentum is conserved since no significant external force in the x-y plane acts on the system in the collision. Visualize: Solve: (a) The x-component of the conservation of momentum is ( N W ) fx N ( ix )N W ( ix )W m + m v = m v + m v ( ) ( )( ) f 75 kg 60 kg 0 kg m/s 60 kg 3.5 m/s x . + v = + - f 1.556 m/s x .v = - The y-component of the conservation of momentum is ( ) ( ) ( ) N W fy N iy N W iy W m + m v = m v + m v ( ) ( )( ) f 75 kg 60 kg 75 kg 2.5 m/s 0 kg m/s y . + v = + f 1.389 m/s y .v = ( ) ( )2 2 f f f 2.085 m/s x y .v = v + v = The time to glide to the edge of the rink is f radius of the rink 25 m 12.0 s v 2.085 m/s = = (b) The location is 1 f f tan 42 y x . = - v v = north of west. Assess: A time of 12.0 s in covering a distance of 25 m at a speed of 2 m/s is reasonable. 9.38. Model: This problem deals with a case that is the opposite of a collision. The two ice skaters, heavier and lighter, will be modeled as particles. The skaters (or particles) move apart after pushing off against each other. During the explosion, the total momentum of the system is conserved. Visualize: Solve: The initial momentum is zero. Thus the conservation of momentum equation fx ix p = p is ( ) ( ) H f H L f L 0 kg m/s x x m v + m v = ( )( ) ( )( ) f H f L 75 kg 50 kg 0 kg m/s x x . v + v = Using the observation that the heavier skater takes 20 s to cover a distance of 30 m, we find ( ) f H 30 m 20 s x v = =1.5 m/s. Thus, ( )( ) ( )( ) f L 75 kg 1.5 m/s 50 kg 0 kg m/s x + v = ( ) f L 2.25 m/s x . v = - Thus, the time for the lighter skater to reach the edge is ( ) f L 30 m 30 m 13.3 s 2.25 m/s x v = = Assess: Conservation of momentum leads to a higher speed for the lighter skater, and hence a shorter time to reach the edge of the ice rink. 9.39. Model: This problem deals with a case that is the opposite of a collision. Our system is comprised of three coconut pieces that are modeled as particles. During the blow up or explosion, the total momentum of the system is conserved in the x-direction and the y-direction. Visualize: Solve: The initial momentum is zero. From pfx = pix , we get ( ) ( ) 1 f 1 3 f 3 cos 0 kg m/s x +m v + m v . = ( ) ( ) ( ) 1 f 1 f 3 3 20 m/s cos 10 m/s 2 x m v m v m m . - - - . = = = From f i, x x p = p we get ( ) ( ) 2 f 3 f3 2 sin 0 kg m/s y +m v + m v . = ( ) 2 ( f )2 ( ) f 3 3 20 m/s sin 10 m/s 2 y m v m v m m . - - - . = = = ( ) ( )2 ( )2 f 3 . v = 10 m/s + 10 m/s =14.1 m/s . = tan-1 (1) = 45 The velocity is 14.1 m/s at 45 east of north. 9.40. Model: The billiard balls will be modeled as particles. The two balls, 1 m (moving east) and 2 m (moving west), together are our system. This is an isolated system because any frictional force during the brief collision period is going to be insignificant. Within the impulse approximation, the momentum of our system will be conserved in the collision. Visualize: Note that m1 = m2 = m. Solve: The equation fx ix p = p yields: ( ) ( ) ( ) ( ) 1 fx 1 2 fx 2 1 ix 1 2 ix 2 m v + m v = m v + m v ( ) ( ) ( ) 1 f 1 1 i 1 2 i 2 cos 0 kg m/s x x .m v . + = m v + m v ( ) ( ) ( ) f 1 i 1 i 2 cos x x . v . = v + v = 2.0 m/s -1.0 m/s =1.0 m/s The equation fy iy p = p yields: ( ) ( ) 1 f 1 2 f 2 sin 0 kg m/s y y +m v . + m v = ( ) ( ) f 1 f 2 sin 1.41 m/s y . v . = - v = - ( ) ( )2 ( )2 f 1 . v = 1.0 m/s + -1.41 m/s =1.73 m/s tan 1 1.41 m/s 55 1.0 m/s . = - .. .. = . . The angle is below +x axis, or south of east. 9.41. Model: This is a two-part problem. First, we have an inelastic collision between the wood block and the bullet. The bullet and the wood block are an isolated system. Since any external force acting during the collision is not going to be significant (the impulse approximation), the momentum of the system will be conserved. The second part involves the dynamics of the block + bullet sliding on the wood table. We treat the block and the bullet as particles. Visualize: Solve: The equation fx ix p = p gives ( ) ( ) ( ) B W fx B ix B W ix W m + m v = m v + m v ( ) ( )( ) ( )( ) f iB 0.010 kg 10 kg 0.010 kg 10 kg 0 m/s x x . + v = v + ( ) f iB 1 1001 x x .v = v From the model of kinetic friction, ( ) ( ) k k k B W B W x f = - n = - m + m g = m + m a x k .a = - g Using the kinematic equation 2 2 ( ) 1 2 1 0 2 , x x x v = v + a x - x 2 2 ( ) 1 2 k 1 0 2 x x v = v - g x - x 2 2 2 f k 1 0 m s 2 x. = v - g x ( ) 2 2 i B k 1 1 2 1001 x... .. v = g x . . ( ) ( )( 2 )( ) i B k 1 1001 2 1001 2 0.20 9.8 m/s 0.050 m 443 m/s x. v = g x = = The bullets speed is 4.4102 m/s. Assess: The bullets speed is reasonable (900 mph). 9.42. Model: This is a two-part problem. First, we have an inelastic collision between Fred (F) and Brutus (B). Fred and Brutus are an isolated system. The momentum of the system during collision is conserved since no significant external force acts on the system. The second part involves the dynamics of the Fred + Brutus system sliding on the ground. Visualize: Note that the collision is head-on and therefore one-dimensional. Solve: The equation fx ix p = p is ( ) ( ) ( ) F B fx F ix F B ix B m + m v = m v + m v ( ) ( )( ) ( )( ) f 60 kg 120 kg 60 kg 6.0 m/s 120 kg 4.0 m/s x . + v = - + f 0.667 m/s x .v = The positive value indicates that the motion is in the direction of Brutus. The model of kinetic friction yields: ( ) ( ) k k k F B F B x x k f = - n = - m + m g = m + m a .a = - g Using the kinematic equation 2 2 ( ) 1 0 1 0 2 , x x x v = v + a x - x we get ( )( ) ( ) ( ) 2 2 2 2 2 2 1 0 k 1 f 1 2 2 2 2 1 1 2 0 m s 2 0.30 9.8 m/s 0 m s 0.667 m/s 5.9 m/s 7.6 cm x x x v v gx v x x x = - . = - . = - . = They slide 7.6 cm in the direction Brutus was running. Assess: After the collision, Fred and Brutus slide with a small speed but with a good amount of kinetic friction. A stopping distance of 7.6 cm is reasonable. 9.43. Model: Model the package and the rocket as particles. This is a two-part problem. First we have an inelastic collision between the rocket (R) and the package (P). During the collision, momentum is conserved since no significant external force acts on the rocket and the package. However, as soon as the package + rocket system leaves the cliff they become a projectile motion problem. Visualize: Solve: The minimum velocity after collision that the package + rocket must have to reach the explorer is 0 , x v which can be found as follows: ( ) 1 ( )2 1 0 0 y 1 0 2 y 1 0 y = y + v t - t + a t - t 1 ( 2 ) 2 2 1 .-200 m = 0 m+ 0 m+ -9.8 m/s t 1 .t = 6.389 s With this time, we can now find 0x v using ( ) 1 ( )2 1 0 0 1 0 2 1 0 . x x x = x + v t - t + a t - t We obtain ( ) 0 30 m 0 m 6.389 s 0 m x = + v + 0 f 4.696 m/s x x .v = = v We now use the momentum conservation equation fx ix p = p which can be written ( ) ( ) ( ) R P fx R ix R P ix P m + m v = m v + m v ( )( ) ( )( ) ( )( ) i R 1.0 kg 5.0 kg 4.696 m/s 1.0 kg 5.0 kg 0 m/s x . + = v + ( ) i R 28 m/s x . v = 9.44. Model: The clay balls undergo a perfectly inelastic collision in the impulse approximation. Visualize: Solve: (a) The total momentum in reference frame S is ( ) ( ) ( ) ( ) ( )( ) 1 2 1 1 2 2 0.020 kg 12 m/s 0 kg m/s 0.24 kg m/s x x x x P = p + p = m v + m v = + = (b) In the S' frame, the total momentum is zero. ( ) ( ) ( ) ( ) 1 2 1 1 2 2 0 x x x x P p p m v m v ' = ' + ' = . ' = - ' Equation 4.24 relates velocities in the two reference frames. ( ) ( ) ( ) ( ) 1 1 2 2 x x x x x x v v V v v V ' = - ' = - Thus (( ) ) (( ) ) ( ) ( ) 1 1 2 2 1 1 2 2 1 2 4.0 m/s x x x x x x x m v V m v V m v m v V m m - =- - + . = = + (c) Since the total momentum is zero, the final velocity of the resulting 60 g clay ball must be ( )f 0 m/s. x v ' = (d) Using Equation 4.24 again, ( ) ( ) f f 0 m/s 4.0 m/s 4.0 m/s x x x v = v ' +V = + = Assess: The final velocity of both balls in the S frame is less than the original velocity of the smaller ball, and is in the same direction. The S' frame is called the center of mass frame. Some collisions are more easily studied in the center of mass frame, such as high energy particle collisions at particle accelerator facilities. 9.45. Model: This is a two-part problem. First, we have an explosion that creates two particles. The momentum of the system, comprised of two fragments, is conserved in the explosion. Second, we will use kinematic equations and the model of kinetic friction to find the displacement of the lighter fragment. Visualize: Solve: The initial momentum is zero. Using momentum conservation fx ix p = p during the explosion, ( ) ( ) ( ) ( ) H 1x H L 1x L H 0x H L 0x L m v + m v = m v + m v ( ) ( ) 1 H 1 L 7 0 kg m/s x x . m v + m v = . ( ) ( 1 )( ) 1x H 7 1x L v = - v Because H m slides to 2H x = -8.2 m before stopping, we have k k H k H k H H H f = n = w = m g = m a H k .a = g Using kinematics, ( )2 ( )2 ( ) 2 H 1 H H 2H 1H 2 x x v = v + a x - x ( ) ( ) ( ) 2 2 1 2 2 7 1 L k 0 m s 2 8.2 m 0 m x. = v + g - - ( )1 L k 88.74 m/s x . v = - How far does L m slide? Using the information obtained above in the following kinematic equation, ( )2 ( )2 ( ) 2 L 1 L L 2L 1L 2 x x v = v + a x - x 2 2 ( )2 k k2L .0 m s = 88.74 - 2 gx 2 2L . x = 4.010 m Assess: Note that H a is positive, but L a is negative, and both are equal in magnitude to k g. Also, 2H x is negative but 2L x is positive. 9.46. Model: We will model the two fragments of the rocket after the explosion as particles. We assume the explosion separates the two parts in a vertical manner. This is a three-part problem. In the first part, we will use kinematic equations to find the vertical position where the rocket breaks into two pieces. In the second part, we will apply conservation of momentum to the system (that is, the two fragments) in the explosion. In the third part, we will again use kinematic equations to find the velocity of the heavier fragment just after the explosion. Visualize: Solve: The rocket accelerates for 2.0 s from rest, so ( ) ( 2 )( ) 1 0 1 0 0 m/s 10 m/s 2 s 0 s 20 m/s y y y v = v + a t - t = + - = ( ) 1 ( )2 1 ( 2 )( )2 1 0 0 1 0 2 1 0 2 0 m 0 m 10 m/s 2 s 20 m y y y = y + v t - t + a t - t = + + = At the explosion the equation fy iy p = p is ( ) ( ) ( ) L 2 y L H 2 y H L H 1y m v + m v = m + m v ( )( ) ( )( ) ( )( ) 2 L 2 H 500 kg 1000 kg 1500 kg 20 m/s y y . v + v = To find 2 H ( )y v we must first find 2 L ( ) , y v the velocity after the explosion of the upper section. Using kinematics, ( ) ( ) ( )( ) 2 2 2 3 L 2 L 3L 2L 2 9.8 m/s y y v = v + - y - y ( ) ( 2 )( ) 2 L 2 9.8 m/s 530 m 20 m 99.98 m/s y . v = - = Now, going back to the momentum conservation equation we get ( )( ) ( )( ) ( )( ) 2 H 500 kg 99.98 m/s 1000 kg 1500 kg 20 m/s y + v = ( ) 2 H 20 m/s y . v = - The negative sign indicates downward motion. 9.47. Model: Let the system be bullet + target. No external horizontal forces act on this system, so the horizontal momentum is conserved. Model the bullet and the target as particles. Since the target is much more massive than the bullet, it is reasonable to assume that the target undergoes no significant motion during the brief interval in which the bullet passes through. Visualize: Solve: (a) By assuming that the target has negligible motion during the interval in which the bullet passes through, the time is that needed to slow from 1200 m/s to 900 m/s in a distance of 30 cm. Well use kinematics to first find the acceleration, then the time. 2 2 1 B 0 B 2 2 2 2 1 B 0 B 6 2 1 B 0 B 1 B 0 B 4 6 2 ( ) ( ) 2 ( ) ( ) (900 m/s) (1200 m/s) 1.05 10 m/s 2 2(0.30 m) ( ) ( ) ( ) ( ) 900 m/s 1200 m/s 2.86 10 s 286 s 1.05 10 m/s x x x x x x x x x x x x v v a x a v v x v v a t t v v a - = + . - - . = = =- . = + . - - . . = = = = - The average force on the bullet is avg 26 kN. x F = m a = (b) Now we can use the conservation of momentum equation 1x 0x p = p to find ( ) T 1 T B 1 B T 0 T B 0 B B 0 B B 0 B B 1 B 1 T T ( ) ( ) ( ) ( ) 0 ( ) ( ) ( ) ( ) 0.025 kg (1200 m/s) (900 m/s) 0.021 m/s 350 kg x x x x x x x x m v m v m v m v m v v m v m v m + = + = + - . = = - = 9.48. Model: Model the two blocks (A and B) and the bullet (L) as particles. This is a two-part problem. First, we have a collision between the bullet and the first block (A). Momentum is conserved since no external force acts on the system (bullet + block A). The second part of the problem involves a perfectly inelastic collision between the bullet and block B. Momentum is again conserved for this system (bullet + block B). Visualize: Solve: For the first collision the equation fx ix p = p is ( ) ( ) ( ) ( ) L 1x L A 1x A L 0x L A 0x A m v + m v = m v + m v ( )( ) ( )( ) ( )( ) 1 L 0.010 kg 0.500 kg 6.0 m/s 0.010 kg 400 m/s 0 kg m/s x . v + = + ( ) 1 L 100 m/s x . v = The bullet emerges from the first block at 100 m/s. For the second collision the equation fx ix p = p is ( ) ( ) L B 2 L 1L x x m + m v = m v ( ) ( )( ) 2 0.010 kg 0.500 kg 0.010 kg 100 m/s x . + v = 2 1.96 m/s x .v = 9.49. Model: Model Brian (B) along with his wooden skis as a particle. The collision between Brian and Ashley lasts for a short time, and during this time no significant external forces act on the Brian + Ashley system. Within the impulse approximation, we can then assume momentum conservation for our system. After finding the velocity of the system immediately after the collision, we will apply constant-acceleration kinematic equations and the model of kinetic friction to find the final speed at the bottom of the slope. Visualize: Solve: Brian skiing down for 100 m: ( )2 ( )2 ( ) 1 B 0 B 1B 0B 2 x x x v = v + a x - x 0 m2 s2 2 (100 m 0 m) x = + a - ( ) ( ) 1 B 200 m x x . v = a To obtain , x a we apply Newtons second law to Brian in the x and y directions as follows: ( ) on B B k B sin x x S F = w . - f = m a ( ) on B B cos 0 N y S F = n - w . = .n = wcos. From the model of kinetic friction, k k k Bf = n = w cos. . The x-equation thus becomes B kB B sin cos x w . - w . = m a ( ) k sin cos x .a = g . - . = (9.8 m/s2 )..sin 20 - (0.060)cos20.. = 2.80 m/s2 Using this value of , x a ( ) ( )( 2 ) 1 B 200 m 2.80 m/s 23.7 m/s. x v = = In the collision with Ashley the conservation of momentum equation fx ix p = p is ( ) ( ) B A 2 B 1B x x m + m v = m v B ( ) ( ) 2 1B B A 80 kg 23.66 m/s 80 kg 50 kg x x v m v m m . = = + + =14.56 m/s Brian + Ashley skiing down the slope: 2 2 ( ) 3 2 3 2 2 x x x v = v + a x - x = (14.56 m/s)2 + 2(2.80 m/s2 )(100 m) 3 27.8 m/s x .v = That is, Brian + Ashley arrive at the bottom of the slope with a speed of 27.8 m/s. Note that we have used the same value of x a in the first and the last parts of this problem. This is because x a is independent of mass. Assess: A speed of approximately 60 mph on a ski slope of 200 m length and 20 slope is reasonable. 9.50. Model: Model the spy plane (P) and the rocket (R) as particles. The plane and the rocket undergo a perfectly inelastic collision. During the brief collision time, momentum for the plane + rocket system will be conserved because no significant external forces act on this system during the collision. After finding the velocity of the system immediately after the collision, we will apply projectile equations to find where the system will hit the ground. Visualize: Solve: For the collision we have fx ix p = p and f i. y y p = p This means the magnitude of the final momentum is ( ) ( ) ( ) ( ) 2 2 2 2 f fx fy ix iy p = p + p = p + p ( ) ( ) ( ) 2 2 R P f R ix R P iy P . m + m v = ..m v .. + ..m v .. ( ) ( )( ) ( )( ) 2 2 f 1 1280 kg 725 m/s 575 kg 450 m/s 1280 kg 575 kg .v = .. .. + .. .. + = 519.4 m/s ( )( ) ( )( ) 1 f 1 f 575 kg 450 m/s tan tan 15.58 1280 kg 725 m/s y x p p . - -. . . . = . . = .. .. = . . . . north of east We call this direction the x' axis in the x-y plane. Let us now look at the falling plane + rocket system in the x'-z plane where z is the vertical axis perpendicular to the x-y plane. We have 0 1 0 0 z = 2700 m z = 0 m x' = 0 m t = 0 s 0 0 0 m/s 519.4 m/s z x v v' = = Using the kinematic equation ( ) 1 ( )2 1 0 0 1 0 2 1 0 , z z z = z + v t - t + a t - t we can find the time to fall: 1 ( 2 ) 2 2 1 1 0 m = 2700 m+ 0 m+ -9.8 m/s t .t = 23.47 s In this time 1t , the wreckage travels horizontally to ( ) 1 0 0x 1 0 x x v ' t t ' = ' + - = 0 m + (519.4 m/s)(23.47 s) =12,192 m =12.19 km The enmeshed plane + rocket system lands 12.19 km from the collision point at an angle of 15.58 north of east. 9.51. Model: This is an isolated system, so momentum is conserved in the explosion. Momentum is a vector quantity, so the direction of the initial velocity vector 1 v.. establishes the direction of the momentum vector. The final momentum vector, after the explosion, must still point in the +x-direction. The two known pieces continue to move along this line and have no y-components of momentum. The missing third piece cannot have a ycomponent of momentum if momentum is to be conserved, so it must move along the x-axiseither straight forward or straight backward. We can use conservation laws to find out. Visualize: Solve: From the conservation of mass, the mass of piece 3 is 5 m3 = mtotal - m1 - m2 = 7.010 kg To conserve momentum along the x-axis, we require [ ] [ ] i total i f 1f 2f 3f 1 1f 2 2f 3f p = m v = p = p + p + p = m v + m v + p 13 3f total i 1 1f 2 2f . p = m v - m v - m v = +1.0210 kg m/s Because 3f p > 0, the third piece moves in the +x-direction, that is, straight forward. Because we know the mass 3m , we can find the velocity of the third piece as follows: 13 3f 7 3f 5 3 1.02 10 kg m/s 1.46 10 m/s 7.0 10 kg v p m = = = 9.52. Model: The two railcars make up a system. The impulse approximation is used while the spring is expanding, so friction can be ignored. The spring is massless. Visualize: Solve: Since the cars are at rest initially, the total momentum of the system is zero. Conservation of momentum holds. 1 ( f )1 2 ( f )2 0 x x = m v + m v We are only told that the relative velocity of the two cars after the spring expands is 4.0 m/s, so ( ) ( ) f 2 f 1 4.0 m/s x x v - v = The positive-going car 2 velocity is written first to ensure that the relative velocity is positive. Substitute ( ) ( ) f 2 f 1 4.0 m/s x x v = v + into the conservation of momentum equation, then solve for ( ) f 1 . x v ( ) (( ) ) ( ) ( ) ( ) ( )( ) ( ) 1 f 1 2 f 1 2 f 1 1 2 0 4.0 m/s 4.0 m/s 90 tons 4.0 m/s 3.0 m/s 30 tons 90 tons x x x m v m v m v m m = + + . =- =- =- + + Assess: The other more massive railcar has a velocity ( ) ( ) f 2 f 1 4.0 m/s 1.0 m/s. x x v = v + = A slower speed for the more massive car makes sense. 9.53. Model: This is a three-part problem. In the first part, the shell, treated as a particle, is launched as a projectile and reaches its highest point. We will use constant-acceleration kinematic equations for this part. The shell, which is our system, then explodes at the highest point. During this brief explosion time, momentum is conserved. In the third part, we will again use the kinematic equations to find the horizontal distance between the landing of the lighter fragment and the origin. Visualize: Solve: The initial velocity is v0x = vcos. = (125 m/s)cos55 = 71.7 m/s ( ) 0 sin 125 m/s sin55 102.4 m/s y v = v . = = At the highest point, 1 0 y v = m/s and 1 71.7 m/s. x v = The conservation of momentum equation fx ix p = p is ( ) ( ) ( ) L 1x L H 1x H L H 1x m v + m v = m + m v The heavier particle falls straight down, so 1 H ( ) 0 m/s. x v = Thus, ( )( ) ( )( ) ( ) 1 L 1 L 15 kg 0 kg m s 15 kg 60 kg 71.7 m/s 358 m/s x x v + = + . v = That is, the velocity of the smaller fragment immediately after the explosion is 358 m/s and this velocity is in the horizontal x-direction. Note that 1 L ( ) 0 m/s. y v = To find 2 x , we will first find the displacement 1 0 x - x and then 2 1x - x . For 1 0x - x , ( ) ( ) ( 2 )( ) 1 0 1 0 1 1 0 m s 102.4 m/s 9.8 m/s 0 s 10.45 s y y y v = v + a t - t . = + - t - .t = ( ) 1 ( )2 ( )( ) 1 0 0 1 0 2 1 0 1 0 71.7 m/s 10.45 s 0 m 749 m x x x = x + v t - t + a t - t . x - x = + = For 2 1x - x : ( ) ( ) 1 ( )2 ( )( ) 2 1 1 2 1 2 2 1 2 1 358 m/s 10.45 s 0 m=3741 m x L x x = x + v t - t + a t - t . x - x = + ( ) ( ) 2 2 1 1 0 . x = x - x + x - x = 3741 m+ 749 m = 4490 m = 4.5 km Assess: Note that the time of ascent to the highest point is equal to the time of descent to the ground, that is, 1 0 2 1 . t - t = t - t 9.54. Model: Model the proton (P) and the gold atom (G) as particles. The two constitute our system, and momentum is conserved in the collision between the proton and the gold atom. Visualize: Solve: The conservation of momentum equation fx ix p = p is ( ) ( ) ( ) ( ) G fx G P fx P P ix P G ix G m v + m v = m v + m v ( )( ) ( )( 7 ) ( )( 7 ) f G 197 u 1 u 0.90 5.0 10 m/s 1 u 5.0 10 m/s 0 u m/s x . v + - = + ( ) 5 f G 4.8 10 m/s x . v = 9.55. Model: Model the proton (P) and the target nucleus (T) as particles. The proton and the target nucleus make our system and in the collision between them momentum is conserved. This is due to the impulse approximation because the collision lasts a very short time and the external forces acting on the system during this time are not significant. Visualize: Solve: The conservation of momentum equation fx ix p = p is ( ) ( ) ( ) ( ) T fx T P fx P T ix T P ix P m v + m v = m v + m v ( 5 ) ( )( 6 ) ( )( 6 ) T .m 3.1210 m/s + 1 u -0.75 2.510 m/s = 0 u m/s + 1 u 2.510 m/s T .m =14.0 u Assess: This is the mass of the nucleus of a nitrogen atom. 9.56. Model: This problem deals with a case that is the opposite of a collision. It is a case of an explosion in which a 214Po nucleus (P) decays into an alpha-particle (A) and a daughter nucleus (N). During the explosion or decay, the total momentum of the system is conserved. Visualize: Solve: Conservation of mass requires the daughter nucleus to have mass mN = 214 u - 4 u = 210 u. The conservation of momentum equation fx ix p = p is ( ) ( ) ( ) N fx N A fx A N A P m v + m v = m + m v ( )( ) ( )( 7 ) f N 210 u 4 u 1.92 10 m/ s 0 u m/s x . v + - = ( ) 5 f N 3.66 10 m/s x . v = 9.57. Model: The neutrons decay is an explosion of the neutron into several pieces. The neutron is an isolated system, so its momentum should be conserved. The observed decay products, the electron and proton, move in opposite directions. Visualize: Solve: (a) The initial momentum is i 0 kg p x = m/s. The final momentum fx e e p p p = m v + m v is 2.7310-23 kg m/s -1.6710-22 kg m/s = -1.4010-22 kg m/s No, momentum does not seem to be conserved. (b) and (c) If the neutrino is needed to conserve momentum, then e P neutrino p + p + p = 0 kg m/s. This requires ( ) 22 neutrino e P p = - p + p = +1.4010- kg m/s The neutrino must carry away 1.4010-22 kg m/s of momentum in the same direction as the electron. 9.58. Model: Model the two balls of clay as particles. Our system comprises these two balls. Momentum is conserved in the perfectly inelastic collision. Visualize: Solve: The x-component of the final momentum is fx ix 1 ( ix )1 2 ( ix )2 p = p = m v + m v = (0.020 kg)(2.0 m/s) - (0.030 kg)(1.0 m/s)cos30 = 0.0140 kg m/s The y-component of the final momentum is ( ) ( ) fy iy 1 iy 1 2 iy 2 p = p = m v + m v = (0.02 kg)(0 m/s) - (0.03 kg)(1.0 m/s)sin30 = -0.0150 kg m/s ( )2 ( )2 f . p = 0.014 kg m/s + -0.015 kg m/s = 0.0205 kg m/s Since ( ) f 1 2 f p = m + m v = 0.0205 kg m/s, the final speed is f ( ) 0.0205 kg m/s 0.41 m/s 0.02 0.03 kg v= = + and the direction is 1 f 1 f tan tan 0.015 47 0.014 y x p p . = - = - = south of east 9.59. Model: Model the three balls of clay as particle 1 (moving north), particle 2 (moving west), and particle 3 (moving southeast). The three stick together during their collision, which is perfectly inelastic. The momentum of the system is conserved. Visualize: Solve: The three initial momenta are p..i1 = m1v..i1 = (0.020 kg)(2.0 m/s) j = 0.040j kg m/s ( )( ) i2 2 i2 p.. = m v.. = 0.030 kg -3.0 m/s i = -0.090i kg m/s ( ) ( ) ( ) i3 3 i3 p = m v = 0.040 kg .. 4.0 m/s cos45i - 4.0 m/s sin 45 j.. .. .. = (0.113i - 0.113 j) kg m/s Since f i i1 i2 i3 p.. = p.. = p + p.. + p.. , we have ( ) ( ) 1 2 3 f m + m + m v.. = 0.023i - 0.073 j kg m/s . ( ) f v.. = 0.256i - 0.811j m/s ( )2 ( )2 f .v = 0.256 m/s + -0.811 m/s = 0.85 m/s 1 f 1 f tan tan 0.811 72 0.256 y x v v . = - = - = below +x 9.60. Model: Model the truck (T) and the two cars (C and C ) ' as particles. The three forming our system stick together during their collision, which is perfectly inelastic. Since no significant external forces act on the system during the brief collision time, the momentum of the system is conserved. Visualize: Solve: The three momenta are p..iT = mTv..iT = (2100 kg)(2.0 m/s)i = 4200i kg m/s ( )( ) iC C iC p.. = m v.. = 1200 kg 5.0 m/s j = 6000j kg m/s ( )( ) iC C iC p m v 1500 kg 10 m/s i 15,000i kg m/s ' ' ' .. = .. = = f i iT iC iC p p p p p' .. = .. = .. + .. + .. = (19,200i + 6000j) kg m/s ( ) ( )2 ( )2 f T C C f p m m m v 19,200 kg m/s 6000 kg m/s ' . = + + = + f .v = 4.2 m/s tan 1 tan 1 6000 17.4 19,200 y x p p . = - = - = above +x Assess: A speed of 4.2 m/s for the entangled three vehicles is reasonable since the individual speeds of the cars and the truck before entanglement were of the same order of magnitude. 9.61. Model: The 14C atom undergoes an explosion and decays into a nucleus, an electron, and a neutrino. Momentum is conserved in the process of explosion or decay. Visualize: Solve: The conservation of momentum equation p..f = p..i = 0 kg m/s is e n N p.. + p.. + p.. = 0 N . ( ) N e n p.. = - p.. + p.. e e n n = -m v.. - m v.. = -(9.1110-31 kg)(5.0107 m/s)i - (8.010-24 kg m/s) j = -(45.5510-24i + 8.010-24 j) kg m/s ( ) ( ) 24 2 24 2 N NN . p = m v = 45.5510- + 8.010- kg m/s . ( 26 ) 23 N 2.3410- kg v = 4.6210- kg m/s 3 N .v =1.9710 m/s 9.62. (a) A 100 g ball traveling to the left at 30 m/s is batted back to the right at 40 m/s. The force curve for the force of the bat on the ball can be modeled as a triangle with a maximum force of 1400 N. How long is the ball in contact with the bat? (b) (c) The solution is .t = 0.100 s =10 ms. 9.63. (a) A 200 g ball of clay traveling to the right overtakes and collides with a 400 g ball of clay traveling to the right at 3.0 m/s. The balls stick and move forward at 4.0 m/s. What was the speed of the 200 g ball of clay? (b) (c) The solution is ( ) i 2 6.0 m/s. x v = 9.64. (a) A 2000 kg auto traveling east at 5.0 m/s suffers a head-on collision with a small 1000 kg auto traveling west at 4.0 m/s. They lock bumpers and stick together after the collision. What will be the speed and direction of the combined wreckage after the collision? (b) (c) The solution is f 2.0 m/s x v = along +x direction. 9.65. (a) A 150 g spring-loaded toy is sliding across a frictionless floor at 1.0 m/s. It suddenly explodes into two pieces. One piece, which has twice the mass of the second piece, continues to slide in the forward direction at 7.5 m/s. What is the speed and direction of the second piece? (b) (c) The solution is f 1 ( ) 12 m/s. x v = - The minus sign tells us that the second piece moves backward at 12 m/s. 9.66. Model: The cart + man (C +M) is our system. It is an isolated system, and momentum is conserved. Visualize: Solve: The conservation of momentum equation fx ix p = p is ( ) ( ) ( ) ( ) M fx M C fx C M ix M C ix C m v + m v = m v + m v Note that f M ( ) x v and f C ( ) x v are the final velocities of the man and the cart relative to the ground. What is given in this problem is the velocity of the man relative to the moving cart. The mans velocity relative to the ground is ( ) ( ) f M f C 10 m/s x x v = v - With this form for f M, ( ) x v we rewrite the momentum conservation equation as ( ) ( ) ( ) ( ) M f C C f C M C 10 m/s 5.0 m/s 5.0 m/s x x m .. v - .. + m v = m + m ( ) ( ) ( )( ) ( )( ) f C f C 70 kg 10 m/s 1000 kg 1000 kg 70 kg 5.0 m/s x x . .. v - .. + v = + ( ) [ ] ( )( ) ( )( ) f C 1000 kg 70 kg 1070 kg 5.0 m/s 70 kg 10 m/s x . v + = + ( ) f C 5.7 m/s x . v = 9.67. Model: Model Ann and cart as particles. The initial momentum is pi = 0 kg m/s in a coordinate system attached to the ground. As Ann begins running to the right, the cart will have to recoil to the left to conserve momentum. Visualize: Solve: The difficulty with this problem is that we are given Anns velocity of 5.0 m/s relative to the cart. If the cart is also moving with velocity cart v then Anns velocity relative to the ground is not 5.0 m/s. Using the Galilean transformation equation for velocity, Anns velocity relative to the ground is ( ) ( ) f Ann f cart 5.0 m/s x x v = v + Now, the momentum conservation equation ix fx p = p is ( ) ( ) Ann f Ann cart f cart 0 kg m/s x x = m v + m v ( ) ( ) ( )( ) f cart f cart 0 kg m/s 50 kg 5.0 m/s 500 kg x x . = .. v + .. + v ( ) f cart 0.45 m/s x . v = - Using the recoil velocity f cart ( ) x v relative to the ground, we find Anns velocity relative to the ground to be f Ann ( ) 5.00 m/s 0.45 m/s 4.55 m/s x v = - = The distance Ann runs relative to the ground is f Ann ( ) , x .x = v .t where .t is the time it takes to reach the end of the cart. Relative to the cart, which is 15 m long, Anns velocity is 5 m/s. Thus .t = (15 m) (5 m/s) = 3.00 s. Her distance over the ground during this interval is ( ) fx Ann .x = v .t = (4.55 m/s)(3.00 s) =13.6 m 9.68. Model: The projectile + wood ball are our system. In the collision, momentum is conserved. Visualize: Solve: The momentum conservation equation fx ix p = p is ( ) ( ) ( ) P B fx P ix P B ix B m + m v = m v + m v ( ) ( )( ) f iP 1.0 kg 20 kg 1.0 kg 0 kg m/s x x . + v = v + ( )i P f 21 x x . v = v We therefore need to determine fx v . Newtons second law for circular motion is ( ) ( ) 2 P B f G P B x m m v T F T m m g r + - = - + = Using max T = 400 N, this equation gives ( )( ) ( ) 2 f 1.0 kg 20 kg 400 N 1.0 kg 20 kg 9.8 m/s 2.0 m x + v - + = ( ) f max 4.3 m/s x . v = Going back to the momentum conservation equation, ( )i f 21 x P x v = v = (21)(4.3 m/s) = 90 m/s That is, the largest speed this projectile can have without causing the cable to break is 90 m/s. 9.69. Model: This is an explosion problem and momentum is conserved. The two-stage rocket is our system. Visualize: Solve: Relative to the ground, the conservation of momentum equation fx ix p = p is ( ) ( ) ( ) 1 fx 1 2 fx 2 1 2 1x m v + m v = m + m v ( ) ( ) ( )( ) 2 f 1 2 f 2 2 3 m 4 m 1200 m/s x x . v + m v = ( ) ( ) f 1 f 2 3 4800 kg m/s x x . v + v = The fact that the first stage is pushed backward at 35 m/s relative to the second can be written ( ) ( ) f 1 f 2 35 m/s x x v = - + v Substituting this form of f 1 ( ) x v in the conservation of momentum equation, ( ) ( ) f 2 f 2 3 35 m/s 4800 kg m/s x x ..- + v .. + v = ( ) f 2 1226 m/s x . v = 9.70. Model: Model the bullet and the vehicle as particles, and use the impulse-momentum theorem to find the impulse provided to the vehicle by bullet(s). Because the final speed of the vehicle is small compared to the bullet speed, and the mass of the bullet is so much smaller than the mass of the vehicle, we will assume that each bullet exerts the same impulse on the vehicle. Visualize: Solve: For one bullet, the impulse-momentum theorem .px = Jx allows us to find that the impulse exerted on the bullet by the vehicles sail is ( ) B B B ( ) ( ) (0.020 kg) ( 200 m/s) (400 m/s) 12.0 kg m/s xJ = m .v = - - = - From Newtons third law, the impulse that the bullet exerts on the vehicle is V B ( ) ( ) 12.0 x x J = - J = kg m/s. A second use of the impulse momentum theorem allows us to find the vehicles increase in velocity due to one bullet: V V V ( ) ( ) 12.0 kg m/s 0.120 m/s 100 kg x v J m . = = = This increase in velocity is independent of the vehicles speed, so as long as the impulse per bullet stays essentially constant (which it does because the bullet speeds are so much larger than the vehicle speed), the number of impacts needed to increase the vehicle speed to 12 m/s is bullets 12 m/s 100 bullets 0.12 m/s per bullet N = = To reach this speed in 20 s requires the bullets to be fired at a rate of 5 bullets per second. 9.71. Model: Let the system be rocket + bullet. This is an isolated system, so momentum is conserved. Visualize: The fact that the bullets velocity relative to the rocket is 139,000 can be written f B (v ) = f R (v ) +139,000m/s. Solve: Consider the firing of one bullet when the rocket has mass M and velocity i v . The conservation of momentum equation f i p = p is f f i f i ( 5kg) (5 kg)( 139,000 m/s) 5 kg (139,000 m/s) M v v Mv v v v M - + + = . . = - = - The rocket starts with mass M = 2000 kg, which is much larger than 5 kg. If only a few bullets are needed, M will not change significantly as the rocket slows. If we assume that M remains constant at 2000 kg, the loss of speed per bullet is .v = -347.5 m/s = -1250 km/h. Thus exactly 8 bullets will reduce the speed by 10,000 km/h, from 25,000 km/h to 15,000 km/h. If youre not sure that treating M as a constant is valid, you can calculate .v for each bullet and reduce M by 5 kg after each shot. The loss of mass causes .v to increase slightly for each bullet. An eight-step calculation then finds that 8 bullets will slow the rocket to 14,900 km/h. Seven bullets wouldnt be enough, and 9 would slow the rocket far too much. 9-1 9.72. Visualize: Solve: Ladies and gentlemen of the jury, how far would the chair slide if it was struck with a bullet from my clients gun? We know the bullets velocity as it leaves the gun is 450 m/s. The bullet travels only a small distance to the chair, so we will neglect any speed loss due to air resistance. The bullet and chair can be considered an isolated system during the brief interval of the collision. The bullet embedded itself in the chair, so this was a perfectly inelastic collision. Momentum conservation allows us to calculate the velocity of the chair immediately after the collision as follows: ix fx p = p ( ) ( ) B i B B C f .m v = m + m v ( ) ( )( ) B i B f B C 450 m/s 0.010 kg 0.225 m/s 20.01 kg m v v m m . = = = + This is the velocity immediately after the collision when the chair starts to slide but before it covers any distance. For the purpose of the problem in dynamics, call this the initial velocity 0v . The free-body diagram of the chair shows three forces. Newtons second law applied to the chair (with the embedded bullet) is ( ) net k k tot tot tot x x F f n a a m m m - = = = = - ( ) 2 net tot tot tot 0 m/s y y F n m g a m m - = = = where weve used the friction model in the x-equation. The y-equation yields tot n = m g, and the x-equation yields 2 k a = - g = -1.96 m/s . We know the coefficient of kinetic friction because it is a wood chair sliding on a wood floor. Finally, we have to determine the stopping distance of the chair. The motion of the chair ends with 1 v = 0 m/s after sliding a distance .x, so 2 2 2 2 1 0 v = 0 m /s = v + 2a.x ( ) ( ) 2 2 0 2 0.225 m/s 0.013 m 1.3 cm 2 2 1.96 m/s x v a . . = - = - = = - If the bullet lost any speed in the air before hitting the chair, the sliding distance would be even less. So you can see that the most the chair could slide if it had been struck by a bullet from my clients gun, would be 1.3 cm. But in actuality, the chair slid 3 cm, more than twice as far. The murder weapon, ladies and gentlemen, was a much more powerful gun than the one possessed by my client. I rest my case. 10.1. Model: We will use the particle model for the bullet (B) and the running student (S). Visualize: Solve: For the bullet, 2 2 B BB 1 1(0.010 kg)(500 m/s) 1250 J 2 2 K = m v = = For the running student, 2 2 S SS 1 1(75 kg)(5.5 m/s) 206 J 2 2 K = m v = = Thus, the bullet has the larger kinetic energy. Assess: Kinetic energy depends not only on mass but also on the square of the velocity. The above calculation shows this dependence. Although the mass of the bullet is 7500 times smaller than the mass of the student, its speed is more than 90 times larger. 10.2. Model: Model the hiker as a particle. Visualize: The origin of the coordinate system chosen for this problem is at sea level so that the hikers position in Death Valley is y0 = -8.5 m. Solve: The hikers change in potential energy from the bottom of Death Valley to the top of Mt. Whitney is gf gi f i f i 2 6 ( ) (65 kg)(9.8 m/s )[4420 m ( 85 m)] 2.9 10 J .U =U -U = mgy - mgy = mg y - y = -- = Assess: Note that .U is independent of the origin of the coordinate system. 10.3. Model: Model the compact car (C) and the truck (T) as particles. Visualize: Solve: For the kinetic energy of the compact car and the kinetic energy of the truck to be equal, 2 2 T C T CC TT C T C 1 1 20,000 kg (25 km/hr) 112 km/hr 2 2 1000 kg K K m v m v v m v m = . = . = = = Assess: A smaller mass needs a greater velocity for its kinetic energy to be the same as that of a larger mass. 10.4. Model: Model the car (C) as a particle. This is an example of free fall, and therefore the sum of kinetic and potential energy does not change as the car falls. Visualize: Solve: (a) The kinetic energy of the car is 2 2 5 C CC 1 1(1500 kg)(30 m/s) 6.75 10 J 2 2 K = m v = = The cars kinetic energy is 6.8105 J. (b) Let us relabel C K as f K and place our coordinate system at f y = 0 m so that the cars potential energy gf U is zero, its velocity is f v , and its kinetic energy is f K . At position i y , i i v = 0 m/s or K = 0 J, and the only energy the car has is gi i U = mgy . Since the sum g K +U is unchanged by motion, f gf i gi . K +U = K +U This means f f i i f i i 5 f i i 2 0 ( ) (6.7510J 0J) 46 m (1500 kg)(9.8 m/ s ) K mgy K mgy K K mgy y K K mg + = + . + = + - - . = = = (c) From part (b), ( ) 2 2 2 2 f i f i f i i 1 1 ( ) 2 2 2 K K mv mv v v y mg mg g - - - = = = Free fall does not depend upon the mass. 10.5. Model: This is a case of free fall, so the sum of the kinetic and gravitational potential energy does not change as the ball rises and falls. Visualize: The figure shows a balls before-and-after pictorial representation for the three situations in parts (a), (b) and (c). Solve: The quantity K +Ug is the same during free fall: f gf i gi . K +U = K +U We have (a) ( ) 2 2 1 1 0 0 2 2 2 2 2 1 0 1 1 1 2 2 2 [(10 m/s) (0 m/s) ]/(2 9.8m/s ) 5.10 m mv mgy mv mgy y v v g + = + . = - = - = 5.1 m is therefore the maximum height of the ball above the window. This is 25.1 m above the ground. (b) 2 2 2 2 0 0 1 1 2 2 mv + mgy = mv + mgy Since 2 0 y = y = 0, we get for the magnitudes 2 0 v = v =10 m/ s. (c) 2 2 2 2 22 3 3 0 0 3 3 0 0 3 0 0 3 1 1 2 2 2( ) 2 2 mv + mgy = mv + mgy .v + gy = v + gy .v = v + g y - y 2 2 2 22 3 .v = (10 m/ s) + 2(9.8 m/ s )[0 m- (-20 m)] = 492 m / s This means the magnitude of 3 v is equal to 22 m/s. Assess: Note that the balls speed as it passes the window on its way down is the same as the speed with which it was tossed up, but in the opposite direction. 10.6. Model: This is a problem of free fall. The sum of the kinetic and gravitational potential energy for the ball, considered as a particle, does not change during its motion. Visualize: The figure shows the balls before-and-after pictorial representation for the two situations described in parts (a) and (b). Solve: The quantity K +Ug is the same during free fall. Thus, f gf i gi . K +U = K +U (a) 2 2 2 2 1 1 0 0 0 1 1 0 2 2 2 22 0 0 1 1 2 ( ) 2 2 (0 m/s) 2(9.8 m/s )(10 m 1.5 m) 166.6 m /s 12.9 m/s mv mgy mv mgy v v g y y v v + = + . = + - . = + - = . = (b) 2 2 22 2 2 0 0 2 0 0 2 2 22 2 2 2 1 1 2 ( ) 2 2 166.6 m /s 2(9.8 m/s )(1.5 m 0 m) 14.0 m/s mv mgy mv mgy v v g y y v v + = + . = + - . = + - . = Assess: An increase in speed from 12.9 m/s to 14.0 m/s as the ball falls through a distance of 1.5 m is reasonable. Also, note that mass does not appear in the calculations that involve free fall. 10.7. Model: Model the oxygen and the helium atoms as particles. Visualize: We denote the oxygen and helium atoms by O and He, respectively. Note that the oxygen atom is four times heavier than the helium atom, so mO = 4 mHe. Solve: The energy conservation equation O He K = K is 2 2 2 2 He O O He He He O He He O 1 1 (4 ) 2.0 2 2 m v m v m v m v v v = . = . = Assess: The result He O v = 2v , combined with the fact that 1 He 4 Om = m , is a consequence of the way kinetic energy is defined: It is directly proportional to the mass and to the square of the speed. 10.8. Model: Model the ball as a particle undergoing rolling motion with zero rolling friction. The sum of the balls kinetic and gravitational potential energy, therefore, does not change during the rolling motion. Visualize: Solve: Since the quantity K +Ug does not change during rolling motion, the energy conservation equations apply. For the linear segment the energy conservation equation 0 g0 1 g1 K +U = K +U is 2 1 1 1 2 mv + mgy = 2 0 0 1 2 mv + mgy 2 2 2 1 010 1 (0 m) 1 (0 m/s) 1 2 2 2 . m v + mg = m + mgy . mv = mgy For the parabolic part of the track, 1 g1 2 g2 K +U = K +U is 2 2 2 2 2 1 1 2 2 1 2 1 2 1 1 1 (0 m) 1 m(0 m/ s) 1 2 2 2 2 2 mv + mgy = mv + mgy . mv + mg = + mgy . mv = mgy Since from the linear segment we have 1 2 2 1 0 0 2 2 0 m v = mgy , we get mgy = mgy or y = y =1.0 m. Thus, the ball rolls up to exactly the same height as it started from. Assess: Note that this result is independent of the shape of the path followed by the ball, provided there is no rolling friction. This result is an important consequence of energy conservation. 10.9. Model: Model the skateboarder as a particle. Assuming that the track offers no rolling friction, the sum of the skateboarders kinetic and gravitational potential energy does not change during his rolling motion. Visualize: The vertical displacement of the skateboarder is equal to the radius of the track. Solve: The quantity K +Ug is the same at the upper edge of the quarter-pipe track as it was at the bottom. The energy conservation equation f gf i gi K +U = K +U is 2 2 22 f f i i i f f i 2 2 2 22 i i 1 1 2 ( ) 2 2 (0 m/s) 2(9.8 m/s )(3.0 m 0 m) 58.8 m /s 7.7 m/s mv mgy mv mgy v v g y y v v + = + . = + - = + - = . = Assess: Note that we did not need to know the skateboarders mass, as is the case with free-fall motion. 10.10. Model: Model the puck as a particle. Since the ramp is frictionless, the sum of the pucks kinetic and gravitational potential energy does not change during its sliding motion. Visualize: Solve: The quantity K +Ug is the same at the top of the ramp as it was at the bottom. The energy conservation equation f gf i gi K +U = K +U is 2 2 22 f f i i i f f i 2 2 2 22 i i 1 1 2 ( ) 2 2 (0 m/s) 2(9.8 m/s )(1.03m 0 m) 20.2 m /s 4.5 m/s mv mgy mv mgy v v g y y v v + = + . = + - . = + - = . = Assess: An initial push with a speed of 4.5 m/s 10 mph to cover a distance of 3.0 m up a 20 ramp seems reasonable. 10.11. Model: In the absence of frictional and air-drag effects, the sum of the kinetic and gravitational potential energy does not change as the pendulum swings from one side to the other. Visualize: The figure shows the pendulums before-and-after pictorial representation for the two situations described in parts (a) and (b). Solve: (a) The quantity K +Ug is the same at the lowest point of the trajectory as it was at the highest point. Thus, 1 g1 0 g0 K +U = K +U means 2 2 2 2 1 1 0 0 1 1 0 0 2 2 1 01 0 1 1 2 2 2 2 2 (0 m) (0 m/s) 2 2 mv mgy mv mgy v gy v gy v g gy v gy + = + . + = + . + = + . = From the pictorial representation, we find that 0 y = L - Lcos30. Thus, 2 1 v = 2gL(1- cos30) = 2(9.8 m/s )(0.75 m)(1- cos30) =1.403 m/s The speed at the lowest point is 1.40 m/s. (b) Since the quantity g K +U does not change, 2 2 1 1. We have g g K +U = K +U 2 2 ( 2 2 ) 2 2 1 1 2 1 2 2 2 2 2 1 1 2 2 2 [(1.403 m/s) (0 m/s) ]/(2 9.8 m/s ) 0.100 m mv mgy mv mgy y v v g y + = + . = - . = - = Since 2 y = L - Lcos. , we obtain 2 1 cos (0.75 m) (0.10 m) 0.8667 cos (0.8667) 30 (0.75 m) L y L . . - - - = = = . = = That is, the pendulum swings to the other side by 30. Assess: The swing angle is the same on either side of the rest position. This result is a consequence of the fact that the sum of the kinetic and gravitational potential energy does not change. This is shown as well in the energy bar chart in the figure. 10.12. Model: Model the child and swing as a particle, and assume the chain to be massless. In the absence of frictional and air-drag effects, the sum of the kinetic and gravitational potential energy does not change during the swings motion. Visualize: Solve: The quantity K +Ug is the same at the highest point of the swing as it is at the lowest point. That is, 0 g0 1 g1 . K +U = K +U It is clear from this equation that maximum kinetic energy occurs where the gravitational potential energy is the least. This is the case at the lowest position of the swing. At this position, the speed of the swing and child will also be maximum. The above equation is 2 2 22 0 0 1 1 1 0 0 1 2 2 1 0 1 0 1 1 2 ( ) 2 2 (0 m/s) 2 ( 0 m) 2 mv mgy mv mgy v v g y y v gy v gy + = + . = + - . = + - . = We see from the pictorial representation that 0 2 1 0 cos45 (3.0 m) (3.0 m)cos45 0.879 m 2 2(9.8 m/s )(0.879 m) 4.2 m/s y L L v gy = - = - = . = = = Assess: We did not need to know the swings or the childs mass. Also, a maximum speed of 4.2 m/s is reasonable. 10.13. Model: Model the car as a particle with zero rolling friction. The sum of the kinetic and gravitational potential energy, therefore, does not change during the cars motion. Visualize: Solve: The initial energy of the car is 2 2 2 5 0 g0 0 0 1 1(1500 kg)(10.0 m/s) (1500 kg)(9.8 m/s )(10 m) 2.22 10 J 2 2 K +U = mv + mgy = + = The car increases its height to 15 m at the gas station. The conservation of energy equation 0 g0 1 g1 K +U = K +U is 5 2 5 2 2 1 1 1 1 2.22 10 J 1 2.22 10 J 1 (1500 kg) (1500 kg)(9.8 m/s )(15 m) 2 2 1.41m/s mv mgy v v = + . = + . = Assess: A lower speed at the gas station is reasonable because the car has decreased its kinetic energy and increased its potential energy compared to its starting values. 10.14. Model: Assume that the spring is ideal and obeys Hookes law. Also model the sled as a particle. Visualize: The only horizontal force acting on the sled is Fsp. .. Solve: Applying Newtons second law to the sled gives on sled sp ( )x x x S F = F = ma .k.y = ma / (150 N/m)(0.20 m)/20 kg 1.50 m/s2 x .a = k.x m = = 10.15. Model: Assume that the spring is ideal and obeys Hookes law. Visualize: According to Hookes law, the spring force acting on a mass (m) attached to the end of a spring is given as Fsp = k.x, where .x is the change in length of the spring. If the mass m is at rest, then sp F is also equal to the gravitational force G F = mg. Solve: sp We have F = k.x = mg. We want a 0.100 kg mass to give .x =0.010 m. This means k = mg /.x = (0.100 kg)(9.8 N/m)/(0.010 m) = 98 N/m 10.16. Model: Assume an ideal spring that obeys Hookes law. Visualize: Solve: (a) The spring force on the 2.0 kg mass is Fsp = -k.y. Notice that .y is negative, so sp F is positive. This force is equal to mg, because the 2.0 kg mass is at rest. We have - k.y = mg. Solving for k: k = -(mg /.y) = -(2.0 kg)(9.8 m/s2 )/(-0.15 m- (-0.10 m)) = 392 N/m The spring constant is 3.9102 N/m. (b) Again using -k.y = mg: 2 e e / (3.0 kg)(9.8 m/s )/(392 N/m) 0.075 m 0.075 m 0.10 m 0.075 m 0.175 m 17.5 cm y mgk y y y y . = - = - ' - = - . ' = - = - - = - = - The length of the spring is 17.5 cm when a mass of 3.0 kg is attached to the spring. The position of the end of the spring is negative because it is below the origin, but length must be a positive number. 10.17. Model: Assume that the spring is ideal and obeys Hookes law. We also model the 5.0 kg mass as a particle. Visualize: We will use the subscript s for the scale and sp for the spring. Solve: (a) The scale reads the upward force s on m F that it applies to the mass. Newtons second law gives 2 on m s on m G s on m G ( ) 0 (5.0kg)(S F y = F - F = . F = F = mg = 9.8 m/s ) = 49 N (b) In this case, the force is on m s on m sp G ( ) 0 20N 0 ( 20 N)/ (49 N 20 N)/0.02 m 1450 N/m y F F F F ky mg k mg y = + - = . + . - = . = - . = - = S The spring constant for the lower spring is 1.45103 N/m. (c) In this case, the force is on m sp G ( ) 0 0 / (49 N)/(1450 N/m) 0.0338 m 3.4 cm y F F F ky mg y mg k = - = . . - = . . = = = = S 10.18. Model: Model the student (S) as a particle and the spring as obeying Hookes law. Visualize: Solve: According to Newtons second law the force on the student is on S spring on S G 2 2 spring on S G ( ) (60 kg)(9.8 m/s 3.0 m/s ) 768 N y y y y F F F ma F F ma mg ma = - = . = + = + = + = S spring on S S on spring Since , 768 N. This F = F = k.y k.y = means .y = (768 N)/(2500 N/m) = 0.31m. 10.19. Model: Assume an ideal spring that obeys Hookes law. Solve: The elastic potential energy of a spring is defined as 1 2 s 2 U = k(.s) , where .s is the magnitude of the stretching or compression relative to the unstretched or uncompressed length. We have .s = 20 cm = 0.20 m and k = 500 N/m. This means 2 s 1 ( ) 1 (500 N/m)(0.20 m) 10 J 2 2 U = k .s = = Assess: Since .s is squared, s U is positive for a spring that is either compressed or stretched. s U is zero when the spring is in its equilibrium position. 10.20. Model: Assume an ideal spring that obeys Hookes law. Solve: The elastic potential energy of a spring is defined as 1 2 s 2 U = k(.s) , where .s is the magnitude of the stretching or compression relative to the unstretched or uncompressed length. s .U = 0 when the spring is at its equilibrium length and .s = 0. s We have U = 200 J and k =1000 N/m. Solving for .s : s .s = 2U /k = 2(200 J) / 1000 N/m = 0.632 m 10.21. Model: Assume an ideal spring that obeys Hookes law. There is no friction, so the mechanical energy K +Us is conserved. Also model the book as a particle. Visualize: The figure shows a before-and-after pictorial representation. The compressed spring will push on the book until the spring has returned to its equilibrium length. We put the origin of our coordinate system at the equilibrium position of the free end of the spring. The energy bar chart shows that the potential energy of the compressed spring is entirely transformed into the kinetic energy of the book. Solve: The conservation of energy equation 2 s2 1 s1 K +U = K +U is 2 2 2 2 2 2 e 1 1 e 1 1 ( ) 1 1 ( ) 2 2 2 2 mv + k x - x = mv + k x - x Using 2 e x = x = 0 m and 1 v = 0 m/s, this simplifies to 2 2 2 2 1 2 1 2 1 1 ( 0 m) (1250 N/m)(0.040 m) 2.0 m/s 2 2 (0.500 kg) mv k x v kx m = - . = = = Assess: This problem cannot be solved using constant-acceleration kinematic equations. The acceleration is not a constant in this problem, since the spring force, given as s F = -k.x, is directly proportional to .x or e |x - x |. 10.22. Model: Assume an ideal spring that obeys Hookes law. Since there is no friction, the mechanical energy K +Us is conserved. Also, model the block as a particle. Visualize: The figure shows a before-and-after pictorial representation. We have put the origin of our coordinate system at the equilibrium position of the free end of the spring. This gives us 1 e x = x = 0 cm and 2 x = 2.0 cm. Solve: The conservation of energy equation 2 s2 1 s1 K +U = K +U is 2 2 2 2 2 2 e 1 1 e 1 1 ( ) 1 1 ( ) 2 2 2 2 mv + k x - x = mv + k x - x Using 2 1 e 2 e v = 0 m/s, x = x = 0 m, and x - x = 0.020 m, we get 2 2 2 e 1 2 e 1 1 ( ) 1 ( ) 2 2 k x x mv x x x mv k - = .. = - = That is, the compression is directly proportional to the velocity 1v . When the block collides with the spring with twice the earlier velocity 1 (2v ), the compression will also be doubled to 2 e 2(x - x ) = 2(2.0 cm) = 4.0 cm. Assess: This problem shows the power of using energy conservation over using Newtons laws in solving problems involving nonconstant acceleration. 10.23. Model: Model the grocery cart as a particle and the spring as an ideal that obeys Hookes law. We will also assume zero rolling friction during the compression of the spring, so that mechanical energy is conserved. Visualize: The figure shows a before-and-after pictorial representation. The before situation is when the cart hits the spring in its equilibrium position. We put the origin of our coordinate system at this equilibrium position of the free end of the spring. This give x1 = xe = 0 and (x2 - xe ) = 60 cm. Solve: The conservation of energy equation 2 s2 1 s1 K +U = K +U is 2 2 2 2 2 2 e 1 1 e 1 1 ( ) 1 1 ( ) 2 2 2 2 mv + k x - x = mv + k x - x Using 2 2 e 1 e v = 0 m/s,(x - x ) = 0.60 m, and x = x = 0 m gives: 2 2 2 e 1 1 2 e 1 ( ) 1 ( ) 250 N/m (0.60 m) 3.0 m/s 2 2 10 kg k x x m v v k x x m - = . = - = = 10.24. Model: Model the jet plane as a particle, and the spring as an ideal that obeys Hookes law. We will also assume zero rolling friction during the stretching of the spring, so that mechanical energy is conserved. Visualize: The figure shows a before-and-after pictorial representation. The before situation occurs just as the jet plane lands on the aircraft carrier and the spring is in its equilibrium position. We put the origin of our coordinate system at the right free end of the spring. This gives x1 = xe = 0 m. Since the spring stretches 30 m to stop the plane, 2 e x - x = 30 m. Solve: The conservation of energy equation 2 s2 1 s1 K +U = K +U for the spring-jet plane system is 2 2 2 2 2 2 e 1 1 e 1 1 ( ) 1 1 ( ) 2 2 2 2 mv + k x - x = mv + k x - x Using 2 1 e 2 e v = 0 m/s, x = x = 0 m, and x - x = 30 m yields 2 2 2 e 1 1 2 1 1 ( ) 1 ( ) 60,000 N/m(30 m) 60 m/s 2 2 15,000 kg k x x mv v k x x m - = . = - = = Assess: A landing speed of 60 m/s or 120 mph is reasonable. 10.25. Model: We assume this is a one-dimensional collision that obeys the conservation laws of momentum and mechanical energy. Visualize: Note that momentum conservation alone is not sufficient to solve this problem because the two final velocities f 1 ( ) x v and f 2 ( ) x v are unknowns and can not be determined from one equation. Solve: 1 i 1 2 i 2 1 f 1 2 f 2 2 2 2 2 1 i 1 2 i 2 1 f 1 2 f 2 Momentum conservation: ( ) ( ) ( ) ( ) Energy conservation: 1 ( ) 1 ( ) 1 ( ) 1 ( ) 2 2 2 2 x x x x x x x x m v m v m v m v m v m v m v m v + = + + = + These two equations can be solved for f 1 ( ) x v and f 2 ( ) , x v as shown by Equations 10.39 through 10.43, to give 1 2 f 1 i 1 1 2 1 f 2 i 1 1 2 ( ) ( ) 50 g 20 g (2.0 m/s) 0.86 m/s 50 g 20 g ( ) 2 ( ) 2(50 g) (2.0 m/s) 2.9 m/s 50 g 20 g x x x x v m m v m m v m v m m - - = = = + + = = = + + Assess: These velocities are of a reasonable magnitude. Since both these velocities are positive, both balls move along the +x-direction. 10.26. Model: This is a case of a perfectly elastic collision between a proton and a carbon atom. The collision obeys the momentum as well as the energy conservation law. Visualize: Solve: P i P C i C P f P C f C 2 2 2 2 P i P C i C P f P C f C Momentum conservation: ( ) ( ) ( ) ( ) Energy conservation: 1 ( ) 1 ( ) 1 ( ) 1 ( ) 2 2 2 2 x x x x x x x x m v m v m v m v m v m v m v m v + = + + = + These two equations can be solved, as described in the text through Equations 10.39 to 10.43: P C P P 7 7 f P P C i P P C P P P P 7 6 f C i P P C P P ( ) ( ) 12 (2.0 10 m/s) 1.69 10 m/s 12 ( ) 2 ( ) 2 (2.0 10 m/s) 3.1 10 m/s 12 x x x x v m m m m v m m m m m m v m v m m m m m - . - . = + = . . = - + . + . . . = = . . = + . + . After the elastic collision the proton rebounds at 1.69107 m/s and the carbon atom moves forward at 3.08106 m/s. 10.27. Model: This is the case of a perfectly inelastic collision. Momentum is conserved because no external force acts on the system (clay + brick). We also represent our system as a particle. Visualize: Solve: (a) The conservation of momentum equation fx ix p = p is 1 2 f 1 i 1 2 i 2 ( ) ( ) ( ) x x x m + m v = m v + m v Using i 1 0 i 2 ( ) and ( ) 0, x x v = v v = we get 1 f i1 i1 i1 0 1 2 ( ) 0.050 kg ( ) 0.0476( ) 0.0476 (1.0 kg 0.050 kg) x x x x v m v v v v m m = = = = + + The brick is moving with speed 0 0.048v . (b) The initial and final kinetic energies are given by 2 2 2 ( )2 2 i 1 i1 2 i 2 0 0 2 22 2 f 1 2 f 0 0 1 ( ) 1 ( ) 1 (0.050 kg) 1 (1.0 kg) 0 m/s (0.025 kg) 2 2 2 2 1 ( ) 1 (1.0 kg 0.050 kg)(0.0476) 0.00119 2 2 x x x K m v m v v v K m mv v v = + = + = = + = + = The percent of energy lost i f i 100% 1 0.00119 100% 95% 0.025 K K K . - . . . = . . = . - . = . . . . 10.28. Model: In this case of a one-dimensional collision, the momentum conservation law is obeyed whether the collision is perfectly elastic or perfectly inelastic. Visualize: Solve: In the case of a perfectly elastic collision, the two velocities f 1 ( ) x v and f 2 ( ) x v can be determined by combining the conservation equations of momentum and mechanical energy. By contrast, a perfectly inelastic collision involves only one final velocity fx v and can be determined from just the momentum conservation equation. (a) 1 i 1 2 i 2 1 f 1 2 f 2 2 2 2 2 1 i 1 2 i 2 1 f 1 2 f 2 Momentum conservation: ( ) ( ) ( ) ( ) Energy conservation: 1 ( ) 1 ( ) 1 ( ) 1 ( ) 2 2 2 2 x x x x x x x x m v m v m v m v m v m v m v m v + = + + = + These two equations can be solved as shown in Equations 10.39 through 10.43: 1 2 f 1 i 1 1 2 1 f 2 i 1 1 2 ( ) ( ) (100 g) (300 g) (10 m/s) 5.0 m/s (100 g) (300 g) ( ) 2 ( ) 2(100 g) (10 m/s) 5.0 m/s (100 g) (300 g) x x x x v m m v m m v m v m m - - = = =- + + = = =+ + + (b) For the inelastic collision, both balls travel with the same final speed f . x v The momentum conservation equation f i is x x p = p 1 2 f 1 i 1 2 i 2 f ( ) ( ) ( ) 100 g (10 m/s) 0 m/s 2.5 m/s 100 g 300 g x x x x m m v m v m v v + = + . . . = . . + = . + . Assess: In the case of the perfectly elastic collision, the two balls bounce off each other with a speed of 5.0 m/s. In the case of the perfectly inelastic collision, the balls stick together and move together at 2.5 m/s. 10.29. Model: For an energy diagram, the sum of the kinetic and potential energy is a constant. Visualize: The particle is released from rest at x =1.0 m. That is, K = 0 at x =1.0 m. Since the total energy is given by E = K +U, we can draw a horizontal total energy (TE) line through the point of intersection of the potential energy curve (PE) and the x =1.0 m line. The distance from the PE curve to the TE line is the particles kinetic energy. These values are transformed as the position changes, causing the particle to speed up or slow down, but the sum K + U does not change. Solve: (a) We have E = 4.0 J and this energy is a constant. For x <1.0, U > 4.0 J and, therefore, K must be negative to keep E the same (note that K = E -U or K = 4.0 J -U). Since negative kinetic energy is unphysical, the particle can not move to the left. That is, the particle will move to the right of x =1.0 m. (b) The expression for the kinetic energy is E -U. This means the particle has maximum speed or maximum kinetic energy when U is minimum. This happens at x = 4.0 m. Thus, 2 max min max max (4.0 J) (1.0 J) 3.0 J 1 3.0 J 2(3.0 J) 8.0 J 17.3m/s 2 0.020 kg K E U mv v m = - = - = = . = = = The particle possesses this speed at x = 4.0 m. (c) The total energy (TE) line intersects the potential energy (PE) curve at x =1.0 m and x = 6.0 m. These are the turning points of the motion. 10.30. Model: For an energy diagram, the sum of the kinetic and potential energy is a constant. Visualize: The particle with a mass of 500 g is released from rest at A. That is, K = 0 at A. Since E = K +U = 0 J +U, we can draw a horizontal TE line through U = 5.0 J. The distance from the PE curve to the TE line is the particles kinetic energy. These values are transformed as the position changes, causing the particle to speed up or slow down, but the sum K +U does not change. Solve: The kinetic energy is given by E -U, so we have 1 2 2( )/ 2 mv = E -U .v = E -U m Using B C D U = 2.0 J,U = 3.0 J, andU = 0 J, we get B C D 2(5.0 J 2.0 J)/0.500 kg 3.5 m/s 2(5.0 J 3.0 J)/0.500 kg 2.8 m/s 2(5.0 J 0 J)/0.500 kg 4.5 m/s v v v = - = = - = = - = 10.31. Model: For an energy diagram, the sum of the kinetic and potential energy is a constant. Visualize: Since the particle oscillates between x = 2.0 mm and x = 8.0 mm, the speed of the particle is zero at these points. That is, for these values of x, E =U = 5.0 J, which defines the total energy (TE) line. The distance from the potential energy (PE) curve to the TE line is the particles kinetic energy. These values are transformed as the position changes, but the sum K +U does not change. Solve: The equation for total energy E =U + K means K = E -U, so that K is maximum when U is minimum. We have 2 max max min max min 1 5.0 J 2 2(5.0 J )/ 2(5.0 J 1.0 J)/0.0020 kg 63 m/s K mv U v U m = = - . = - = - = 10.32. Model: For an energy diagram, the sum of the kinetic and potential energy is a constant. Visualize: For the speed of the particle at A that is needed to reach B to be a minimum, the particles kinetic energy as it reaches the top must be zero. Similarly, the minimum speed at B for the particle to reach A obtains when the particle just makes it to the top with zero kinetic energy. Solve: (a) The energy equation KA +UA = Ktop +Utop is 2 A A top A top A 1 0 J 2 2( )/ 2(5.0 J 2.0 J)/0.100 kg 7.7 m/s mv U U v U U m + = + . = - = - = (b) To go from point B to point A, B B top top K +U = K +U is 2 B B top B top B 1 0 J 2 2( )/ 2(5.0 J 0 J)/0.100 kg 10.0 m/s mv U U v U U m + = + . = - = - = Assess: The particle requires a higher kinetic energy to reach A from B than to reach B from A. 10.33. Model: Model your vehicle as a particle. Assume zero rolling friction, so that the sum of your kinetic and gravitational potential energy does not change as the vehicle coasts down the hill. Visualize: The figure shows a before-and-after pictorial representation. Note that neither the shape of the hill nor the angle of the downward slope is given, since these are not needed to solve the problem. All we need is the change in potential energy as you and your vehicle descend to the bottom of the hill. Also note that 35 km/ hr = (35,000 m/3600 s) = 9.722 m/ s Solve: Using f y = 0 and the equation i gi f gf K +U = K +U we get 2 2 2 2 i i f f i i f 2 2 2 f i i 1 1 2 2 2 2 (9.722 m/s) 2(9.8 m/s )(15 m) 19.7 m/s 71 km/h mv mgy mv mgy v gy v v v gy + = + . + = . = + = + = = You are driving over the speed limit. Yes, you will get a ticket. Assess: A speed of 19.7 m/s or 71 km/h at the bottom of the hill, when your speed at the top of the hill was 35 km/s, is reasonable. From the energy bar chart, we see that the initial potential energy is completely transformed into the final kinetic energy. 10.34. Model: This is case of free fall, so the sum of the kinetic and gravitational potential energy does not change as the cannon ball falls. Visualize: The figure shows a before-and-after pictorial representation. To express the gravitational potential energy, we put the origin of our coordinate system on the ground below the fortress. Solve: Using yf = 0 and the equation i gi f gf K +U = K +U we get 2 2 2 2 i i f f i i f 2 2 2 f i i 1 1 2 2 2 2 (80 m/s) 2(9.8 m/s )(10 m) 81 m/s mv mgy mv mgy v gy v v v gy + = + . + = = + = + = Assess: Note that we did not need to use the tilt angle of the cannon, because kinetic energy is a scalar. Also note that using the energy conservation equation, we can find only the magnitude of the final velocity, not the direction of the velocity vector. 10.35. Model: This is a case of free fall, so the sum of the kinetic and gravitational potential energy does not change as the rock is thrown. Model the Frisbee and the rock as particles. Visualize: The coordinate system is put on the ground for this system, so that yf =16 m. The rocks final velocity f v must be at least 5.0 m/s to dislodge the Frisbee. Solve: (a) The energy conservation equation for the rock f gf i gi K +U = K +U is 2 2 f f i i 1 1 2 2 mv + mgy = mv + mgy This equation involves only the velocity magnitudes and not the angle at which the rock is to be thrown to dislodge the Frisbee. This equation is true for all angles that will take the rock to the Frisbee. (b) Using the above equation we get 2 2 2 2 2 f f i i i f f i v + 2gy = v + 2gy v = v + 2g( y - y ) = (5.0 m/s) + 2(9.8 m/s )(16 m- 2.0 m) =17.3 m/s Assess: Kinetic energy is defined as 1 2 2 K = mv and is a scalar quantity. Scalar quantities do not have directional properties. 10.36. Model: For the ice cube sliding around the inside of a smooth pipe, the sum of the kinetic and gravitational potential energy does not change. Visualize: We use a coordinate system with the origin at the bottom of the pipe, that is, y1 = 0. The radius (R) of the pipe is 10 cm, and therefore top 2 y = y = 2R = 0.20 m. At an arbitrary angle . , measured counterclockwise from the bottom of the circle, y = R - Rcos. . Solve: (a) The energy conservation equation 2 g2 1 g1 K +U = K +U is 2 2 2 2 1 1 2 2 2 2 1 1 2 1 1 2 2 2 ( ) (3.0 m/s) 2(9.8 m/s )(0 m 0.20 m) 2.25 m/s mv mgy mv mgy v v gy y . + = + . = + - = + - = (b) Expressing the energy conservation equation as a function of . : ( ) 2 2 g 1 g1 1 2 2 1 1 ( ) ( ) 1 ( ) ( ) 1 0 J 2 2 2 ( ) 2 (1 cos ) K U K U mv mgy mv v v gy v gR . . . . . . . + = + . + = + . = - = - - Using 2 1 v = 3.0 m/s, g = 9.8 m/s , and R = 0.10 m, we get v(. ) = 9 -1.96(1- cos. ) (m/s) (c) The accompanying figure shows a graph of v for a complete revolution (0 =. = 360). Assess: Beginning with a speed of 3.0 m/s at the bottom, the marbles potential energy increases and kinetic energy decreases as it gets toward the top of the circle. At the top, its speed is 2.25 m/s. This is reasonable since some of the kinetic energy has been transformed into the marbles potential energy. 10.37. Model: Assume that the rubber band behaves similar to a spring. Also, model the rock as a particle. Visualize: Please refer to Figure P10.37. Solve: (a) The rubber band is stretched to the left since a positive spring force on the rock due to the rubber band results from a negative displacement of the rock. That is, (Fsp )x = -kx, where x is the rocks displacement from the equilibrium position due to the spring force sp F . (b) Since the sp F versus x graph is linear with a negative slope and can be expressed as sp F = -kx, the rubber band obeys Hookes law. (c) From the graph, sp |.F | = 20 N for |.x| =10 cm. Thus, sp 2 | | 20 N 200 N/m 2.0 10 N/m | | 0.10m F k x . = = = = . (d) The conservation of energy equation f sf i si K +U = K +U for the rock is 2 2 2 2 2 2 2 2 f f i i f i f i 1 1 1 1 1 1 (0 m) 1 (0 m/s) 1 2 2 2 2 2 2 2 2 200 N/m(0.30 m) 19.0 m/s 0.050 kg mv kx mv kx mv k m kx v k x m + = + . + = + = = = Assess: Note that i x is .x, which is the displacement relative to the equilibrium position, and f x is the equilibrium position of the rubber band, which is equal to zero. 10.38. Model: Model the block as a particle and the springs as ideal springs obeying Hookes law. There is no friction, hence the mechanical energy K +Us is conserved. Visualize: Note that f e i e x = x and x - x = .x. The before-and-after pictorial representations show that we put the origin of the coordinate system at the equilibrium position of the free end of the springs. Solve: The conservation of energy equation f sf i si K +U = K +U for the single spring is 2 2 2 2 f f e i i e 1 1 ( ) 1 1 ( ) 2 2 2 2 mv + k x - x = mv + k x - x Using the value for f v given in the problem, we get 2 2 2 2 0 0 1 0 J 0 J 1 ( ) 1 1 ( ) 2 2 2 2 mv + = + k .x . mv = k .x Conservation of energy for the two-spring case: 2 2 2 f ie ie 1 0 J 0 J 1 ( ) 1 ( ) 2 2 2 mV + = + k x - x + k x - x 2 2 f 1 ( ) 2 mV = k .x Using the result of the single-spring case, 2 2 f 0 f 0 1 2 2 mV = mv .V = v Assess: The block separates from the spring at the equilibrium position of the spring. 10.39. Model: Model the block as a particle and the springs as ideal springs obeying Hookes law. There is no friction, hence the mechanical energy K +Us is conserved. Visualize: The springs in both cases have the same compression .x. We put the origin of the coordinate system at the equilibrium position of the free end of the spring for the single-spring case (a), and at the free end of the two connected springs for the two-spring case (b). Solve: The conservation of energy for the single-spring case: 2 2 2 2 f sf i si f f e i i e 1 1 ( ) 1 1 ( ) 2 2 2 2 K +U = K +U . mv + k x - x = mv + k x - x Using f e i f 0, x = x = 0 m, v = 0 m/s, and v = v this equation simplifies to 2 2 0 1 1( ) 2 2 mv = k .x Conservation of energy in the case of the two springs in series, where each spring compresses by .x/2, is 2 2 2 2 f sf i si f i 1 0 1 1 ( /2) 1 ( /2) 2 2 2 2 K +U = K +U . mV + = mv + k .x + k .x Using f e x = x' = 0 m and i v = 0 m/s, we get 2 2 f 1 11( ) 2 22 mV = .. k .x .. . . Comparing the two results we see that f 0V = v / 2. Assess: The block pushes on the spring until the spring has returned to its equilibrium length. 10.40. Model: Model the ball and earth as particles. An elastic collision between the ball and earth conserves both momentum and mechanical energy. Visualize: The before-and-after pictorial representation of the collision is shown in the figure. The ball as it is dropped from a height of 10 m has zero velocity. Its speed just before it hits earth can be found using kinematics. We can then apply the conservation laws to find earths recoil velocity. Solve: (a) To get the speed of the ball at collision: 2 2 2 1 B 0 B 1 0 B 2 1 B ( ) ( ) 2 ( ) (0 m/s) 2(9.8m/s )( 10.0 m) ( ) 2(9.8 m/s )(10 m) 14.0 m/s v v gy y v = - - = - - . = = In an elastic collision, the velocity of the object that is struck is B 24 2 E 1 B 24 E B ( ) 2 ( ) 2(0.500 kg) (14.0 m/ s) 2.3 10 m/s 5.98 10 kg v m v m m = = = - + (b) The time it would take earth to move 1.0 mm at this speed is given by 3 20 13 24 speed distance 1.0 10 m 4.3 10 s 1.36 10 years velocity 2.3 10 m/s - - = = = = 10.41. Model: Assume an ideal spring that obeys Hookes law. There is no friction, and therefore the mechanical energy K +Us +Ug is conserved. Visualize: The figure shows a before-and-after pictorial representation. We have chosen to place the origin of the coordinate system at the position where the ice cube has compressed the spring 10 cm. That is, 0 y = 0. Solve: The energy conservation equation 2 s2 g2 0 s0 g0 K +U +U = K +U +U is 2 2 2 2 2 e e 0 0 e 0 1 1 ( ) 1 1 ( ) 2 2 2 2 mv + k x - x + mgy = mv + k x - x + mgy 2 0 0 Using v = 0 m/ s, y = 0 m, and v = 0 m/ s, 2 2 2 e 2 e 2 2 1 ( ) ( ) (25 N/m)(0.10 m) 25.5 cm 2 2 2(0.050 kg)(9.8 m/s ) mgy k x x y k x x mg - = - . = = = The distance traveled along the incline is 2y / sin 30 = 25.5 cm/ sin30 = 51 cm. Assess: The net effect of the launch is to transform the potential energy stored in the spring into gravitational potential energy. The block has kinetic energy as it comes off the spring, but we did not need to know this energy to solve the problem. 10.42. Model: Model the two packages as particles. Momentum is conserved in both inelastic and elastic collisions. Kinetic energy is conserved only in a perfectly elastic collision. Visualize: Solve: For a package with mass m the conservation of energy equation is 2 2 1 g1 0 g0 1 1 0 0 1 ( ) 1 ( ) 2 2 m m K +U = K +U . m v + mgy = m v + mgy 0 1 Using ( ) 0 m/ s and 0 m, mv = y = 2 2 1 0 1 0 1 ( ) ( ) 2 2(9.8 m/s )(3.0 m) 7.668 m/s 2 m m m v = mgy . v = gy = = (a) For the perfectly inelastic collision the conservation of momentum equation is f i 23 1 12 ( 2 )( ) ( ) (2 )( ) x x m m m p = p . m + m v = m v + m v 1 2 Using ( ) 0 m / s, we get m v = 2 3 1 ( ) ( ) /3 2.56 m/s m m v = v = The packages move off together at a speed of 2.6 m/s. (b) For the elastic collision, the mass m package rebounds with velocity ( ) 3 1 ( ) 2 ( ) 1 7.668 m/s 2.56 m/s 2 3 m m v m m v m m - = =- =- + The negative sign with 3 ( )m v shows that the package with mass m rebounds and goes to the position 4y . We can determine 4 y by applying the conservation of energy equation as follows. For a package of mass m: 2 2 f gf i gi 4 4 3 3 1 ( ) 1 ( ) 2 2 m m K +U = K +U . m v + mgy = m v + mgy 3 3 4 Using ( ) 2.55 m/s, 0 m, and ( ) 0 m/s, m m v = - y = v = we get 2 4 4 1 ( 2.56m/s) 33cm 2 mgy = m - . y = 10.43. Model: Model the marble and the steel ball as particles. We will assume an elastic collision between the marble and the ball, and apply the conservation of momentum and the conservation of energy equations. We will also assume zero rolling friction between the marble and the incline. Visualize: This is a two-part problem. In the first part, we will apply the conservation of energy equation to find the marbles speed as it exits onto a horizontal surface. We have put the origin of our coordinate system on the horizontal surface just where the marble exits the incline. In the second part, we will consider the elastic collision between the marble and the steel ball. Solve: The conservation of energy equation K1 +Ug1 = K0 +Ug0 gives us: 2 2 M 1 M M 1 M 0 M M 0 1 ( ) 1 ( ) 2 2 m v + m gy = m v + m gy 2 0 M 1 1 M 0 1 M 0 Using ( ) 0 m/ s and 0 m, we get 1 ( ) ( ) 2 . 2 v = y = v = gy . v = gy When the marble collides with the steel ball, the elastic collision gives the ball velocity M M 2 S 1 M 0 M S M S (v ) 2m (v ) 2m 2gy m m m m = = + + Solving for 0 y gives 2 M S 0 2S M 1 ( ) 0.258 m 25.8 cm 2 2 y m m v g m . + . = . . = = . . 10.44. Model: Assume an ideal spring that obeys Hookes law. Since this is a free-fall problem, the mechanical energy K +Ug +Us is conserved. Also, model the safe as a particle. Visualize: We have chosen to place the origin of our coordinate system at the free end of the spring, which is neither stretched nor compressed. The safe gains kinetic energy as it falls. The energy is then converted into elastic potential energy as the safe compresses the spring. The only two forces are gravity and the spring force, which are both conservative, so energy is conserved throughout the process. This means that the initial energyas the safe is releasedequals the final energywhen the safe is at rest and the spring is fully compressed. Solve: The conservation of energy equation 1 g1 s1 0 g0 s0 K +U +U = K +U +U is 2 2 2 2 1 1 e 1 e 0 0 e e e 1 ( ) 1 ( ) 1 ( ) 1 ( ) 2 2 2 2 mv + mg y - y + k y - y = mv + mg y - y + k y - y Using 1 0 v = v = 0 m/s and e y = 0 m, the above equation simplifies to 2 1 1 0 1 2 mgy + ky = mgy 2 0 1 5 2 2 1 2 ( ) 2(1000 kg)(9.8 m/s )(2.0 m ( 0.50 m)) 1.96 10 N/m ( 0.50m) k mg y y y - -- . = = = - Assess: By equating energy at these two points, we do not need to find how fast the safe was moving when it hit the spring. 10.45. Model: Assume an ideal spring that obeys Hookes law. There is no friction and hence the mechanical energy K +Us +Ug is conserved. Visualize: Solve: (a) When releasing the block suddenly, 2 s2 g2 1 s1 g1 K +U +U = K +U +U 2 2 2 2 2 2 e 2 1 1 e 1 1 1 ( ) 1 1 ( ) 2 2 2 2 mv + k y - y + mgy = mv + k y - y + mgy Using 2 1 v = 0 m/s, v = 0 m/s, and 1 ey = y , we get 2 2 2 1 2 1 2 1 1 2 2 2 1 2 1 2 1 2 0 J 1 (490 N/m)( ) 0 J 0 J (245 N/m)( ) ( ) 2 (245N/m)( ) (5.0kg)(9.8m/s)( ) ( ) 0.20m y y mgy mgy y y mg y y y y y y y y + - + = + + . - = - . - = - . - = (b) When lowering the block gently until it rests on the spring, the block reaches a point of static equilibrium. 2 net 0 (5.0 kg)(9.8 m/s ) 0.10 m 490 N/m F k y mg y mg k = . - = . . = = = (c) In part (b), at a point 0.10 m down, the forces balance. But in part (a) the block has kinetic energy as it reaches 0.10 m. So the block continues on past the equilibrium point until all the gravitational potential energy is stored in the spring. 10.46. Model: Assume an ideal spring that obeys Hookes law. There is no friction, and thus the mechanical energy K +Us +Ug is conserved. Visualize: We place the origin of our coordinate system at the springs compressed position 1 y = 0. The rock leaves the spring with velocity 2 v as the spring reaches its equilibrium position. Solve: (a) The conservation of mechanical energy equation is 2 2 2 2 2 s2 g2 1 s1 g1 2 2 e 2 1 1 e 1 1 1 ( ) 1 1 ( ) 2 2 2 2 K +U +U = K +U +U mv + k y - y + mgy = mv + k y - y + mgy Using 2 e 1 1 y = y , y = 0 m, and v = 0 m/s, this simplifies to 2 2 2 2 1 e 2 2 2 2 2 1 0 J 0 J 1 ( ) 0 2 2 1 (0.400 kg) (0.400 kg)(9.8 m/s )(0.30 m) 1 (1000 N/m)(0.30 m) 14.8 m/s 2 2 mv mgy k y y v v + + = + - + + = .= (b) Let us use the conservation of mechanical energy equation once again to find the highest position 3 ( y ) of the rock where its speed 3 (v ) is zero: 2 2 3 g3 2 g2 3 3 2 2 2 2 2 2 3 2 2 3 2 2 1 1 2 2 0 ( ) 1 ( ) (14.8 m/s) 11.2 m 2 2 2(9.8 m/s ) K U K U mv mgy mv mgy g y y v y y v g + = + . + = + . + - = . - = = = If we assume the springs length to be 0.5 m, then the distance between ground and fruit is 11.2 m+ 0.5 m =11.7 m. This is much smaller than the distance of 15 m between fruit and ground. So, the rock does not reach the fruit, and the contestants go hungry. 10.47. Model: Assume an ideal spring that obeys Hookes law. There is no friction, so the mechanical energy K +Ug +Us is conserved. Visualize: Place the origin of the coordinate system at the end of the unstretched spring, making e y = 0 m. Solve: The clay is in static equilibrium while resting in the pan. The net force on it is zero. We can start by using this to find the spring constant. 2 sp G 1 e 1 1 ( ) (0.10 kg)(9.8 m/s ) 9.8 N/m 0.10 m F F k y y ky mg k mg y = . - - = - = . = - = - = - Now apply conservation of energy. Initially, the spring is unstretched and the clay ball is at the ceiling. At the end, the spring has maximum stretch and the clay is instantaneously at rest. Thus 1 2 1 2 1 2 2 g 2 s 2 0 g 0 s 0 2 2 2 2 2 2 0 0 K + (U ) + (U ) = K + (U ) + (U ) . mv + mgy + ky = mv + mgy + 0 J Since 0 v = 0 m/s and 2 v = 0 m/s, this equation becomes 1 2 2 0 2 2 2 0 2 2 2 2 2 2 2 0 0.20 0.10 0 mgy ky mgy y mg y mgy k k y y + = . + - = + - = The numerical values were found using known values of m, g, k, and 0y . The two solutions to this quadratic equation are 2 y = 0.231 m and 2 y = -0.432 m. The point were looking for is below the origin, so we need the negative root. The distance of the pan from the ceiling is 2 L = y + 50 cm = 93 cm 10.48. Model: Assume an ideal spring that obeys Hookes law. Also assume zero rolling friction between the roller coaster and the track, and a particle model for the roller coaster. Since no friction is involved, the mechanical energy K +Us +Ug is conserved. Visualize: We have chosen to place the origin of the coordinate system on the end of the spring that is compressed and touches the roller coaster car. Solve: (a) The energy conservation equation for the car going to the top of the hill is 2 2 2 2 2 g2 s2 0 g0 s0 2 2 e e 0 0 0 e 1 1 ( ) 1 1 ( ) 2 2 2 2 K +U +U = K +U +U mv + mgy + k x - x = mv + mgy + k x - x 0 2 1 0 e Noting that y = 0 m,v = 0 m/s, v = 0 m/s, and x - x = 2.0 m, we obtain 2 2 2 2 4 2 2 0 J 0 J 0 J 0 J 1 (2.0 m) 2 2 2(400 kg)(9.8 m/s )(10 m) 1.96 10 N/m (2.0 m) (2.0 m) mgy k k mgy + + = + + . = = = We now increase this value for k by 10% for safety, giving a value of 2.156104 N/m 2.2104 N/m. (b) The energy conservation equation 3 g3 s3 0 g0 s0 K +U +U = K +U +U is 2 2 2 2 3 3 e e 0 0 0 e 1 1 ( ) 1 1 ( ) 2 2 2 2 mv + mgy + k x - x = mv + mgy + k x - x 3 0 0 0 e Using y = -5.0 m, v = 0 m/ s, y = 0 m, and |x - x | = 2.0 m, we get 2 2 3 0e 2 2 4 2 3 3 1 ( 5.0 m) 0 J 0 J 0 J 1 ( ) 2 2 1 (400 kg) (400 kg)(9.8 m/s )(5.0 m) 1 (2.156 10 N/m)(2.0 m) 2 2 17.7 m/s mv mg k x x v v + - + = + + - - = . = 10.49. Model: Since there is no friction, the sum of the kinetic and gravitational potential energy does not change. Model Julie as a particle. Visualize: We place the coordinate system at the bottom of the ramp directly below Julies starting position. From geometry, Julie launches off the end of the ramp at a 30 angle. Solve: Energy conservation: K1 +Ug1 = K0 +Ug0 2 2 1 1 0 0 1 1 2 2 . mv + mgy = mv + mgy Using 0 0 1 v = 0 m/s, y = 25 m, and y = 3m, the above equation simplifies to 2 2 1 1 0 1 0 1 1 2 ( ) 2(9.8 m/s )(25 m 3m) 20.77 m/s 2 mv + mgy = mgy .v = g y - y = - = We can now use kinematic equations to find the touchdown point from the base of the ramp. First well consider the vertical motion: 2 2 2 2 1 1 2 1 2 1 1 2 1 2 1 2 2 1 2 2 1 2 2 2 2 1 2 1 2 1 ( ) 1 ( ) 0 m 3m ( sin30 )( ) 1 ( 9.8 m/ s )( ) 2 2 ( ) (20.77 m/s)sin30 ( ) (3m) 0 (4.9 m/s ) (4.9 m/s ) ( ) (2.119 s)( ) (0.6122 s ) 0 ( ) 2.377 s y y y y v t t a t t v t t t t t t t t t t t t t t = + - + - = + - + - - . - - - - = - - - - = . - = For the horizontal motion: 2 2 1 1 2 1 2 1 ( ) 1 ( ) 2 x x x = x + v t - t + a t - t 2 1 1 2 1 x - x = (v cos30)(t - t ) + 0 m = (20.77 m/s)(cos30)(2.377 s) = 43 m Assess: Note that we did not have to make use of the information about the circular arc at the bottom that carries Julie through a 90 turn. 10.50. Model: We assume the spring to be ideal and to obey Hookes law. We also treat the block (B) and the ball (b) as particles. In the case of an elastic collision, both the momentum and kinetic energy equations apply. On the other hand, for a perfectly inelastic collision only the equation of momentum conservation is valid. Visualize: Place the origin of the coordinate system on the block that is attached to one end of the spring. The before-andafter pictorial representations of the elastic and perfectly inelastic collision are shown in figures (a) and (b), respectively. Solve: (a) For an elastic collision, the balls rebound velocity is b B f b i b b B ( ) ( ) 80 g (5.0 m/s) 3.33 m/s 120 g v m m v m m - - = = =- + The balls speed is 3.3 m/s. (b) An elastic collision gives the block speed B f B i b b B ( ) 2 ( ) 40 g (5.0 m/s) 1.667 m/s 120 g v m v m m = = = + To find the maximum compression of the spring, we use the conservation equation of mechanical energy for the block + spring system. 1 s1 0 s0 That is K +U = K +U : 2 2 2 2 2 2 B f B 1 0 B f B 0 0 1 0 B f B 2 1 0 1 ( ) 1 ( ) 1 ( ) 1 ( ) 0 ( ) ( ) 0 2 2 2 2 ( ) (0.100 kg )(1.667 m/s ) /(20 N/m) 11.8 cm m v k x x m v k x x k x x m v x x ' + - = + - + - = + - = = (c) Momentum conservation f i p = p for the perfectly inelastic collision means B B f b ib B iB f f ( ) ( ) ( ) (0.100 kg 0.020 kg) (0.020 kg)(5.0 m/s) 0 m/v 0.833 m/s m m v m v m v v v + = + + = + . = The maximum compression in this case can now be obtained using the conservation of energy equation 1 s1 K +U = 0 s0 : K +U 2 2 B b f B b f 0 J (1/ 2) ( ) (1/ 2)( ) 0 J 0.120 kg (0.833m/s) 0.0645 m 6.5 cm 20N/m k x m m v x m m v k + . = + + + . . = = = = 10.51. Model: Assume an ideal spring that obeys Hookes law. We treat the bullet and the block in the particle model. For a perfectly inelastic collision, the momentum is conserved. Furthermore, since there is no friction, the mechanical energy of the system (bullet + block + spring) is conserved. Visualize: We place the origin of our coordinate system at the end of the spring that is not anchored to the wall. Solve: (a) Momentum conservation for perfectly inelastic collision states f ip = p . This means f im iM f B f B (m M)v m(v ) M(v ) (m M)v mv 0 kg m/s v m v m M + = + . + = + . = . . . + . . . where we have used B v for the initial speed of the bullet. The mechanical energy conservation equation 1 s1 K +U = e se K +U as the bullet embedded block compresses the spring is: 2 2 2 2 1 e f e e 2 2 2 2 B B 2 1 ( ) 1 ( ) 1 ( )( ) 1 ( ) 2 2 2 2 0 J 1 1 ( ) 0 J ( ) 2 2 f m v k x x m M v k x x kd m M m v v m M kd m M m ' + - = + + - . . + + = + . . + . = . + . (b) Using the above formula with m = 5.0 g, M = 2.0 kg, k = 50 N/m, and d =10 cm, 2 2 2 B v = (0.0050 kg + 2.0 kg)(50 N/m)(0.10 m) /(0.0050) = 2.010 m/s (c) The fraction of energy lost is 2 2 2 2 B f f 2 B B 1 1 ( ) + 2 2 1 1 1 2 1 1 0.0050 kg 0.9975 (0.0050 kg 2.0 kg) mv m M v m M v m M m mv m v m m M m m M - + . . + . . = - . . = - . . . . . + . = - = - = + + That is, during the perfectly inelastic collision 99.75% of the bullets energy is lost. The energy is dissipated inside the block. Although it is common to say, The energy is lost to heat, in the next chapter well see that it is more accurate to say, The energy is transformed to thermal energy. 10.52. Model: Assume an ideal spring that obeys Hookes law. There is no friction, hence the mechanical energy K +Ug +Us is conserved. Visualize: We have chosen to place the origin of the coordinate system on the free end of the spring that is neither stretched nor compressed, that is, at the equilibrium position of the end of the unstretched spring. The bullets mass is m and the blocks mass is M. Solve: (a) The energy conservation equation 2 s2 g2 1 s1 g1 K +U +U = K +U +U becomes 2 2 2 2 2 2 e 2 e 1 1 e 1 e 1 ( ) 1 ( ) ( ) ( ) 1 ( ) 1 ( ) ( ) ( ) 2 2 2 2 m+M v + k y - y + m+M g y - y = m+M v + k y - y - m+M g y - y Noting 2 v = 0 m/s, we can rewrite the above equation as 2 22 2 2 1 1 1 k(.y ) + 2(m + M)g(.y + .y ) = (m + M)v + k(.y ) Let us express 1 v in terms of the bullets initial speed B v by using the momentum conservation equation f i p = p which is 1 B block (m + M)v = mv + Mv . Since block v = 0 m/s, we have 1 B v m v m M = . . . + . . . We can also find the magnitude of 1 y from the equilibrium condition 1 e k( y - y ) = Mg. 1 y Mg k . = With these substitutions for 1 v and 1.y , the energy conservation equation simplifies to 2 2 2 2 B 2 1 2 2 2 2 2 2 B 1 2 2 2 2 ( ) 2( ) ( ) ( ) 2 ( ) ( ) ( ) k y m M g y y m v k Mg m M k v m M g y y m M M g k m M y m m k m . + + . + . = + . . + . . . . . + . + . + . . = . . . + . - + . . . . . . . We still need to include the springs maximum compression (d) into this equation. We assume that 1 2d = .y + .y , that is, maximum compression is measured from the initial position 1 ( y ) of the block. Thus, using 2 1 .y = d - .y = (d -Mg /k), we have 2 1 2 2 2 2 B 2 2 v 2 m M gd m M M g k m M (d Mg /k) m m k m . . + . . + . . + . . = . . . - . . + . . - . .. . . . . . . .. (b) Using m = 0.010 kg, M = 2.0 kg, k = 50 N/m, and d = 0.45 m, 2 2 2 2 22 B 2 2 2 B 2 2.010 kg (9.8 m/s )(0.45 m) (2.010 kg) 2.0 kg (9.8 m/s ) /(50 N/m) 0.010kg 0.010 kg (50 N/m)(2.010 kg) 1 [0.45 m (2.0 kg) (9.8 m/s )/50 N/m] (0.010 kg) 453m/s v v . . . . = . . - . . . . . . + - . = The bullet has a speed of 4.5102 m/s. Assess: This is a reasonable speed for the bullet. 10.53. Model: Because the track is frictionless, the sum of the kinetic and gravitational potential energy does not change during the cars motion. Visualize: We place the origin of the coordinate system at the ground level directly below the cars starting position. This is a two-part problem. If we first find the maximum speed at the top of the hill, we can use energy conservation to find the maximum initial height. Solve: Because its motion is circular, at the top of the hill the car has a downward-pointing centripetal acceleration 2 ac = -(mv /r) j. .. Newtons second law at the top of the hill is ( ) 2 2 2 net G c ( ) ( ) y y y y F n F n mg m a mv n mg mv m g v R R R . . = + = - = = - . = - = . - . . . If v = 0 m/ s, n = mg as expected in static equilibrium. As v increases, n gets smallerthe weight of the car and passengers decreases as they go over the top. But n has to remain positive for the car to be on the track, so the maximum speed max v occurs when n.0. We see that max v = gR. Now we can use energy conservation to relate the top of the hill to the starting height: 2 2 f f i i f f 1 1 max 1 1 1 0 J 2 2 2 K +U = K +U . mv + mgy = mv + mgy . mgR + mgR = + mgh where we used f max v = v and f y = R. Solving for max h gives 3 max 2 h = R. (b) If R =10 m, then max h =15 m. 10.54. Model: This is a two-part problem. In the first part, we will find the critical velocity for the block to go over the top of the loop without falling off. Since there is no friction, the sum of the kinetic and gravitational potential energy is conserved during the blocks motion. We will use this conservation equation in the second part to find the minimum height the block must start from to make it around the loop. Visualize: We place the origin of our coordinate system directly below the blocks starting position on the frictionless track. Solve: The free-body diagram on the block implies 2 c G F n mv R + = For the block to just stay on track, n = 0. Thus the critical velocity c v is 2 c 2 G c F mg mv v gR R = = . = The block needs kinetic energy 1 2 2 c mv = 1 2 mgR to go over the top of the loop. We can now use the conservation of mechanical energy equation to find the minimum height h. 2 2 f gf i gi f f i i 1 1 2 2 K +U = K +U . mv + mgy = mv + mgy f c f i Using v = v = gR, y = 2R, v = 0 m/s, and i y = h, we obtain 1 (2 ) 0 2.5 2 gR + g R = + gh.h = R 10.55. Model: Model Lisa (L) and the bobsled (B) as particles. We will assume the ramp to be frictionless, so that the mechanical energy of the system (Lisa + bobsled + spring) is conserved. Furthermore, during the collision, as Lisa leaps onto the bobsled, the momentum of the Lisa + bobsled system is conserved. We will also assume the spring to be an ideal one that obeys Hookes law. Visualize: We place the origin of our coordinate system directly below the bobsleds initial position. Solve: (a) Momentum conservation in Lisas collision with bobsled states 1 0p = p , or L B 1 L 0L B 0B L B 1 L 0L L 1 0L L B ( ) ( ) ( ) ( ) ( ) 0 ( ) 40 kg (12 m/s) 8.0 m/s 40 kg 20 kg m m v m v m v m m v m v v m v m m + = + . + = + . . . . . = . . = . . = . + . . + . The energy conservation equation: 2 s2 g2 1 s1 g1 K +U +U = K +U +U is 2 2 2 2 L B 2 2 e L B 2 L B 1 e e L B 1 1 ( ) 1 ( ) ( ) 1 ( ) 1 ( ) ( ) 2 2 2 2 m + m v + k x - x + m + m gy = m + m v + k x - x + m + m gy Using 2 2 1 1 v = 0 m/s, k = 2000 N/m, y = 0 m, y = (50 m)sin 20 =17.1 m, v = 8.0 m/s, and L B (m m ) = 60 kg, we get 2 2 2 2 e 0 J 1 (2000 N/m)( ) 0 J 1 (60 kg)(8.0 m/ s) 0 J (60 kg)(9.8 m/ s )(17.1m) 2 2 + x - x + = + + 2 e Solving this equation yields (x - x ) = 3.5 m. (b) As long as the ice is slippery enough to be considered frictionless, we know from conservation of mechanical energy that the speed at the bottom depends only on the vertical descent .y. Only the ramps height h is important, not its shape or angle. 10.56. Model: We can divide this problem into two parts. First, we have an elastic collision between the 20 g ball (m) and the 100 g ball (M). Second, the 100 g ball swings up as a pendulum. Visualize: The figure shows three distinct moments of time: the time before the collision, the time after the collision but before the two balls move, and the time the 100 g ball reaches its highest point. We place the origin of our coordinate system on the 100 g ball when it is hanging motionless. Solve: For a perfectly elastic collision, the ball moves forward with speed 1 0 0 ( ) 2 ( ) 1 ( ) 3 m M m m m M v m v v m m = = + In the second part, the sum of the kinetic and gravitational potential energy is conserved as the 100 g ball swings up after the collision. That is, 2 g2 1 g1 . K +U = K +U We have 2 2 2 2 1 1 1 ( ) 1 ( ) 2 2 M M M v + Mgy = M v + Mgy 0 2 1 1 2 Using ( ) 0 J, ( ) ( ) , 0 m, and cos , 3 m M M v = v = v y = y = L - L . the energy equation simplifies to 2 0 2 0 ( cos ) 1 ( ) 2 9 ( ) 18 g (1 cos ) 18(9.8 m/s )(1.0 m)(1 cos50 ) 7.9 m/s m m g L L v v L . . - = . = - = - = 10.57. Model: Model the balls as particles. We will use the Galilean transformation of velocities (Equation 10.44) to analyze the problem of elastic collisions. We will transform velocities from the lab frame S to a frame S' in which one ball is at rest. This allows us to apply Equations 10.43 to the case of a perfectly elastic collision in S , ' find the final velocities of the balls in S , ' and then transform these velocities back to the lab frame S. Visualize: Let S' be the frame of the 200 g ball. Denoting masses as m1 =100 g and 2 m = 200 g, the initial velocities in the S frame are i 1 ( ) 4 m/s x v = and i 2 ( ) 3 m/s. x v = - Figure (a) shows the before-collision situation as seen in frame S, and figure (b) shows the before-collision situation as seen in frame S'. The after-collision velocities in S' are shown in figure (c), and figure (d) indicates velocities in S after they have been transformed to frame S from S'. Solve: (a) In the S frame, i 1 ( ) 4 m/s x v = and i 2 ( ) 3 m/s. x v = - S' is the reference frame of the 200 g ball, so V = -3 m/s. The velocities of the two balls in this frame can be obtained using the Galilean transformation of velocities v' = v -V. So, i 1 i 1 ( ) ( ) 4m/s ( 3m/s) 7m/s x x v ' = v -V = - - = i 2 i 2 ( ) ( ) 3m/s ( 3m/s) 0m/s x x v ' = v -V = - - - = Figure (b) shows the before situation, where ball 2 is at rest. Now we can use Equations 10.43 to find the after-collision velocities in frame S': 1 2 f 1 i 1 1 2 1 f 2 i 1 1 2 ( ) ( ) 100 g 200 g (7 m/s) 7 m/s 100 g 200 g 3 ( ) 2 ( ) 2(100) g (7 m/s) 14 m/s 100 g 200 g 3 x x x x v m m v m m v m v m m ' - ' - = = =- + + ' = ' = = + + Finally, we need to apply the reverse Galilean transformation v = v' +V, with the same V, to transform the aftercollision velocities back to the lab frame S: f 1 f 1 f 2 f 2 ( ) ( ) 7m/s 3m/s 5.33m/s 3 ( ) ( ) 14m/s 3m/s 1.67 m/s 3 x x x x v v V v v V = ' + = - - = - = ' + = - = Figure (d) shows the after situation in the lab frame. The 100 g ball is moving left at 5.3 m/s; the 200 g ball is moving right at 1.7 m/s. (b) The momentum conservation equation fx ix p = p for a perfectly inelastic collision is 1 2 f 1 i 1 2 i 2 f f ( ) ( ) ( ) (0.100 kg 0.200 kg) (0.100 kg)(4.0 m/s) (0.200 kg)( 3.0 m/s) 0.667 m/s x x x x x m m v m v m v v v + = + + = + - . =- Both balls are moving left at 0.67 m/s. 10.58. Model: Model the balls as particles. We will use the Galilean transformation of velocities (Equation 10.44) to analyze the problem of elastic collisions. We will transform velocities from the lab frame S to a frame S' in which one ball is at rest. This allows us to apply Equations 10.43 to a perfectly elastic collision in S'. After finding the final velocities of the balls in S', we can then transform these velocities back to the lab frame S. Visualize: Let S' be the frame of the 400 g ball. Denoting masses as m1 =100 g and 2 m = 400 g, the initial velocities in the S frame are i 1 ( ) 4.0 m/s x v = + and i 2 ( ) 1.0 m/s. x v = + Figures (a) and (b) show the before-collision situations in frames S and S', respectively. The after-collision velocities in S' are shown in figure (c). Figure (d) indicates velocities in S after they have been transformed to S from S'. Solve: In frame S, i 1 ( ) 4.0 m/s x v = and i 2 ( ) 1.0 m/s. x v = Because S' is the reference frame of the 400 g ball, V =1.0 m/s. The velocities of the two balls in this frame can be obtained using the Galilean transformation of velocities v' = v -V.So, i 1 i 1 ( ) ( ) 4.0 m/s 1.0 m/s 3.0 m/s x x v ' = v -V = - = i 2 i 2 ( ) ( ) 1.0m/s 1.0m/s 0m/s x x v ' = v -V = - = Figure (b) shows the before situation in frame S' where the ball 2 is at rest. Now we can use Equations 10.43 to find the after-collision velocities in frame S': 1 2 f 1 i 1 1 2 1 f 2 i 1 1 2 ( ) ( ) 100 g 400 g (3.0 m/s) 1.80 m/s 100 g 400 g ( ) 2 ( ) 2(100 g) (3.0 m/s) 1.20 m/s 100 g 400 g x x x x v m m v m m v m v m m ' - ' - = = =- + + ' = ' = = + + Finally, we need to apply the reverse Galilean transformation v = v' +V, with the same V, to transform the aftercollision velocities back to the lab frame S. f 1 f 1 f 2 f 2 ( ) ( ) 1.80 m/s 1.0m/s 0.80m/s ( ) ( ) 1.20 m/s 1.0m/s 2.20m/s x x x x v v V v v V = ' + = - + = - = ' + = + = Figure (d) shows the after situation in frame S. The 100 g ball moves left at 0.80 m/s, the 400 g ball right at 2.2 m/s. Assess: The magnitudes of the after-collision velocities are similar to the magnitudes of the before-collision velocities. 10.59. Model: Use the model of the conservation of mechanical energy. Visualize: Solve: (a) The turning points occur where the total energy line crosses the potential energy curve. For E =12 J, this occurs at the points x =1 m and x = 7 m. (b) The equation for kinetic energy K = E -U gives the distance between the potential energy curve and total energy line. U = 8 J at x = 2 m, so K =12 J - 8 J = 4 J. The speed corresponding to this kinetic energy is 2 2(4J) 4.0 m/s 0.5 kg v K m = = = (c) Maximum speed occurs for minimum U. This occurs at x =1 m and x = 4 m, where U = 0 J and K =12 J. The speed at these two points is 2 2(12J) 6.9 m/s 0.500 kg v K m = = = (d) The particle leaves x =1 m with v = 6.9 m/s. It gradually slows down, reaching x = 2 m with a speed of 4.0 m/s. After x = 2 m, it speeds up again, returning to a speed of 6.9 m/s as it crosses x = 4 m. Then it slows again, coming instantaneously to a halt (v = 0 m/s) at the x = 7 m turning point. Now it will reverse direction and move back to the left. (e) If the particle has E = 4 J it cannot cross the 8 J potential energy mountain in the center. It can either oscillate back and forth over the range 1.0 m = x =1.5 m or over the range 3 m = x = 5 m. 10.60. Model: We will use the conservation of mechanical energy. Visualize: The potential energy (U) of the nitrogen atom as a function of z exhibits a double-minimum behavior; the two minima correspond to the nitrogen atoms position on both sides of the plane containing the three hydrogens. Solve: (a) At room temperature, the total energy line is below the hill in the center of the potential energy curve. That is, the nitrogen atom does not have sufficient energy to pass from one side of the molecule to the other. Theres a stable equilibrium position on either side of the hydrogen-atom plane at the points where U = 0. Since E > 0, the nitrogen atom will be on one side of the plane and will make small vibrations back and forth along the z-axisthat is, toward and away from the hydrogen-atom plane. In the figure above, the atom oscillates between points A and B. (b) The total energy line is now well above the hill, and the turning points of the nitrogen atoms motion (where the total energy line crosses the potential curve) are at points C and D. In other words, the atom oscillates from one side of the 3 H plane to the other. It slows down a little as it passes through the plane of hydrogen atoms, but it has sufficient energy to get through. 10.61. Solve: (a) The equilibrium positions are located at points where dU 0. dx = ( ) ( ) 1 0 1 2cos 2 cos 2 1 2 1 cos 1 2 2 dU x x dx x -
= = + . = - . = .- . . . . . Note that 1 2 - is in radians and x is in meters. The function cos 1 1 2 - .. - .. . . may have values 2 3 p and 4 . 3 p Thus there are two values of x, 1 2 and 2 3 3 x x p p = = within the interval 0 m = x =p m. (b) A point of stable equilibrium corresponds to a local minimum, while a point of unstable equilibrium corresponds to a local maximum. Compute the concavity of U(x) at the equilibrium positions to determine their stability. ( ) 2 2 d U 4sin 2x dx = - At 1 , 3 x p = ( ) 2 2 1 4 3 2 3. 2 d U x dx . . = - .. .. = - . . Since ( ) 2 2 1 1 0, 3 d Ux x dx p < = is a local maximum, so 1 3 x p = is a point of unstable equilibrium. At ( ) 2 2 2 2 2 , 4 3 2 3. 3 2 x d U x dx p . . = - = - .. - .. = + . . Since 2 2 2 0, 2 3 d U x dx p > = is a local minimum, so 2 2 3 x p = is a point of stable equilibrium. 10.62. Model: The potential energy of two nucleons interacting via the strong force is / 0 0U =U [1- e-x x ] where x is the distance between the centers of the two nucleons, 15 0 x = 2.010- m, and 11 0 U = 6.010- J. Visualize: Nucleons are protons and neutrons, and they are held together in the nucleus by a force called the strong force. This force exists between nucleons at very small separations. Solve: (a) (b) For x = 5.010-15 m, U = 55.110-12 J. This energy is represented by a total energy line. (c) Due to conservation of total energy, the potential energy when x = 5.010-15 J is transformed into kinetic energy until x = twice the radius =1.010-15 m. At this separation, u =15.710-12 J. Thus, 1 2 1 2 15.7 10 12 J 55.1 10 12 J 1.53 108 m/s 2 2 mv + mv + - = - .v = Assess: A speed of 1.53108 m/s is approximately 0.5 c where c is the speed of light. This speed is understandable for the present model. 10.63. A 2.5 kg ball is thrown upward at a speed of 4.0 m/s from a height of 82 cm above a vertical spring. When the ball comes down it lands on and compresses the spring. If the spring has a spring constant of k = 600 N/m, by how much is it compressed? 10.64. (a) A 1500 kg auto coasts up a 10.0 m high hill and reaches the top with a speed of 5.0 m/s. What initial speed must the auto have had at the bottom of the hill? (b) We place the origin of our coordinate system at the bottom of the hill. (c) The solution of the equation is vi =14.9 m/s. This is approximately 30 mph and is a reasonable value for the speed at the bottom of the hill. 10.65. (a) A spring gun is compressed 15 cm to launch a 200 g ball on a horizontal, frictionless surface. The ball has a speed of 2.0 m/s as it loses contact with the spring. Find the spring constant of the gun. (b) We place the origin of our coordinate system on the free end of the spring in the equilibrium position. Because the surface is frictionless, the mechanical energy for the system (ball + spring) is conserved. (c) The conservation of energy equation is f sf i si 2 2 2 2 1 2 2 1 1 (0 m) 1 (0 m/s) 1 ( 0.15 m) 2 2 2 2 (0.200 kg)(2.0 m/s) ( 0.15 m) 36 N/m K U K U mv k m k k k + = + + = + - = - = 10.66. (a) A 100 g lump of clay traveling at 3.0 m/s strikes and sticks to a 200 g lump of clay at rest on a frictionless surface. The combined lumps smash into a horizontal spring with k = 3.0 N/m. The other end of the spring is firmly anchored to a fixed post on the surface. How far will the spring compress? (b) (c) Solving the conservation of momentum equation we get 1 1.0 m/s. x v = Substituting this value into the conservation of energy equation yields 2 .x = 32 cm. 10.67. (a) A spring with spring constant 400 N/m is anchored at the bottom of a frictionless 30 incline. A 500 g block is pressed against the spring, compressing the spring by 10 cm, then released. What is the speed with which the block is launched up the incline? (b) The origin is placed at the end of the uncompressed spring. This is the point from which the block is launched as the spring expands. (c) Solving the energy conservation equation, we get vf = 2.6 m/s. 10.68. Model: Assume an ideal spring that obeys Hookes law. The mechanical energy K +Us +Ug is conserved during the launch of the ball. Visualize: This is a two-part problem. In the first part, we use projectile equations to find the balls velocity 2 v as it leaves the spring. This will yield the balls kinetic energy as it leaves the spring. Solve: Using the equations of kinematics, 2 3 2 2 3 2 3 2 2 3 2 3 3 2 2 3 2 2 3 2 3 2 2 2 2 3 3 ( ) 1 ( ) 5.0 m 0 m ( cos30 )( 0 s) 0 m 2 ( cos30 ) 5.0 m (5.0 m/ cos30 ) ( ) 1 ( ) 2 1.5 m 0 ( sin30 )( 0 s) 1 (9.8 m/ s )( 0 s) 2 x x y y x x v t t a t t v t v t t v y y v t t a t t v t t = + - + - . = + - + = . = = + - + - - = + - + - Substituting the value for 3t , ( ) 2 2 2 2 2 ( 1.5 m) ( sin30 ) 5.0 m 4.9 m/s 5.0 m cos30 cos30 v v v . . . . - = . . - . . . . . . 2 2 2 ( 1.5 m) (2.887 m) 163.33 v 6.102 m/s v . - = + - . = The conservation of energy equation 2 s2 g2 1 s1 g1 K +U +U = K +U +U is 2 2 2 2 2 2 1 1 1 1 (0 m) 1 1 ( ) 2 2 2 2 mv + k + mgy = mv + k .s + mgy Using 2 y = 0 m, 1 v = 0 m/s, .s = 0.20 m, and 1 y = -(.s)sin g30, we get 2 2 2 2 2 2 2 2 2 1 0 J 0 J 0 J 1 ( ) ( )sin30 ( ) 2 ( )sin30 2 2 (0.20 m) (0.020 kg)(6.102 m/s) 2(0.020 kg)(9.8 m/s )(0.20)(0.5) 19.6 N/m mv k s mg s s k mv mg s k k + + = + . - . . = + . = + .= Assess: Note that 1 y = -(.s)sin30 is with a minus sign and hence the gravitational potential energy 1 mgy is -mg(.s)sin30. 10.69. Model: This is a two-part problem. In the first part, we will find the critical velocity for the ball to go over the top of the peg without the string going slack. Using the energy conservation equation, we will then obtain the gravitational potential energy that gets transformed into the critical kinetic energy of the ball, thus determining the angle . . Visualize: We place the origin of our coordinate system on the peg. This choice will provide a reference to measure gravitational potential energy. For . to be minimum, the ball will just go over the top of the peg. Solve: The two forces in the free-body force diagram provide the centripetal acceleration at the top of the circle. Newtons second law at this point is 2 G F T mv r + = where T is the tension in the string. The critical speed c v at which the string goes slack is found when T .0. In this case, 2 C 2 C mg mv v gr gL 3 r = . = = The ball should have kinetic energy at least equal to 2C 1 1 2 2 3 mv = mg . L . . . . . for the ball to go over the top of the peg. We will now use the conservation of mechanical energy equation to get the minimum angle . . The equation for the conservation of energy is 2 2 f gf i gi f f i i 1 1 2 2 K +U = K +U . mv + mgy = mv + mgy 1 2 f c f 3 i C Using v = v , y = L, v = 0, and the above value for v , we get i i 1 2 3 3 2 mg L + mg L = mgy . y = L That is, the ball is a vertical distance 1 2 L above the pegs location or a distance of 2 3 2 6 . L - L . = L . . . . below the point of suspension of the pendulum, as shown in the figure on the right. Thus, cos /6 1 80.4 6 L L . = = .. = 10.70. Model: Choose yourself + spring + earth as the system. There are no forces from outside this system, so it is an isolated system. The interaction forces within the system are the spring force of the bungee cord and the gravitational force. These are both conservative forces, so mechanical energy is conserved. Visualize: We can equate the systems initial energy, as you step off the bridge, to its final energy when you reach the lowest point. We do not need to compute your speed at the point where the cord starts to stretch. We do, however, need to note that the end of the unstretched cord is at 1 2 0 1 2s 2 2 0 y = y - 30 m = 70 m, so U = k( y - y ) . Also note that 1s U = 0, since the cord is not stretched. The energy conservation equation is 2 2 2g 2s 1 1g 1s 2 2 0 1 0 J 1 ( ) 0 J 0 J 2 K +U +U = K +U +U . + mgy + k y - y = + mgy + Multiply out the square of the binomial and rearrange: 2 2 2 2 02 0 1 2 2 1 2 2 02 0 2 2 1 1 2 2 2 2 2 100.8 980 0 mgy ky ky y ky mgy y mg y y y mgy y y k k + - + = . + . - . + . - . = - + = . . . . . . . . This is a quadratic equation with roots 89.9 m and 10.9 m. The first is not physically meaningful because it is a height above the point where the cord started to stretch. So we find that your distance from the water when the bungee cord stops stretching is 10.9 m. 10.71. Model: Assume an ideal spring that obeys Hookes law. There is no friction, hence the mechanical energy K +Ug +Us is conserved. Visualize: We have chosen to place the origin of the coordinate system at the point of maximum compression. We will use lengths along the ramp with the variable s rather than x. Solve: (a) The conservation of energy equation 2 g2 s2 1 g1 s1 K +U +U = K +U +U is 2 2 2 2 2 2 1 1 2 2 2 2 2 1 1 ( ) 1 (0 m) 2 2 2 1 (0 m/s) (0 m) 1 ( ) 1 (0 m/s) (4.0 m )sin30 0 J 2 2 2 1 (250 N/m)( ) (10 kg)(9.8 m/s )(4.0 m ) 1 2 2 mv mgy k s mv mgy k m mg ks m mg s s s + + . = + + + + . = + +. + . = + . . . . . . . This gives the quadratic equation: (125 N/m)( )2 (49 kg m/s2 ) 196 kg m2 /s2 0 1.46 m and 1.07 m (unphysical) s s s . - . - = . . = - The maximum compression is 1.46 m. (b) We will now apply the conservation of mechanical energy to a point where the vertical position is y and the blocks velocity is v. We place the origin of our coordinate system on the free end of the spring when the spring is neither compressed nor stretched. 2 2 2 2 1 1 2 2 1 1 ( ) 1 1 (0 m) 2 2 2 2 1 ( sin30 ) 1 ( ) 0 J (4.0 m sin30 ) 0 J 2 2 mv mgy k s mv mgy k mv mg s k s mg + + . = + + + -. + . = + + 1 ( )2 ( sin30 ) 1 2 sin30 (4.0 m) 0 2 2 k .s - mg .s + mv - mg = To find the compression where v is maximum, take the derivative of this equation with respect to .s: 1 2( ) ( sin30 ) 1 2 0 0 2 2 k s mg m v dv d s . - + - = . Since dv 0 at the maximum, we have d s = . .s = (mg sin30)/k = (10 kg)(9.8 m/s2 )(0.5)/(250 N/m) =19.6 cm 10.72. Model: Assume an ideal, massless spring that obeys Hookes law. Let us also assume that the cannon (C) fires balls (B) horizontally and that the spring is directly behind the cannon to absorb all motion. Visualize: The before-and-after pictorial representation is shown, with the origin of the coordinate system located at the springs free end when the spring is neither compressed nor stretched. This free end of the spring is just behind the cannon. Solve: The momentum conservation equation fx ix p = p is B f B C f C B i B C i C ( ) ( ) ( ) ( ) x x x x m v + m v = m v + m v Since the initial momentum is zero, C f B f C f C f C B ( ) ( ) 200 kg ( ) 20( ) 10 kg x x x x v m v v v m . . = - = -. . = - . . The mechanical energy conservation equation for the cannon + spring f sf i si K +U = K +U is 2 2 2 2 2 f C i C f C f C 1 ( ) 1 ( ) 1 ( ) 0 J 0 J 1 ( ) 1 ( ) 2 2 2 2 2 ( ) (20,000 N/m) (0.50 m) 5.0 m/s 200 kg x x mv k x mv k x mv v k x m + . = + . + . = . = . = = To make this velocity physically correct, we retain the minus sign with f C ( ) . x v Substituting into the momentum conservation equation yields: f B ( ) 20( 5.0m/s) 100m/s x v = - - = 10.73. Model: This is a collision between two objects, and momentum is conserved in the collision. In addition, because the interaction force is a spring force and the surface is frictionless, energy is also conserved. Visualize: Let part 1 refer to the time before the collision starts, part 2 refer to the instant when the spring is at maximum compression, and part 3 refer to the time after the collision. Notice that just for an instant, when the spring is at maximum compression, the two blocks are moving side by side and have equal velocities: vA2 = vB2 = v2. This is an important observation. Solve: First relate part 1 to part 2. Conservation of energy is 2 2 2 2 2 2 A A1 A A2 B B2 max A B 2 max 1 1 1 1 ( ) 1 ( ) 1 ( ) 2 2 2 2 2 2 m v = m v + m v + k .x = m + m v + k .x where max .x is the springs compression. g U is not in the equation because there are no elevation changes. Also note that 2 K is the sum of the kinetic energies of all moving objects. Both 2 v and max .x are unknowns. Now add the conservation of momentum: A A1 A A1 A A2 B B2 A B 2 2 A B m v m v m v (m m )v v m v 2.667 m/s m m = + = + . = = + Substitute this result for 2 v into the energy equation to find: 2 2 2 max A A1 A B 2 2 2 A A1 A B 2 max 1 ( ) 1 1 ( ) 2 2 2 ( ) 0.046 m 4.6 cm k x mv m m v x m v m m v k . = - + - + . . = = = Notice how both conservation laws were needed to solve this problem. (b) Again, both energy and momentum are conserved between before and after. Energy is 2 2 2 2 2 A A1 A A3 B B3 A3 B3 1 1 1 16 1 2 2 2 2 m v = m v + m v . = v + v The spring is no longer compressed, so the energies are purely kinetic. Momentum is A A1 A A3 B B3 A3 B3 m v = m v + m v .8 = 2v + v We have two equations in two unknowns. From the momentum equation, we can write B3 A3 v = 2(4 - v ) and use this in the energy equation to obtain: ( )2 2 2 2 2 A3 A3 A3 A3 A3 A3 16 1 4 4 3 16 32 3 16 16 0 2 v = + - v = v - v + . v - v + = This is a quadratic equation for A3 v with roots A3 v = (4 m/s, 1.33 m/s). Using B3 A3 v = 2(4 - v ), these two roots give B3 v = (0 m/s, 5.333 m/s). The first pair of roots corresponds to a collision in which A misses B, so each keeps its initial speed. Thats not the situation here. We want the second pair of roots, from which we learn that the blocks speeds after the collision are A3 B3 v =1.33m/s and v = 5.3 m/s. 10.74. Model: Mechanical energy and momentum are conserved during the expansion of the spring. Visualize: Please refer to Figure CP10.74. Solve: Example 10.8 is a very similar problem, except that the objects are initially at rest. We can use the solution from Example 10.8 for this problem in a reference frame S' in which the two carts are initially at rest, then transform the answer to the frame S in which the carts are initially moving. Thus in the S' frame, ( ) ( ) ( ) 2 i f 2 2 2 1 1 x k x v m m m ' . = + ( ) 2 ( ) f 1 f 2 1 x x v m v m ' = - ' Let the 100 g cart be Cart 1 and the 300 g cart be Cart 2. With k =120 N/m and i .x = 4.0cm, ( ) ( ) f 2 f 1 0.40 m/s, 1.2 m/s x x v ' = v ' = - An object at rest in the S' frame is traveling to the right at 1.0 m/s in the S frame. The equation of transformation is therefore 1.0 m/s x x v = v ' + In the S frame, the velocities of the carts are ( ) f 1 1.2 m/s 1.0 m/s 0.2 m/s x v = - + = - ( ) f 2 0.40 m/s 1.0 m/s 1.4 m/s x v = + = Assess: Cart 1 is moving slowly to the left while the heavier Cart 2 is moving quickly to the right. 10.75. Model: Model the balls as particles, and assume a perfectly elastic collision. After the collision is over, the balls swing out as pendulums. The sum of the kinetic energy and gravitational potential energy does not change as the balls swing out. Visualize: In the pictorial representation we have identified before-and-after quantities for both the collision and the pendulum swing. We have chosen to place the origin of the coordinate system at a point where the two balls at rest barely touch each other. Solve: As the ball with mass m1, whose string makes an angle il . with the vertical, swings through its equilibrium position, it lowers its gravitational energy from 1 0 1 i1 m gy = m g(L - Lcos. ) to zero. This change in potential energy transforms into a change in kinetic energy. That is, 2 1 i1 111 11 i1 ( cos ) 1 ( ) ( ) 2 (1 cos )
2 m g L - L . = m v . v = gL - . Similarly, 1 2 i2 i1 i2 1 1 1 2 (v ) = 2gL(1- cos. ). Using . =. = 45, we get (v ) = 2.396 m/s = (v ) . Both balls are moving at the point where they have an elastic collision. Since our analysis of elastic collisions was for a situation in which ball 2 is initially at rest, we need to use the Galilean transformation to change to a frame S' in which ball 2 is at rest. Ball 2 is at rest in a frame that moves with ball 2, so choose S' to have V = -2.396 m/s, with the minus sign because this frame (like ball 2) is moving to the left. In this frame, ball 1 has velocity 1 1 1 1 (v') = (v ) -V = 2.396 m/s + 2.396 m/s = 4.792 m/s and ball 2 is at rest. The elastic collision causes the balls to move with velocities 1 2 2 1 1 1 1 2 1 2 2 1 1 1 2 ( ) ( ) 1 (4.792 m/s)= 1.597 m/s 3 ( ) 2 ( ) 2 (2.396 m/s)=3.194 m/s 3 v m m v m m v m v m m ' - ' = = - + ' = ' = + We can now use v = v' +V to transform these back into the laboratory frame: 2 1 2 2 ( ) 1.597 m/s 2.396 m/s 3.99 m/s ( ) 3.195 m/s 2.396 m/s 0.799 m/s v v = - - = - = - = Having determined the velocities of the two balls after collision, we will once again use the conservation equation f gf i gi K +U = K +U for each ball to solve for the f1 . and f2. . 2 2 1 3 1 1 f1 1 2 1 1 ( ) (1 cos ) 1 ( ) 0 J 2 2 m v + m gL - . = m v + Using 3 1 (v ) = 0, this equation simplifies to 2 2 f1 f1 f1 (1 cos ) 1 ( 3.99 m/s) cos 1 1 ( 3.99 m/s) 79.3 2 2 gL gL - . = - . . = - - .. = The 100 g ball rebounds to 79. Similarly, for the other ball: 2 2 2 3 2 2 f2 2 2 2 1 ( ) (1 cos ) 1 ( ) 0 J 2 2 m v + m gL - . = m v + Using 3 2 (v ) = 0, this equation becomes 2 f 2 f 2 cos 1 1 (0.799) 14.7 2gL . . . . = - . . . = . . The 200 g ball rebounds to 14.7. 10.76. Model: Model the sled as a particle. Because there is no friction, the sum of the kinetic and gravitational potential energy is conserved during motion. Visualize: Place the origin of the coordinate system at the center of the hemisphere. Then y0 = R and, from geometry, 1 y = Rcosf . Solve: The energy conservation equation 1 1 0 0 K +U = K +U is 2 2 2 1 1 0 0 1 1 1 1 1 cos 2 (1 cos ) 2 2 2 mv + mgy = mv + mgy . mv + mgR f = mgR.v = gR - f (b) If the sled is on the hill, it is moving in a circle and the r-component of net F .. has to point to the center with magnitude 2 net F = mv /R. Eventually the speed gets so large that there is not enough force to keep it in a circular trajectory, and that is the point where it flies off the hill. Consider the sled at angle f . Establish an r-axis pointing toward the center of the circle, as we usually do for circular motion problems. Newtons second law along this axis requires: 2 net G 2 2 ( ) cos cos cos cos r r F F n mg n ma mv R n mg mv m g v R R f f f f = - = - = = . . . = - = . - . . . The normal force decreases as v increases. But n cant be negative, so the fastest speed at which the sled stays on the hill is the speed max v that makes n.0. We can see that max v = gRcosf . (c) We now know the sleds speed at angle f , and we know the maximum speed it can have while remaining on the hill. The angle at which v reaches max v is the angle max f at which the sled will fly off the hill. Combining the two expressions for 1 v and max v gives: max max 1 max max 2 (1 cos ) cos 2 (1 cos ) cos cos 2 cos 2 48 3 3 gR f gR f R f R f f f - - = . - = . = . = . . = . . . . 11.1. Visualize: Please refer to Figure EX11.1. Solve: (a) A B = ABcosa = (4)(5)cos40 =15.3. .. .. (b) C D = CDcosa = (2)(4)cos120 = -4.0. .. .. (c) E F = EF cosa = (3)(4)cos90 = 0. .. .. 11.2. Visualize: Please refer to Figure EX11.2. Solve: (a) A B = ABcosa = (3)(4)cos110 = -4.1. .. .. (b) C D = CDcosa = (4)(5)cos180 = -20. .. .. (c) E F = EF cosa = (4)(3)cos30 =10.4. .. .. 11.3. Solve: (a) (3 A B = AxBx + AyBy = )(-2) + (-4)(6) = -30. .. .. (b) (2)(6) (3)( 4) 0. x x y y A B = A B + A B = + - = .. .. 11.4. Solve: (a) A B = AxBx + AyBy = (4)(-3) + (-2)(-2) = -16. .. .. (b) ( 4)( 1) (2)( 2) 0. x x y y A B = A B + A B = - - + - = .. .. 11.5. Solve: (a) The length of A .. is ( )2 ( )2 A = A = A A = 3 + -4 = 25 = 5. .. .. .. The length of B .. is ( )2 ( )2 B = -2 + 6 = = 40 = 2 10.Using the answer A B = -30 .. .. from Ex11.3(a), ( )( ) 1 ( ) cos 30 5 2 10 cos cos 0.9487 162 A B AB a a a - = . - = . = - = .. .. (b) From EX11.3 (b), A B = 0. .. .. Thus cos 0 90 a a = = Assess: In part (a) the vectors are nearly pointing in opposite directions, while in part (b) the vectors are perpendicular. 11.6. Solve: (a) The length of A .. is ( )2 ( )2 A = A = A A = 4 + 2 = 20. .. .. .. The length of B .. is ( )2 ( )2 B = -3 + -2 = 13. Using the answer A B = -16 .. .. from EX11.4(a), ( )( ) 1 ( ) cos 16 20 13 cos cos 0.9923 173 A B AB a a a - = . - = . = - = .. .. (b) From EX11.4(b), A B = 0. .. .. Thus cos 0 90 a a = = Assess: In part (a) the vectors are nearly pointing in opposite directions, while in part (b) the vectors are perpendicular. 11.7. Solve: (a) W = F .r = (6.0i - 3.0 j) (2.0i) Nm = (12.0i i - 3.0 j i) J =12.0 J. .. .. (b) W = F .r = (6.0i - 3.0 j) (2.0 j) Nm = (12.0i j - 6.0 j j) J = -6.0 J. .. .. 11.8. Solve: (a) W = F .r = (-4.0i - 6.0 j) N (3.0i) m = (-12.0i i +12.0 j i) J= -12.0 J. .. .. .. (b) W = F .r = (-4.0i - 6.0 j) N (-3.0i + 2.0 j) m = (12.0 -12.0) J=0 J. .. .. .. 11.9. Model: Use the work-kinetic energy theorem to find the net work done on the particle. Visualize: Solve: From the work-kinetic energy theorem, 2 2 ( 2 2 ) 2 2 1 0 1 0 1 1 1 1( 0.020 kg)[(30 m/s) ( 30 m/s) ] 0 J 2 2 2 2 W = .K = mv - mv = m v - v = - - = Assess: Negative work is done in slowing down the particle to rest, and an equal amount of positive work is done in bringing the particle to the original speed but in the opposite direction. 11.10. Model: Work done by a force F .. on a particle is defined as W = F .r, .. .. .. where r ... is the particles displacement. Visualize: Solve: (a) The work done by gravity is 2 g G W = F .r = (-mgj) N (2.25 - 0.75) j m = -(2.0 kg)(9.8 m/s )(1.50 m) J = -29 J .. .. (b) The work done by hand is H hand on book W = F .r. .. .. As long as the book does not accelerate, hand on book earth on book 2 H ( ) ( ) (2.25 0.75) m (2.0 kg)(9.8 m/s )(1.50 m)=29 J F F mgj mgj W mgj j = - = - - = . = - = .. .. 11.11. Model: Model the piano as a particle and use W = F .r, .. .. where W is the work done by the force F .. through the displacement .r... Visualize: Solve: For the force GF : .. 4 G ( )( )cos0 (2500 N)(5.00 m)(1) 1.250 10 J g W = F .r = F .r = F .r = = .. .. .. .. For the tension 1 T : .. 3 1 1W = T .r = (T )(.r)cos(150) = (1830 N)(5.00 m)(-0.8660) = -7.9210 J .. .. For the tension 2 T : .. 3 2 2W = T .r = (T )(.r)cos(135) = (1295 N)(5.00 m)(-0.7071) = -4.5810 J .. .. Assess: Note that the displacement .r.. in all the above cases is directed downwards along - j. 11.12. Model: Model the crate as a particle and use W = F .r, .. .. where W is the work done by a force F .. on a particle and .r.. is the particles displacement. Visualize: Solve: For the force k f : .. k kW = f .r = f (.r)cos(180) = (650 N)(3.0 m)(-1) = -1.95 kJ .. .. For the tension 1 T : .. 1 1W = T .r = (T )(.r)cos20 = (600 N)(3.0 m)(0.9397) =1.69 kJ .. .. For the tension 2T : .. 2 2W = T .r = (T )(.r)cos30 = (410 N)(3.0 m)(0.866) =1.07 kJ .. .. Assess: Negative work done by the force of kinetic friction k ( f ) .. means that 1.95 kJ of energy has been transferred out of the crate. 11.13. Model: Model the 2.0 kg object as a particle, and use the work-kinetic energy theorem. Visualize: Please refer to Figure EX11.13. For each of the five intervals the velocity-versus-time graph gives the initial and final velocities. The mass of the object is 2.0 kg. Solve: According to the work-kinetic energy theorem: 2 2 ( 2 2 ) f i f i 2 2 i f 2 2 i f 1 1 1 2 2 2 Interval AB: 2 m/s, 2 m/s 1 (2.0 kg)[( 2 m/s) (2 m/s) ] 0 J 2 Interval BC: 2 m/s, 2 m/s 1 (2.0 kg)[( 2 m/s) ( 2 m/s) ] 0 J 2 Inte W K mv mv mv v v v W v v W = . = = = - = =- . = - - = = - = - . = - - - = 2 2 i f 2 2 i f 2 2 i f rval CD: 2 m/s, 0 m/s 1 (2.0 kg)[(0 m/s) ( 2 m/s) ] 4.0 J 2 Interval DE: 0 m/s, 2 m/s 1 (2.0 kg)[(2 m/s) (0 m/s) ] 4.0 J 2 Interval EF: 2 m/s, 1m/s 1 (2.0 kg)[(1 m/s) (2 m/s) ] 3.0 J 2 v v W v v W v v W = - = . = - - = - = = . = - =+ = = . = - =- Assess: The work done is zero in intervals AB and BC. In the interval CD + DE the total work done is zero. It is not whether v is positive or negative that counts because K . v2. What is important is the magnitude of v and how v changes. 11.14. Model: Use the definition of work. Visualize: Please refer to Figure EX11.14. Solve: Work is defined as the area under the force-versus-position graph: f i area under the force curve s s s W = . F ds = Interval 01 m: (4.0 N)(1.0 m 0.0 m) 4.0 J Interval 12 m: (4.0 N)(0.5 m) ( 4.0 N)(0.5 m) 0 J Interval 23 m: 1 ( 4.0 N)(1 m) 2.0 J 2 W W W = - = = +- = = - =- 11.15. Model: Use the work-kinetic energy theorem to find velocities. Visualize: Please refer to Figure EX11.15. Solve: The work-kinetic energy theorem is f i f x 2 2 f i i f x 2 2 2 f f 0 m 1 1 area under the force curve from to 2 2 1 1 (0.500 kg)(2.0m/s) 1 1.0 J area from 0 to 2 2 2 x x x K mv mv W F dx x x mv mv F dx x . = - = = = . - = - = = . . 2 f f 2 f f 2 f f At =1 m: 1 (0.500 kg) 1.0 J 12.5 J 7.35 m/s 2 At 2 m: 1 (0.500 kg) 1.0 J 20 J 9.17 m/s 2 At 3 m: 1 (0.500 kg) 1.0 J 22.5 J 9.70 m/s 2 x v v x v v x v v - = . = = - = . = = - = . = 11.16. Model: Use the work-kinetic energy theorem. Visualize: Please refer to Figure EX11.16. Solve: The work-kinetc energy theorem is f i f x 2 2 f i i f x 2 2 2 f f 0 m 1 1 area under the force curve from to 2 2 1 1 (2.0 kg)(4.0 m/s) 1 16.0 J area from 0 to 2 2 2 x x x K mv mv W F dx x x mv mv F dx x . = - = = = . - = - = = . . 2 f f 2 f f At 2 m: 1 (2.0 kg) 16.0 J 1 (10 N)(2 m) 10 J 5.1 m/s 2 2 At 4 m: 1 (2.0 kg) 16.0 J 0 J 4.0 m/s 2 x v v x v v = - = = . = = - = . = 11.17. Model: Use the work-kinetic energy theorem. Visualize: Please refer to Figure EX11.17. Solve: The work-kinetic energy theorem is f i i f 2 2 f i max area of the -versus- graph between and 1 1 1( )(2 m) 2 2 2 x x x x K W Fdx F x x x mv mv F . = = = - = . Using m = 0.500 kg, vf = 6.0 m/s, and i v = 2.0 m/s, the above equation yields max F = 8.0 N. Assess: Problems in which the force is not a constant can not be solved using constant-acceleration kinematic equations. 11.18. Model: Use the definition Fs = -dU /ds. Visualize: Please refer to Figure EX11.18. Solve: x F is the negative of the slope of the potential energy graph at position x. Between x = 0 cm and x =10 cm the slope is slope = f i f i (U -U ) / (x - x ) = (0 J -10 J) / (0.10 m- 0.0 m) = -100 N Thus, 100 N x F = at x = 5 cm. The slope between x =10 cm and x = 20 cm is zero, so 0 N x F = at x =15 cm. Between 20 cm and 40 cm, slope = (10 J - 0 J) / (0.40 m - 0.20 m) = 50 N At x = 25 cm and x = 35 cm, therefore, 50 N. x F = - 11.19. Model: Use the definition Fs = -dU ds. Visualize: Please refer to Figure EX11.19. Solve: x F is the negative of the slope of the potential energy graph at position x. x F dU dx = -. . . . . . Between x = 0 m and x = 3 m, the slope is f i f i slope = (U -U ) / (x - x ) = (60 J - 0 J) / (3 m- 0 m) = 20 N Thus, 20 N x F = - at x =1 m. Between x = 3 m and x = 5 m, the slope is slope ( ) ( ) ( ) ( ) f i f i = U -U / x - x = 0 J - 60 J / 5 m- 3 m = -30 N Thus, 30 N at 4 m. xF = x = 11.20. Model: Use the negative derivative of the potential energy to determine the force acting on a particle. Solve: (a) The graph of the potential energy U = 4y3 is shown. (b) The y-component of the force is (4 3) 12 2 y F dU d y y dy dy = - = - = - At y = 0 m, 0 N; y F = at y =1 m, 12 N; y F = - and at y = 2 m, 48 N. y F = - 11.21. Model: Use the negative derivative of the potential energy to determine the force acting on a particle. Solve: (a) The graph of the potential energy U =10/ x is shown. (b) The x-component of the force is 2 2 m 2 5 m 2 8 m 2 2m 5m 8m 10 10 10 2.5 N 10 0.40 N 10 0.156 N x x x x x x x F dU d dx dx x x F F F x x x = = = = = = = - = - . . = . . . . = = = = = = 11.22. Model: Assume the carbon-carbon bond acts like an ideal spring that obeys Hookes law. Visualize: The quantity (x - xe ) is the stretching relative to the springs equilibrium length. In the present case, bond stretching is analogous to spring stretching. Solve: (a) The kinetic energy of the carbon atom is 1 2 1 (2.0 10 26 kg)(500 m/s)2 2.5 10 21 J 2 2 K = mv = - = - (b) The energy of the spring is given by 2 s e 21 2 92 e 1 ( ) 2 2 2(2.5 10 J) 2.0 N ( ) (0.050 10 m) m U kx x K k K x x - - = - = . = = = - 11.23. Visualize: One mole of helium atoms in the gas phase contains 23 NA = 6.0210 atoms. Solve: If each atom moves with the same speed v, the microscopic total kinetic energy will be 2 micro micro A 27 23 A 1 3700 J 2 2(3700 J) 1360 m/s 2 (6.68 10 kg)(6.02 10 ) K N mv v K mN - = . . = . = = = . . . . 11.24. Solve: (a) The car has an initial kinetic energy Ki . That energy is transformed into thermal energy of the car and the road surface. The gravitational potential energy does not change and no work is done by external forces, so during the skid i K transforms entirely into thermal energy th E . This energy transfer and transformation is shown on the energy bar chart. Note that 2 2 5 i 1 1(1500 kg)(20 m/s) 3.0 10 J 2 2 K = mv = = (b) The change in the thermal energy of the car and the road surface is 3.0105 J. 11.25. Visualize: Solve: (a) 2 2 2 i 0 0 i g0 0 2 2 ext f 1 1 f g1 1 1 0 J (20 kg)(9.8 m/s )(3.0 m) 5.9 10 J 2 0 J 1 1 (20 kg)(2.0 m/s) 40 J 0 J 2 2 K K mv U U mgy W K K mv U U mgy = = = = = = = = = = = = = = = At the top of the slide, the child has gravitational potential energy of 5.9102 J. This energy is transformed into thermal energy of the childs pants and the slide and the kinetic energy of the child. This energy transfer and transformation is shown on the energy bar chart. (b) The change in the thermal energy of the slide and of the childs pants is 5.9102 J - 40 J = 5.5102 J. 11.26. Visualize: The system loses 400 J of potential energy. In the process of losing this energy, it does 400 J of work on the environment, which means Wext = -400 J. Since the thermal energy increases 100 J, we have th .E =100 J, which must have been 100 J of kinetic energy originally. This is shown in the energy bar chart. 11.27. Visualize: Note that the conservation of energy equation Ki +Ui +Wext = Kf +Uf + .Eth requires that ext W be equal to +400 J. 11.28. Solve: Please refer to Figure EX11.28. The energy conservation equation yields i i ext f f K +U +W = K +U + .Eth .4 J +1 J +Wext =1 J + 2 J +1 J. Wext = -1 J Thus, the work done to the environment is -1 J. In other words, 1 J of energy is transferred from the system into the environment. This is shown in the energy bar chart. 11.29. Visualize: The tension of 20.0 N in the cable is an external force that does work on the block ext W = (20.0 N)(2.00 m) = 40.0 J, increasing the gravitational potential energy of the block. We placed the origin of our coordinate system on the initial resting position of the block, so we have i U = 0 J and f f U = mgy = (1.02 kg)(9.8 m/s2 )(2.00 m) = 20.0 J. Also, i K = 0 J, and th .E = 0 J. The energy bar chart shows the energy transfers and transformations. Solve: The conservation of energy equation is 2 i i ext f f th f f 0 J 0 J 40.0 J 1 20.0 J 0 J 2 (20.0 J)(2) /1.02 kg 6.26 m/s K U W K U E mv v + + = + +. . + + = + + . = = 11.30. Model: Model the elevator as a particle, and apply the conservation of energy equation. Solve: The tension in the cable does work on the elevator to lift it. Because the cable is pulled by the motor, we say that the motor does the work of lifting the elevator. (a) The energy conservation equation is i i ext f f th . K +U +W = K +U + .E Using i f K = 0 J, K = 0 J, and th .E = 0 J gives 2 5 ext f i f i W = (U -U ) = mg( y - y ) = (1000 kg)(9.8 m/s )(100 m) = 9.8010 J (b) The power required to give the elevator this much energy in a time of 50 s is 5 ext 4 9.8 10 J 1.96 10 W 50 s P W t = = = . Assess: Since 1 horsepower (hp) is 746 W, the power of the motor is 26 hp. This is a reasonable amount of power to lift a mass of 1000 kg to a height of 100 m in 50 s. 11.31. Model: Model the steel block as a particle subject to the force of kinetic friction and use the energy conservation equation. Visualize: Solve: (a) The work done on the block is Wnet = Fnet .r .. .. where r ... is the displacement. We will find the displacement using kinematic equations and the force using Newtons second law of motion. The displacement in the x-direction is 2 1 0 0 1 0 1 0 ( ) 1 ( ) 0 m (1.0 m/s)(3.0 s 0 s) 0 m 3.0 m 2 x x .x = x = x + v t - t + a t - t = + - + = Thus .r.. = 3.0i m. The equations for Newtons second law along the x and y components are ( ) ( ) 2 net G G net k k k net net 0 N (10 kg)(9.8 m/s ) 98.0 N 0 N (0.6)(98.0 N) 58.8 N cos0 (58.8 N)(3.0 m)(1) 176 J y x F n F n F mg F F f F f n W F r F x = - = . = = = = = - = . = = = = . = . = . = = .. .. .. .. (b) The power required to do this much work in 3.0 s is 176 J 59 W 3.0 s P W t = = = 11.32. Solve: The power of the solar collector is the solar energy collected divided by time. The intensity of the solar energy striking the earth is the power divided by area. We have 6 2 2 2 150 10 J 41,667 W and intensity 1000W 3600 s m Area of solar collector 41,667 W 41.7 m 1000 W/m P E t . = = = = . . = = 11.33. Solve: The night light consumes more energy than the hair dryer. The calculations are 3 5 5 1.2 kW 10 min 1.2 10 10 60 J 7.2 10 J 10 W 24 hours 10 24 60 60 J 8.64 10 J = = = = 11.34. Solve: (a) A kilowatt hour is a kilowatt multiplied by 3600 seconds. It has the dimensions of energy. (b) One kilowatt hour is energy 1 kwh = (1000 J/s)(3600 s) = 3.6106 J Thus 6 500 kwh (500 kwh) 3.6 10 J 1.8 109 J 1 kwh . . = . . = . . 11.35. Model: Model the sprinter as a particle, and use the constant-acceleration kinematic equations and the definition of power in terms of velocity. Visualize: Solve: (a) We can find the acceleration from the kinematic equations and the horizontal force from Newtons second law. We have 2 2 2 0 0 1 0 1 0 2 ( ) 1 ( ) 50 m 0 m 0 m 1 (7.0 s 0 s) 2.04 m/s 2 2 (50 kg)(2.04 m/s ) 102 N x x x x x x x x v t t a t t a a F ma = + - + - . = + + - . = . = = = (b) We obtain the sprinters power output by using P = F v, .. .. where v.. is the sprinters velocity. At 2.0 s t = the power is 2 0 0 ( )[ ( )] (102 N)[0 m/s (2.04 m/s )(2.0 s 0 s)] 0.42 kW x x x P = F v + a t - t = + - = The power at t = 4.0 s is 0.83 kW, and at t = 6.0 s the power is 1.25 kW. 11.36. Model: Use the definition of work for a constant force F, .. W = F .s, .. .. where s ... is the displacement. Visualize: Please refer to Figure P11.36. The force F = (6i + 8 j) N .. on the particle is a constant. Solve: (a) ABD AB BD AB BD ( ) ( ) (6 8 ) N (3) m (6 8 ) N (4 ) m 18 J 32 J 50 J W W W F s F s i j i i j j = + = . + . = + + + = + = .. .. .. .. (b) ACD AC CD AC CD ( ) ( ) (6 8 ) N (4 ) m (6 8 ) N (3 ) m 32 J 18 J 50 J W W W F s F s i j j i j j = + = . + . = + + + = + = .. .. .. .. (c) AD AD W = F (.s) = (6i + 8 j) N (3i + 4 j) m =18 J + 32 J = 50 J .. .. The force is conservative because the work done is independent of the path. 11.37. Model: The force is conservative, so it has a potential energy. Visualize: Please refer to Figure P11.37 for the graph of the force. Solve: The definition of potential energy is .U = -W(i.f ). In addition, work is the area under the force-versusdisplacement graph. Thus f i .U =U -U = - (area under the force curve). Since i U = 0 at x = 0 m, the potential energy at position x is U(x) = - (area under the force curve from 0 to x). From 0 m to 3 m, the area increases linearly from 0 N m to -60 N m, so U increases from 0 J to 60 J. At x = 4 m, the area is -70 J. Thus U = 70 J at x = 4 m, and U doesnt change after that since the force is then zero. Between 3 m and 4 m, where F changes linearly, U must have a quadratic dependence on x (i.e., the potential energy curve is a parabola). This information is shown on the potential energy graph below. (b) Mechanical energy is E = K +U. From the graph, U = 20 J at x =1.0 m. The kinetic energy is 1 2 1 2 2 2K = mv = (0.100 kg) (25 m/s) = 31.25 J. Thus E = 51.25 J. (c) The total energy line at 51.25 J is shown on the graph above. (d) The turning point occurs where the total energy line crosses the potential energy curve. We can see from the graph that this is at approximately 2.5 m. For a more accurate value, the potential energy function is U = 20x J. The TE line crosses at the point where 20x = 51.25, which is x = 2.56 m. 11.38. Model: Use the relationship between force and potential energy and the work-kinetic energy theorem. Visualize: Please refer to Figure P11.38. We will find the slope in the following x regions: 0 cm < x <1 cm, 1< x < 3 cm, 3 < x < 5 cm, 5 < x < 7 cm, and 7 < x < 8 cm. Solve: (a) Fx is the negative slope of the U-versus-x graph, for example, for 0 m < x < 2 m 4 J 400 N 400 N 0.01 m x dU F dx - = =- . =+ Calculating the values of Fx in this way, we can draw the force-versus-position graph as shown. (b) Since f i x x x W = . F dx = area of the Fx-versus-x graph between xi and xf, the work done by the force as the particle moves from i x = 2 cm to f x = 6 cm is -2 J. (c) The conservation of energy equation is f f i i . K +U = K +U We can see from the graph that i U = 0 J and f U = 2 J in moving from x = 2 cm to x = 6 cm. The final speed is f v =10 m/s, so 1 2 1 2 2 2 i i 2 J + (0.010 kg)(10.0 m/s) = 0 J + (0.010 kg)v . v = 22.4 m/s 11.39. Model: Use the relationship between a conservative force and potential energy. Visualize: Please refer to Figure P11.39. We will obtain U as a function of x and x F as a function of x by using the calculus techniques of integration and differentiation. Solve: (a) For the interval 0 m < x < 0.5 m, Fx = (4x) N, where x is in meters. This means 2 2 1 4 2 2 x dU F x U x C x dx = - = - . = - + = - where we have used U = 0 J at x = 0 m to obtain 1 C = 0. For the interval 0.5 m < x <1 m, ( 4 4) N. xF = - x + Likewise, 2 2 dU 4x 4 U 2x 4x C dx = - . = - + Since U should be continuous at the junction, we have the continuity condition 2 2 0.5 m 2 0.5 m 2 2 ( 2 ) (2 4 ) 0.5 0.5 2 1 x x x x xC C C = = - = - + .- = - + . = U remains constant for x =1 m. (b) For the interval 0 m < x < 0.5 m, U = +4x, and for the interval 0.5 m < x <1.0 m, U = -4x + 4, where x is in meters. The derivatives give 4 N and 4 N, x x F = - F = + respectively. The slope is zero for x =1 m. 11.40. Model: Use x , x a dv dt = , x x = . v dt 1 2 2 , x K = mv and . x F = ma Visualize: Please refer to Figure P11.40. We know slope x a = of the vx-versus-t graph and x = area under the vx-versus-x graph between 0 and x. Solve: Using the above definitions and methodology, we can generate the following table: t(s) ax (m/s2) x(m) K(J) F(N) 0 0.5 1.0 10 10 10 0 1.25 5 0 6.25 25 5 5 5 1.5 2.0 2.5 10 +10 or -10 -10 11.25 20 28.75 56.25 100 56.25 5 -5 or +5 -5 3.0 3.5 4.0 -10 or 0 0 0 35 40 45 25 25 25 -5 or 0 0 0 (e) Let J1 be the impulse from t = 0 s to t = 2 s and 2 J be the impulse from t = 2 s to t = 4 s. We have 2 s 4 s 1 2 0 s 0 s (5 N)(2 s) 10 N s and ( 5 N)(1 s) 5 N s x x J = . F dt = = J = . F dt = - = - (f) f i f i / At 2 s, 0 m/s (10 N s) / 0.5 kg 0 m/s 20 m/s 20 m/s At 4 s, 20 m/s ( 5 N s) / 0.5 kg 20 m/s 10 m/s 10 m/s x x J p mv mv v v J m t v t v = . = - . = + = = + = + = = = + - = - = The vx-versus-t graph also gives 20 m/s x v = at t = 2 s and 10 m/s x v = at t = 4 s. (g) (h) From 0 s to 2 s, (5 N)(20 m) 100 J From 2 s to 4 s, ( 5 N)(15 m) 75 J x x t t W Fdx t t W Fdx = = = = = = = = = - =- . . (i) At t = 0 s, 0 m/s x v = so the work-kinetic energy theorem for calculating x v at t = 2 s is 2 2 2 2 f i 1 1 100 J 1 (0.5 kg) 1 (0.5 kg)(0 m/s) 20 m/s 2 2 2 2x x W = .K = mv - mv . = v - .v = To calculate x v at t = 4 s, we use x v at t = 2 s as the initial velocity: 75 J 1 (0.5 kg) 2 1 (0.5 kg)(20 m/s)2 10 m/s 2 2x x - = v - .v = Both of these values agree with the values on the velocity graph. 11.41. Model: Model the elevator as a particle. Visualize: Solve: (a) The work done by gravity on the elevator is 2 4 g g 0 1 1 0 W = -.U = mgy - mgy = -mg( y - y ) = -(1000 kg)(9.8 m/s )(10 m) = -9.810 J (b) The work done by the tension in the cable on the elevator is 1 0 ( )cos0 ( ) (10 m) T W = T .y = T y - y = T To find T we write Newtons second law for the elevator: 2 2 G G 4 4 5 ( ) (1000 kg)(9.8 m/s 1.0 m/s ) 1.08 10 N (1.08 10 N)(10 m) 1.08 10 J y y y y T F T F ma T F ma mg a W = - = . = + = + = + = . = = S (c) The work-kinetic energy theorem is 2 2 net g f i f 0 f g 0 4 5 2 4 f 1 1 2 2 ( 9.8 10 J) (1.08 10 J) 1 (1000 kg)(0 m/s) 1.0 10 J 2 T T W W W K K K K mv K W W mv K = + =. = - = - . = + + . = - + + = (d) 2 4 2 f f f 1 1.0 10 J 1 (1000 kg) 4.5 m/s 2 2 f K = mv . = v .v = 11.42. Model: Model the rock as a particle, and apply the work-kinetic energy theorem. Visualize: Solve: (a) The work done by Bob on the rock is 2 2 2 2 2 Bob 1 0 1 1 1 1 1(0.500 kg)(30 m / s) 225 J = 2.3 10 J 2 2 2 2 W = .K = mv - mv = mv = = (b) For a constant force, 2 Bob Bob Bob BobW = F .x. F =W /.x = 2.310 N. (c) Bobs power output is Bob Bob rock P = F v and will be a maximum when the rock has maximum speed. This is just as he releases the rock with rock 1 v = v = 30 m/s. Thus, ( )( ) max Bob 1 P = F v = 225 J 30 m/s = 6750W = 6.8 kW. 11.43. Model: Model the crate as a particle, and use the work-kinetic energy theorem. Visualize: Solve: (a) The work-kinetic energy theorem is 1 2 1 2 1 2 2 1 2 0 2 1 total.K = mv - mv = mv =W . Three forces act on the box, so total grav n push W =W +W +W . The normal force is perpendicular to the motion, so n W = 0 J. The other two forces do the following amount of work: ( ) push push push grav G G cos20 137.4 J ( sin 20 ) 98.0 J x W = F .r = F .x = W = F .r = F . x = -mg . x = - .. .. .. .. Thus, total W = 39.4 J, leading to a speed at the top of the ramp equal to total 1 2 2(39.4 J) 4.0 m/s 5.0 kg v W m = = = (b) The x-component of Newtons second law is net push G push 2 ( ) cos20 sin 20 cos20 sin 20 x 1.35 m/s x F F F F mg a a m m m - - = = = = = Constant-acceleration kinematics with 1 x = h / sin 20 = 5.85 m gives the final speed 2 2 2 1 0 1 0 1 1 1 v = v + 2a(x - x ) = 2ax .v = 2ax = 2(1.35 m/ s )(5.85 m) = 4.0 m/s 11.44. Model: Model Sam strapped with skis as a particle, and apply the law of conservation of energy. Visualize: Solve: (a) The conservation of energy equation is K1 +Ug1 + .Eth = K0 +Ug0 +Wext The snow is frictionless, so th .E = 0 J. However, the wind is an external force doing work on Sam as he moves down the hill. Thus, ext wind 1 g1 0 g0 2 2 2 2 1 1 0 0 1 0 1 0 wind 1 0 ( )( ) 1 1 1 0 J (0 J ) 1 2 2 2 2 2 2 W W K U K U mv mgy mv mgy mv mgy mv mgy v gy W m = = + - + = . + . - . + . = . + . - + = - . . . . . . . . . . . . . = + We compute the work done by the wind as follows: wind wind wind W = F .r = F .r cos160 = (200 N)(146 m)cos160 = -27,400 J .. .. where we have used .r = h / sin 20 =146 m. Now we can compute 2 1 2(9.8 m/ s )(50 m) 2( 27,400 J) 15.7 m/ s 75 kg v - = + = (b) We will use a tilted coordinate system, with the x-axis parallel to the slope. Newtons second law for Sam is net G wind wind 2 2 ( ) sin 20 cos20 sin 20 cos20 (75 kg)(9.8 m/s )sin 20 (200 N)cos20 0.846 m/s 75 kg x x a a F F F mg F m m m - - = = = = - = = Now we can use constant-acceleration kinematics as follows: 2 2 2 1 0 1 0 1 1 1 v = v + 2a(x - x ) = 2ax .v = 2ax = 2(0.846 m/s )(146 m) =15.7 m/s Assess: We used a vertical y-axis for energy analysis, rather than a tilted coordinate system, because g U is determined by its vertical position. 11.45. Model: Model Paul and the mat as a particle, assume the mat to be massless, use the model of kinetic friction, and apply the work-kinetic energy theorem. Visualize: We define the x-axis along the floor and the y-axis perpendicular to the floor. Solve: We need to first determine fk . Newtons second law in the y-direction is G n + T sin30 = F = mg .n = mg -T sin30 = (10 kg)(9.8 m/s2 ) - (30 N)(sin30) = 83.0 N. Using n and the model of kinetic friction, k k f = n = (0.2)(83.0 N) =16.60 N. The net force on Paul and the mat is therefore ( ) net k F = T cos30 - f = 30 N cos30 -16.6 N = 9.4 N . Thus, net net W = F .r = (9.4 N)(3.0 m) = 28 J The other forces G n and F .. .. make an angle of 90 with .r.. and do zero work. We can now use the work-kinetic energy theorem to find the final velocity as follows: 2 net f i f f f net 0 J 1 2 / 2(28 J)/10 kg 2.4 m/s 2 W = K - K = K - = mv .v = W m = = Assess: A speed of 2.4 m/s or 5.4 mph is reasonable for the present problem. 11.46. Model: Assume an ideal spring that obeys Hookes law. Model the box as a particle and use the model of kinetic friction. Visualize: Solve: When the horizontal surface is frictionless, conservation of energy means 2 2 2 0 e 1 1 1 1 ( ) 1 1 (100 N/m)(0.20 m 0 m) 2.0 J 2 2 2 x k x - x = mv = K . K = - = That is, the box is launched with 2.0 J of kinetic energy. It will lose 2.0 J of kinetic energy on the rough surface. The net force on the box is net . k k F = - f = - mgi .. .. The work-kinetic energy theorem is net net 2 1 2 1 2 1 2 k 0 J 2.0 J 2.0 J ( )( ) 2.0 J ( ) 2.0 J 2.0 J 54 cm ( )( ) (0.15)(2.5 kg)(9.8 m/s ) k W F r K K mg x x x x mg = . = - = - =- . - - = - . - = = = .. .. Assess: Because the force of friction transforms kinetic energy into thermal energy, energy is transferred out of the box into the environment. In response, the box slows down and comes to rest. 11.47. Model: Model the suitcase as a particle, use the model of kinetic friction, and use the work-kinetic energy theorem. Visualize: The net force on the suitcase is Fnet = fk . .. .. Solve: The work-kinetic energy theorem is 2 2 2 2 net 1 0 net k 0 k 1 0 0 2 2 2 0 k 1 0 0 k 2 1 0 1 1 0 J 1 ( )( )cos180 1 2 2 2 2 ( ) 1 (1.2 m/s) 0.037 2 2 ( ) 2(9.8 m/s )(2.0 m 0 m) W K mv mv F r f r mv f x x mv mg x x mv v g x x = . = - . . = . = - . - = - . - - = - . = = = - - .. .. .. .. Assess: Friction transforms kinetic energy of the suitcase into thermal energy. In response, the suitcase slows down and comes to rest. 11.48. Model: Identify the truck and the loose gravel as the system. We need the gravel inside the system because friction increases the temperature of the truck and the gravel. We will also use the model of kinetic friction and the conservation of energy equation. Visualize: We place the origin of our coordinate system at the base of the ramp in such a way that the x-axis is along the ramp and the y-axis is vertical so that we can calculate potential energy. The free-body diagram of forces on the truck is shown. Solve: The conservation of energy equation is 1 g1 th 0 g0 ext . K +U + .E = K +U +W In the present case, ext W = 0 J, 1 0 m/s, x v = 0 0 J, g U = 0 35 m/s. x v = The thermal energy created by friction is th k 1 0 k 1 0 k 1 0 2 1 0 1 0 ( )( ) ( )( ) ( cos6.0)( ) (0.40)(15,000 kg)(9.8 m/s )(cos6.0 )( ) (58,478 J/m)( ) E f x x n x x mg x x xx xx . = - = - = - = -= - Thus, the energy conservation equation simplifies to 2 1 10 0 2 2 1 0 1 0 1 0 0 J (58,478 J/m)( ) 1 0 J 0 J 2 (15,000 kg)(9.8 m/s )[( )sin 6.0 ] (58,478 J/m)( ) 1 (15,000 kg)(35 m/s) 2 ( ) 124 m x mgy x x mv x x x x x x + + - = + + - + - = . - = Assess: A length of 124 m at a slope of 6 seems reasonable. 11.49. Model: We will use the spring, the package, and the ramp as the system. We will model the package as a particle. Visualize: We place the origin of our coordinate system on the end of the spring when it is compressed and is in contact with the package to be shot. Model: (a) The energy conservation equation is 1 g1 s1 th 0 g0 s0 ext 2 2 2 2 1 1 e e th 0 0 ext 1 1 ( ) 1 1 ( ) 2 2 2 2 K U U E K U U W mv mgy k x x E mv mgy k x W + + +. = + + + + + - +. = + + . + Using 1 y =1 m, th .E = 0 J (note the frictionless ramp), 0 v = 0 m/s, 0 y = 0 m, .x = 30 cm, and ext W = 0 J, we get 2 2 1 2 2 2 1 1 1 (1 m) 0 J 0 J 0 J 0 J 1 (0.30 m) 0 J 2 2 1 (2.0 kg) (2.0 kg)(9.8 m/s )(1 m) 1 (500 N/m)(0.30 m) 2 2 1.70 m/s mv mg k v v + + + = + + + + = . = (b) How high can the package go after crossing the sticky spot? If the package can reach 1 y =1.0 m before stopping 1 (v = 0), then it makes it. But if 1 y <1.0 m when 1 v = 0, it does not. The friction of the sticky spot generates thermal energy 2 th k .E = ( mg).x = (0.30)(2.0 kg)(9.8 m/s )(0.50 m) = 2.94 J The energy conservation equation is now 1 2 1 2 2 1 1 th 2 mv + mgy + .E = k(.x) If we set 1 v = 0 m/s to find the highest point the package can reach, we get (1 2 ) (1 2 ) 2 1 2 th 2 y = k(.x) - .E /mg = (500 N/m)(0.30 m) - 2.94 J /(2.0 kg)(9.8 m/s ) = 0.998 m The package does not make it. It just barely misses. 11.50. Model: Model the two blocks as particles. The two blocks make our system. Visualize: We place the origin of our coordinate system at the location of the 3.0 kg block. Solve: (a) The conservation of energy equation is f gf th i gi ext . K +U + .E = K +U +W Using th .E = 0 J and ext W = 0 J we get 2 2 2 2 2 f2 3 f3 2 f 2 i2 3 i3 2 i 1 ( ) 1 ( ) ( ) 1 ( ) 1 ( ) ( ) 2 2 2 2 m v + m v + m g y = m v + m v + m g y Noting that f 2 f 3 f (v ) = (v ) = v and i 2 i 3 (v ) = (v ) = 0 m/ s, this becomes 2 2 3 f 2 f i 2 2 i f f 2 3 1( ) ( ) 2 2 ( ) 2(2.0 kg)(9.8 m/s )(1.50 m) 3.4 m/s (2.0 kg 3.0 kg) m m v mg y y v m g y y m m + =- - - = = = + + (b) We will use the same energy conservation equation. However, this time 2 th k 3 .E = (m g)(.x) = (0.15)(3.0 kg)(9.8 m/s )(1.50 m) = 6.615 J The energy conservation equation is now 2 2 2 2 2 f 3 f 2 f 2 i 2 3 1 3 2 i 1 1 6.615 J 1 ( ) 1 ( ) 0 J 2 2 2 2 m v + m v + m gy + = m v + m v + m gy + 2 2 3 f 2 i f f 2 i f 2 3 2 1 ( ) 6.615 J ( ) 2 [ ( ) 6.615 J] 2 2 [(2.0 kg)(9.8 m/s )(1.50 m) 6.615 J] 3.0 m/s 5.0 kg m m v mg y y v mg y y m m . . + + = - . = . . - - . + . . . = . . - = . . Assess: A reduced speed when friction is present compared to when there is no friction is reasonable. 11.51. Model: Use the particle model, the definition of work W = F .s, .. .. and the model of kinetic friction. Visualize: We place the coordinate frame on the incline so that its x-axis is along the incline. Solve: (a) T W = T .r = (T)(.x)cos18 = (120 N)(5.0 m)cos18 = 0.57 kJ .. .. 2 g G W = (F .r ) = mg(.x)cos(120) = (8 kg)(9.8 m/s )(5.0 m)cos120 = -196 J .. .. n W = (n.. .r..) = (n)(.x)cos90 = 0 J (b) The amount of energy transformed into thermal energy is th k k .E = ( f )(.x) = ( n)(.x). To find n, we write Newtons second law as follows: G G 2 cos30 sin18 0 N cos30 sin18 cos30 sin18 (8.0 kg)(9.8 m/s )cos30 (120 N)sin18 30.814 N y F n F T n F T n mg T = - + = . = - . = - = - = S Thus, th .E = (0.25)(30.814 N)(5.0 m) = 38.5 J. Assess: Any force that acts perpendicular to the displacement does no work. 11.52. Model: Assume the spring is ideal so that Hookes law is obeyed, and model the weather rocket as a particle. Visualize: The origin of the coordinate system is placed on the free end of the spring. Note that the bottom of the spring is anchored to the ground. Solve: (a) The rocket is initially at rest. The free-body diagram on the rocket helps us write Newtons second law as ( ) sp G 2 0 N (10.2 kg)(9.8 m/s ) 20.0 cm (500 N/m) yF F F mg kymg y mg k = . = = . . = . . = = = S (b) The thrust does work. Using the energy conservation equation when 2 0 y - y = 40 cm = 0.40 m: 2 g2 sp2 1 g1 sp1 ext 2 2 2 2 ext 2 1 2 2 2 e 1 1 1 e 2 2 2 ( ) 1 1 ( ) 1 1 ( ) (200 N)(0.60 m) 2 2 2 2 (5.10 kg) 40.0 J 40.0 J 0 20.0 J 10.0 J 120 J 2.43 m/s K U U K U U W W F y y mv mgy k y y mv mgy k y y v v + + = + + + = - + + - = + + - + + + = - + + . = If the rocket were not attached to the spring, the energy conservation equation would not involve the spring energy term sp2U . That is, 2 g2 1 g1 sp1 ext 2 2 2 2 2 1 (10.2 kg) (10.2 kg)(9.8 m/s )(0.40 m) 0 J (10.2 kg)(9.8 m/s )(0.20 m) 2 1 (500 N/m)(0.20 m) (200 N)(0.60 m) 2 K U K U U W v + = + + + + = - + + 2 2 2 .(5.10 kg)v = 70.0 J.v = 3.70 m/s 11.53. Model: Use the particle model for the ice skater, the model of kinetic/static friction, and the workkinetic energy theorem. Visualize: Solve: (a) The work-kinetic energy theorem is 2 2 1 0 net wind 1 1 2 2 .K = mv - mv =W =W There is no kinetic friction along her direction of motion. Static friction acts to prevent her skates from slipping sideways on the ice, but this force is perpendicular to the motion and does not contribute to a change in thermal energy. The angle between wind F .. and .r.. is . =135, so wind wind wind W = F .r = F .y cos135 = (4 N)(100 m)cos135 = -282.8 J .. .. Thus, her final speed is 2 wind 1 0 v v 2W 2.16 m/s m = + = (b) If the skates dont slip, she has no acceleration in the x-direction and so net ( ) 0 N. x F = That is: s wind s wind f - F cos45 = 0 N. f = F cos45 = 2.83 N Now there is an upper limit to the static friction: s smax s f = ( f ) = mg. To not slip requires s s 2 2.83 N 0.0058 (50 kg)(9.8 m/s ) f mg = = = Thus, the minimum value of s is 0.0058. Assess: The work done by the wind on the ice skater is negative, because the wind slows the skater down. 11.54. Model: Model the ice cube as a particle, the spring as an ideal that obeys Hookes law, and the law of conservation of energy. Visualize: Solve: (a) The normal force does no work and the slope is frictionless, so mechanical energy is conserved. Weve drawn two separate axes: a vertical y-axis to measure potential energy and a tilted s-axis to measure distance along the slope. Both have the same origin which is at the point where the spring is not compressed. Thus, the two axes are related by y = ssin. . Also, this choice of origin makes the elastic potential energy simply 1 2 s 2 0 U = k(s - s ) = 1 2 2 ks . Because energy is conserved, we can relate the initial pointwith the spring compressedto the final point where the ice cube is at maximum height. We do not need to find the speed with which it leaves the spring. We have 2 g2 s2 1 g1 s1 2 2 2 2 2 2 0 1 1 1 1 1 1 1 2 2 2 2 K U U K U U mv mgy ks mv mgy ks + + = + + + + = + + It is important to note that at the final point, when the ice cube is at y2, the end of the spring is only at s0. The spring does not stretch to 2 s , so s2 U is not 1 2 2 2 ks . Three of the terms are zero, leaving 2 2 1 2 1 1 2 1 1 height gained 0.255 m 25.5 cm 2 2 mgy mgy ks y y y ks mg =+ + . - =. = = = = The distance traveled is .s = .y / sin30 = 51.0 cm. (b) Using the energy equation and the expression for thermal energy: 2 g2 s2 th 1 g1 s1 ext th k k K +U +U + .E = K +U +U +W .E = f .s = n.s From the free-body diagram, net ( ) 0 N cos30 cos30 y F = = n - mg .n = mg .. Now, having found th k.E = (mg cos30).s, the energy equation can be written 2 2 k 1 1 2 2 1 1 k 0 J 0 J ( cos30 ) 0 J 1 0 J 2 ( ) 1 cos30 0 2 mgy mg s mgy ks mg y y ks mg s + + + . = + + + . - - + . = Using .y = (.s)sin30, the above equation simplifies to 2 2 1 k 1 k sin30 cos30 1 0.379 m 37.9 cm 2 2(sin30 cos30) mg s mg s ks s ks mg . + . = .. = = = + 11.55. Model: Assume an ideal spring, so Hookes law is obeyed. Treat the box as a particle and apply the energy conservation law. Box, spring, and the ground make our system, and we also use the model of kinetic friction. Visualize: We place the origin of the coordinate system on the ground directly below the boxs starting position. Solve: (a) The energy conservation equation is 1 g1 s1 th 0 g0 s0 ext 2 2 2 1 1 0 0 1 0 2 1 0 1 0 J 0 J 1 0 J 0 J 1 0 J 0 J 2 2 2 2 2(9.8m/s )(5.0 m) 9.9 m/s K U U E K U U W mv mgy mv mgy mv mgy v gy + + +. = + + + + + + = + + + . + = + . = = = (b) The friction creates thermal energy. The energy conservation equation for this part of the problem is 2 2 2 g2 s2 th 1 g1 s1 ext 2 k 2 1 1 2 2 2 2 2 k 2 1 1 2 k 2 1 1 1 0 J 0 J ( ) 1 0 J 0 J 0 J 2 2 1 ( ) 1 1 ( ) 1 2 2 2 2 K U U E K U U W mv mg x x mv mv nx x mv mv mgx x mv + + +. = + + + + + + - = + + + + - = . + - = 2 2 2 2 1 k 2 1 .v = v - 2 g(x - x ) = (9.9 m/s) - 2(0.25)(9.8 m/s )(2.0 m) = 9.4 m/s (c) To find how much the spring is compressed, we apply the energy conservation once again: 2 2 3 g3 s3 th 2 g2 s2 ext 3 2 2 0 J 0 J 1 ( ) 0 J 1 0 J 0 J 0 J 2 2 K +U +U + .E = K +U +U +W + + k x - x + = mv + + + Using 2 v = 9.4 m/s, k = 500 N/m and m = 5.0 kg, the above equation yields 3 2 (x - x ) = .x = 94 cm. (d) The initial energy 2 0 = mgy = (5.0 kg)(9.8 m/ s )(5.0 m) = 254 J. The energy transformed to thermal energy during each passage is 2 k 2 1 mg(x - x ) = (0.25)(5.0 kg)(9.8 m/ s )(2.0 m) = 24.5 J The number of passages is equal to 245 J / 24.5 J or 10. 11.56. Model: Assume an ideal spring, so Hookes law is obeyed. Treat the physics student as a particle and apply the law of conservation of energy. Our system is comprised of the spring, the student, and the ground. We also use the model of kinetic friction. Visualize: We place the origin of the coordinate system on the ground directly below the end of the compressed spring that is in contact with the student. Solve: (a) The energy conservation equation is 1 g1 s1 th 0 g0 s0 ext 2 2 2 2 1 1 1 e 0 0 1 0 1 1 ( ) 0 J 1 1 ( ) 0 J 2 2 2 2 K U U E K U U W mv mgy k x x mv mgy k x x + + +. = + + + + + - + = + + - + Since 1 0 y = y =10 m, 1 ex = x , 0 v = 0 m/s, k = 80,000 N/m, m =100 kg, and 1 0 (x - x ) = 0.5 m, 2 2 1 1 0 1 1 0 1 1( ) ( ) 14.14 m/s 2 2 mv k x x v k x x m = - . = - = (b) Friction creates thermal energy. Applying the conservation of energy equation once again: 2 g2 s2 th 0 g0 s0 ext 2 2 2 2 k 0 1 0 1 0 J 0 J 1 ( ) 0 J 2 2 K U U E K U U W mv mgy f s mgy k x x + + +. = + + + + + + . = + + - + With 2 v = 0 m/s and 2 y = (.s)sin30, the above equation is simplified to 2 k 0 1 0 ( )sin30 1 ( ) 2 mg .s + n.s = mgy + k x - x From the free-body diagram for the physics student, we see that G n = F cos30 = mg cos30. Thus, the conservation of energy equation gives 2 k 0 10 ( sin30 cos30 ) 1 ( ) 2 .s mg + mg = mgy + k x - x Using m =100 kg, k = 80,000 N/m, 1 0 (x - x ) = 0.50 m, 0 y =10 m, and k = 0.15, we get 2 0 1 0 k 1 ( ) 2 32.1 m (sin30 cos30 ) mgy k x x s mg + - . = = + Assess: 2 y = (.s)sin30 =16.05 m, which is greater than 0 y =10 m. The higher value is due to the transformation of the spring energy into gravitational potential energy. 11.57. Model: Treat the block as a particle, use the model of kinetic friction, and apply the energy conservation law. The block and the incline comprise our system. Visualize: We place the origin of the coordinate system directly below the blocks starting position at the same level as the horizontal surface. On the horizontal surface the model of kinetic friction applies. Solve: (a) For the first incline, the conservation of energy equation gives 2 1 g1 th 0 g0 ext 1 0 1 0 1 0 J 0 J 0 J 0 J 2 2 2 K +U + .E = K +U +W mv + + = + mgy + .v = gy = gh (b) The friction creates thermal energy. Applying once again the conservation of energy equation, we have 2 2 3 g3 th 1 g1 ext 3 3 2 1 1 1 ext 1 ( ) 1 2 2 k K +U + .E = K +U +W mv + mgy + mg x - x = mv + mgy +W Using 3 v = 0 m/s, 1 y = 0 m, ext W = 0 J, 1 v = 2gh, and 2 1 (x - x ) = L, we get 3 k 3 k 1 (2 ) 2 mgy + mgL = m gh . y = h - L Assess: For k = 0, 3y = h which is predicted by the law of the conservation of energy. 11.58. Model: Assume an ideal spring, so Hookes law is obeyed. Visualize: Solve: For a conservative force the work done on a particle as it moves from an initial to a final position is independent of the path. We will show that A C B A B W W . . . = for the spring force. Work done by a spring force F = -kx is given by f i x x W = . Fdx = - . kxdx This means ( ) ( ) ( ) B C B A A C 2 2 2 2 2 2 A B B A A C C A C B B C , , and 2 2 2 x x x x x x W kxdx k x x W kx dx k x x W kx dx k x x . . . = - . = - - = - . = - - = - . = - - Adding the last two: ( 2 2 2 2 ) A C B A C C B C A B A B 2 C W W W k x x x x W . . . . . = + =- - + - = 11.59. Model: A sprong that obeys the force law 3 Fx = -q(x - xe ) , where q is the sprong constant and e x is the equilibrium position. Visualize: We place the origin of the coordinate system on the free end of the sprong, that is, e f x = x = 0 m. Solve: (a) The units of q are N/m3. (b) A cubic curve rises more steeply than a parabola. The force increases by a factor of 8 every time x increases by a factor of 2. (c) 4 3 0 Since , we have ( ) ( ) . 4 x x x F dU U x F dx qx dx qx dx = - = -. = -. - = (d) Applying the energy conservation equation to the ball and sprong system: f f i i 4 2 i f 4 3 4 f 1 0 J 0 J 2 4 (40,000 N/m ) ( 0.10 m) 10 m/s 2 (0.020 kg) 2 K U K U mv qx v q x m + = + + = + - . = = = 11.60. Solve: (a) Because sin (cx) is dimensionless, 0 F must have units of force in newtons. (b) The product cx is an angle because we are taking the sine of it. An angle has no real physical units. If x has units of m and the product cx is unitless, then c has to have units of m-1. (c) The force is a maximum when sin(cx) =1. This occurs when cx =p /2, or for max x =p /2c. (d) The graph is the first quarter of a sine curve. (e) We can find the velocity f v at f max x = x from the work-kinetic energy theorem: 2 2 2 2 2 f i f 0 f 0 1 1 1 1 2 2 2 2 2 K mv mv mv mv W v v W m . = - = - = . = + This is a variable force. As the particle moves from i x = 0 m to f max x = x =p /2c, the work done on it is f i / 2 0 / 2 0 0 0 0 0 ( ) sin( ) cos( )| cos cos0 2 x c c x W F x dx F cx dx F cx F F c c c p p . p . = = = - = - . - . = . . . . Thus, the particles speed at f max x = x =p /2c is 2 f 0 0 v = v + 2F /mc. 11.61. Visualize: We place the origin of the coordinate system at the base of the stairs on the first floor. Solve: (a) We might estimate y2 - y1 4.0 m 12 ft y3 - y2 , thus, 3 1 y - y 8.0 m. (b) We might estimate the time to run up these two flights of stairs to be 20 s. (c) Estimate your mass as m 70 kg 150 lb. Your power output while running up the stairs is 3 1 2 work done by you change in potential energy ( ) time time time (70 kg)(9.8 m/s )(8.0 m) 270 W (270 W) 1 hp 0.35 hp 20 s 746 W mg y - y = = = = . . . . . . Assess: Your estimate may vary, depending on your mass and how fast you run. 11.62. Solve: Power output during the push-off period is equal to the work done by the cat divided by the time the cat applied the force. Since the force on the floor by the cat is equal in magnitude to the force on the cat by the floor, work done by the cat can be found using the work-kinetic energy theorem during the push-off period,Wnet =Wfloor = .K. We do not need to explicitly calculate cat W , since we know that the cats kinetic energy is transformed into its potential energy during the leap. That is, 2 g 2 1 .U = mg( y - y ) = (5.0 kg)(9.8 m/ s )(0.95 m) = 46.55 J Thus, the average power output during the push-off period is net 46.55 J 0.23 kW 0.20 s P W t = = = 11.63. Solve: Using the conversion 746 W =1 hp, we have a power of 1492 J/s. This means W = Pt = (1492 J/s)(1 h) = 5.3712106 J is the total work done by the electric motor in one hour. Furthermore, motor g gf gi f i 6 motor 4 4 4 2 ( ) (10 m) 5.3712 10 J 5.481 10 kg 5.481 10 kg 1 liter 5.5 10 liters (10 m) (9.8 m/s )(10 m) 1 kg W W U U mgy y mg m W g = - = - = - = = = = = = 11.64. Solve: (a) The change in the potential energy of 1.0 kg of water in falling 25 m is .Ug = -mgh = -(1.0 kg)(9.8 m/s2 )(25 m) = -245 J -0.25 kJ (b) The power required of the dam is 50 106 Watts 50 106 J 1 s P W W W t = = = . = That is, 50106 J of energy is required per second for the dam. Out of the 245 J of lost potential energy, (245 J)(0.80) =196 J is converted to electrical energy. Thus, the amount of water needed per second is (50106 J)(1 kg /196 J) = 255,000 kg 2.6105 kg. 11.65. Solve: The force required to tow a water skier at a speed v is Ftow = Av. Since power P = Fv, the power required to tow the water skier is 2 tow tow P = F v = Av . We can find the constant A by noting that a speed of v = 2.5 mph requires a power of 2 hp. Thus, 2 2 (2 hp) (2.5 mph) 0.32 hp (mph) = A . A = Now, the power required to tow a water skier at 7.5 mph is 2 2 tow 2 0.32 hp (7.5 mph) 18 hp (mph) P = Av = = Assess: Since P . v2 , a three-fold increase in velocity leads to a nine-fold increase in power. 11.66. Solve: By definition, the maximum power output of a horse is P 1 hp = 746 W. At maximum speed, when the horse is running at constant speed, net F = 0. .. .. The propulsion force horse F , provided by the horse pushing against the ground, is balanced by the drag force of air resistance: horse F = D. We learned in Chapter 6 that a reasonable model for drag is 1 2 4 D Av , where A is the cross section area. Since the power needed for force F to push an object at velocity v is P = Fv, we have (1 2 ) 1 3 horse 4 4 1/ 3 1/ 3 1 hp 746 W 4 4(746 W) 15 m/s (0.5 m)(1.8 m) P Fv AvvAv v P A = = = = . . . . . . = . . = . . . . Assess: 15 m/s 30 mph is a reasonable top speed for a well-trained horse. 11.67. Solve: The net force on a car moving at a steady speed is zero. The motion is opposed both by rolling friction and by air resistance. Thus the propulsion force provided by the drive wheels must be 1 2 car r 4 F = mg + Av , where r is the rolling friction, m is the mass, A is the cross-section area, and v is the cars velocity. The power required to move the car at speed v is 3 car r 1 4 P = F v = mgv + Av Since the maximum power output is 200 hp and 75% of the power reaches the drive wheels, P = (200 hp)(0.75) =150 hp. Thus, 2 3 3 (150 hp) 746 W (0.02)(1500 kg)(9.8 m/s ) 1 (1.6 m)(1.4 m) 1 hp 4 0.56 294 111,900 0 55.5 m/s v v v v v . . . . = + . . . + - = . = The easiest way to solve this equation is through iterations by trial and error. Assess: A speed of 55.5 m/s 110 mph is very reasonable. 11.68. Model: Use the model of static friction, kinematic equations, and the definition of power. Solve: (a) The rated power of the Porsche is 217 hp =161,882 W and the gravitational force on the car is (1480 kg)(9.8 m/s2 ) = 14,504 N. The amount of that force on the drive wheels is (14,504)(2/3) = 9670 N. Because the static friction of the tires on road pushes the car forward, max s max s s max 2 max (1.00)(9670 N) 9670 N 6.53 m/s 1480 kg F f n mg ma a = = = = = . = = (b) Only 70% of the power generated by the motor is applied at the wheels. max max (0.70)(161,882 W) 11.7 m/s 9670 N P Fv v P F = . = = = (c) Using the kinematic equation, max 0 max min 0 v = v + a (t - t ) with 0 v = 0 m/s and 0 t = 0 s, we obtain max min 2 max 11.7 m/s 1.79 s 6.53 m/s t v a = = = Assess: An acceleration time of 1.79 s for the Porsche to reach a speed of 26 mph from rest is reasonable. 11.69. (a) A student uses a string to pull her 2.0 kg physics book, starting from rest, across a 2.0-m-long lab bench. The coefficient of kinetic friction between the book and the lab bench is 0.15. If the books final speed is 4.0 m/s, what is the tension in the string? (b) (c) The tension does external work Wext . This work increases the books kinetic energy and also causes an increase th .E in the thermal energy of the book and the lab bench. Solving the equation gives T =10.9 N. 11.70. (a) A 20 kg chicken crate slides down a 2.5-m-high, 40 ramp from the back of a truck to the ground. The coefficient of kinetic friction between the crate and the ramp bench is 0.15. How fast are the chickens going at the bottom of the ramp? (b) (c) v1 = 6.34 m/s. 11.71. (a) If you expend 75 W of power to push a 30 kg sled on a surface where the coefficient of kinetic friction between the sled and the surface is k = 0.20, what speed will you be able to maintain? (b) (c) 2 push (0.20)(30 kg)(9.8 m/s ) 58.8 N 75 W (58.8 N) 75 W 1.28 m/s 58.8 N F = = . = v.v = = 11.72. (a) A 1500 kg object is being accelerated upward at 1.0 m/s2 by a rope. How much power must the motor supply at the instant when the velocity is 2.0 m/s? (b) (c) (1500 kg)(9.8 m/s2 ) 1500 kg(1.0 m/s2 ) 16,200 N=16.2 kN (2m/s) (16,200 N)(2.0 m/s) 32,400 W 32 kW T P T = + = = = = = 11.73. Model: Model the water skier as a particle, apply the law of conservation of mechanical energy, and use the constant-acceleration kinematic equations. Visualize: We placed the origin of the coordinate system at the base of the frictionless ramp. Solve: Well start by finding the smallest speed v1 at the top of the ramp that allows her to clear the shark tank. From the vertical motion for jumping the shark tank, 2 2 1 1 2 1 2 1 2 2 2 1 2 1 ( ) 1 ( ) 2 0 m (2.0 m) 0 m 1 ( 9.8 m/s )( ) ( ) 0.639 s 2 y y y y v t t a t t t t t t = + - + - . = + + - - . - = From the horizontal motion, 2 2 1 1 2 1 2 1 1 1 1 1 ( ) 1 ( ) 2 ( 5.0 m) (0.639 s) 0 m 5.0 m 7.825 m/s 0.639 s x x x x v t t a t t x xv v = + - + - . + = + + . = = Having found the v1 that will take the skier to the other side of the tank, we now use the energy equation to find the minimum speed v0. We have 2 2 1 g1 0 g0 1 1 0 0 2 2 2 0 1 1 0 1 1 2 2 2 ( ) (7.825 m/s) 2(9.8 m/s )(2.0 m) 10.0 m/s K U K U mv mgy mv mgy v v gy y + = + . + = + = + - = + = 11.74. Model: Assume the spring to be ideal that obeys Hookes law, and model the block as a particle. Visualize: We placed the origin of the coordinate system on the free end of the compressed spring which is in contact with the block. Because the horizontal surface at the bottom of the ramp is frictionless, the spring energy appears as kinetic energy of the block until the block begins to climb up the incline. Solve: Although we could find the speed v1 of the block as it leaves the spring, we dont need to. We can use energy conservation to relate the initial potential energy of the spring to the energy of the block as it begins projectile motion at point 2. However, friction requires us to calculate the increase in thermal energy. The energy equation is 1 2 1 2 2 g2 th 0 g0 ext 2 2 2 k 2 0 e K +U + .E = K +U +W . mv + mgy + f .s = k(x - x ) The distance along the slope is 2.s = y / sin 45. The friction force is k k f = n, and we can see from the freebody diagram that n = mg cos45. Thus 1/ 2 2 2 0 e 2 k 2 v k (x x ) 2gy 2 gy cos 45 / sin45 m = .. - - - .. . . 1/ 2 1000 N/m (0.15 m)2 2(9.8 m/s2 )(2.0 m) 2(0.20)(9.8 m/s2 )(2.0 m) cos 45 / sin45 8.091 m/s 0.20 kg . . = . - - . = . . Having found the velocity 2 v , we can now find 3 2 (x - x ) = d using the kinematic equations of projectile motion: 2 3 2 2 3 2 2 3 2 2 2 2 3 2 3 2 ( ) 1 ( ) 2 2.0 m 2.0 m ( sin 45 )( ) 1 ( 9.8 m/ s )( ) 2 y y y y v t t a t t v t t t t = + - + - = + - + - - 3 2 .(t - t ) = 0 s and 1.168 s Finally, 2 3 2 2 3 2 2 3 2 3 2 2 ( ) 1 ( ) 2 ( ) ( cos45 )(1.168 s) 0 m 6.68 m x x x x v t t a t t d x x v = + - + - = - = + = 11.75. Model: Assume an ideal spring that obeys Hookes law, the particle model for the ball, and the model of kinetic friction. The ball, the spring, and the barrel comprise our system. Visualize: We placed the origin of the coordinate system on the free end of the spring which is in contact with the ball. Solve: Force F .. does external work Wext . This work compresses the spring, gives the ball kinetic energy, and is partially dissipated to thermal energy by friction. (a) The energy equation for our system is 1 s1 th 0 s0 ext 2 2 2 2 1 1 0 1 0 0 e e 1 0 2 2 1 1 ( ) ( ) 1 1 ( ) ( ) 2 2 2 2 1 (1.0 kg)(2.0 m/s) 1 (3000 N/m)(0.3 m) (0.30 m) 0 J 0 J (0.30 m) 2 2 k k K U E K U W mv k x x mg x x mv k x x F x x mg F + + . = + + + - + - = + - + - + + = + + With k = 0.30 and m =1.0 kg, we can solve this equation to obtain F = 460 N. (b) Using the energy equation for our system once again, we have 2 2 2 2 2 s2 th 1 s1 ext 2 e e 2 1 1 1 0 2 2 2 2 1 1 0 2 2 2 2 2 2 1 1 ( ) ( ) 1 1 ( ) 0 J 2 2 2 2 1 ( ) 0 J+ 1 ( ) 2 2 1 (1.0 kg) (0.30)(1.0 kg)(9.8 m/s )(1.5 m) 1 (3000 N/m)(0.30 m) 2 2 (0.5 kg) 4.41 J k k K U E K U W mv k x x mg x x mv k x x mv mg x x k x x v v + +. = + + + - + - = + - + + - = - + = + 2 =135 J.v =16.2 m/s 11.76. Solve: (a) The graph is a hyperbola. (b) The separation for zero potential energy is r =8, since U Gm1m2 0 J as r r =- . .8 This makes sense because two masses dont interact at all if they are infinitely far apart. (c) Due to the absence of nonconservative forces in our system of two particles, the mechanical energy is conserved. The equations of energy and momentum conservation are 2 2 1 2 2 2 1 2 f gf i gi 1 1f 2 2f 1 1i 2 2i f i 2 2 1 1f 2 2f 1 2 f i 1 1 1 1 2 2 2 2 1 1 1 1 2 2 K U K U m v m v Gm m m v m v Gm m r r mv m v Gmm r r . . . . + = + + + . - . = + + .- . . . . . . . + = . - . . . 1 f i 1 1f 2 2f 2f 1f 2 p p mv m v 0 kg m/ s v m v m = . + = . =- Substituting this expression for v2f into the energy equation, we get 2 2 1 2 2 1 1f 2 1f 1 2 1f 2 fi 12fi 1 1 1 1 2 1 1 2 2 (1 / ) m v m m v Gm m v Gm m rr mmrr . . . . . . + . . = . - .. = . - . . . . . + . . With 11 2 8 14 30 30 f 1 2 i 1 2 G = 6.6710- Nm(kg)- , r = R + R =1810 m, r =1.010 m, m = 8.010 kg, and m = 2.010 kg, the above equation can be simplified to yield ( ) 30 5 1 5 5 1f 2f 1f 30 2 1.72 10 m/s, and 8.0 10 kg 1.72 10 m/s 6.89 10 m/s 2.0 10 kg v v m v m . . = = - = . . = . . The speed of the heavier star is 1.7105 m/s. That of the lighter star is 6.9105m/s. 11-1 11.77. Model: Model the lawnmower as a particle and use the model of kinetic friction. Visualize: We placed the origin of our coordinate system on the lawnmower and drew the free-body diagram of forces. Solve: The normal force n.., which is related to the frictional force, is not equal to GF . .. This is due to the presence of F. .. The rolling friction is r r r r f = n, or n = f . The lawnmower moves at constant velocity, so net F = 0. .. .. The two components of Newtons second law are ( ) ( ) G rr rr r r r r 2 r r sin37 0 N / sin37 0 N sin37 cos37 0 N cos37 sin37 0 N (0.15)(12 kg)(9.8 m/s ) 29.4 N cos37 sin37 (0.7986) (0.15)(0.6018) y y x F n F F ma f mg F f mg F F F f F mg F F mg = - - = = . - - = . = + = - = . - - = . = = = - - S S Thus, the power supplied by the gardener in pushing the lawnmower at a constant speed of 1.2 m/s is P = F v = .. .. Fvcos. = (24.9 N)(1.2 m/s)cos37 = 24 W. 12.1. Model: A spinning skater, whose arms are outstretched, is a rigid rotating body. Visualize: Solve: The speed v = r., where r =140 cm/2 = 0.70 m. Also, 180 rpm = (180)2p /60 rad/s = 6p rad/s. Thus, v = (0.70 m)(6p rad/s) =13.2 m/s. Assess: A speed of 13.2 m/s 26 mph for the hands is a little high, but reasonable. 12.2. Model: Assume constant angular acceleration. Solve: (a) The final angular velocity is ( ) f 2000 rpm 2 rad min 209.4 rad/s. rev 60 s p . = .. .... .. = . .. . The definition of angular acceleration gives us f i 209.4 rad/s 0 rad/s 419 rad/s t t 0.50 s . . . a . - - = = = = . . The angular acceleration of the drill is 4.2102 rad/s. (b) ( )2 ( )( )2 f i 1 0 rad 0 rad 1 419 rad/s 0.50 s 52.4 rad 2 2 i. =. +. .t + a .t = + + = The drill makes (52.4 rad) rev 8.3 revolutions. 2p rad . . . . = . . 12.3. Model: Assume constant angular acceleration. Visualize: Solve: (a) Since at = ra , find a first. With 90 rpm = 9.43 rad/s and 60 rpm = 6.28 rad/s, 9.43 rad 6.28 rad/s 0.314 rad/s2 t 10 s . a . - = = = . The angular acceleration of the sprocket and pedal are the same. So (0.18 m)(0.314 rad/s2 ) 0.057 m/s2 t a = ra = = (b) The length of chain that passes over the sprocket during this time is L = r.. . Find .. : ( ) ( )( ) ( )( ) 2 f i i 2 2 f i 1 2 6.28 rad/s 10 s 1 0.314 rad/s 10 s 78.5 rad 2 . . . t a t . . . = + . + . - = . = + = The length of chain which has passed over the top of the sprocket is L = (0.10 m)(78.5 rad) = 7.9 m 12.4. Model: Assume constant angular acceleration. Visualize: Solve: The initial angular velocity is ( ) i 60 rpm 2 rad min 2 rad/s. rev 60 s p . = .. .... .. = p . .. . The angular acceleration is f i 2 0 rad/s 2 rad/s 0.251 rad/s t 25 s . . p a - - = = =- . The angular velocity of the fan blade after 10 s is ( ) ( 2 )( ) f i 0 . =. +a t - t = 2p rad/s+ -0.251 rad/s 10 s - 0 s = 3.77 rad/s The tangential speed of the tip of the fan blade is (0.40 m)(3.77 rad/s) 1.51 m/s tv = r. = = (b) ( )2 ( )( ) ( 2 )( )2 f i i 1 0 rad 2 rad/s 25 s 1 0.251 rad/s 25 s 78.6 rad 2 2 . =. +. .t + a .t = + p + - = The fan turns 78.6 radians =12.5 revolutions while coming to a stop. 12.5. Model: The earth and moon are particles. Visualize: Choosing 0 m E x = sets the coordinate origin at the center of the earth so that the center of mass location is the distance from the center of the earth. Solve: ( 24 )( ) ( 22 )( 8 ) E E M M cm 24 22 E M 6 5.98 10 kg 0 m 7.36 10 kg 3.84 10 m 5.98 10 kg 7.36 10 kg 4.67 10 m x m x m x m m + + = = + + = Assess: The center of mass of the earth-moon system is called the barycenter, and is located beneath the surface of the earth. Even though E x = 0 m the earth influences the center of mass location because E m is in the denominator of the expression for cmx . 12.6. Visualize: Please refer to Figure EX12.6. The coordinates of the three masses mA, mB, and C m are (0 cm, 0 cm), (0 cm, 10 cm), and (10 cm, 0 cm), respectively. Solve: The coordinates of the center of mass are A A B B C C cm A B C A A B B C C cm A B C (100 g)(0 cm) (200 g)(0 cm) (300 g)(10 cm) 5.0 cm (100 g 200 g 300 g) (100 g)(0 cm) (200 g)(10 cm) (300 g)(0 cm) 3.3 cm (100 g 200 g 300 g) x m x m x m x m m m y m y m y m y m m m + + + + = = = + + + + + + + + = = = + + + + 12.7. Visualize: Please refer to Figure EX12.7. The coordinates of the three masses mA, mB, and C m are (0 cm, 10 cm), (10 cm, 10 cm), and (10 cm, 0 cm), respectively. Solve: The coordinates of the center of mass are A A B B C C cm A B C A A B B C C cm A B C (200 g)(0 cm) (300 g)(10 cm) (100 g)(10 cm) 6.7 cm (200 g 300 g 100 g) (200 g)(10 cm) (300 g)(10 cm) (100 g)(0 cm) 8.3 cm (200 g 300 g 100 g) x m x m x m x m m m y m y m y m y m m m + + + + = = = + + + + + + + + = = = + + + + 12.8. Model: The balls are particles located at the balls respective centers. Visualize: Solve: The center of mass of the two balls measured from the left hand ball is ( )( ) ( )( ) cm 100 g 0 cm 200 g 30 cm 20 cm 100 g 200 g x + = = + The linear speed of the 100 g ball is ( )( ) 1 cm 0.20 m 120 rev/min 2 rad min 2.5 m/s rev 60 s v r x p = . = . = .. .... .. = . .. . 12.9. Model: The earth is a rigid, spherical rotating body. Solve: The rotational kinetic energy of the earth is 1 2 rot 2 K = I. . The moment of inertia of a sphere about its diameter (see Table 12.2) is 2 2 5 earth I = M R and the angular velocity of the earth is 2 rad 7.27 10 5 rad/s 24 3600 s p . = = - Thus, the rotational kinetic energy is 2 2 rot earth 24 6 2 5 2 29 1 2 2 5 1 (5.98 10 kg)(6.37 10 m) (7.27 10 rad/s) 2.57 10 J 5 K M R. - = . . . . . . = = 12.10. Model: The triangle is a rigid body rotating about an axis through the center. Visualize: Please refer to Figure EX12.10. Each 200 g mass is a distance r away from the axis of rotation, where r is given by 0.20 m cos30 0.20 m 0.2309 m cos30 r r = . = = Solve: The moment of inertia of the triangle is I = 3mr2 = 3(0.200 kg)(0.2309 m)2 = 0.0320 kg m2. The frequency of rotation is given as 5.0 revolutions per s or 10p rad/s. The rotational kinetic energy is 2 2 2 rot. 1 1(0.0320 kg m )(10.0 rad/s) 15.8 J 2 2 K = I. = p = 12.11. Model: The disk is a rigid body rotating about an axis through its center. Visualize: Solve: The speed of the point on the rim is given by vrim = R.. The angular velocity . of the disk can be determined from its rotational kinetic energy which is 1 2 2 K = I. = 0.15 J. The moment of inertia I of the disk about its center and perpendicular to the plane of the disk is given by 2 2 52 2 5 2 1 1(0.10 kg)(0.040 m) 8.0 10 kg m 2 2 2(0.15 J) 0.30 J 61.237 rad/s 8.0 10 kg m I MR I . . - - = = = . = = . = Now, we can go back to the first equation to find rim v . We get rim v = R. = (0.040 m)(61.237 rad/s) = 2.4 m/s. 12.12. Model: The baton is a thin rod rotating about a perpendicular axis through its center of mass. Solve: The moment of inertia of a thin rod rotating about its center is 1 2. 12 I = ML For the baton, 1 (0.400 kg)(0.96 m)2 0.031 kg m2 12 I= = The rotational kinetic energy of the baton is ( )( ) 2 2 2 rot 1 1 0.031 kg m 100 rev/min 2 rad min 1.68 J 2 2 rev 60 s K I p . . . .. .. = = . . .. .. = . . .. .. 12.13. Model: The structure is a rigid body rotating about its center of mass. Visualize: We placed the origin of the coordinate system on the 300 g ball. Solve: First, we calculate the center of mass: cm (300 g)(0 cm) (600 g)(40 cm) 26.67 cm 300 g 600 g x + = = + Next, we will calculate the moment of inertia about the structures center of mass: 2 2 cm cm 2 2 2 (300 g)( ) (600 g)(40 cm ) (0.300 kg)(0.2667 m) (0.600 kg)(0.1333 m) 0.032 kg m I = x + - x = + = Finally, we calculate the rotational kinetic energy: 2 2 2 rot 1 1 (0.032 kg m ) 100 2 rad/s 1.75 J 2 2 60 K I p . . . = = . . = . . 12.14. Model: The moment of inertia of any object depends on the axis of rotation. In the present case, the rotation axis passes through mass A and is perpendicular to the page. Visualize: Please refer to Figure EX12.14. Solve: (a) A A B B C C D D cm A B C D A A B B C C D D cm A B C D (100 g)(0 m) (200 g)(0 m) (200 g)(0.10 m) (200 g)(0.10 m) 0.057 m 100 g 200 g 200 g 200 g (100 g)(0 m) (200 g)(0.10 m) (200 g)(0.10 cm) (200 g i i i m x x m x m x m x m x m m m m m y m y m y m y m y m m m m + + + = = + + + + + + = = + + + + + + = + + + + + + = S S )(0 m) 0.057 m 700 g = (b) The distance from the axis to mass C is 14.14 cm. The moment of inertia through A and perpendicular to the page is 2 2 2 2 2 A AA BB CC DD (0.100 kg)(0 m)2 (0.200 kg)(0.10 m)2 (0.200 kg)(0.1414 m)2 (0.200 kg)(0.10 m)2 0.0080 kg m2 i i i I = m r = m r + m r + m r + m r = + + + = S 12.15. Model: The moment of inertia of any object depends on the axis of rotation. Visualize: Solve: (a) A A B B C C D D cm A B C D A A B B C C D D cm A B C D (100 g)(0 m) (200 g)(0 m) (200 g)(0.10 m) (200 g)(0.10 m) 0.057 m 100 g 200 g 200 g 200 g (100 g)(0 m) (200 g)(0.10 m) (200 g)(0.10 cm) (200 g i i i m x x m x m x m x m x m m m m m y m y m y m y m y m m m m + + + = = + + + + + + = = + + + + + + = + + + + + + = S S )(0 m) 0.057 m 700 g = (b) The moment of inertia about a diagonal that passes through B and D is 2 2 BD A A C C I = m r + m r where A C r = r = (0.10 m)cos45 = 7.07 cm and are the distances from the diagonal. Thus, 2 2 2 BD A C I = (0.100 kg)r + (0.200 kg)r = 0.0015 kg m Assess: Note that the masses B and D, being on the axis of rotation, do not contribute to the moment of inertia. 12.16. Model: The three masses connected by massless rigid rods is a rigid body. Visualize: Please refer to Figure EX12.16. Solve: (a) cm (0.100 kg)(0 m) (0.200 kg)(0.06 m) (0.100 kg)(0.12 m) 0.060 m 0.100 kg 0.200 kg 0.100 kg i i i m x x m + + = = = + + S S ( 2 2 ) cm (0.100 kg)(0 m) (0.200 kg) (0.10 m) (0.06 m) (0.100 kg)(0 m) 0.040 m 0.100 kg 0.200 kg 0.100 kg i i i m y y m + - + = = = + + S S (b) The moment of inertia about an axis through A and perpendicular to the page is 2 2 2 2 2 2 A B C (0.10 m) (0.10 m) (0.100 kg)[(0.10 m) (0.10 m) ] 0.0020 kg m i i I =Smr = m + m = + = (c) The moment of inertia about an axis that passes through B and C is ( )2 2 2 2 BC A I = m (0.10 m) - (0.06 m) = 0.00128 kg m Assess: Note that mass A m does not contribute to AI , and the masses B m and C m do not contribute to BCI . 12.17. Model: The door is a slab of uniform density. Solve: (a) The hinges are at the edge of the door, so from Table 12.2, 1 (25 kg)(0.91 m)2 6.9 kg m2 3 I= = (b) The distance from the axis through the center of mass along the height of the door is 0.91 m 0.15 m 0.305 m. 2 d = .. - .. = . . Using the parallelaxis theorem, 2 ( )( )2 ( )( )2 2 cm 1 25 kg 0.91 m 25 kg 0.305 cm 4.1 kg m 12 I = I + Md = + = Assess: The moment of inertia is less for a parallel axis through a point closer to the center of mass. 12.18. Model: The CD is a disk of uniform density. Solve: (a) The center of the CD is its center of mass. Using Table 12.2, 2 ( )( )2 5 2 cm 1 10.021 kg 0.060 m 3.8 10 kg m 2 2 I = MR = = - (b) Using the parallelaxis theorem with d = 0.060 m, 2 5 2 ( )( )2 4 2 cm I = I + Md = 3.810- kg m + 0.021 kg 0.060 m =1.1410- kg m 12.19. Visualize: Solve: Torque by a force is defined as t = Fr sinf where f is measured counterclockwise from the r.. vector to the F .. vector. The net torque on the pulley about the axle is the torque due to the 30 N force plus the torque due to the 20 N force: 1 1 2 2 (30 N) sin (20 N) sin (30 N)(0.02 m) sin ( 90 ) (20 N)(0.02 m) sin (90 ) ( 0.60 N m) (0.40 N m) 0.20 N m r f + r f = - + = - + = - Assess: A negative torque causes a clockwise acceleration of the pulley. 12.20. Visualize: The two equal but opposite 50 N forces, one acting at point P and the other at point Q, make a couple that causes a net torque. Solve: The distance between the lines of action is l = d cos30. The net torque is given by t = lF = (d cos30)F = (0.10 m)(0.866)(50 N) = 4.3 N m 12.21. Visualize: Solve: The net torque on the spark plug is t = Fr sinf = -38 N m = F(0.25 m)sin(-120). F =176 N That is, you must pull with a force of 176 N to tighten the spark plug. Assess: The force applied on the wrench leads to its clockwise motion. That is why we have used a negative sign for the net torque. 12.22. Model: The disk is a rotating rigid body. Visualize: The radius of the disk is 10 cm and the disk rotates on an axle through its center. Solve: The net torque on the axle is A A A B B B C C C D D D sin sin sin sin (30 N)(0.10 m)sin( 90 ) (20 N)(0.050 m)sin90 (30 N)(0.050 m)sin135 (20 N)(0.10 m)sin 0 3 N m 1 N m 1.0607 N m 0.94 N m t = F r f + F r f + F r f + F r f = -+ + + = - + + = - Assess: A negative torque means a clockwise rotation of the disk. 12.23. Model: The beam is a solid rigid body. Visualize: The steel beam experiences a torque due to the gravitational force on the construction worker( ) G C F .. and the gravitational force on the beam ( ) G B F . .. The normal force exerts no torque since the net torque is calculated about the point where the beam is bolted into place. Solve: The net torque on the steel beam about point O is the sum of the torque due to ( ) G C F .. and the torque due to ( ) G B F . .. The gravitational force on the beam acts at the center of mass. G C G B 2 2 (( ) )(4.0 m)sin( 90 ) (( ) )(2.0 m)sin( 90 ) (70 kg)(9.80 m/s )(4.0 m) (500 kg)(9.80 m/s )(2.0 m) 12.5 kN m t = F - + F - = - - = - The negative torque means these forces would cause the beam to rotate clockwise. The magnitude of the torque is 12.5 kN m. 12.24. Model: Model the arm as a uniform rigid rod. Its mass acts at the center of mass. Visualize: Solve: (a) The torque is due both to the gravitational force on the ball and the gravitational force on the arm: ball arm b b a a 2 2 ( ) sin90 ( ) sin90 (3.0 kg)(9.8 m/s )(0.70 m)+(4.0 kg)(9.8 m/s )(0.35 m) 34 N m t =t +t = m g r + m g r = = (b) The torque is reduced because the moment arms are reduced. Both forces act at f = 45 from the radial line, so ball arm b b a a 2 2 ( ) sin45 ( ) sin45 (3.0 kg)(9.8 m/s )(0.70 m)(0.707) (4.0 kg)(9.8 m/s )(0.35 m)(0.707) 24 N m t =t +t = m g r + m g r = + = 12.25. Solve: t = Ia is the rotational analog of Newtons second law F = ma. We have t = (2.0 kg m2 )(4.0 rad/s2 ) = 8.0 kg m2/s2 = 8.0 N m. 12.26. Visualize: Since a =t /I, a graph of the angular acceleration looks just like the torque graph with the numerical values divided by I = 4.0 kg m2. Solve: From the discussion about Figure 4.47 f i . =. + area under the angular acceleration a curve between i f t and t The area under the curve between t = 0 s and t = 3 s is 0.75 rad/s. With 1 . = 0 rad/s, we have f . = 0 rad/s + 0.75 rad/s = 0.75 rad/s 12.27. Model: Two balls connected by a rigid, massless rod are a rigid body rotating about an axis through the center of mass. Assume that the size of the balls is small compared to 1 m. Visualize: We placed the origin of the coordinate system on the 1.0 kg ball. Solve: The center of mass and the moment of inertia are cm cm 2 2 2 2 about cm (1.0 kg)(0 m) (2.0 kg)(1.0 m) 0.667 m and 0 m (1.0 kg 2.0 kg) (1.0 kg)(0.667 m) (2.0 kg)(0.333 m) 0.667 kg m i i x y I mr + = = = + =S = + = We have f . = 0 rad/s, f i t - t = 5.0 s, and 2 i 3 . = -20 rpm = -20(2p rad/60 s) = - p rad/s, so f i f i . =. +a (t - t ) becomes 0 rad/s 2 rad/s (5.0 s) 2 rad/s2 3 15 p p = .. - .. +a .a = . . Having found I and a , we can now find the torque t that will bring the balls to a halt in 5.0 s: 2 2 about cm 2 kg m 2 rad/s 4 N m 0.28 N m 3 15 45 I p p t = a = .. .... .. = = . .. . The magnitude of the torque is 0.28 N m, applied in the counterclockwise direction. 12.28. Model: A circular plastic disk rotating on an axle through its center is a rigid body. Assume axis is perpendicular to the disk. Solve: To determine the torque (t) needed to take the plastic disk from .i = 0 rad/s to f . =1800 rpm = (1800)(2p ) / 60 rad/s = 60p rad/s in f i t - t = 4.0 s, we need to determine the angular acceleration (a ) and the disks moment of inertia (I) about the axle in its center. The radius of the disk is R =10.0 cm. We have 2 2 32 f i 2 f i f i f i 1 1(0.200 kg)(0.10 m) 1.0 10 kg m 2 2 ( ) 60 rad/s 0 rad/s 15 rad/s 4.0 s I MR t t t t . . p . . a a p = = = - - - = + - . = = = - Thus, t = Ia = (1.010-3 kg m2 )(15p rad/s2 ) = 0.047 N m. 12.29. Model: The compact disk is a rigid body rotating about its center. Visualize: Solve: (a) The rotational kinematic equation .1 =.0 +a (t1 - t0 ) gives (2000 rpm) 2 rad/s 0 rad (3.0 s 0 s) 200 rad/s2 60 9 p p .. .. = +a - .a = . . The torque needed to obtain this operating angular velocity is (2.5 10 5 kg m2 ) 200 rad/s2 1.75 10 3 N m 9 I p t = a = - .. .. = - . . (b) From the rotational kinematic equation, 2 2 ( )2 1 0 0 1 0 1 0 ( ) 1 ( ) 0 rad 0 rad 1 200 rad/s 3.0 s 0 s 2 29 100 rad 100 revolutions 50 rev 2 t t t t p . . . a p p p = + - + - = + + . . - . . . . = = = Assess: Fifty revolutions in 3 seconds is a reasonable value. 12.30. Model: The rocket attached to the end of a rigid rod is a rotating rigid body. Assume the rocket is small compared to 60 cm. Visualize: Please refer to Figure EX12.30. Solve: We can determine the rockets angular acceleration from the relationship t = Ia. The torque t can be found from the thrust (F) using t = Fr sinf . The moment of inertia (I) can be calculated from equations given in Table 12.2. Specifically, rod about one end rocket I = I + I becomes 2 2 2 2 rod 2 2 2 1 1(0.100 kg)(0.60 m) (0.200 kg)(0.60 m) 3 3 0.012 kg m 0.072 kg m 0.0840 kg m M L + ML = + = + = 2 2 sin (4.0 N)(0.60 m)sin(45 ) 20 rads/s 0.0840 kg m Fr I I t f a . = = = = Assess: The rocket will accelerate counterclockwise since a is positive. 12.31. Model: The rod is in rotational equilibrium, which means that t net = 0. Visualize: As the gravitational force on the rod and the hanging mass pull down (the rotation of the rod is exaggerated in the figure), the rod touches the pin at two points. The piece of the pin at the very end pushes down on the rod; the right end of the pin pushes up on the rod. To understand this, hold a pen or pencil between your thumb and forefinger, with your thumb on top (pushing down) and your forefinger underneath (pushing up). Solve: Calculate the torque about the left end of the rod. The downward force exerted by the pin acts through this point, so it exerts no torque. To prevent rotation, the pins normal force pin n.. exerts a positive torque (ccw about the left end) to balance the negative torques (cw) of the gravitational force on the mass and rod. The gravitational force on the rod acts at the center of mass, so 2 2 net pin pin 0 N m (0.40 m)(2.0 kg)(9.8 m/s ) (0.80 m)(0.50 kg)(9.8 m/s ) 11.8 N m t t t = = - - . = 12.32. Model: The massless rod is a rigid body. Visualize: Solve: To be in equilibrium, the object must be in both translational equilibrium (Fnet = 0 N) .. and rotational equilibrium net (t = 0 Nm). We have ( ) net (40 N) (100 N) (60 N) 0 N, y F = - + = so the object is in translational equilibrium. Measuring net t about the left end, net t = (60 N)(3.0 m)sin(+90) + (100 N)(2.0 m)sin(-90) = -20 N m The object is not in equilibrium. 12.33. Model: The object balanced on the pivot is a rigid body. Visualize: Since the object is balanced on the pivot, it is in both translational equilibrium and rotational equilibrium. Solve: There are three forces acting on the object: the gravitational force ( ) G 1 F .. acting through the center of mass of the long rod, the gravitational force ( ) G 2 F .. acting through the center of mass of the short rod, and the normal force P .. on the object applied by the pivot. The translational equilibrium equation ( ) net 0 N y F = is ( ) ( ) ( ) ( ) 2 2 G 1 G 2 G 1 G 2 - F - F + P = 0 N. P = F + F = (1.0 kg)(9.8 m/s ) + (4.0 kg)(9.8 m/s ) = 49 N Measuring torques about the left end, the equation for rotational equilibrium net t = 0 Nm is 1 2 2 2 (1.0 m) (1.5 m) 0 Nm (49 N) (1.0 kg)(9.8 m/s )(1.0 m) (4.0 kg)(9.8 m/s )(1.5 m) 0 N 1.40 m Pd w w d d - - = . - - = . = Thus, the pivot is 1.40 m from the left end. 12.34. Model: The see-saw is a rigid body. The cats and bowl are particles. Visualize: Solve: The see-saw is in rotational equilibrium. Calculate the net torque about the pivot point. ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )( ) ( )( ) net G 1 G 2 G 2 1 1 2 0 2.0 m 2.0 m 2.0 m 2.0 m 2.0 m 5.0 kg 2.0 kg 2.0 m 1.5 m 4.0 kg B B B F F d F m gd mg m g m m d m t = = - - = - - - = = = Assess: The smaller cat is close but not all the way to the end by the bowl, which makes sense since the combined mass of the smaller cat and bowl of tuna is greater than the mass of the larger cat. 12.35. Solve: (a) According to Equation 12.35, the speed of the center of mass of the tire is cm ( ) 2 cm 20 m/s 20 m/s 66.67 rad/s 66.7 60 rpm 6.4 10 rpm 0.30 m 2 v R v R . . p = = . = = = = . . = . . . . (b) The speed at the top edge of the tire relative to the ground is top cm v = 2v = 2(20 m/s) = 40 m/s. (c) The speed at the bottom edge of the tire relative to ground is bottom v = 0 m/s. 12.36. Model: The can is a rigid body rolling across the floor. Assume that the can has uniform mass distribution. Solve: The rolling motion of the can is a translation of its center of mass plus a rotation about the center of mass. The moment of inertia of the can about the center of mass is 1 2 2 MR , where R is the radius of the can. Also cm v = R., where . is the angular velocity of the can. The total kinetic energy of the can is 2 2 2 2 2 cm cm rot cm cm cm 2 2 cm 1 1 1 11 2 2 2 22 3 3(0.50 kg)(1.0 m/s) 0.38 J 4 4 K K K Mv I Mv MR v R Mv = + = + . = + .. .... .. . .. . = = = 12.37. Model: The sphere is a rigid body rolling down the incline without slipping. Visualize: The initial gravitational potential energy of the sphere is transformed into kinetic energy as it rolls down. Solve: (a) If we choose the bottom of the incline as the zero of potential energy, the energy conservation equation will be Kf =Ui . The kinetic energy consists of both translational and rotational energy. This means 2 2 2 2 2 f cm cm 2 2 10 10 7 7 2 2 1 1 12 1 ( ) 2 2 25 2 7 (2.1 m)sin 25 10 (2.1 m)(sin 25 ) (2.1 m)(sin 25 ) 88 rad/s (0.04 m) K I Mv Mgh MR M R Mgh MR Mg g g R . . . . . = + = . . . + = . . . . . = . = = = (b) From part (a) 2 2 22 2 2 2 22 total cm cm rot cm 1 2 2 rot 5 7 2 2 total 10 1 1 7 and 1 1 2 1 2 2 10 2 25 5 1 10 2 5 7 7 K I Mv MR K I MR MR K MR K MR . . . . . . . = + = = = . . = . . . . . = = = 12.38. Visualize: Please refer to Figure EX12.38. To determine angle a , put the tails of the vectors together. Solve: (a) The magnitude of A B is ABsina = (6)(4)sin 45 =17. .. .. The direction of A B, .. .. using the right hand rule, is out of the page. Thus, A B = (17, out of the page). .. .. (b) The magnitude of C D is CDsina = (6)(4)sin180 = 0. .. .. Thus C D = 0. .. .. .. 12.39. Visualize: Please refer to Figure EX12.39. Solve: (a) The magnitude of A B .. .. is ABsina = (6)(4)sin 60 = 20.78. The direction of A B .. .. is given by the right hand rule. To curl our fingers from A .. to , B .. we have to point our thumb out of the page. Thus, A B = .. .. (21, out of the page). (b) C D = ((6)(4)sin90, into the page) = (24, into the page). .. .. 12.40. Solve: (a) (i j)i = ki = j (b) i ( j i) = i (-k) = -i k = -(- j) = j 12.41. Solve: (a) i (i j) = i k = - j (b) ( ) 0 i j k k k = = .. 12.42. Solve: (a) (3 ) (3 2 2 ) 9 6 6 3 2 2 0 6 6( ) 3( ) 0 2 2 6 9 = + - + = - + + - + = - + - + - - + = - - .. .. A B i j i j k i i i j i k j i j j j k k j k i i j k (b) 12.43. Solve: (a) C D = 0 .. .. implies that D .. must also be in the same or opposite direction as the C .. vector or zero, because ii = 0. Thus D = ni, .. where n could be any real number. (b) C E = 6k .. .. implies that E .. must be along the j vector, because i j = k. Thus E = 2 j. .. (c) C F = -3 j .. .. implies that F .. must be along the k vector, because i k = - j. Thus F =1k. .. 12.44. Solve: (5 5 ) ( 10 ) N m [ 50( ) 50( )] N m [ 50( ) 0] N m 50 N m t = = + - = - - = - + - = - .. .. .. .. r F i j j i j j j k k 12.45. Solve: (5 ) ( 10 10 ) N m ( 5 50 ) N m [ 50( ) 0] N m 50 N m t = = - + = - + = - - + = .. .. .. .. r F j i j j i j j k k 12.46. Visualize: Please refer to Figure EX12.46. Solve: ( ) 2 2 2 (1.0 2.0 ) m (0.200 kg)(3.0 m/s) cos45 sin 45 (0.42 0.42 0.85 0.85 ) kg m /s (1.27 ) kg m /s or (1.27 kg m /s, into page) = = + - = - + - =- .. .. .. L r mv i j i j i i i j j i j j k 12.47. Visualize: Please refer to Figure EX12.47. Solve: 2 2 2 2 2 2 (3.0 2.0 ) m (0.1 kg)(4.0) m/s 1.20( ) kg m /s 0.8( ) kg m /s 1.20 kg m /s 0 kg m /s 1.20 kg m /s or (1.20 kg m /s, out of page) = = + = + = + = .. .. .. L r mv i j j i j j j k k 12.48. Model: The bar is a rotating rigid body. Assume that the bar is thin. Visualize: Please refer to Figure EX12.48. Solve: The angular velocity 120 . = rpm = (120)(2p ) / 60 rad/s = 4p rad/s. From Table 12.2, the moment of inertial of a rod about its center is 1 2 12 I = ML . The angular momentum is 1 (0.50 kg)(2.0 m)2 (4 rad/s) 2.1 kg m2/s 12 L = I. = .. .. p = . . If we wrap our fingers in the direction of the rods rotation, our thumb will point in the z direction or out of the page. Consequently, L = (2.1 kg m2/s, out of the page) .. 12.49. Model: The disk is a rotating rigid body. Visualize: Please refer to Figure EX12.49. Solve: From Table 12.2, the moment of inertial of the disk about its center is 1 2 1 (2.0 kg)(0.020 m)2 4.0 10 4 kg m2 2 2 I = MR = = - The angular velocity . is 600 rpm = 600 2p /60 rad/s = 20p rad/s. Thus, L = I. = (4.010-4 kg m2 )(20p rad/s) = 0.025 kg m2/s. If we wrap our right fingers in the direction of the disks rotation, our thumb will point in the -x direction. Consequently, L = -0.025 i kg m2/s = (0.025 kg m2/s, into page) .. 12.50. Model: The beach ball is a spherical shell. Solve: From Table 12.2, the moment of inertia about a diameter of a spherical shell is 2 2 2 (0.100 kg)(0.50 m)2 1 kg m2 3 3 60 I = MR = = Require ( ) 0.10 kg m2/s 1 kg m2 60 60 rad/s L I. . . = = = . . . . . . . = In rpm, this is (60 rad/s) rev 60 s 57 rpm. 2p rad min . .. . = . .. . . .. . 12.51. Model: The wheel is a rigid rolling body. Visualize: Solve: The front of the disk is moving forward at velocity vcm. Also, because of rotation the point is moving downward at velocity rel cmv = R. = v . So, this point has a speed 2 2 cm cm cm v = v + v = 2v = 2(20 m/s) = 28 m/s Assess: The speed v is independent of the radius of the wheel. 12.52 Model: The triangle is a rigid body rotating about its center of mass perpendicular to the plane of the triangle. The center of mass of any symmetrical object of uniform density is at the physical center of the object. Visualize: The distance to one tip of the triangle from the center of mass is given by r cos30 = (5.0/2) cm, which yields r = (2.5 cm)/ cos30 = 2.9 cm. Solve: The speed of the tip is v = r. = (2.9 cm)(120 rpm) = (2.9 cm)(4p rad/s) = 36 cm/s 12.53. Visualize: Solve: We will consider a vertical strip of width dx and of mass dm at a position x from the origin. The formula for the x component of the center of mass is cm x 1 x dm M = . The area of the steel plate is 1 2 2 A = (0.2 m)(0.3 m) = 0.030 m . Mass dm in the strip is the same fraction of M as dA is of A. Thus 2 2 0.800 kg (26.67 kg/m ) 0.030 m dm dA dm M dA dA ldx M A A = . = = . . = . . . . The relationship between l and x is 2 0.20 m 0.30 m 3 l = x .l = x Therefore, 2 3 0.3 m 2 2 2 2 cm 0 m 1 (26.67 kg/m ) 2 (17.78 kg/m ) (17.78 kg/m ) (0.3 m) 20 cm 3 3 0.8 kg 3 x x dx x M M = . . = = = . . . . . Due to symmetry cm y = 0 cm. 12.54. Visualize: Solve: Build the plate from the three shapes 1, 2, and 3. The center of mass of the complete shape is the center of mass of the center of masses of the three smaller shapes. ( ) ( ) ( ) ( ) ( ) ( ) 1 cm 1 2 cm 2 3 cm 3 cm 1 2 3 1 cm 1 2 cm 2 3 cm 3 cm 1 2 3 m x m x m x x m m m m y m y m y y m m m + + = + + + + = + + The masses of each of the smaller shapes are a fraction of the larger shapes mass by the ratio of areas. The following table will be useful in the solution. Shape A mass cm x cm y All 32 cm2 M cm x cm y 1 24 cm2 24 32 M -1.0 cm 0 cm 2 2.0 cm2 2.0 32 M 2.0 cm -2.5 cm 3 6.0 cm2 6.0 32 M 2.0 cm 1.5 cm Thus, ( ) ( ) ( ) cm 24 M 1.0 cm 2.0 M 2.0 cm 6.0 M 2.0 cm 32 32 32 0.25 cm M x . . - + . . + . . . . . . . . = . . . . . . = - Similarly, cm y = 0.125 cm. Assess: The plates center of mass of (-0.25 cm, 0.125 cm) is reasonable. The cutout means more of the mass is left of center and above the center. 12.55. Model: The disk is a rigid rotating body. The axis is perpendicular to the plane of the disk. Visualize: Solve: (a) From Table 12.2, the moment of inertia of a disk about its center is 1 2 1 (2.0 kg)(0.10 m)2 0.010 kg m2 2 2 I = MR = = (b) To find the moment of inertia of the disk through the edge, we can make use of the parallel axis theorem: 2 2 2 2 center I = I + Mh = (0.010 kg m ) + (2.0 kg)(0.10 m) = 0.030 kg m Assess: The larger moment of inertia about the edge means there is more inertia to rotational motion about the edge than about the center. 12.56. Model: The object is a rigid rotating body. Assume the masses 1 m and 2 m are small and the rod is thin. Visualize: Please refer to P12.56. Solve: The moment of inertia of the object is the sum of the moment of inertia of the rod, mass m1, and mass 2m . Using Table 12.2 for the moment of inertia of the rod, we get 1 2 2 2 2 rod rod about center m m 1 2 2 2 2 2 2 1 2 1 1 12 2 4 1 1 1 12 4 16 4 3 4 I I I I ML m L m L ML m L m L L M m m = + + = + . . + . . . . . . . . . . = + + = . + + . . . . . Assess: With 1 2 1 2 rod12 m = m = 0 kg, I ML , as expected. 12.57. Visualize: We chose the origin of the coordinate system to be on the axis of rotation, that is, at a distance d from one end of the rod. Solve: The moment of inertia can be calculated as follows: 2 1 2 3 2 33 33 and 1 [( ) ( ) ] [( ) ] 3 3 3 x x L d L d d d I x dm dm dx dm M dx M L L I M x dx M x M L d d M L d d L L L L - - - - = = . = . = = . . = . . - - - = - + . . . . . . . . . . For d = 0 m, 1 2 3 I = ML , and for 1 2 d = L, 3 3 1 2 3 2 2 12 I M L L ML L .. . . . . = .. . + . . . = ... . . . .. Assess: The special cases d = 0 m and d = L/2 of the general formula give the same results that are found in Table 12.2. 12.58. Visualize: Solve: We solve this problem by dividing the disk between radii 1 r and 2 r into narrow rings of mass dm. Let dA = 2p rdr be the area of a ring of radius r. The mass dm in this ring is the same fraction of the total mass M as dA is of the total area A. (a) The moment of inertia can be calculated as follows: ( )( ) ( ) ( ) ( ) ( )( ) ( ) 2 2 2 1 1 1 2 disk 2 2 2 1 4 2 3 disk 2 2 2 2 2 2 2 1 2 1 2 1 4 4 2 2 2 2 2 1 2 1 2 1 and 2 (2 ) 2 2 4 2 4 2 r r r r r r I r dm dm M dA M r dr A r r I M r r dr M r dr M r r r r r r r M r r M r r r r p p p p = = =- . = = = - - - = - = + - . . . Replacing 1 r with r and 2 r with R, the moment of inertia of the disk through its center is 1 2 2 disk 2 I = M(R + r ). (b) For r = 0 m, 1 2 disk 2 I = MR . This is the moment of inertia for a solid disk or cylinder about the center. Additionally, for r . R, we have I = MR2. This is the expression for the moment of inertia of a cylindrical hoop or ring about the center. (c) The initial gravitational potential energy of the disk is transformed into kinetic energy as it rolls down. If we choose the bottom of the incline as the zero of potential energy, and use cm v =. R, the energy conservation equation f i K =U is 2 2 2 2 2 cm 2 cm 2 cm i 2 2 2 2 2 2 22 cm 2 cm cm 2 2 2 22 cm 2 cm 1 1 1 ( ) 1 (0.50 m)sin 20 2 2 2 2 2 1 11 1.6759 m /s 4 2 2 4 4 3 (0.015 m) 1.6759 m /s 1.37 m/s 4 4(0.020 m) I Mv Mgh M R r v Mv Mgy Mg R v R r v v r R R v v . + = . .. .. + + = = . . . + . . . . . . + = . + + . = . . . . . . . + . = . = . . For a sliding particle on a frictionless surface f iK =U , so 2 cm f i f i f 1 2 2 (0.50 m)sin 20 1.83 m/s 0.75 2 mv mgy v gy g v v = . = = = . = That is, cm v is 75% of the speed of a particle sliding down a frictionless ramp. 12.59. Model: The plate has uniform density. Visualize: Solve: The moment of inertia is I = .r2dm. Let the mass of the plate be M. Its area is L2. A region of area dA located at (x,y) has mass 2 dm M dA M dx dy. A L = = The distance from the axis of rotation to the point (x, y) is r = x2 + y2 . With 2 2 - L = x = L and , 2 2 - L = y = L ( ) 2 2 2 2 3 2 2 2 2 2 2 2 2 2 2 3 2 3 2 2 3 4 3 2 2 2 2 2 2 2 2 4 3 3 2 3 24 2 24 2 12 12 3 1 12 24 24 6 L L L L L L L L L L L L L L I x y M dxdy M x y x dy L L M L y L L y L dy M L Ly dy M L L y L L L M L L L L ML L - - - - - - - . . . . = + . . = . + . . . . . . . . . . - . . . . . = .. + - . - ... = . + . = . + . . . .. . . . . . . . . .. = .. + . + ... = . . .. . . . . . 2 12.60. Solve: From Equation 12.16, I = .(x2 + y2 )dm A small region of area dA has mass dm, and dm M dA M dx dy A A = = The area of the plate is 1 2 2 (0.20 m)(0.30 m) = 0.030 m . So 2 2 0.800 kg 26.67 kg m 0.030 m M A = . . = . . . . The limits for x are 0 = x = 30 cm. For a particular value of x, 1 1 . 3 3 - x = y = x Note that 1 3 is the slope of the top and bottom edges of the triangle. Therefore, ( )( ) ( ) ( ) ( ) 1 30 cm 3 2 2 2 0 1 3 1 30 cm 3 3 2 2 0 1 3 30 cm 3 2 3 0 30 cm 2 4 2 0 26.67 kg/m 26.67 kg/m 3 26.67 kg/m 2 2 3 81 26.67 kg/m 56 1 0.037 kg m 81 4 x x x x I x y dxdy x y y dx x x dx x - - = + . . = . + . . . . . = . + . . . = . .. . = . .. . . .. . . . . . 12.61. Model: The ladder is a rigid rod of length L. To not slip, it must be in both translational equilibrium (Fnet = 0 N) .. .. and rotational equilibrium net (t = 0 N m). We also apply the model of static friction. Visualize: Since the wall is frictionless, the only force from the wall on the ladder is the normal force 2n.. . On the other hand, the floor exerts both the normal force 1 n.. and the static frictional force s f . .. The gravitational force G F .. on the ladder acts through the center of mass of the ladder. Solve: The x- and y-components of net F = 0 N .. .. are 2 s s 2 1 G 1 G 0 N 0 N x y SF = n - f = . f = n SF = n - F = .n = F The minimum angle occurs when the static friction is at its maximum value s max s 1f = n . Thus we have 2 s s 1 s . n = f = n = mg We choose the bottom corner of the ladder as a pivot point to obtain net t , because two forces pass through this point and have no torque about it. The net torque about the bottom corner is net 1 2 2 min min s t = d mg - d n = (0.5Lcos. )mg - (Lsin. ) mg = 0 N m min s min min min s 0.5cos sin tan 0.5 0.5 1.25 51 0.4 . . . . . = . = = = . = 12.62. Model: The beam is a rigid body of length 3.0 m and the student is a particle. Visualize: Solve: To stay in place, the beam must be in both translational equilibrium (Fnet = 0 N) .. .. and rotational equilibrium net (t = 0 Nm). The first condition is ( ) ( ) G beam G student 1 2 0 N ySF = - F - F + F + F = 2 1 2 G beam G student .F + F = (F ) + (F ) = (100 kg + 80 kg)(9.80 m/s ) =1764 N Taking the torques about the left end of the beam, the second condition is G beam G student 2 -(F ) (1.5 m) - (F ) (2.0 m) + F (3.0 m) = 0 N m 2 2 2 -(100 kg)(9.8 m/s )(1.5 m) - (80 kg)(9.8 m/s )(2.0 m) + F (3.0 m) = 0 N m 2 . F =1013 N From 1 2 F + F =1764 N, we get 1 F =1764 N -1013 N = 0.75 kN. Assess: To establish rotational equilibrium, the choice for the pivot is arbitrary. We can take torques about any point on the body of interest. 12.63. Model: The structure is a rigid body. Visualize: Solve: We pick the left end of the beam as our pivot point. We dont need to know the forces h F and v F because the pivot point passes through the line of application of h F and v F and therefore these forces do not exert a torque. For the beam to stay in equilibrium, the net torque about this point is zero. We can write about left end G B G W ( ) (3.0 m) t = - F - (F ) (4.0 m) + (T sin150)(6.0 m) = 0 N m Using 2 G B (F ) = (1450 kg)(9.8 m/s ) and 2 G W (F ) = (80 kg)(9.8 m/s ), the torque equation can be solved to yield T =15,300 N. The tension in the cable is slightly more than the cable rating. The worker should be worried. 12.64. Model: Model the beam as a rigid body. For the beam not to fall over, it must be both in translational equilibrium (Fnet = 0 N) .. .. and rotational equilibrium net (t = 0 N m). Visualize: The boy walks along the beam a distance x, measured from the left end of the beam. There are four forces acting on the beam. 1 F and 2 F are from the two supports, ( ) G b F .. is the gravitational force on the beam, and ( ) G B F .. is the gravitational force on the boy. Solve: We pick our pivot point on the left end through the first support. The equation for rotational equilibrium is G b 2 G B -(F ) (2.5 m) + F (3.0 m) - (F ) x = 0 N m 2 2 2 -(40 kg)(9.80 m/s )(2.5 m) + F (3.0 m) - (20 kg)(9.80 m/s )x = 0 N m The equation for translation equilibrium is ( ) ( ) ( ) ( ) y 1 2 G b G B 2 1 2 G b G B 0 N (40 kg 20 kg)(9.8 m/s ) 588 N F F F F F F F F F = = + - - . + = + = + = S Just when the boy is at the point where the beam tips, 1 F = 0 N. Thus 2 F = 588 N. With this value of 2 F , we can simplify the torque equation to: -(40 kg)(9.80 m/s2 )(2.5 m) + (588 N)(3.0 m) - (20 kg)(9.80 m/s2 )x = 0 N m . x = 4.0 m Thus, the distance from the right end is 5.0 m- 4.0 m =1.0 m. 12.65. Visualize: Please refer to Figure P12.65. Solve: The bricks are stable when the net gravitational torque on each individual brick or combination of bricks is zero. This is true as long as the center of gravity of each individual brick and any combination is over a base of support. To determine the relative positions of the bricks, work from the top down. The top brick can extend past the second brick by L 2. For maximum extension, their combined center of gravity will be at the edge of the third brick, and the combined center of gravity of the three upper bricks will be at the edge of the fourth brick. The combined center of gravity of all four bricks will be over the edge of the table. Measuring from the left edge of the brick 2, the center of gravity of the top two bricks is 1 1 2 2 12 com 1 2 ( ) 2 3 . 2 4 m L mL x m x m x L m m m . . + + . . = = . . = + Thus the top two bricks can extend L 4 past the edge of the third brick. The top three bricks have a center of mass ( ) 1 1 2 2 3 3 123 com 1 2 3 3 5 2 4 4 5 . 3 6 m L m L m L x m x m x m x L m m m m . . + . . + . . + + . . . . . . = = . . . . . . = + + Thus the top three bricks can extend past the edge of the fourth brick by L 6. Finally, the four bricks have a combined center of mass at ( ) 1234 com 4 11 177 2 6 12 12 . 4 8 m L m L m L m L x L m . . + . . + . . + . . . . . . . . . . = . . . . . . . . = The center of gravity of all four bricks combined is 7 L 8 from the left edge of the bottom brick, so brick 4 can extend L 8 past the table edge. Thus the maximum distance to the right edge of the top brick from the table edge is max 25 . 8 6 4 2 24 d = L + L + L + L = L Thus, yes, it is possible that no part of the top brick is directly over the table because max d > L. Assess: As crazy as this seems, the center of gravity of all four bricks is stably supported, so the net gravitational torque is zero, and the bricks do not fall over. 12.66. Model: The pole is a uniform rod. The sign is also uniform. Visualize: Solve: The geometry of the rod and cable give the angle that the cable makes with the rod. tan 1 250 51.3 200 . = - .. .. = . . The rod is in rotational equilibrium about its left-hand end. ( )( ) ( ) ( ) ( ) ( ) ( ) ( )( )( ) ( )( ) ( ) net G P G S G S 2 2 S 0 100 cm 80 cm 1 200 cm 1 200 cm sin51.3 2 2 100 cm 5.0 kg 9.8 m/s 9.8 m/s 140 cm 156 cm F F F T m T t = = - - .. .. - .. .. + . . . . = - - + With S T = 300 N, m = 30.6 kg. Assess: A mass of 30.6 kg is reasonable for a sign. 12.67. Model: The hollow cylinder is a rigid rotating body. Visualize: We placed the origin of the coordinate system on the ground. Solve: (a) Newtons second law for the block is -(FG )B + T = mBay , where T is the tension in the string, G B B (F ) = m g is the gravitational force on the block, and y a is the acceleration of block. The string tension exerts a negative (cw) torque on the cylinder, so the rotational form of Newtons second law for the hollow cylinder is 2 y y a Ia TR I I T R R t = - = a = . = - where we used the acceleration constraint . ya =a R With this expression for T, Newtons second law for the block becomes B B 2 B 2 B ( / ) y y m g Ia m a a m g R m I R - - - = . = + The moment of inertia of a hollow cylinder is 2 C I = m R , so the equation for y a simplifies to 2 B 2 B C (3.0 kg)(9.8 m/s ) 5.88 m/s 3.0 kg 2.0 kg y a m g m m - - = = =- + + The speed of the block just before it hits the ground can now be found using kinematics: 2 2 2 2 2 1 0 1 0 1 2 ( ) 0 m /s 2( 5.88 m/s )(0 m 1.0 m) 3.4 m/s y v = v + a y - y = + - - .v = (b) The conservation of energy equation 1 g1 0 g0 K +U = K +U for the system (block + cylinder + earth) is ( ) 2 2 2 2 B 1 1 B 1 B 0 0 B 0 2 2 2 1 2 B C B 1 C 2 B 0 1 B 0 2 2 B 0 2 2 1 1 B C 1 1 1 1 2 2 2 2 1 1 0 J 0 J 0 J 2 2 2 2 2 2(3.0 kg)(9.8 m/s )(1.0 m) 11.76 m /s 3.4 m/s (3.0 kg) (2.0 kg) m v I m gy m v I m gy m v m R v m gy v m m m gy R v m gy v m m + . + = + . + + + = + + . . + . = . . . . . = = = . = + + Assess: Newtons second law and the conservation of energy method give the same result for the blocks final velocity. 12.68. Model: The bar is a solid body rotating through its center. Visualize: Solve: (a) The two forces form a couple. The net torque on the bar about its center is net LF I F I L a t = = a . = where F is the force produced by one of the air jets. We can find I and a as follows: 2 2 2 2 1 0 1 0 2 2 1 1(0.50 kg)(0.60 m) 0.015 kg m 12 12 ( ) 150 rpm 5.0 rad/s 0 rad (10 s 0 s) 0.50 rad/s (0.015 kg m )(0.5 rad/s ) 0.0393 N (0.60 m) I ML t t F . . a p a a p p = = = = + - . = = + - . = . = = The force F = 39 mN. (b) The torque of a couple is the same about any point. It is still net t = LF. However, the moment of inertia has changed. 2 2 2 net 2 2 where 1 1 (0.500 kg)(0.6 m) 0.060 kg m 3 3 (0.0393 N) (0.60 m) 0.393 rad/s 0.060 kg m LF I LF I ML I t a a a = = . = = = = . = = Finally, 2 1 0 1 0 ( ) 0 rad/s (0.393 rad/s )(10 s 0 s) 3.93 rad/s (3.93)(60) rpm 37.5 rpm 2 . . a t t p = + - = + - = = = The angular speed is 38 rpm. Assess: Note that . .a and a .1/ I. Thus, . .1/ I. I about the center of the rod is 4 times smaller than I about one end of the rod. Consequently, . is 4 times larger. 12.69. Model: The flywheel is a rigid body rotating about its central axis. Visualize: Solve: (a) The radius of the flywheel is R = 0.75 m and its mass is M = 250 kg. The moment of inertia about the axis of rotation is that of a disk: 1 2 1 (250 kg)(0.75 m)2 70.31 kg m2 2 2 I = MR = = The angular acceleration is calculated as follows: 2 2 net net t = Ia .a =t / I = (50 N m)/(70.31 kg m ) = 0.711 rad/s Using the kinematic equation for angular velocity gives 2 1 0 1 0 1 1 ( ) 1200 rpm 40 rad/s 0 rad/s 0.711 rad/s ( 0 s) 177 s t t t t . =. +a - = = p = + - . = (b) The energy stored in the flywheel is rotational kinetic energy: 2 2 2 5 rot 1 1 1(70.31 kg m )(40 rad/s) 5.55 10 J 2 2 K = I. = p = The energy stored is 5.6105 J. (c) 5 Average power delivered energy delivered (5.55 10 J)/2 1.39 105 W 139 kW time interval 2.0 s = = = = (d) Because I , I I full energy half energy . t t . . . t a t . . - . = . = = . . . . . . full energy 1 . =. (from part (a)) = 40p rad/s. half energy . can be obtained as: 5 2 rot half energy rot half energy 2 1 1 5.55 10 J 88.85 rad/s 2 2 70.31 kg m I K K I . . = . = = = Thus (70.31 kg m2 ) 40 rad/s 88.85 rad/s 1.30 kN m 2.0 s p t . - . = . . = . . 12.70. Model: The pulley is a rigid rotating body. We also assume that the pulley has the mass distribution of a disk and that the string does not slip. Visualize: Because the pulley is not massless and frictionless, tension in the rope on both sides of the pulley is not the same. Solve: Applying Newtons second law to m1, 2 m , and the pulley yields the three equations: ( ) ( ) 1 G 1 1 1 G 2 2 2 2 2 1 T - F = m a - F + T = m a T R -T R - 0.50 N m = Ia Noting that 2 1 -a = a = a, 1 2 2 p I = m R , and a = a/R, the above equations simplify to 2 1 1 1 2 2 2 2 1 p p 1 1 0.50 N m 1 0.50 N m 2 2 0.060 m T m g m a m g T m a T T m R a m a R R R - = - = - = . .. . + = + . .. . . .. . Adding these three equations, 2 1 1 2 p 2 2 1 2 1 1 2 2 p ( ) 1 8.333 N 2 ( ) 8.333 N (4.0 kg 2.0 kg)(9.8 m/s ) 8.333 N 1.610 m/s 2.0 kg 4.0 kg (2.0 kg/2) m m g a m m m a m m g m m m - = . + + . + . . . . - - - - . = = = + + + + We can now use kinematics to find the time taken by the 4.0 kg block to reach the floor: 2 2 2 1 0 0 1 0 2 1 0 1 1 2 ( ) 1 ( ) 0 1.0 m 0 1 ( 1.610 m/s )( 0 s) 2 2 2(1.0 m) 1.11 s (1.610 m/s ) y y v t t a t t t t = + - + - . = + + - - . = = 12.71. Model: Assume the string does not slip on the pulley. Visualize: The free-body diagrams for the two blocks and the pulley are shown. The tension in the string exerts an upward force on the block m2 , but a downward force on the outer edge of the pulley. Similarly the string exerts a force on block 1 m to the right, but a leftward force on the outer edge of the pulley. Solve: (a) Newtons second law for 1 m and 2 m is 1 1 T = m a and 2 22 . T - m g = m a Using the constraint 2 1 -a = +a = a, we have 1 T = m a and 2 2-T + m g = m a. Adding these equations, we get 2 1 2 m g = (m + m )a, or 2 12 1 1 2 1 2 a m g T m a m m g m m m m = . = = + + (b) When the pulley has mass m, the tensions 1 2 (T and T ) in the upper and lower portions of the string are different. Newtons second law for 1 m and the pulley are: 1 1 1 2 T = m a and T R -T R = -Ia We are using the minus sign with a because the pulley accelerates clockwise. Also, a = Ra. Thus, 1 1 T = m a and 2 1 2 T T I a aI RR R - = = Adding these two equations gives 2 1 2 T a m I R = . + . . . . . Newtons second law for 2 m is 2 2 2 2 2 .T - m g = m a = -m a Using the above expression for 2 T , 2 1 2 2 2 2 1 2 / a m I m a m g a m g R m m I R . + . + = . = . . + + . . Since 1 2 2 p I = m R for a disk about its center, 2 1 1 2 2 p a m g m m m = + + With this value for a we can now find 1 T and 2 T : ( ) ( 1 ) 1 2 2 2 2 1 2 p 1 1 1 2 1 1 1 p 1 1 2 2 p 1 2 2 p 1 2 2 p ( / ) 1 2 m m g m g m m m g T ma T am I R m m m m m m m m m m m . . + = = = + = . + . = + + + + . . + + Assess: For m = 0 kg, the equations for a, 1T , and 2 T of part (b) simplify to 2 12 12 1 2 1 2 1 2 1 2 a m g and T m m g and T m m g m m m m m m = = = + + + These agree with the results of part (a). 12.72. Model: The disk is a rigid spinning body. Visualize: Please refer to Figure P12.72. The initial angular velocity is 300 rpm or (300)(2p )/60 =10p rad/s. After 3.0 s the disk stops. Solve: Using the kinematic equation for angular velocity, 1 0 2 1 0 1 0 1 0 ( ) (0 rad/s 10 rad/s) 10 rad/s (3.0 s 0 s) 3 t t t t . . p p . . a a - - - = + - . = = = - - Thus, the torque due to the force of friction that brings the disk to rest is (1 2 ) 2 1 ( ) 1 (2.0 kg)(0.15 m) 10 rad/s2 1.57 N 2 2 3 I mR I fR f mR R R a a p t = a = - . = - = - = - a = - ..- .. = . . The minus sign with t = - fR indicates that the torque due to friction acts clockwise. 12.73. Model: The entire structure is a rigid rotating body. The two thrust forces are a couple that exerts a torque on the structure about its center of mass. We will assume the thrust forces are perpendicular to the connecting tunnel. Visualize: Please refer to Figure 12.30. We chose a coordinate system in which 1 m and 2 m are on the x-axis and m1 = 100,000 kg is at the origin. m3 is the mass of the tunnel, whose center is at x3 = 45 m. Solve: (a) Assuming the center of mass of the tunnel is at the center of the tunnel, we get 5 5 4 1 1 2 2 3 3 cm cm 5 5 4 1 2 3 (1.0 10 kg)(0 m) (2.0 10 kg)(90 m) (5.0 10 kg)(45 m) 57.9 m 1.0 10 kg 2.0 10 kg 5.0 10 kg x m x m x m x x m m m + + + + = = = + + + + The center of mass of the entire structure is 58 m from the 100,000 kg rocket. (b) Initially, the angular velocity is zero. The structures angular velocity after 30 s is 1 0 1 0 . =. +a (t - t ) = 0 rad/s +a (30 s - 0 s) =a (30 s) The angular acceleration a can be found from t = Ia , where t is the net torque on the structure and I is its moment of inertia. The two thrusts form a couple with torque t = lF = (90 m)(50,000 N) = 4.50106 N m ( ) ( ) ( )( ) 1 2 2 2 2 2 m m tunnel 1 1 2 2 3 3 5 2 5 2 4 2 4 2 8 2 1 90 m 58 m 45 m 12 (1.0 10 kg)(57.9 m) (2.0 10 kg)(32.1 m) 1 (5 10 kg)(90 m) 5 10 kg 13 m 12 5.83 10 kg m I = I + I + I = m x + m x + .. m + m - .. . . = + + + = 6 3 2 8 2 3 2 1 4.5 10 N m 7.71 10 rad/s 5.83 10 kg m (30 s)(7.71 10 rad/s ) 0.23 rad/s I t a . - - . = = = . = = Assess: Note that the parallel axis theorem was used in finding the moment of inertia of the tunnel. 12.74. Model: Assume that the hollow sphere is a rigid rolling body and that the sphere rolls up the incline without slipping. We also assume that the coefficient of rolling friction is zero. Visualize: The initial kinetic energy, which is a combination of rotational and translational energy, is transformed in gravitational potential energy. We chose the bottom of the incline as the zero of the gravitational potential energy. Solve: The conservation of energy equation Kf +Ugf = Ki +Ugi is 2 2 2 2 1 cm cm 1 1 0 cm cm 0 0 2 2 2 2 2 20 cm 1 0cm 0cm 1 0cm 2 5 2 2 2 6 0 cm 1 0cm 1 2 1 ( ) 1 ( ) 1 ( ) 1 ( ) 2 2 2 2 0 J 0 J 1 ( ) 1 2 ( ) 0 J 1 ( ) 1 ( ) 2 23 2 3 5 ( ) 5 (5.0 m/s) ( ) 2.126 m 6 69.8 m/s M v I Mgy M v I Mgy Mgy M v MR Mgy M v MR v R gy v y v g . . . + + = + + + + = + . . + . = + . . . . . = . = = = The distance traveled along the incline is 1 2.126 m 4.3 m sin30 0.5 s = y = = Assess: This is a reasonable stopping distance for an object rolling up an incline when its speed at the bottom of the incline is approximately 10 mph. 12.75. Model: The masses are particles. Visualize: Solve: (a) The moment of inertia of the barbell is 2 ( )2 M m I = I + I = Mx + m L - x The rotational kinetic energy is therefore 2 ( 2 ( )2 ) 2 rot 1 1 2 2 K = I. = Mx + m L - x . To find x such that rot K is a minimum, set rot ( ( )) 2 0 1 2 2 2 dK Mx m L x x m L dx m M = = - - . . = + (b) This is the center of mass location measured from the mass M. 12.76. Model: The disk is a rigid body rotating on an axle passing through one edge. The gravitational potential energy is transformed into rotational kinetic energy as the disk is released. Visualize: We placed the origin of the coordinate system at a distance R just below the axle. In the initial position, the center of mass of the disk is at the same level as the axle. The center of mass of the disk in the final position is coincident with the origin of the coordinate system. Solve: (a) The torque is due to the gravitational force on the disk acting at the center of mass. Thus t = (mg)R = (5.0 kg)(9.8 m/s2 )(0.30 m) =14.7 N m The moment of inertia about the disks edge is obtained using the parallel-axis theorem: 2 2 2 2 2 2 cm 2 2 1 3 3(5.0 kg)(0.30 m) 0.675 kgm 2 2 2 14.7 N m 22 rad/s 0.675 kg m I I mR mR mR mR I t a = + = + = = . . = . . . . . = = = (b) The energy conservation equation f gf i gi K +U = K =U is 2 2 2 1 1 0 0 1 2 1 2 1 1 1 0 J 0 J 2 2 2 2 2(5.0 kg)(9.8 m/s )(0.30 m) 6.6 rad/s 0.675 kg m I mgy I mgy I mgR mgR I . . . . + = + . + = + = = = Assess: An angular velocity of 6.6 rad/s (or 1.05 revolutions/s) as the center of mass of the disk reaches below the axle is reasonable. 12.77. Model: The hoop is a rigid body rotating about an axle at the edge of the hoop. The gravitational torque on the hoop causes it to rotate, transforming the gravitational potential energy of the hoops center of mass into rotational kinetic energy. Visualize: We placed the origin of the coordinate system at the hoops edge on the axle. In the initial position, the center of mass is a distance R above the origin, but it is a distance R below the origin in the final position. Solve: (a) Applying the parallel-axis theorem, 2 2 2 2 Iedge = Icm + mR = mR + mR = 2mR . Using this expression in the energy conservation equation f gf i gi K +U = K +U yields: 2 2 22 edge 1 1 edge 0 0 1 1 1 1 1 (2 ) 0 J 2 2 2 2 I mgy I mgy mR mgR mgR g R . + = . + . - = + .. = (b) The speed of the lowest point on the hoop is 1 v ( )(2R) 2g (2R) 8gR R = . = = Assess: Note that the speed of the lowest point on the loop involves a distance of 2R instead of R. 12.78. Model: The long, thin rod is a rigid body rotating about a frictionless pivot on the end of the rod. The gravitational torque on the rod causes it to rotate, transforming the gravitational potential energy of the rods center of mass into rotational kinetic energy. Visualize: We placed the origin of the coordinate system at the pivot point. In the initial position, the center of mass is a distance 1 2 L above the origin. In the final position, the center of mass is at y = 0 m and thus has zero gravitational potential energy. Solve: (a) The energy conservation equation for the rod f gf i gi K +U = K +U is 2 2 22 1 1 0 0 1 1 1 1 1 1 0 J 0 J ( /2) 3 / 2 2 23 I. + mgy = I. + mgy .. mL ... + = + mg L .. = g L . . (b) The speed at the tip of the rod is tip 1 v = (. )L = 3gL. 12.79. Model: The sphere attached to a thin rod is a rigid body rotating about the rod. Assume the rod is vertical and the sphere solid. Visualize: Please refer to Figure P12.79. The sphere rotates because the string wrapped around the rod exerts a torque t. Solve: The torque exerted by the string on the rod is t = Tr. From the parallel-axis theorem, the moment of inertia of the sphere about the rods axis is 2 2 2 2 off center cm 2 13 2 5 4 20 I = I + M . R . = MR + MR = MR . . . . From Newtons second law, 2 2 20 (13 /20) 13 Tr Tr I MR MR t a = = = 12.80. Model: The pulley is a rigid rotating body. Visualize: We placed the origin of the coordinate system on the floor. The pulley rotates about its center. Solve: Using kinematics for the physics book (mass = m1), 2 2 2 1 0 0 1 0 1 0 ( ) 1 ( ) 0 m 1.0 m 0 m 1 (0.71 s 0 s) 3.967 m/s 2 2 y = y + v t - t + a t - t = + + a - .a = - Since the torque acting on the pulley is t = -TR = Ia , we have 2 / I TR TR TR a a R a = - = - = - We can compare the measured value of I for the pulley with the theoretical value. We first must find the tension T. From the free-body diagram, Newtons second law of motion is 1 1T - m g = m a. This means 2 2 1T = m (g + a) = (1.0 kg)(9.80 m/s - 3.967 m/s ) = 5.833 N With these values of T and a, we can now find I as: 2 2 3 2 2 (5.833 N)(0.06 m) 5.3 10 kg m ( 3.967 m/s ) I TR a = - = - = - - Let us now calculate the theoretical values of I: 2 2 3 2 2 2 3 2 disk hoop about center: (2.0 kg)(0.06 m) 7.2 10 kg m disk about center: 1 1 (2.0 kg)(0.06 m) 3.6 10 kg m 2 2 I MR I MR - - = = = = = = Since disk hoop I < I < I , the mass of the disk is not uniformly distributed. The mass is concentrated near the rim. 12.81. Model: The angular momentum of the satellite in the elliptical orbit is a constant. Visualize: Solve: (a) Because the gravitational force is always along the same direction as the direction of the moment arm vector, the torquet = r Fg .. .. .. is zero at all points on the orbit. (b) The angular momentum of the satellite at any point on the elliptical trajectory is conserved. The velocity is perpendicular to r.. at points a and b, so = 90 and L = mvr. Thus a b a bb aa b a b a b b 30,000 km 9000 km 6000 km and 30,000 km 9000 km 24,000 km 2 2 6000 km (8000 m/s) 2000 m/s 24,000 km L L mv r mv r v r v r r r v . . = . = . =. . . . = - = = + = . . . = . . = . . (c) Using the conservation of angular momentum c aL = L , we get a 2 2 7 c c c a a c a c c c mv r sin mv r v r v /sin r (9000 km) (12,000 km) 1.5 10 m r . . = . = . . = + = . . From the figure, we see that c sin =12,000 15,000 = 0.80. Thus c 6000 km (8000 m/s) 4000 m/s 15,000 km 0.80 v . . = . . = . . 12.82. Model: The clay balls are particles and undergo a totally inelastic collision. Linear momentum is conserved during the collision. Visualize: Solve: (a) The angle is measured counterclockwise from r.. to v... From geometry, 1 ( ) 1 =180 + tan- 1 = 225, and 1 2 tan 1 26.6 . 2 = - .. .. = . . So ( )( )( ) ( )( )( ) 1 2 111 1 22 2 2 2 sin sin 0.015 kg 2 m 2 m/s sin 225 0.025 kg 5 m 2.0 m/s sin 26.6 0.020 kg m /s. L = L + L + m rv + m r v = + = Note that the signs of 1 L and 2 L agree with those determined by the right-hand rule. (b) At the instant before the clay balls collide they are located at (1.5 m, 1.0 m). Here, 1 2 1 2 tan 2 33.7 180 146.3 3 = - .. .. = = + = . . So ( )( ( ) ( ) )( ) ( )( ( ) ( ) )( ) 2 2 2 2 2 0.015 kg 1.0 m 1.5 m 2.0 m/s sin 213.7 0.025 kg 1.0 m 1.5 m 2.0 m/s sin33.7 0.020 kg m /s L= + + + = (c) The clay balls have a final speed v after the collision. Linear momentum is conserved. ( )( ) ( )( ) ( ) 1 2 (12) 0.015 kg 2.0 m/s 0.025 kg 2.0 m/s 0.015 kg 0.025 kg 0.50 m/s. i i f p p p v v + + = + - = + . = - The balls are moving to the left. The angle 2 = from part (b). The angular momentum after the collision is L = (0.040 kg)( (1.0 m)2 + (1.5 m)2 )(0.50 m/s)sin33.7 = 0.020 kg m2/s Assess: Angular momentum is also conserved since net t.. = 0. 12.83. Model: For the (bullet + block + rod) system, angular momentum is conserved. After the bullet is stuck in the block, the mechanical energy of the system is conserved. Assume the block is small. Visualize: The origin of the coordinate system was placed at the center-of-mass of the block as it freely hangs from the bottom of the rod. Solve: The initial angular momentum of the system about the pivot is due only to the bullet: i bb b b L = m v L = (0.010 kg)v (1.0 m) = (0.010 kgm)v The angular momentum immediately after the bullet hits and sticks in the block is equal to I.. The moment of inertia is 2 2 2 2 rod block bullet R B b R B b 2 2 1 1 3 3 1 (1.0 kg) 2.0 kg 0.010 kg (1.0 m) 2.343 kg m 3 I = I + I + I = m L + m L + m L = .. m + m + m ..L . . = . + + . = .. .. Angular momentum is conserved in the collision: 2 f i b L = (2.343 kg m ). = L = (0.010 kg m)v We need to determine . before we can find bv . To find . we use the conservation of mechanical energy equation f gf i gi K +U = K +U as the pendulum swings out, which is 2 b B B R R 2 2 22 0 J ( ) 1 0 J 2 (0.010 kg 2.0 kg)(9.8 m/s ) (1 cos30 ) (1.0 kg)(9.8 m/s ) (1 cos30 ) 1 (2.343 kg m ) 2 2 m m gy mgy I L L . . + + + = + + - + - = The energy equation can be further simplified to 2.6390 kg m2/s2 + 0.6565 kg m2/s2 = (1.1715 kg m2 ). 2 .. =1.677 rad/s Finally, we can use the conservation of angular momentum equation to obtain the speed of the bullet: 2 2 b (2.343 kg m )(1.677 rad/s) 3.9 10 m/s 0.010 kg m v= = Assess: A speed of 390 m/s for a bullet is reasonable. 12.84. Model: For the (bullet + door) system, the angular momentum is conserved in the collision. Visualize: Solve: As the bullet hits the door, its velocity v.. is perpendicular to r... Thus the initial angular momentum about the rotation axis, with r = L, is 2 i BB L = m v L = (0.010 kg)(400 m/s)(1.0 m) = 4.0 kg m /s After the collision, with the bullet in the door, the moment of inertia about the hinges is door bullet I = I + I 2 2 D B 1 3 = m L + m L 1 (10.0 kg)(1.0 m)2 (0.010 kg)(1.0 m)2 3.343 kg m2 3 = + = Therefore, 2 f L = I. = (3.343 kg m ).. Using the angular momentum conservation equation 2 f iL = L (3.343 kg m ). = 4.0 kg m2/s and thus . =1.20 rad/s. 12.85. Model: The mechanical energy of both the hoop (h) and the sphere (s) is conserved. The initial gravitational potential energy is transformed into kinetic energy as the objects roll down the slope. The kinetic energy is a combination of translational and rotational kinetic energy. We also assume no slipping of the hoop or of the sphere. Visualize: The zero of gravitational potential energy is chosen at the bottom of the slope. Solve: (a) The energy conservation equation for the sphere or hoop Kf +Ugf = Ki +Ugi is 2 2 2 2 1 1 1 0 0 0 1 ( ) 1 ( ) 1 ( ) 1 ( ) 2 2 2 2 I . + m v + mgy = I . + m v + mgy For the sphere, this becomes 2 2 1 s 2 2 1 s s 2 2 1 s 1 s 1 2 ( ) 1 ( ) 0 J 0 J 0 J 2 5 2 7 ( ) ( ) 10 /7 10(9.8 m/s )(0.30 m)/7 2.05 m/s 10 mR v m v mgh R v gh v gh . . + + = + + . . . . . = . = = = For the hoop, this becomes 2 2 1 h 2 2 1 h hoop 2 1 h hoop 1 ( ) ( ) 1 ( ) 0 J 0 J 0 J 2 2 ( ) mR v m v mgh R h v g + + = + + . = For the hoop to have the same velocity as that of the sphere, 2 2 1 s hoop 2 ( ) (2.05 m/s) 42.9 cm 9.8 m/s h v g = = = The hoop should be released from a height of 43 cm. (b) As we see in part (a), the speed of a hoop at the bottom depends only on the starting height and not on the mass or radius. So the answer is No. 12.86. Model: Model the turntable as a rigid disk rotating on frictionless bearings. As the blocks fall from above and stick on the turntable, the turntable slows down due to increased rotational inertia of the (turntable + blocks) system. Any torques between the turntable and the blocks are internal to the system, so angular momentum of the system is conserved. Visualize: The initial moment of inertia is 1 I and the final moment of inertia is 2I . Solve: The initial moment of inertia is 1 2 1 2 2 1 disk 2 2I = I = mR = (2.0 kg)(0.10 m) = 0.010 kg m and the final moment of inertia is 2 2 2 2 2 2 2 1 I = I + 2mR = 0.010 kg m + 2(0.500 kg) (0.10 m) = 0.010 kg m + 0.010 kg m = 0.020 kg m Let 1 . and 2 . be the initial and final angular velocities. Then 2 1 1 f i 22 11 2 2 2 (0.010 kg m )(100 rpm) 50 rpm 0.020 kg m L L I I I I . = .. =. .. = = = 12.87. Model: Model the turntable as a rigid disk rotating on frictionless bearings. For the (turntable + block) system, no external torques act as the block moves outward towards the outer edge. Angular momentum is thus conserved. Visualize: The initial moment of inertia of the turntable is 1 I and the final moment of inertia is 2I . Solve: The initial moment of inertia is 1 2 1 2 2 1 disk 2 2I = I = mR = (0.2 kg)(0.2 m) = 0.0040 kg m . As the block reaches the outer edge, the final moment of inertia is 2 2 2 2 1 B 2 2 2 0.0040 kg m (0.020 kg)(0.20 m) 0.0040 kg m 0.0008 kg m 0.0048 kg m I = I + m R = + = + = Let 1 . and 2 . be the initial and final angular velocities, then the conservation of angular momentum equation is 2 1 1 f i 22 11 2 2 2 (0.0040 kg m )(60 rpm) 50 rpm (0.0048 kg m ) L L I I I I . = .. =. .. = = = Assess: A change of angular velocity from 60 rpm to 50 rpm with an increase in the value of the moment of inertia is reasonable. 12.88. Model: Model the merry-go-round as a rigid disk rotating on frictionless bearings about an axle in the center and John as a particle. For the (merry-go-round + John) system, no external torques act as John jumps on the merry-go-round. Angular momentum is thus conserved. Visualize: The initial angular momentum is the sum of the angular momentum of the merry-go-round and the angular momentum of John. The final angular momentum as John jumps on the merry-go-round is equal to final finalI . . Solve: Johns initial angular momentum is that of a particle: J JJ JJ L = m v Rsin = m v R. The angle = 90 since John runs tangent to the disk. The conservation of angular momentum equation f i L = L is 2 final final disk J i J J 2 2 1 2 1 (250 kg)(1.5 m) (20 rpm) 2 rad (30 kg)(5.0 m/s)(1.5 m) 814 kg m /s 2 60rpm I . L L MR . m v R p = + = . . + . . . . . . . . = . . . . + = . . . . 2 final final 2 2 2 2 2 final disk J J 2 final 2 814 kg m /s 1 1(250 kg)(1.5 m) (30 kg)(1.5 m) 349 kgm 2 2 814 kg m /s 2.33 rad/s 22 rpm 349 kg m I I I I MR mR . . . = = + = + = + = = = = 12.89. Model: The toy car is a particle located at the rim of the track. The track is a cylindrical hoop rotating about its center, which is an axis of symmetry. No net torques are present on the track, so the angular momentum of the car and track is conserved. Visualize: Solve: The toy cars steady speed of 0.75 m/s relative to the track means that vc - vt = 0.75 m/s.vc = vt + 0.75 m/s, where t v is the velocity of a point on the track at the same radius as the car. Conservation of angular momentum implies that ( ) ( ) i f 2 2 c c t t c t c t 0 L L I . I. mr . Mr . m. M. = = + = + = + The initial and final states refer to before and after the toy car was turned on. Table 12.2 was used for the track. Since c c v , r . = t t v , r . = we have ( ) ( ) ( ) ( )( ) c t t t t 0 0.75 m/s 0 0.200 kg 0.75 m/s 0.75 m/s 0.125 m/s 0.200 kg 1.0 kg mv Mv m v Mv v M m M = + . + + = . =- = - = - + + The minus sign indicates that the track is moving in the opposite direction of the car. The angular velocity of the track is ( ) t t 0.125 m/s 0.417 rad/s clockwise. 0.30 m v r . = = = In rpm, ( ) t 0.417 rad/s rev 60 s 2 rad min 4.0 rpm . p = . .. . . .. . . .. . = Assess: The speed of the track is less than that of the car because it is more massive. 12.90. Model: Model the skater as a cylindrical torso with two rod-like arms that are perpendicular to the axis of the torso in the initial position and collapse into the torso in the final position. Visualize: Solve: For the initial position, the moment of inertia is I1 = ITorso + 2IArm. The moment of inertia of each arm is that of a 66 cm long rod rotating about a point 10 cm from its end, and can be found using the parallel-axis theorem. In the final position, the moment of inertia is 1 2 2 2 I = MR . The equation for the conservation of angular momentum f i L = L can be written 2 2 1 1 I . = I. . 2 1 2 1 . = (I / I ). . Calculating 1 I and 2 I , ( ) 2 2 2 1 T AA A 2 2 2 2 2 2 2 2 2 2 2 1 2 1 2 12 1 (40 kg)(0.10 m) 2 1 (2.5 kg)(0.66 m) (2.5 kg) 0.33 m 0.10 m kg m 1.306 2 12 1 1 (45 kg)(0.10 m) 0.225 kg m (1.306 kg m ) ( rev/s) rev 2 2 (0.225 kg m ) 1.0 5.8 I MR ML Md I MR . = + . + . .. .. = + . + + . = .. .. = = = . = = /s 12.91. Model: Assume that the marble does not slip as it rolls down the track and around a loop-the-loop. The mechanical energy of the marble is conserved. Visualize: Solve: The balls center of mass moves in a circle of radius R - r. The free-body diagram on the marble at its highest position shows that Newtons second law for the marble is 2 1 mg n mv R r + = - The minimum height (h) that the track must have for the marble to make it around the loop-the-loop occurs when the normal force of the track on the marble tends to zero. Then the weight will provide the centripetal acceleration needed for the circular motion. For n.0 N, 2 2 1 ( ) ( ) mg mv v g R r R r = . = - - Since rolling motion requires 2 2 2 1 1v = r . , we have 2 2 2 1 1 2 r g(R r) g(R r) r . . - = - . = The conservation of energy equation is 2 2 f gf top of loop i gi initial 1 1 1 0 ( ) ( ) 1 1 2 2 K +U = K +U . mv + I. + mgy = mgy = mgh Using the above expressions and 2 2 5 I = mr the energy equation simplifies to 2 ( ) 2 1 ( ) 1 2 ( ) 2( ) 2.7 2 25 mg R r mr g R r mg R r mgh h R r r . . . - . - + . . . . + - = . = - . . . . 12.92. Model: The Swiss cheese wedge is of uniform densityor at least uniform enough that its center of mass is at the same location as that of a solid piece. To find the angle at which the cheese starts sliding, the cheese will be treated as a particle, and the model of static friction will be used. Visualize: Solve: The angle at which the cheese starts sliding, .S, will be compared to the critical angle c . for stability. Use Newtons second law with the free body diagram. ( ) ( ) net s G s net G s 0 sin 0 cos x y F f F F nF . . = = - = = - With G F = mg, the y-direction equation gives n = mg cos. . The cheese starts sliding when s is at its maximum value. Combining that with the x-direction equation and s s f = n, ( ) ( ) ( ) s s s 1 1 s s 0 cos sin tan tan 0.90 42 mg . mg . . - - = - . = = = The cheese will start sliding at an angle of 42. The center of mass of the cheese wedge can be found using the result of problem 12.53. There, the center of mass of a triangle with the same proportions as the cheese wedge was found. So cm x is at the center of the cheese wedge (by symmetry). The cm y can be found by proportional reasoning. ( ) cm cm 30 cm 20 cm 4.0 cm 12 cm 30 cm y y - = . = Note that here we have measured cm y from the base of the wedge. Stability considerations require that the center of mass be no further than the left corner of the wedge. At the critical angle geometry shown in the figure above, the right triangle formed by the wedges center of mass, lower left corner, and center point of the base is a 45-45-90 triangle. So c . = 45. Assess: The cheese will slide first as the incline reaches 42. It would not topple until the angle reaches 45. So Emily is correct. 12.93. Model: Define the system as the rod and cube. Energy and angular momentum are conserved in a perfectly elastic collision in the absence of a net external torque. The rod is uniform. Visualize: Please refer to Figure CP12.93. Solve: Let the final speed of the cube be vf , and the final angular velocity of the rod be .. Energy is conserved, and angular momentum around the rods pivot point is conserved. 2 2 2 i f 0 f rod i f 0 f rod 1 1 1 2 2 2 2 2 E E mv mv I L L mv d mv d I . . = . = + = . . . = . . + . . . . . . . . This is two equations in the two unknowns f v and .. From Table 12.2, 2 ( ) 2 2 rod 1 1 2 1 12 12 6 I = Md = m d = md From the angular momentum equation, ( ) 0 f 0 f 3 3 v v d v v d = + . .. = - Substituting into the energy equation, ( ) ( ) 2 2 2 2 0 f 2 0 f 2 2 2 0 f 0 f 2 2 f 0f 0 1 1 11 9 2 2 26 3 2 0 6 1 5 5 mv mv md v v d v v v v v vv v = + . .. . - . .. . . .. . = + - = - + This is a quadratic equation in f v . The roots are 2 2 0 0 0 f 00 0 0 6 6 4 5 5 5 3 2 2 5 5 1 5 v v v v vv v v . . . . . . - . . . . . . = = .. =... The answer f 0 v = v means the ice cube missed the rod. So f 0 1 5 v = v to the right. 12.94. Model: The clay ball is a particle. The rod is a uniform thin rod rotating about its center. Angular momentum is conserved in the collision. Visualize: Solve: This is a two-part problem. Angular momentum is conserved in the collision, and energy is conserved as the ball rises like a pendulum. The angular momentum conservation equation about the rods pivot point is Li = Lf ( ) 0 ball+rod .mv r = I . Note 15 cm. 2 r = L = The rod and ball are a composite object. From Table 12.2, 2 rod 1 , 12 I = ML so 2 2 2 2 2 ball+rod ball rod 1 1 12 4 12 4 3 I = I + I = mr + ML = m L + ML = L .m + M . . . . . If f v is the final velocity of the clay ball, f f v 2v r L . = = since the ball sticks to the rod. Thus ( )( ) ( ) ( ) 2 0 f 0 f 2 2 4 3 0.010 kg 2.5 m/s 0.714 m/s 0.075 kg 3 0.010 kg 3 mv L L m M v L v mvm M = . + .. . . .. . . .. . . = = = + + Energy is conserved as the clay ball rises. Compare the energy of the ball-rod system just after the collision to when the ball reaches the maximum height. Note that the center of mass of the rod does not change position. 2 i f rod + ball 1( ) 2 E = E . I . = mgh Thus ( ) 2 2 f 2 f 2 f 1 2 1 cos 2 4 3 3 1 cos 3 L m M v mgh v m M mgL L v m M mgL . . . + .. . = . . + . = - . .. . . . . .. . . . . - . + . = . . . . Using the various values, cos. = 0.393.. = 67. Assess: The clay ball rises (1 cos ) 9.1 cm. 2 h = L - . = This is about 2/3 of the height of the pivot point, and is reasonable. 12.95. Model: Because no external torque acts on the star during gravitational collapse, its angular momentum is conserved. Model the star as a solid rotating sphere. Solve: (a) The equation for the conservation of angular momentum is 2 2 i f i i f f i i f f i f i f 2 2 5 5 L L I I mR mR R R . . . . . . = . = .. . = . . . . . .
. . . . . = The angular velocity is inversely proportional to the period T. We can write f ( 8 ) 5 f i 6 i 7.0 10 m 0.10 s 1.3749 10 m 137 km 2.592 10 s R R T T = = = = (b) A point on the equator rotates with f r = R . Its speed is f 6 2 2 (137,490 m) 8.6 10 m/s 0.10 s v R T p p = = = 12-1 12.96. Model: For the (turntable + bicycle wheel + professor) system the angular momentum is conserved because the turntable is frictionless and no external torques act on the system. Solve: (a) Nothing happens. The bicycle wheel already has an angular momentum and nothing changes for the wheel when it is handed to the professor. So, nothing happens to the professor. (b) The initial angular momentum is 2 2 2 wheel i ( ) ( ) (4.0 kg)(0.32 m) 180 2 rad 7.72 kg m /s 60 s L I mR p . . . . = = = . . = . . When the wheel is turned upside down, the angular momentum of the wheel becomes 2 wheel f (L ) = -7.72 kg m /s Since the initial and final total angular momentum should be equal, the professor must acquire some angular momentum. From the angular momentum conservation equation, 2 2 2 prof wheel f wheel i prof wheel i wheel f L + (L ) = (L ) . L = (L ) - (L ) = 7.72 kg m /s - (-7.72 kg m /s) =15.44 kg m /s Let us now calculate the angular velocity of the turntable by using prof total L = I ., where total turntable body arms I = I + I + I + wheel in handI . We assume that the axis of the professor and turntable also become the axis of the arms and wheel (i.e., no leaning). 2 2 2 turntable T T 2 2 2 body body 2 2 2 arms arms 2 2 wheel in hand wheel 1 1(5.0 kg)(0.25 m) 0.156 kg m 2 2 1 (70 kg)(0.125 m) 0.547 kg m 2 2 1 2 (2.5 kg)(0.45 m) 0.338 kg m 3 3 (4.0 kg)(0.45 m) 0.810 kg I mR I m R I mR I m r = = = = = = = . . = = . . . . = = = m2 Thus, 2 2 total I = (0.156 + 0.547 + 0.338 + 0.810) kg m =1.851 kg m . Using this value for total I , we can now find . to be 2 prof 2 total 15.44 kg m /s 8.34 rad/s 79.7 rpm 1.851 kg m L I . = = = = 13.1. Model: Model the sun (s) and the earth (e) as spherical masses. Due to the large difference between your size and mass and that of either the sun or the earth, a human body can be treated as a particle. Solve: s y s on you 2 s e GM M F r - = and e y e on you 2 e GM M F r = Dividing these two equations gives 2 30 6 2 s on y s e 4 24 11 e on y e s e 1.99 10 kg 6.37 10 m 6.00 10 5.98 10 kg 1.50 10 m F M r F M r - - . .. . . .. . = . .. . = . .. . = . .. . . .. . 13.2. Model: Assume the two lead balls are spherical masses. Solve: (a) 11 2 2 1 2 9 1on 2 2 on12 2 (6.67 10 N m /kg )(10 kg)(0.100 kg) 6.7 10 N (0.10 m) F F Gm m r - - = = = = (b) The ratio of the above gravitational force to the gravitational force on the 100 g ball is 9 9 2 6.67 10 N 6.81 10 (0.100 kg)(9.8 m/s ) - - = Assess: The answer in part (b) shows the smallness of the gravitational force between two lead balls separated by 10 cm compared to the gravitational force on the 100 g ball. 13.3. Model: Model the sun (s), the moon (m), and the earth (e) as spherical masses. Solve: s m s on m 2 s m F GM M r - = and e m e on m 2 e m F GM M r - = Dividing the two equations and using the astronomical data from Table 13.2, 2 30 8 2 s on m s e m 24 11 e on m e s m 1.99 10 kg 3.84 10 m 2.18 5.98 10 kg 1.50 10 m F M r F M r - - . .. . . .. . = . .. . = . .. . = . .. . . .. . Note that the sun-moon distance is not noticeably different from the tabulated sun-earth distance. 13.4. Solve: s p sphere on particle 2 s p GM M F r - = and e p earth on particle 2 e GM M F r = Dividing the two equations, 2 6 2 sphere on particle s e 7 24 earth on particle e s p 5900 kg 6.37 10 m 1.60 10 5.98 10 kg 0.50 m F M r F M r - - . .. . . .. . = . ... .. = . .. . = . .. . . .. . 13.5. Model: Model the woman (w) and the man (m) as spherical masses or particles. Solve: 11 2 2 w m 7 w on m m on w2 2 m w (6.67 10 N m /kg )(50 kg)(70 kg) 2.3 10 N (1.0 m) F F GM M r - - - = = = = 13.6. Model: Model the earth (e) as a sphere. Visualize: The space shuttle or a 1.0 kg sphere (s) in the space shuttle is 6 6 Re + rs = 6.3710 m+ 0.3010 m = 6.67106m away from the center of the earth. Solve: (a) 11 2 2 24 e s e on s 2 62 e s (6.67 10 N m /kg )(5.98 10 kg)(1.0 kg) 9.0 N ( ) (6.6710 m) F GM M R r - = = = + (b) Because the sphere and the shuttle are in free fall with the same acceleration around the earth, there cannot be any relative motion between them. That is why the sphere floats around inside the space shuttle. 13.7. Model: Model the sun (s) as a spherical mass. Solve: (a) 11 2 2 30 s 2 sun surface2 82 s (6.67 10 N m /kg )(1.99 10 kg) 274 m/s (6.96 10 m) g GM R - = = = (b) 11 2 2 30 s 3 2 sun at earth2 112 s e (6.67 10 N m /kg )(1.99 10 kg) 5.90 10 m/s (1.50 10 m) g GM r - - - = = = 13.8. Model: Model the moon (m) and Jupiter (J) as spherical masses. Solve: (a) 11 2 2 22 m 2 moon surface2 62 m (6.67 10 N m /kg )(7.36 10 kg) 1.62 m/s (1.74 10 m) g GM R - = = = (b) 11 2 2 27 J 2 Jupiter surface2 72 J (6.67 10 N m /kg )(1.90 10 kg) 25.9 m/s (6.99 10 m) g GM R - = = = 13.9. Model: Model the earth (e) as a spherical mass. Visualize: The acceleration due to gravity at sea level is 9.83 m/s2 (see Table 13.1) and 6 e R = 6.3710 m (see Table 13.2). Solve: e e earth 2 observatory 2 2 2 e 2 e e e (9.83 0.0075) m/s ( ) 1 1 g GM GM g R h h h R R R = = = = - + . . . . . + . . + . . . . . Here 2 earth e e g = GM R is the acceleration due to gravity on a non-rotating earth, which is why weve used the value 9.83 m/s2. Solving for h, e 9.83 1 2.43 km 9.8225 h R . . = .. - .. = . . 13.10. Model: Model the earth (e) as a spherical mass. Solve: Let the acceleration due to gravity be surface 3g when the earth is shrunk to a radius of x. Then, e surface 2 e g GM R = and e surface 2 3g GM x = e e e 2 2 e e 3 0.577 3 GM GM x R R R x . = . = = The earths radius would need to be 0.577 times its present value. 13.11. Model: Model Planet Z as a spherical mass. Solve: (a) 11 2 2 Z 2 Z 24 Z surface 2 6 2 Z Z 8.0 m/s (6.67 10 N m /kg ) 3.0 10 kg (5.0 10 m) g GM M M R - = . = . = (b) Let h be the height above the north pole. Thus, 2 Z Z Z surface 2 above N pole 2 2 2 6 2 Z 2 Z 6 Z Z 8.0 m/s 0.89 m/s ( ) 10.0 10 m 1 1 1 5.0 10 m g GM GM g R h h h R R R = = = = = + . . . . . . . + . . + . . + . . . . . . . 13.12. Model: Model Mars (M) as a spherical mass. Ignore air resistance. Also consider Mars and the ball as an isolated system, so mechanical energy is conserved. Visualize: Solve: A height of 15 m is very small in comparison with the radius of earth or Mars. We can use flat-earth gravitational potential energy to find the speed with which the astronaut can throw the ball. On earth, with yi = 0 m and f v = 0 m/s, energy conservation is 1 2 1 2 2 2 f f 2 i i i e f mv + mgy = mv + mgy .v = 2g y = 2(9.80 m/s )(15 m) =17.1 m/s Energy is also conserved on Mars, but the acceleration due to gravity is different. 11 2 2 23 M 2 Mars surface2 62 M (6.67 10 N m /kg )(6.42 10 kg) 3.77 m/s (3.37 10 m) g GM R - = = = On Mars, with i y = 0 m and f v = 0 m/s, energy conservation is 2 2 1 2 1 2 i 2 f f 2 i i f 2 m (17.1 m/s) 39 m 2 2(3.77 m/s) mv mgy mv mgy y v g + = + . = = = 13.13. Model: Model Jupiter as a spherical mass and the object as a point particle. The object and Jupiter form an isolated system, so mechanical energy is conserved. The minimum launch speed for escape, which is called the escape speed, causes an object to stop only as the distance approaches infinity. Visualize: Solve: The energy conservation equation K2 +U2 = K1 +U1 is 2 J o 2 J o o 2 o 1 2 J 1 1 2 2 m v GM m m v GM m r R - = - where J R and J M are the radius and mass of Jupiter. Using the asymptotic condition 2 v = 0 m/s as 2 r .8, 11 2 2 27 2 J o J 4 o 1 1 7 J J 0 J 1 2 2(6.67 10 N m /kg )(1.90 10 kg) 6.02 10 m/s 2 6.9910 m m v GM m v GM R R - = - . = = = Thus, the escape velocity from Jupiter is 60.2 km/s. 13.14. Model: Model the earth (e) as a spherical mass. Compared to the earths size and mass, the rocket (r) is modeled as a particle. This is an isolated system, so mechanical energy is conserved. Visualize: Solve: The energy conservation equation K2 +U2 = K1 +U1 is 2 e r 2 e r r 2 r 1 2 e 1 1 2 2 m v GM m m v GM m r R - = - In the present case, 2 r .8, so 2 2 e r 2 2 e r 2 r 1 2 1 e e 1 1 2 2 2 m v m v GM m v v GM R R = - . = - 11 2 2 24 4 2 2 6 3 (1.50 10 m/s) 2(6.67 10 N m /kg )(5.98 10 kg) 6.37 10 m 9.99 10 m/s v - . = - = 13.15. Model: The probe and the sun form an isolated system, so mechanical energy is conserved. The minimum launch speed for escape, which is called the escape speed, causes an object to stop only as the distance approaches infinity. Visualize: We denote by p m the mass of the probe. S M is the suns mass, and S p R - is the separation between the centers of the sun and the probe. Solve: The conservation of energy equation K2 +U2 = K1 +U1 is 2 S p 2 S p p 2 p 1 2 Sp 1 1 2 () 2 GM m GM m m v m v r R- - = - Using the condition 2 2 v = 0 m/s asymptotically as r .8, 11 2 2 30 2 S p S 4 p escape escape 11 S p S p 1 2 2(6.67 10 N m /kg )(1.99 10 kg) 4.21 10 m/s 2 (1.50 10 m) GM m GM m v v R R - - - = . = = = 13.16. Model: Model the distant planet (p) and the earth (e) as spherical masses. Because both are isolated, the mechanical energy of the object on both the planet and the earth is conserved. Visualize: Let us denote the mass of the planet by Mp and that of the earth by Me. Your mass is 0m . Also, acceleration due to gravity on the surface of the planet is p g and on the surface of the earth is eg . p R and e R are the radii of the planet and the earth, respectively. Solve: (a) We are given that 1 P e p 4 e M = 2M and g = g . Since p p 2 p GM g R = and e e 2 e g GM , R = we have p e p 6 7 2 2 2 p e p e e 1 1 ( /2) 8 8(6.37 10 m) 1.80 10 m 4 4 GM GM G M R R R R R = = . = = = (b) The conservation of energy equation 2 2 1 1 K +U = K +U is 2 p o 2 p o o 2 o 1 2 p 1 1 2 2 GM m GM m m v m v r R - = - Using 2 v = 0 m/s as 2 r .8, we have 2 p o o escape p 11 2 2 24 p e escape 7 p p 1 2 2 2 (2 ) 4(6.67 10 N m /kg )(5.98 10 kg) 9.41 km/s 1.80 10 m GM m m v R GM G M v R R - = . = = = = 13.17. Model: Model the sun (s) as a spherical mass and the asteroid (a) as a point particle. Visualize: The asteroid, having mass ma and velocity va , orbits the sun in a circle of radius a r . The asteroids time period is a T = 5.0 earth years =1.5779108 s. Solve: The gravitational force between the sun S (mass = M ) and the asteroid provides the centripetal acceleration required for circular motion. 2 2 2 1/ 3 s a a a s a s a 2 a 2 a a a a 2 4 GM m m v GM r r GM T r r r T p p . . . . = . = . . . = . . . . . . Substituting G = 6.6710-11 Nm2 /kg2 , 30 s M =1.9910 kg, and the time period of the asteroid, we obtain 11 a r = 4.3710 m. The velocity of the asteroid in its orbit will therefore be 11 a 4 a 8 a 2 (2)(4.3710 m) 1.74 10 m/s 1.5779 10 s v r T p p = = = 13.18. Model: Model the sun (s) and the earth (e) as spherical masses. Visualize: The earth orbits the sun with velocity ve in a circular path with a radius denoted by rs-e. The suns and the earths masses are denoted by s M and em . Solve: The gravitational force provides the centripetal acceleration required for circular motion. 2 2 s e e e e s e 2 2 s e s e s e e GM m m v m (2 r ) r r rT p - - - - = = 2 3 2 11 3 s e 30 s 2 11 2 2 2 e 4 4(1.5010 m) 2.01 10 kg (6.67 10 N m /kg )(365 24 3600 s) M r GT p p - - . = = = Assess: The tabulated value is 1.991030 kg. The slight difference can be ascribed to the fact that the earths orbit isnt exactly circular. 13.19. Model: Model the star (s) and the planet (p) as spherical masses. Solve: A planets acceleration due to gravity is p p 2 p 2 2 6 2 p p 25 p 11 2 2 (12.2 m/s )(9.0 10 m) 1.48 10 kg 6.67 10 N m /kg GM g R g R M G - = . = = = (b) A planets orbital period is 2 2 3 s 2 3 2 11 3 30 s 2 11 2 2 2 4 4 4(2.210 m) 5.2 10 kg (6.67 10 N m /kg )(402 24 3600 s) T r GM M r GT p p p - . . =. . . . . = = = 13.20. Model: Model the planet and satellites as spherical masses. Visualize: Please refer to Figure EX13.20. Solve: (a) The period of a satellite in a circular orbit is T = [(4p 2/GM)r3]1/2. This is independent of the satellites mass, so we can find the ratio of the periods of two satellites a and b: 3 a a b b T r T r . . = . . . . Satellite 2 has 2 1r = r , so 2 1 T = T = 250 min. Satellite 3 has 3 1 r = (3/2)r , so 3/ 2 3 1 T = (3/2) T = 459 min. (b) The force on a satellite is F = GMm/r2. Thus the ratio of the forces on two satellites a and b is 2 a b a b a b F r m F r m . . . . = . . . . . . . . Satellite 2 has 2 1 r = r and 2 1 m = 2m , so 2 2 1 F = (1) (2)F = 20,000 N. Similarly, satellite 3 has 3 1 r = (3/2)r and 3 1m = m , so 2 3 1 F = (2/3) (1)F = 4440 N. (c) The speed of a satellite in a circular orbit is v = (GM/r)2 , so its kinetic energy is 1 2 2 K = mv = GMm/2r.Thus the ratio of the kinetic energy of two satellites a and b is a b a b a b K r m K r m . .. . = . .. . . .. . Satellite 3 has 3 1 r = (3/2)r and 3 1m = m , so 1 3 K /K = (3/2)(1/1) = 3/2 =1.50. 13.21. Model: Model the sun (S) as a spherical mass and the satellite (s) as a point particle. Visualize: The satellite, having mass ms and velocity vs , orbits the sun with a mass S M in a circle of radius s r . Solve: The gravitational force between the sun and the satellite provides the necessary centripetal acceleration for circular motion. Newtons second law is 2 S s s s 2 s s GM m m v r r = Because s s s v = 2p r /T where s T is the period of the satellite, this equation simplifies to 2 2 11 2 2 30 2 S s 3 S s 9 2 2 s 2 2 s s ss (2 ) (6.67 10 N m /kg )(1.99 10 kg)(24 3600 s) 2.9 10 m 4 4 GM r r GM T r r Tr p p p - = . = = . = 13.22. Model: Model the earth (e) as a spherical mass and the shuttle (s) as a point particle. Visualize: The shuttle, having mass ms and velocity vs , orbits the earth in a circle of radius s r . We will denote the earths mass by eM . Solve: The gravitational force between the earth and the shuttle provides the necessary centripetal acceleration for circular motion. Newtons second law is 2 e s s s 2 e e 2 s s s s s s GM m m v v GM v GM r r r r = . = . = Because 6 s e r = R + 350 km = 6.7210 m, 11 2 2 24 s 6 6 s s 3 s (6.67 10 N m /kg )(5.98 10 kg) 7704 m/s (6.72 10 m) 2 2(6.7210 m)5 580 s 91.3 minutes 7.704 10 m/s v T r v p p - = = = = = = 13.23. Model: Model the earth (e) as a spherical mass and the satellite (s) as a point particle. Visualize: The satellite has a mass is ms and orbits the earth with a velocity vs . The radius of the circular orbit is denoted by s r and the mass of the earth by eM . Solve: The satellite experiences a gravitational force that provides the centripetal acceleration required for circular motion: 2 1122 247 e s s s e 2 s 2 2 s s s 7 s 4 s s (6.67 10 N m /kg )(5.98 10 kg) 1.32 10 m (5500 m/s) 2 (2)(1.3210 m) 1.51 10 s 4.2 hr (5500 m/s) GM m m v r GM r r v T R v p p - = . = = = . = = = = 13.24. Model: Model Mars (m) as a spherical mass and the satellite (s) as a point particle. Visualize: The geosynchronous satellite whose mass is s m and velocity is s v orbits in a circle of radius s r around Mars. Let us denote mass of Mars by Mm. Solve: The gravitational force between the satellite and Mars provides the centripetal acceleration needed for circular motion: 2 2 2 1/ 3 m s s s s s m s 2 2 s 2 s s ss (2 ) ( ) 4 GM m m v m r r GM T r r rT p p . . = = . =. . . . Using G = 6.6710-11 Nm2 /kg2 , 6.42 1023 kg, m M = and s T = (24.8 hrs) = (24.8)(3600) s = 89,280 s, we obtain 7 s r = 2.0510 m. Thus, altitude 7 s m = r - R =1.7210 m. 13.25. Solve: We are givenM1 + M2 =150 kg which means M1 =150 kg -M2. We also have ( ) 6 2 1 2 6 2 2 1 2 11 2 2 8.00 10 N (8.00 10 N)(0.20 m) 4798 kg 0.20 m 6.67 10 N m /kg GM M M M - - - = . = = Thus, 2 2 2 2 2 2 2 (150 kg -M )M = 4798 kg or M - (150 kg)M + (4798 kg ) = 0. Solving this equation gives 2 M =103.75 kg and 46.25 kg. The two masses are 104 kg and 46 kg. 13.26. Visualize: Because of the gravitational force of attraction between the lead spheres, the cables will make an angle of . with the vertical. The distance between the sphere centers is therefore going to be less than 1 m. The free-body diagram shows the forces acting on the lead sphere. Solve: We can see from the diagram that the distance between the centers is d =1.000 m - 2Lsin. . Each sphere is in static equilibrium, so Newtons second law is grav grav sin 0 sin cos 0 cos x y SF = F -T . = .T . = F SF = T . - mg = .T . = mg Dividing these two equations to eliminate the tension T yields 2 grav 2 sin tan / cos F Gmm d Gm mg mg d g . . . = = = = We know that d is going to be only very, very slightly less than 1.00 m. The very slight difference is not going to be enough to affect the value of grav F , the gravitational attraction between the two masses, so well evaluate the right side of this equation by using 1.00 m for d. This gives 11 2 2 10 8 2 2 tan (6.67 10 N m /kg )(100 kg) 6.81 10 (3.90 10 ) (1.00 m) (9.80 m/s ) . . - - - = = .= This small angle causes the two spheres to move closer by 2Lsin. =1.410-7 m = 0.00000014 m. Consequently, the distance between their centers is d = 0.99999986 m. 13.27. Visualize: We placed the origin of the coordinate system on the 20 kg sphere (m1). The sphere 2 (m ) with a mass of 10 kg is 20 cm away on the x-axis. The point at which the net gravitational force is zero must lie between the masses 1 m and 2m . This is because on such a point, the gravitational forces due to 1 m and 2 m are in opposite directions. As the gravitational force is directly proportional to the two masses and inversely proportional to the square of distance between them, the mass m must be closer to the 10-kg mass. The small mass m, if placed either to the left of 1 m or to the right of 2 m , will experience gravitational forces from 1 m and 2 m pointing in the same direction, thus always leading to a nonzero force. Solve: 1 2 1 2 on on 2 2 2 2 2 20 10 (0.20 ) (0.20 ) 10 8 0.8 0 0.683 m and 0.117 m m m m m F F G m m G m m x x x x x x x = . = . = - - . - + = . = The value x = 68.3 cm is unphysical in the current situation, since this point is not between 1 m and 2m . Thus, the point (x, y) = (11.7 cm, 0 cm) is where a small mass is to be placed for a zero gravitational force. 13.28. Visualize: We placed the origin of the coordinate system on the 20.0 kg mass (m1) so that the 5.0 kg mass 3 (m ) is on the x-axis and the 10.0 kg mass 2 (m ) is on the y-axis. Solve: (a) The forces acting on the 20 kg mass 1 (m ) are 2 1 3 1 1 11 2 2 1 2 7 on 2 2 12 7 2 2 1 3 7 on 2 2 13 7 on (6.67 10 N m /kg )(20.0 kg)(10.0 kg) 3.335 10 N (0.20 m) (6.67 10 N m /kg )(20.0 kg)(5.0 kg) 6.67 10 N (0.10 m) 6.67 10 N 3.33 m m m m m F Gm m j j j r F Gm m i i i r F i - - - - - = = = = = = = + .. .. .. 1 3 1 2 1 7 7 on 7 1 on 1 7 on 5 10 N 7.46 10 N tan tan 6.67 10 N 63.4 3.335 10 N m m m m m j F F F . - - - - - - . = . . . . = .. .. = . . = . . . . Thus the force is ( ) 1 7 on 7.5 10 N, 63 cw from the -axis . mF = - y .. (b) The forces acting on the 5 kg mass 3 (m ) are 1 3 3 1 2 3 2 3 7 on on 11 2 2 2 3 8 on 2 2 2 23 8 8 on 8 6.67 10 N (6.67 10 N m /kg )(10.0 kg)(5.0 kg) 6.67 10 N [(0.20) (0.10) ] m (6.67 10 N)cos (6.67 10 N)sin (6.67 10 N) 10 cm 22. m m m m m m m m F F i F Gm m r F fi f j - - - - - - = - = - = = = + = - + = - .. .. .. 3 3 8 8 8 7 8 on 7 2 8 2 7 on 7 1 (6.67 10 N) 20 cm 36 cm 22.36 cm (2.983 10 N) (5.966 10 N) 6.968 10 N 5.966 10 N ( 6.968 10 N) (5.966 10 N) 6.99 10 N tan 6.968 10 N 5.966 10 m m i j i j F i j F . - - - - - - - - - - . . + . . . . . . . . . . = - + = - + = - + = ' = .. 8 85.1 - N . . . . = . . Thus ( ) 3 7 on 7.0 10 N, 85 ccw from the -axis . mF = - y .. 13.29. Visualize: Solve: The angle tan 1 5 14.04 . 20 . = - .. .. = . . The distance ( )2 ( )2 1 2 r = r = 0.050 m + 0.200 m = 0.206 m. The forces on the 20.0 kg mass are ( ) ( ) 1 1 2 1 2 2 2 2 sin cos sin cos F G Mm i j r F G Mm i j r . . . . = - + = + .. .. Note 1 2 m = m and 1 2r = r . Thus the net force on the 20.0 kg mass is ( )( )( ) ( ) 1 net 1 2 2 1 11 2 2 2 7 2 cos 2 6.67 10 N m /kg 20.0 kg 5.0 kg cos14.04 0.206 m 3.0 10 N F F F G Mm j r j j . - - = + = = = .. .. .. 13.30. Visualize: Solve: The total gravitational potential energy is the sum of the potential energies due to the interactions of the pairs of masses. 12 13 23 1 2 1 3 2 3 12 13 23 U U U U G m m G m m G m m r r r = + + = - - - With 1 2 3 12 13 m = 20.0 kg, m =10.0 kg, m = 5.0 kg, r = 0.20 m, r = 0.10 m, and 23 r = 0.2236 m, U = -1.4810-7 J Assess: The gravitational potential energy is negative because the masses attract each other. It is a scalar, so there are no vector calculations required. 13.31. Visualize: Solve: The total gravitational potential energy is the sum of the potential energies due to the interactions of the pairs of masses. 12 13 23 1 2 1 3 2 3 12 13 23 U U U U G m m G m m G m m r r r = + + = - - - With 1 2 3 1213 m = 20.0 kg, m = m =10.0 kg, r = r = ( )2 ( )2 0.050 m + 0.200 m = 0.206 m, and 23 r = 0.100 m, U = -1.9610-7 J Assess: The gravitational potential energy is negative because the masses attract each other. It is a scalar, so there are no vector calculations required. 13.32. Model: Model the earth (e) as a spherical mass and the satellite (s) as a point particle. Visualize: Let h be the height from the surface of the earth where the acceleration due to gravity (galtitude ) is 10% of the surface value surface (g ). Solve: (a) Since altitude surface g = (0.10) g , we have e e 2 2 2 2 e e e 6 7 7 e (0.10) ( ) 10 ( ) 2.162 (2.162)(6.37 10 m) 1.377 10 m= 1.38 10 m e GM GM R h R R h R h R h = . + = + . = . = = (b) For a satellite orbiting the earth at a height h above the surface of the earth, the gravitational force between the earth and the satellite provides the centripetal acceleration necessary for circular motion. For a satellite orbiting with velocity s v , 2 1122 24 e s s s e 2 s 6 7 e e e (6.67 10 N m /kg )(5.98 10 kg) 4.45 km/s ( ) ( ) (6.37 10 m 1.377 10 m) GM m m v v GM R h R h R h - = . = = = + + + + 13.33. Model: Model the earth as a spherical mass and the object (o) as a point particle. Ignore air resistance. This is an isolated system, so mechanical energy is conserved. Visualize: Solve: (a) The conservation of energy equation K2 +Ug2 = K1 +Ug1 is 2 e o 2 e o o 2 o 1 e e1 1 1 2 2 ( ) m v GM m m v GM m R R y - = -+ 2 e e e 1 11 2 2 24 6 6 2 1 1 2(6.67 10 N m /kg )(5.98 10 kg) 1 1 3.02 km/s 6.37 10 m 6.87 10 m v GM R R y - . . . = . - . . + . = . - . = . . . . (b) In the flat-earth approximation, g U = mgy. The energy conservation equation thus becomes 2 2 o 2 o 2 o 1 o 1 2 2 5 2 1 1 2 1 1 2 2 2 ( ) 2(9.80 m/s )(5.00 10 m 0 m) 3.13 km/s m v m gy m v m gy v v gy y + = + . = + - = - = (c) The percent error in the flat-earth calculation is 3130 m/s 3020 m/s 3.6% 3020 m/s - 13.34. Model: Model the earth and the projectile as spherical masses. Ignore air resistance. This is an isolated system, so mechanical energy is conserved. Visualize: A pictorial representation of the before-and-after events is shown. Solve: After using v2 = 0 m/s, the energy conservation equation 2 2 1 1 K +U = K +U is e p 2 e p p 1 e e 0 J 1 2 GM m GM m m v R h R - = - + The projectile mass cancels. Solving for h, we find 2 1 1 e e e 5 1 2 4.18 10 m 418 km h v R R GM - . . = . - . - . . = = 13.35. Model: Model the planet (p) as a spherical mass and the projectile as a point mass. This is an isolated system, so mechanical energy is conserved. Visualize: The projectile of mass m was launched on the surface of the planet with an initial velocity v0. Solve: (a) The energy conservation equation 1 1 0 0 K +U = K +U is 2 p 2 p 1 0 p 1 p 1 2 0 6 1 p p p 1 1 2 2 1 2.8 10 m 2 GM m GM m mv mv R y R y v R R GM - - = - + . . . = . - . - = .. .. (b) Using the energy conservation equation 1 1 0 0 K +U = K +U with 6 1 y =1000 km =1.00010 m: 2 p 2 p 1 0 p 1 p 1 1 2 2 GM m GM m mv mv R y R - = - + 1 2 2 1 0 p p 1 p v v 2GM 1 1 R y R . . .. . = . + .. - ... .. . + ... 1 2 2 11 2 2 24 6 6 6 (5000 m/s) 2(6.67 10 N m /kg )(2.6 10 kg) 1 1 (5.0 10 m 1.000 10 m) 5.0 10 m 3.7 km/s - . . .. = . + . - .. . . + .. = 13.36. Model: The object is a particle. The planet is a spherical mass. Solve: Conservation of mechanical energy of the object gives 1 2 2 G Mm mv G Mm Rh R - = - + The objects mass drops out. Solving for the speed as it hits the ground, ( ) ( ) v 2GM 1 1 2GM R h R 2GMh R R h R R h R R h . . . + - . = . - . = .. .. = . + . . + . + Assess: Compare this to 2 v 2gh 2GMh , R = = which is the result if the potential U = mgh is used. 13.37. Model: Model the earth as a spherical mass and the meteoroids as point masses. Visualize: Solve: (a) The energy conservation equation K2 +U2 = K1 +U1 is 1/ 2 1 2 e 1 2 e 2 2 2 2 1 2 1 e e m em 4 2 1 1 1.13 10 m/s=11.3 km/s mv GM m mv GM m v v GM R r Rr . . .. - = - . = . + . - .. .. . ... = The speed does not depend on the meteoroids mass. (b) This part differs in that 7 2 e r = R + 5000 km =1.13710 m. The shape of the meteoroids trajectory is not important for using energy conservation. Thus 1/ 2 1 2 e 1 2 e 2 2 2 2 1 2 1 e e m e m 3 2 1 1 5000 km 5000 km 8.94 10 m/s 8.94 km/s mv GM m mv GM m v v GM R r R r . . .. - = - . = . + . - .. + .. . + ... = = 13.38. Model: Model the two stars as spherical masses, and the comet as a point mass. This is an isolated system, so mechanical energy is conserved. Visualize: In the initial state, the comet is far away from the two stars and thus it has neither kinetic energy nor potential energy. In the final state, as the comet passes through the midpoint connecting the two stars, it possesses both kinetic energy and potential energy. Solve: The conservation of energy equation Kf +Uf = Ki +Ui is 2 f f1 f2 11 2 2 30 f 12 f 1 0 J 0 J 2 4 4(6.67 10 N m /kg )(1.99 10 kg) 32,600 m/s (0.50 10 m) mv GMm GMm r r v GM r - - - = + . = = = Assess: Note that the final velocity of 33 km/s does not depend on the mass of the comet. 13.39. Model: Model the asteroid as a spherical mass and yourself as a point mass. This is an isolated system, so mechanical energy is conserved. Visualize: The radius of the asteroid is Ma and its mass is Ra . Solve: The conservation of energy equation f f i i K +U = K +U for the asteroid is 2 a 2 a f i a a 1 1 2 2 mv GM m mv GM m R r R - = - + The minimum speed for escape is the one that will cause you to stop only when the separation between you and the asteroid becomes very large. Noting that f v .0m/s as r .8, we have 11 2 2 14 2 a i 3 i a 2 2(6.67 10 N m /kg )(1.0 10 kg) 2.58 m/s (2.0 10 m) v GM v R - = = . = That is, you need a speed of 2.58 m/s to escape from the asteroid. We can now calculate your jumping speed on the earth. The conservation of energy equation is 2 2 i i 0 J 1 (0.50 m) 2(9.8 m/s )(0.50 m) 3.13 m/s 2 - mv = -mg .v = = This means you can escape from the asteroid. 13.40. Model: The projectile is a particle. The earth and moon are spherical masses. Solve: The projectile is attracted to both the moon and earth. Its final velocity and potential energy are zero. Since the projectile is fired from the far side of the moon, its initial distance from the center of the earth is the earth-moon distance Re-m plus the radius of the moon Rm. Let the projectile have mass m. The conservation of mechanical energy equation is ( ) 2 e m escape e-m m m 2 e m escape e-m m m 1 0 J 0 J 2 2 mv G M m G M m R R R v G M M R R R - - = + + . . . = . + . . + . From Table 13.2, 8 6 24 e-m m e R = 3.8410 m, R =1.7410 m, M = 5.9810 kg, and 22 m M = 7.3610 kg. So escape v = 2.78 km/s. Assess: The escape velocity does not depend on the mass of the object which is trying to escape. 13.41. Model: The earth and sun are spherical masses. The earth is in a circular orbit around the sun. The projectile is a particle. Visualize: Solve: At the earths distance from the sun, to escape the suns gravitational pull a projectile must have speed vescape. The energy conservation equation for the projectile is ( )( ) ( ) 2 2 escape e s 11 2 2 30 s 4 escape 11 e s 1 0 J 2 2 6.67 10 N m /kg 1.99 10 kg 2 4.207 10 m/s 1.50 10 m - - - - = . = = = mv G M m R v G M R The earths speed in its orbit is found by considering its period T =1 year 365.25 243600 = 3.156107s. ( 11 ) e s 4 e 7 2 2 1.50 10 m 2.987 10 m/s 3.156 10 s p p - = = = v R T The projectiles total speed is the sum of its launch speed and the earths speed. escape e launch 4 launch escape e 1.22 10 m/s v v v v v v = + . = - = Assess: The projectile need only attain a launch speed of 12.2 km/s if launched in the direction of earths motion. This is about a factor of 3.5 times less than what is required from rest. 13.42. Model: Gravity is a conservative force, so we can use conservation of energy. Visualize: The planets will be pulled together by gravity and each will have speed 2 v as they crash and the separation between their centers will be 2R. Solve: The planets begin with only gravitational potential energy. When they crash, they have both potential and kinetic energy. Thus, 2 2 2 2 2 2 1 1 2 1 2 2 1 1 1 0 J 2 2 1 1 K U Mv Mv GMM K U GMM r r v GM r r . .. . . + = + - . = . + = - . . .. . . . . = . - . . . Because the planet is Jupiter-size, well use 27 Jupiter M = M =1.910 kg and 8 2 Jupiter r = 2R =1.410 m. The crash speed of each planet is 4 2 v = 3.010 m/s. Assess: Note that the force is not constant, because it varies with distance, so the motion is not constant acceleration motion. The formulas from constant-acceleration kinematics do not work for problems such as this. 13.43. Model: The two asteroids make an isolated system, so mechanical energy is conserved. We will also use the law of conservation of momentum for our system. Visualize: Solve: The conservation of momentum equation fx ix p = p is f 1 f 2 f 1 f 2 ( ) 2 ( ) 0 kg m/s ( ) 2( ) x x x x M v + M v = . v = - v The equation for mechanical energy conservation f f i i K +U = K +U is 2 2 2 2 f 1 f 2 f 2 f 2 f 2 f 1 f 2 1 ( ) 1 (2 )( ) ( )(2 ) ( )(2 ) 1[2( ) ] ( ) 0.8 2 2 2 10 2 ( ) 0.516 ( ) 2( ) 1.032 x x x x x x x M v M v G M M G M M v v GM R R R v GM v v GM R R + - =- . + = . =- . =- = The heavier asteroid has a speed of 0.516(GM/R)1/ 2 and the lighter one a speed of 1.032(GM/R)1/ 2. 13.44. Model: Model the distant planet (P) as a spherical mass. Solve: The acceleration at the surface of the planet and at the altitude h are P P P surface 2 altitude surface 2 2 2 2 and 1 1 2 ( ) 2 ( ) 2 2 ( 2 1) 0.414 g GM g g G M GM R R h R R h R R h R h R R = = . = + . + = . + = . = - = That is, the starship is orbiting at an altitude of 0.414R. 13.45. Model: The stars are spherical masses. Visualize: Solve: The starts are identical, so their final speeds f v are the same. They collide when their centers are 2R apart. From energy conservation, 2 2 2 9 f 2 f 9 5 f 3 3 1 3 (5.0 10 m) 2 2 2 1 1 2 5.0 10 m 3.71 10 m/s M M G Mv GR v GM R v . . . . . . . - . = . . - . . . . . . . . = . - . . . . . . = 13.46. Model: Model the moon (m) as a spherical mass and the lander (l) as a particle. This is an isolated system, so mechanical energy is conserved. Visualize: The initial position of the lunar lander (mass = m1) is at a distance 1 m r = R + 50 km from the center of the moon. The final position of the lunar lander is the orbit whose distance from the center of the moon is 2 m r = R + 300 km. Solve: The external work done by the thrusters is 1 ext mech 2 g W = .E = .U where we used 1 mech 2 g E = U for a circular orbit. The change in potential energy is from the initial orbit at 1 50 km m r = R + to the final orbit f m r = R + 300 km. Thus m m m ext f i i f 11 2 2 22 6 6 8 1 11 2 2 (6.67 10 N m / kg )(7.36 10 kg)(4000 kg) 1 1 2 1.79 10 m 2.04 10 m 6.72 10 J W GM m GM m GM m r r r r - . - - . . . = . - . = . - . . . . . . . = . - . . . = 13.47. Model: Model the earth (e) as a spherical mass and the space shuttle (s) as a point particle. This is an isolated system, so the mechanical energy is conserved. Visualize: The space shuttle (mass = ms ) is at a distance of 1 e r = R + 250 km. Solve: The external work done by the thrusters is 1 ext mech 2 g W = .E = .U where we used 1 mech 2 g E = U for a circular orbit. The change in potential energy is from the initial orbit at 1 e r = R + 250 km to the final orbit f e r = R + 610 km. Thus e e e ext f i i f 11 2 2 24 6 6 11 1 11 2 2 (6.67 10 N m / kg )(5.98 10 kg)(75,000 kg) 1 1 2 6.62 10 m 6.98 10 m 1.17 10 J W GM m GM m GM m r r r r - . - - . . . = . - . = . - . . . . . . . = . - . . . = This much energy must be supplied by burning the on-board fuel. 13.48. Solve: (a) Using Equation 13.31 for a satellite in a circular orbit, mech g e e e 11 2 2 24 4 1 1 ( ) 1 1 2 2 300 km 500 km 1(6.67 10 N m /kg )(5.98 10 kg)(5.00 10 kg) 2 E U GMm R R - . . . = . = - . - . . + + . - = 10 6 5 6 5 1 1 4.35 10 J 6.37 10 m 3 10 m 6.37 10 m 5 10 m . - . = - . + + . . . The negative sign indicates that 4.351010 J of energy is lost. (b) The shuttle would fire retro-rockets to lose enough energy to go into an elliptical orbit. When halfway around, at min r , the shuttle would fire retro-rockets again to lose the rest of the energy to stay at min r . 13.49. Model: Planet Physics is a spherical mass. The cruise ship is in a circular orbit. Solve: (a) At the surface, the free-fall acceleration is 2 g G M . R = From kinematics, ( )2 ( )( ) ( )2 0 2 2 0 m 0 m 11 m/s 2.5 s 2 2.5 s 2.20 m/s f i y y v t g t g g = + . - . . = + - . = The period of the cruise ships orbit is 230 60 =13,800 s. For the circular orbit of the cruise ship, ( ) ( )( ) 2 2 2 2 3 2 2 2 6 2 4 2 1 32 2.20 m/s 13,800 s 1.327 10 m. 32 T R T R R R GM GM g R p p p . . . . . . = . . . = . . = . . . . . . . . . = = The mass is thus 2 M R g 5.8 1022 kg. G = = (b) From part (a), R =1.33106 m. 13.50. Model: Assume a spherical asteroid and a point mass model for the satellite. This is an isolated system, so mechanical energy is conserved. Visualize: The orbital radius of the satellite is r = Ra + h = 8,800 m+ 5,000 m =13,800 m Solve: (a) The speed of a satellite in a circular orbit is (6.67 10 11 N m2 /kg2 )(1.0 1016 kg) 1 2 7.0 m/s (13,800 m) v GM r . - . = = . . = . . (b) The minimum launch speed for escape i (v ) will cause the satellite to stop asymptotically f (v = 0 m/s) as f r .8. Using the energy conservation equation 2 2 1 1 , K +U = K +U we get 2 a s 2 a s 2 a s f s i escape f a a 11 2 2 16 a escape a 1 1 0 J 0 J 1 2 2 2 2 2(6.67 10 N m /kg )(1.0 10 kg) 12.3 m/s 8800 m m v GM m m v GM m v GM r R R v GM R - - = - . - = - . = = = 13.51. Model: Model the moon as a spherical mass and the satellite as a point mass. Visualize: The rotational period of the satellite is the same as the rotational period of the moon around its own axis. This time happens to be 27.3 days. Solve: The gravitational force between the moon and the satellite provides the centripetal acceleration necessary for circular motion around the moon. Therefore, 2 m 2 2 2 11 2 2 22 2 3 m 2 2 7 2 (6.67 10 N m /kg )(7.36 10 kg)(27.3 24 3600 s) 4 4 8.84 10 m GM m mr mr r T r GM T r p . p p - = = . . . . . . . = = . = Since m r = R + h, then 7 6 7 m h = r - R = 8.8410 m -1.7410 m = 8.6710 m. 13.52. Model: Model the earth as a spherical mass and the satellite as a point mass. Visualize: The satellite is directly over a point on the equator once every two days. Thus, T = 2Te = 2 24 3600 s =1.728105 s. Solve: A satellites period is 2 2 3 e 2 11 2 2 24 52 3 e 2 2 7 4 (6.67 10 N m /kg )(5.98 10 kg)(1.728 10 ) 4 4 6.71 10 m T r GM r GM T r p p p - . . =. . . . . = = . = Assess: The radius of the orbit is larger than the geosynchronous orbit. 13.53. Solve: (a) Taking the logarithm of both sides of v p = Cuq gives [log(v p ) plogv] [log(Cuq ) logC qlogu] logv q logu logC p p = = = + . = + But x = logu and y = logv, so x and y are related by y q x logC p p . . = . . + . . (b) The previous result shows there is a linear relationship between x and y, hence there is a linear relationship between log u and log v. The graph of a linear relationship is a straight line, so the graph of log v-versus-log u will be a straight line. (c) The slope of the straight line represented by the equation y = (q/ p)x + logC/ p is q/p. Thus, the slope of the log v-versus-log u graph will be q/p. (d) From Newtons theory, the period T and radius r of an orbit around the sun are related by 2 T 2 4 r3 GM . p . =. . . . This equation is of the form T p = Crq , with p = 2, q = 3, and C = 4p 2/GM. If the theory is correct, we expect a graph of log T-versus-log r to be a straight line with slope q/p = 3/2 =1.500. The experimental measurements of actual planets yield a straight line graph whose slope is 1.500 to four significant figures. Note that the graph has nothing to do with theoryit is simply a graph of measured values. But the fact that the shape and slope of the graph agree precisely with the prediction of Newtons theory is strong evidence for its correctness. (e) The predicted y-intercept of the graph is log C/p, and the experimentally determined value is 9.264. Equating these, we can solve for M. Because the planets all orbit the sun, the mass we are finding is sun M = M . 2 2 18.528 18.528 sun sun 2 18.528 30 sun 1 log 1 log 4 9.264 4 10 1 2 2 10 4 10 1.996 10 kg C GM GM M G p p p - . . = . . = - . = = . . . = = The tabulated value, to three significant figures, is 30 sum M =1.9910 kg. We have used the orbits of the planets to weigh the sun! 13.54. Visualize: Please refer to Figure P13.54. Solve: The gravitational force on one of the masses is due to the star and the other planet. Thus 2 2 2 2 2 2 2 2 2 2 3 1 2 2 2 4 (2 ) 4 4 4 1 4 ( /4) G Mm Gmm mv m r GM Gm r r r r r T r r T G M m r T r r T GMm p p p p + = = . . . + = . . . . . . . . . + . = . = . . . . . + . 13.55. Solve: (a) Dividing the circumference of the orbit by the period, 4 s 4 2 2(1.010 m) 6.3 10 m/s 1.0 s v R T p p = = = (b) Using the formula for the acceleration at the surface, 11 2 2 30 s 12 2 surface2 42 s (6.67 10 N m /kg )(1.99 10 kg) 1.33 10 m/s (1 10 m) g GM R - = = = (c) The mass of an object on the earth will be the same as its mass on the star. The gravitational force is ( ) 12 G star surface F = mg =1.3310 N (d) The radius of the orbit of the satellite is r =1104 m+1.0103 m =1.1104 m. The period is 2 3 2 4 3 2 4 11 2 2 30 s 4 4 (1.1 10 m) 6.29 10 s (6.67 10 N m /kg )(1.99 10 kg) T r T GM p p - - = = . = This means there are 1589 revolutions per second or 9.5104 orbits per minute. (e) Applying Equation 13.25 for a geosynchronous orbit, 11 2 2 30 3 s 2 6 2 2 (6.67 10 N m /kg )(1.99 10 kg)(1.0 s) 1.50 10 m 4 4 r GM T r p p - = = . = 13.56. Model: Assume the solar system is a point particle. Solve: (a) The radius of the orbit of the solar system in the galaxy is 25,000 light years. This means 8 20 20 15 8 5 25,000 light years 2500(3 10 )(365)(24)(3600) m 2.36 10 m 2 (2)(2.36 10 m) 6.64 10 s 2.05 10 years (2.30 10 m/s) r T r v p p = = = = = = = (b) The number of orbits 9 8 5.0 10 years 24 orbits. 2.05 10 years = = (c) Applying Newtons second law yields 2 2 5 2 20 g center ss ss 41 2 g center 11 2 2 (2.30 10 m/s) (2.36 10 m) 1.87 10 kg 6.67 10 N m /kg GM m m v v r M r r G - = . = = = (d) The number of stars in the center of the galaxy is 41 10 30 1.87 10 kg 9.4 10 1.99 10 kg = 13.57. Model: Assume the two stars are spherical masses. Visualize: The gravitational force between the two stars provides the centripetal acceleration required for circular motion about the center of mass. Solve: Newtons second law is 1 2 2 3 2 gravitation2 2 2 (2 ) 16 F GMM MR MR R GT M R T p . p . . . . = = = . . . = . . . . . . Using T = 90 days = 90 243600 s and 30 sun M = 2M = 3.9810 kg, we get R = 4.6671010 m. Thus the star separation is 2R = 9.331010 m. 13.58. Model: Assume the three stars are spherical masses. Visualize: The stars rotate about the center of mass, which is the center of the triangle and equal distance r from all three stars. The gravitational force between any two stars is the same. On a given star the two forces from the other stars make an angle of 60. Solve: The value of r can be found as follows: 12 /2 cos30 1.0 10 m 0.577 1012 m 2cos30 2cos30 L r L r = . = = = The gravitational force between any two stars is 2 11 2 2 30 2 26 g2 122 (6.67 10 N m /kg )(1.99 10 kg) 2.64 10 N (1.0 10 m) F GM L - = = = The component of this force toward the center is 26 26 c gF = F cos30 = (2.6410 N)cos30 = 2.2910 N The net force on a star toward the center is twice this force, and that force equals MR. 2. This means 2 26 2 2 2 30 12 8 26 26 2 2.29 10 N 2 4 4 (1.99 10 kg)(0.577 10 m) 3.15 10 s 10 years 4.58 10 N 4.58 10 N MR MR T T MR p . p p = = . . . . . . . = = = = 13.59. Model: Angular momentum is conserved for a particle following a trajectory of any shape. Visualize: For a particle in an elliptical orbit, the particles angular momentum is L = mrvt = mrvsin , where v is the velocity tangent to the trajectory and is the angle between r.. and v... Solve: At the distance of closest approach min (r ) and also at the most distant point, = 90. Since there is no tangential force (the only force being the radial force), angular momentum must be conserved: Pluto 1 min Pluto 2 max 3 12 12 2 1 min max ( / ) (6.12 10 m/s)(4.43 10 m/7.30 10 m) 3.71 km/s m vr m vr v v r r = . = = = 13.60. Model: Angular momentum is conserved for a particle following a trajectory of any shape. Visualize: For a particle in an elliptical orbit, the particles angular momentum is L = mrvt = mrvsin , where v is the velocity tangent to the trajectory, and is the angle between r.. and v... Solve: At the distance of closest approach min (r ) and also at the most distant point, = 90. Since there is no tangential force (the only force being the radial force), angular momentum must be conserved: Mercury 1 min Mercury 2 max 10 10 min max 2 1 ( / ) (6.99 10 m)(38.8 km/s) 4.60 10 m 59.0 km/s m vr m vr r r v v = . = = = 13.61. Model: For the sun + comet system, the mechanical energy is conserved. Visualize: Solve: The conservation of energy equation f f i i K +U = K +U is 2 s c 2 s c c 2 c 1 2 1 1 1 2 2 M v GM M M v GM M r r - = - Using G = 6.6710-11 Nm2/kg2 , 30 s M =1.9910 kg, 10 1 r = 8.7910 m, 12 2 r = 4.5010 m, and 1 v = 54.6 km/s, we get 2 v = 4.49 km/s. 13.62. Model: Model the planet (p) as a spherical mass and the spaceship (s) as a point mass. Visualize: Solve: (a) For the circular motion of the spaceship around the planet, 2 p s 0 p 2 0 0 0 0 GM m mv GM v r r r = . = Immediately after the rockets were fired 1 0v = v /2 and 1 0r = r . Therefore, p 1 0 1 2 GM v r = (b) The spaceships maximum distance is max 0r = r . Its minimum distance occurs at the other end of the ellipse. The energy at the firing point is equal to the energy at the other end of the elliptical trajectory. That is, 2 p s 2 p s s 1 s 2 1 2 1 1 2 2 GM m GM m m v m v r r - = - Since the angular momentum at these two ends is conserved, we have 1 1 2 2 2 1 1 2 mv r = mv r .v = v (r /r ) With this expression for 2 v , the energy equation simplifies to 2 p 2 2 p 1 112 1 2 1 1( / ) 2 2 GM GM v vrr r r - = - Using 1 0 r = r and p 1 0 0 /2 1 , 2 GM v v r = = 2 p p p 0 p 0 2 2 0 0 0 2 2 0 0 2 2 0 2 0 2 2 2 0 2 2 0 1 1 1 1 1 1 1 2 4 2 4 8 8 7 1 0 7 0 8 8 8 8 GM GM GM r GM r r r r r r r r r r r r r r r r r r . . . . . . - = . . - . - = - . . . . . . . + - = .. . - + = . . The solutions are 2 0 r = r (the initial distance) and 2 0r = r / 7. Thus the minimum distance is min 0 r = r / 7. 13.63. Solve: (a) The satellite is in a circular orbit if 1 . 2 K = - Ug Calculate K and : g U ( ) ( )( ) 2 3 2 7 24 11 2 2 7 7 1 1 5.5 10 m/s 1.51 10 . 2 2 5.98 10 kg 6.67 10 N m /kg 1.10 10 m 3.63 10 e g K mv m m M m m U G r m - = = = = - = - = - The quantity 1 1.81 107 , 2 g- U = m . K so no, the satellite is not in a circular orbit. (b) The orbit is bound if the total mechanical energy is negative. 7 7 mech 1.51 10 3.63 10 0. g E = K +U = m - m < Yes, the orbit is bound. Assess: Even without knowing the mass of the satellite we could answer questions about its orbit. 13.64. Visualize: Solve: We choose two equal time intervals tb - ta and c bt - t . A constant velocity and equal time intervals means that b a c b . x - x = x - x The area swept from a a t = 0 s to t = t is Rx /2 and the area swept from b b t = 0 s to t = t is Rx /2. Thus, the area swept between a b b a t = t and t = t is R(x - x ) /2. In the same way, the area swept between b t = t and t = t is c b R(x - x )/2. Since c b b ax - x = x - x , the area swept during time b a (t - t ) is the same as the area swept during an equal time c b (t - t ). Keplers second law is obeyed. 13.65. Solve: (a) At what distance from the center of Saturn is the acceleration due to gravity the same as on the surface of the earth? (b) (c) The distance is 6.21107 m. This is Saturn 1.06R 13.66. Solve: (a) A 1000 kg satellite orbits the earth with a speed of 1997 m/s. What is the radius of the orbit? (b) (c) The radius of the orbit is 11 2 2 24 E 8 2 2 payload 6.67 10 N m /kg (5.98 10 kg) 1.00 10 m ( ) (1997 m/s) r GM v - = = = 13.67. Solve: (a) A 100 kg object is released from rest at an altitude above the moon equal to the moons radius. At what speed does it impact the moons surface? (b) (c) The speed is v2 =1680 m/s. 13.68. Solve: (a) A 2.01030 kg star and a 4.01030 kg star, each 1.0109 m in diameter, are at rest 1.01012 m apart. What are their speeds as they crash together? (b) (c) The first equation is the conservation of momentum. It can be used in the conservation of energy equation to give f1 v = 596 km/s and f2 v = -298 km/s. That is, the speeds of the two stars are 6.0102 km/s and 3.0102 km/s. 13.69. Solve: (a) Keplers third law for circular orbits is 32 2 2 T 2 4 r3 T 4 r GM GM . p . p = . . . = . . Letting 4 2 a , GM p = the first satellite obeys 3 T = ar 2 . For the second satellite, which orbits the same mass, ( ) 32 3 3 T T a r r 2 ar 2 1 r r . . . + . = + . = . + . . . Since r 1, r . << we can use the approximation (1 ) 1 . n x nx Thus 32 1 3 2 T T ar r r . . . + . . + . . . Subtracting the equation 32 T = ar for the first satellite from this, 32 3 2 r T arr . . . . = . . . . Dividing this by the equation for the first satellite, 3 2 T r T r . . = (b) The satellites orbit the earth. The fractional difference in their periods is 3 1 km 2.24 10 4 2 6700 km T T - . = = After 4 1 4467 2.24 10- = periods they will meet again. For the inner satellite, ( )( )32 2 6 24 4 6.700 10 m 5.98 10 kg 5456 s 1.52 hrs T G p = = = So the satellites will meet again in 44671.52 hrs = 6770 hrs = 282 days. Assess: A communications satellite has an orbital period of around 1.5 h. The surprising length of time between the two satellites meeting is due to the small differences in their periods. 13.70. Model: Model the earth and sun as spherical masses, the satellite as a point mass. Assume the satellites distance from the earth is very small compared to the earths (and satellites) distance to the sun. Visualize: Solve: The net force on the satellite is the sum of the gravitational force toward the sun and the gravitational force toward the earth. This net force is responsible for circular motion around the sun. We want to chose the distance d to make the period T match the period e T with which the earth orbits the sun. The earths orbital period is given by 2 2 3 Te = (4p /GMs )Re . Thus 2 2 2 s e s net 2 2 centripetal 2 3 e e e F GM m GM m ma mv m 2 r mr 4 mr GM r d r r T T R . p . p = - = = = . . = = . . Using e r = R - d and canceling the Gm term gives s e se 2 2 3 e e ( ) ( ) M M MR d R d d R - = - - This equation cant be solved exactly, but we can make use of the fact that e d .. R to use the binomial approximation. Factor the e R out of the expressions e (R - d) to get s e se 2 2 2 2 e e e (1 / ) (1 / ) M M M d R R dR d R - = - - If we think of ed/R = x ..1, we can simplify the first term by using (1 x)n 1 nx. Here n = -2, so we get s e s e s 2 e 2 2 e 2 2 e e e M (1 ( 2)d / R ) M M (1 d/R ) M 3M R d R d R - - - = - . = Thus 1/3 9 e s e d = (M /3M ) R =1.5010 m. Assess: e d/R = 0.010, so our assumption that e d/R ..1 is justified. 13.71. Model: Model the earth as a spherical mass and the shuttle and payload as point masses. Well assume mpayload .. mshuttle. Visualize: Solve: (a) The payload, in steady state, is undergoing uniform circular motion. This means that the net force is directed toward the center of the earth. There is no tangential force component, since such a force would cause the payload to speed up or slow down and the motion wouldnt be uniform. Since the net force is due to gravity and tension, and the gravitational force is radial, the tension force cannot have any tangential component. Thus the rope is radially outward at angle 0. (b) To move with the shuttle, the payloads period p T in an orbit of radius p r must exactly match the period s T of the shuttle in an orbit of radius s r . The shuttles period around the earth is given by 2 2 3 s es T = (4p /GM ) r , where weve used the assumption p s m .. m to infer that the ropes tension will be too small to have any influence on the shuttles motion. Newtons second law for the payloads motion is 2 2 2 e p p p p net 2 p p p 2 p p p p p 2 4 ( )r r GM m m v m r F T m a m r r r rT T . p . p = - = = = .. .. = . . Using p s T = T and the above expression for s T , this becomes 3 e p e e p p 2 p p 3 2 3 p s p s 3 3 e p p e p e 2 2 p s p e 1 1 290 km 4.04 N 300 km GM m GM GM m r T mr r r r r GM m r GM m R T r r r R - = = . . . . . . + . . . = . - . . . = . - . . . = .. . . .. .. . + . .. Assess: The fact that the tension is so small justifies our assumption that it will have no effect on the motion of the shuttle. 13.72. Model: Model the 400 kg satellite and the 100 kg satellite as point masses and model the earth as a spherical mass. Momentum is conserved during the inelastic collision of the two satellites. Visualize: Solve: For the given orbit, 6 6 r0 = Re +110 m = 7.3710 m. The speed of a satellite in this orbit is e 0 0 v = GM 7357 m/s r = The two satellites collide, stick together, and move with velocity 1v . The equation for momentum conservation for the perfectly inelastic collision is 1 1 (400 kg 100 kg) (400 kg)(7357 m/s) (100 kg)(7357 m/s) 4414 m/s v v + = - . = The new satellites radius immediately after the collision is still 6 1 0 r = r = 7.3710 m. Now it is moving in an elliptical orbit. We need to determine if the minimum distance 2 r is larger or smaller than the earths radius 6 e R = 6.3710 m. The combined satellites will continue moving in an elliptical orbit. The momentum of the combined satellite is L = mrv sin (see Equation 13.26) and is conserved in a trajectory of any shape. The angle is 90 when 1 v = 4414 m/s and when the satellite is at its closest approach to the earth. From the conservation of angular momentum, we have 6 102 1 1 2 2 10 2 2 2 (7.37 10 )(4414 m/s) 3.253 10 m /s 3.253 10 m /s rv r v r v = = = . = Using the conservation of energy equation at positions 1 and 2, 2 e 2 e 6 2 2 1 (500 kg)(4414 m/s) (500 kg) 1 (500 kg) (500 kg) 2 7.3710 m2 GM v GM r - = - Using the above expression for 2 r , we can simplify the energy equation to 2 4 722 22 2 2 2 (2.452 10 m/s) (8.876 10 m /s ) 0 m /s 20,107 m/s and 4414 m/s v v v - + = . = A velocity of 20,107 m/s for 2 v yields 10 2 6 2 3.253 10 m /s 1.62 10 m 20,107 m/s r = = Since 6 2 e r < R = 6.3710 m, the combined mass of the two satellites will crash into the earth. 13.73. Model: The stars are spherical masses. They each rotate about the systems center of mass. Visualize: Solve: (a) The stars rotate about the systems center of mass with the same period: T1 = T2 = T. We can locate the center of mass by letting the origin be at the smaller-mass star. Then 30 30 12 12 1 cm 30 30 (2.0 10 kg)(0 m) (6.0 10 kg)(2.0 10 m) 1.5 10 m 2.0 10 kg 6.0 10 kg r r + = = = + Mass 2 m undergoes uniform circular motion with radius 12 2 r = 0.510 m due to the gravitational force of mass 1 m at distance R = 2.01012 m. The gravitational force is responsible for the centripetal acceleration, so 2 2 2 1 2 2 2 2 2 2 2 grav 2 2 centripetal 2 2 2 2 1/ 2 2 12 12 2 1/ 2 2 8 2 11 2 2 30 1 2 4 4 4 (0.5 10 m)(0.5 10 m) 7.693 10 s = 24 years (6.67 10 N m /kg ) (2.0 10 kg) F Gm m m a m v m r m r R r rT T T rR Gm p p p p - = = = = . . = . . . . . . . . . = . . = . . = . . . . (b) The speed of each star is v = (2p r)/T. Thus 12 1 1 8 12 2 2 8 2 2 (1.5 10 m) 12.3 km/s 7.693 10 s 2 2 (0.5 10 m) 4.1 km/s 7.693 10 s v r T v r T p p p p = = = = = = 13.74. Model: The moon is a spherical mass. The moon lander is originally in a circular orbit. Visualize: Solve: Energy and momentum are conserved between points 1 and 2 in the elliptical orbit. Also, at both points = 90, so L = mrvsin = mrv. Let h =1000 km. The original speed of the lander is 0 m 1338 m/s. m v GM R h = = + Conservation of angular momentum requires 2 1 1 1 2 2 1 2 m . m mrv mr v v r R h v r R + = . = = The energy conservation equation is 2 2 1 2 1 1 2 2 m m m m mv G M m mv G M m Rh R - = - + Substituting 2 1m , m v R h v R . + . =. . . . ( ) ( ) 2 2 2 1 1 2 2 1 2 6 2 2 1 2 1 1 1 2 2 1 2 1 1 2 1 2 1.392 10 m /s 2 1 1180 m/s m m m m m m m m m m m m m m m m m m m m v G M v R h G M R h R R v R h GM R RhR v GM h GM R R R h R h R h R h R v . + . - = . . - + . . . . + . . . . . - . . . = . - . . . . . . + . . . . . = = . . = + . + . + . + . . . - . . = The fractional change in speed required to just graze the moon at point 2 is 1338 m/s 1180 m/s 11.8% 1338 m/s - = Assess: A reduction in speed by almost 12% is reasonable. 13.75. Model: Model the earth as a spherical mass and the satellite as a point mass. This is an isolated system, so mechanical energy is conserved. Also, the angular momentum of the satellite is conserved. Visualize: Please refer to Figure CP13.75. Solve: (a) Angular momentum is L = mrvsin . The angle = 90 at points 1 and 2, so conservation of angular momentum requires 2 1 1 2 2 1 2 1 mrv mr v v r v r . . ' = ' . ' = . . ' . . The energy conservation equation is 2 2 2 2 2 1 2 1 2 1 21 1 ( ) 1 ( ) ( ) ( ) 2 1 1 2 2 m v GMm m v GMm v v GM r r rr . . ' - = ' - . ' - ' = . - . . . Using the angular momentum result for 1 v' gives 2 2 2 2 2 2 1 2 2 1 2 1 2 2 2 2 2 1 12 1 12 1 2 2 2 1 2 1 2 1 2 1 1 2 ( ) ( ) 2 ( ) 2 2 ( / ) and 2 ( / ) v v r GM r r v r r GM r r r rr r rr v GM r r v r v GM r r r r r r r . . . - . . - . . - . ' - ' . . = . .. ' . . = . . . . . . . . . . . . ' = ' = . . ' = + . . + (b) For the circular orbit, 11 2 2 24 1 6 5 1 (6.67 10 N m /kg )(5.98 10 kg) 7730 m/s (6.37 10 m 3 10 m) v GM r - = = = + For the elliptical orbit, 6 5 6 1 e 6 7 7 2 e 2 1 1 1 1 2 300 km 6.37 10 m 3 10 m 6.67 10 m 35,900 km 6.37 10 m 3.59 10 m 4.23 10 m 2 ( / ) 10,160 m/s ( ) r R r R v GM r r v r r = + = + = = + = + = ' = . ' = + (c) From the work-kinetic energy theorem, 2 2 10 1 1 1 1 2.17 10 J 2 2 W = .K = mv' - mv = (d) 1 2 2 1 2 v 2GM(r /r ) r r ' = + Using the same values of 1 r and 2 r as in (b), 2 v' =1600 m/s. For the circular orbit, 2 2 v GM 3070 m/s r = = (e) 2 2 9 2 2 1 1 3.43 10 J 2 2 W = mv - mv' = (f) The total work done is 2.5131010 J. This is the same as in Example 13.6, but here weve learned how the work has to be divided between the two burns. 13.76. Model: The rod is thin and uniform. Visualize: Solve: (a) The rod is not spherical so must be divided into thin sections each dr wide and having mass dm. Since the rod is uniform, dm dr dm M dr M L L = . = The width of the rod is small enough so that all of dm is distance r away from m. The gravitational potential energy of dm and m is m M dr dU G m dm G L r r . . . . = - = - . . The total potential energy of the rod and mass m is found by adding the contributions dU from every point along the rod in an integral: 2 2 ln 2 2 x L x L x L U dU GMm dr GMm L r L x L + - . + . . . = =- =- . . . - . . . . . Note r is increasing with the limits chosen as they are. (b) The force on m when at x is 2 2 ln ln 1 1 2 2 2 2 4 , 4 2 dU GMm d L L GMm F x xL dx L dx L x x L GMm x L x L . . . . . . .. . . = - = . . + . - . - .. = . - . . . . . .. . + - . . . = - . . = . - . . . Assess: The direction of the force is towards the x direction, as expected. The force magnitude approaches 8 as the mass m approaches the end of the rod, but goes to zero like 2 1 x as x gets large. This is expected since from far away the rod looks like a point mass. 13-1 13.77. Model: The ring is uniform and is so thin that every point on it is the same distance r from m. Visualize: Solve: (a) We must determine the gravitational potential (dU) between m and an arbitrary part of the ring dm, then add using an integral all the contributions to U. Since the ring is uniform, 2 2 dm dl dm M dl M p r p r = . = The distance from m to dm is r = x2 + R2 . The gravitational potential between m and dm is 2 2 2 dU G m dm G mM dl r p r x R = - = - + The total gravitational potential is 2 2 2 ring U dU GmM dl p r x R = =- + . . Note that x and R do not change for any location of dm. The integral is just the length of the ring. ring . dl = 2p R Thus 2 2 U GmM x R = - + (b) The force on m when at x is (( ) ) ( ) ( ) ( ) 1 3 2 2 32 2 2 2 2 2 2 1 2 2 x F dU GmM d x R GmM x R x dx dx GmM x x R = - = + - = . - . + - . . . . = - + Thus the magnitude of the force is 2 232 . ( ) F GmM x x R = + Assess: The force is zero at the center of the ring. Elsewhere its direction is toward the origin. As x gets large, the force decreases like 2 1 . x This is expected since from far away the ring looks like a point mass. 14.1. Solve: The frequency generated by a guitar string is 440 Hz. The period is the inverse of the frequency, hence 1 1 2.27 10 3 s 2.27 ms 440 Hz T f = = = - = 14.2. Model: The air-track glider oscillating on a spring is in simple harmonic motion. Solve: The glider completes 10 oscillations in 33 s, and it oscillates between the 10 cm mark and the 60 cm mark. (a) 33 s 3.3 s oscillation 3.3 s 10 oscillations T= = = (b) 1 1 0.303 Hz 0.30 Hz 3.3 s f T = = = (c) . = 2p f = 2p (0.303 Hz) =1.90 rad s (d) The oscillation from one side to the other is equal to 60 cm -10 cm = 50 cm = 0.50 m. Thus, the amplitude is 1 ( ) 2 A = 0.50 m = 0.25 m. (e) The maximum speed is ( )( ) max v A 2 A 1.90 rad s 0.25 m 0.48 m s T p =. = .. .. = = . . 14.3. Model: The air-track glider attached to a spring is in simple harmonic motion. Visualize: The position of the glider can be represented as x(t) = Acos.t. Solve: The glider is pulled to the right and released from rest at t = 0 s. It then oscillates with a period T = 2.0 s and a maximum speed max v = 40 cm s = 0.40 m s. (a) max max and 2 2 rad s 0.40 m s 0.127 m 12.7 cm 2.0 s rad s v A A v T p p . . p . p = = = = . = = = = (b) The gliders position at t = 0.25 s is ( ) ( )( ) 0.25 s x = 0.127 m cos .. p rad s 0.25 s .. = 0.090 m = 9.0 cm 14.4. Model: The oscillation is the result of simple harmonic motion. Visualize: Please refer to Figure EX14.4. Solve: (a) The amplitude A =10 cm. (b) The time to complete one cycle is the period, hence T = 2.0 s and 1 1 0.50 Hz 2.0 s f T = = = (c) The position of an object undergoing simple harmonic motion is ( ) ( ) 0 x t = Acos . t +f . 0 At t = 0 s, x = -5 cm, thus ( ) ( ) 0 1 0 0 5 cm 10 cm cos 0 s cos 5 cm 1 cos 1 2 rad or 120 10 cm 2 2 3 . f p f f - - = .. + .. - . . . = = - . = .- . = . . Since the oscillation is originally moving to the left, 0 f = +120. 14.5. Model: The oscillation is the result of simple harmonic motion. Visualize: Please refer to Figure EX14.5. Solve: (a) The amplitude A = 20 cm. (b) The period T = 4.0 s, thus 1 1 0.25 Hz 4.0 s f T = = = (c) The position of an object undergoing simple harmonic motion is ( ) ( ) 0 x t = Acos . t +f . At 0 t = 0 s, x =10 cm. Thus, ( ) 1 1 0 0 10 cm 20 cm cos cos 10 cm cos 1 rad 60 20 cm 2 3 p = f .f = - .. .. = - .. .. = = . . . . Because the object is moving to the right at t = 0 s, it is in the lower half of the circular motion diagram and thus must have a phase constant between p and 2p radians. Therefore, 0 rad 60 . 3 p f =- =- 14.6. Visualize: The phase constant 2 3p has a plus sign, which implies that the object undergoing simple harmonic motion is in the second quadrant of the circular motion diagram. That is, the object is moving to the left. Solve: The position of the object is ( ) ( ) ( ) ( ) ( ) 2 0 0 x t = Acos .t +f = Acos 2p ft +f = 4.0 cm cos .. 4p rad s t + 3p rad.. The amplitude is A = 4 cm and the period is T =1 f = 0.50 s. A phase constant 0 f = 2p 3 rad =120 (second quadrant) means that x starts at 1 2 - A and is moving to the left (getting more negative). Assess: We can see from the graph that the object starts out moving to the left. 14.7. Visualize: A phase constant of 2 p - implies that the object that undergoes simple harmonic motion is in the lower half of the circular motion diagram. That is, the object is moving to the right. Solve: The position of the object is given by the equation ( ) ( ) ( ) ( ) 0 0 cos cos 2 8.0 cm cos rad s rad 2 2 x t A t A ft t p p . f p f .. . . = + = + = .. . - . .. . . The amplitude is A = 8.0 cm and the period is T =1 f = 4.0 s. With 0 f = -p 2 rad, x starts at 0 cm and is moving to the right (getting more positive). Assess: As we see from the graph, the object starts out moving to the right. 14.8. Solve: The position of the object is given by the equation x(t ) = Acos(.t +f0 ) = Acos(2p ft +f0 ) We can find the phase constant 0 f from the initial condition: ( ) 1 ( ) 1 0 0 0 2 0 cm = 4.0 cm cosf .cosf = 0.f = cos- 0 = p rad Since the object is moving to the right, the object is in the lower half of the circular motion diagram. Hence, 1 0 2 f = - p rad. The final result, with f = 4.0 Hz, is ( ) ( ) ( ) 1 2 x t = 4.0 cm cos .. 8.0p rad s t - p rad.. 14.9. Solve: The position of the object is given by the equation x(t ) = Acos(.t +f0 ) The amplitude A = 8.0 cm. The angular frequency . = 2p f = 2p (0.50 Hz) =p rad/s. Since at t = 0 it has its most negative velocity, it must be at the equilibrium point x = 0 cm and moving to the left, so 0 . 2 p f = Thus ( ) (8.0 cm)cos[( rad/s) rad] 2 x t t p = p + 14.10. Model: The air-track glider is in simple harmonic motion. Solve: (a) We can find the phase constant from the initial conditions for position and velocity: x0 = Acosf0 v0x = -. Asinf0 Dividing the second by the first, we see that 0 0 0 0 0 sin tan cos x v x f f f . = =- The glider starts to the left 0 (x = -5.00 cm) and is moving to the right 0( 36.3 cm/s). x v = + With a period of 1.5 s = 3 2 s, the angular frequency is 4 3 . = 2p /T = p rad/s. Thus 1 1 2 0 3 3 tan 36.3 cm/s rad (60 ) or rad (120 ) (4 /3 rad/s)( 5.00 cm) f p p p -. . = . - . = . - . The tangent function repeats every 180, so there are always two possible values when evaluating the arctan function. We can distinguish between them because an object with a negative position but moving to the right is in the third quadrant of the corresponding circular motion. Thus 2 0 3 f = - p rad, or -120. (b) At time t, the phase is 4 2 0 3 3 ( rad/s) rad. t t f . f p p = + = - This gives 23 f = - p rad, 0 rad, 2 3 p rad, and 4 3 p rad at, respectively, t = 0 s, 0.5 s, 1.0 s, and 1.5 s. This is one period of the motion. 14.11. Model: The block attached to the spring is in simple harmonic motion. Solve: The period of an object attached to a spring is 0 T 2 m T 2.0 s k = p = = where m is the mass and k is the spring constant. (a) For mass = 2m, ( ) 0 T 2 2m 2 T 2.8 s k = p = = (b) For mass 1 2 m, 1 2 0 2 21.41 s T m T k = p = = (c) The period is independent of amplitude. Thus 0 T = T = 2.0 s (d) For a spring constant = 2k, 0 2 2 1.41 s 2 T m T k = p = = 14.12. Model: The air-track glider attached to a spring is in simple harmonic motion. Solve: Experimentally, the period is T = (12.0 s) (10 oscillations) =1.20 s. Using the formula for the period, ( ) 2 2 2 2 2 0.200 kg 5.48 N m 1.20 s T m k m k T p p = p . = .. .. = .. .. = . . . . 14.13. Model: The mass attached to the spring oscillates in simple harmonic motion. Solve: (a) The period T =1 f =1 2.0 Hz = 0.50 s. (b) The angular frequency . = 2p f = 2p (2.0 Hz) = 4p rad/s. (c) Using energy conservation 1 2 1 2 1 2 2 2 0 2 0x kA = kx + mv Using 0 x = 5.0 cm, 0 30 cm/s x v = - and k = m. 2 = (0.200 kg)(4p rad/s)2 , we get A = 5.54 cm. (d) To calculate the phase constant 0 f , 0 0 1 0 cos 5.0 cm cos 5.0 cm 0.45 rad 5.54 cm A f x f - = = . = . . = . . . . (e) The maximum speed is ( )( ) max v =. A = 4p rad s 5.54 cm = 70 cm s. (f) The maximum acceleration is 2 ( ) ( )( ) 2 max a =. A =. . A = 4p rad s 70 cm s = 8.8 m s (g) The total energy is 1 2 1 ( )( )2 2 max 2 E = mv = 0.200 kg 0.70 m s = 0.049 J. (h) The position at t = 0.40 s is ( ) ( )( ) 0.4 s x = 5.54 cm cos .. 4p rad s 0.40 s + 0.45 rad.. = +3.8 cm 14.14. Model: The oscillating mass is in simple harmonic motion. Solve: (a) The amplitude A = 2.0 cm. (b) The period is calculated as follows: 2 10 rad s 2 0.63 s 10 rad s T T p p . = = . = = (c) The spring constant is calculated as follows: k k m 2 (0.050 kg)(10 rad s)2 5.0 N m m . = . = . = = (d) The phase constant 1 0 4 f = - p rad. (e) The initial conditions are obtained from the equations ( ) ( ) ( 1 ) ( ) ( ) ( 1 ) 4 4 2.0 cm cos 10 and 20.0 cm s sin 10 x x t = t - p v t = - t - p At t = 0 s, these equations become ( ) ( 1 ) ( ) ( 1 ) 0 4 0 4 2.0 cm cos 1.41 cm and 20 cm s sin 14.1 cm s x x = - p = v = - - p = In other words, the mass is at +1.41 cm and moving to the right with a velocity of 14.1 cm/s. (f) The maximum speed is ( )( ) max v = A. = 2.0 cm 10 rad s = 20 cm s. (g) The total energy 1 2 1 ( )( )2 3 2 2E = kA = 5.0 N m 0.020 m =1.0010- J. (h) At t = 0.41 s, the velocity is ( ) ( )( ) 1 0 4 20 cm s sin 10 rad s 0.40 s 1.46 cm s x v = - .. - p .. = 14.15. Model: The block attached to the spring is in simple harmonic motion. Visualize: Solve: (a) The conservation of mechanical energy equation Kf +Usf = Ki +Usi is ( ) ( ) 1 2 1 2 1 2 1 2 1 2 2 1 2 2 0 2 2 0 0 0 J 0 J 0 J 1.0 kg 0.40 m s 0.10 m 10.0 cm 16 N m mv k x mv kA mv A mv k + . = + . + = + . = = = = (b) We have to find the velocity at a point where x = A/2. The conservation of mechanical energy equation 2 s2 i si K +U = K +U is ( ) 2 2 2 2 2 2 2 2 2 2 0 2 0 0 0 0 2 0 1 1 1 0 J 1 1 1 1 1 1 1 3 1 2 2 2 2 2 2 42 2 42 42 3 30.40 m s 0.346 m s 4 4 mv k A mv mv mv kA mv mv mv v v + . . = + . = - . . = - . . = . . . . . . . . . . . . . . . . . . . = = = The velocity is 35 cm/s. 14.16. Model: The vertical oscillations constitute simple harmonic motion. Visualize: Solve: (a) At equilibrium, Newtons first law applied to the physics book is ( ) ( )( ) ( ) sp 2 0 N 0 N 0.500 kg 9.8 m s 0.20 m 24.5 N m y F mg ky mg k mg y - = .- . - = . = - . = - - = (b) To calculate the period: 24.5 N m 7.0 rad s and 2 2 rad 0.90 s 0.500 kg 7.0 rad s k T m p p . . = = = = = = (c) The maximum speed is ( )( ) max v = A. = 0.10 m 7.0 rad s = 0.70 m s Maximum speed occurs as the book passes through the equilibrium position. 14.17. Model: The vertical oscillations constitute simple harmonic motion. Solve: To find the oscillation frequency using . = 2p f = k m, we first need to find the spring constant k. In equilibrium, the weight mg of the block and the spring force k.L are equal and opposite. That is, mg = k.L.k = mg .L. The frequency of oscillation f is thus given as 1 1 1 1 9.8 m s2 3.5 Hz 2 2 2 2 0.020 m f k mg L g p m p m p L p . = = = = = . 14.18. Model: The vertical oscillations constitute simple harmonic motion. Visualize: Solve: The period and angular frequency are 20 s 0.6667 s and 2 2 9.425 rad s 30 oscillations 0.6667 s T T p p = = . = = = (a) The mass can be found as follows: 2 ( )2 15 N/m 0.169 kg 9.425 rad s k m k m . . = . = = = (b) The maximum speed ( )( ) max v =. A = 9.425 rad s 0.060 m = 0.57 m/s. 14.19. Model: Assume a small angle of oscillation so there is simple harmonic motion. Solve: The period of the pendulum is 0 0 T 2 L 4.0 s g = p = (a) The period is independent of the mass and depends only on the length. Thus 0 T = T = 4.0 s. (b) For a new length 0 L = 2L , 0 0 T 2 2L 2T 5.7 s g = p = = (c) For a new length 0L = L /2, 0 0 2 2 1 2.8 s 2 T L T g = p = = (d) The period is independent of the amplitude as long as there is simple harmonic motion. Thus T = 4.0 s. 14.20. Model: The pendulum undergoes simple harmonic motion. Solve: (a) The amplitude is 0.10 rad. (b) The frequency of oscillations is 5 Hz 0.796 Hz 2 2 f . p p = = = (c) The phase constant f =p rad. (d) The length can be obtained from the period: ( ) ( ) 2 2 2 1 1 9.8 m s2 0.392 m 2 2 0.796 Hz f g L g L f . p p p . . . . = = . = . . = .. .. = . . . . (e) At t = 0 s, ( ) ( ) 0 . = 0.10 rad cos p = -0.10 rad. To find the initial condition for the angular velocity we take the derivative of the angular position: ( ) (0.10 rad)cos(5 ) ( ) (0.10 rad)(5)sin(5 ) d t t t t dt . . = +p . = - +p At t = 0 s, ( ) ( ) ( ) 0 d. dt = -0.50 rad sin p = 0 rad s. (f) ( ) ( ( ) ) 2.0 At t = 2.0 s, . = 0.10 rad cos 5 2.0 s +p = 0.084 rad. 14.21. Model: Assume the small-angle approximation so there is simple harmonic motion. Solve: The period is T =12 s 10 oscillations =1.20 s and is given by the formula ( ) 2 2 2 1.20 s 9.8 m s2 36 cm 2 2 T L L T g g p p p = . = . . = . . = . . . . . . . . 14.22. Model: Assume a small angle of oscillation so there is simple harmonic motion. Solve: (a) On the earth the period is earth 2 2 2 1.0 m 2.0 s 9.80 m s T L g = p = p = (b) On Venus the acceleration due to gravity is ( )( ) ( ) 11 2 2 24 Venus 2 Venus 2 6 2 Venus Venus 2 Venus 6.67 10 N m kg 4.88 10 kg 8.86 m s 6.06 10 m 2 2 1.0 m 2.1 s 8.86 m s g GM R T L g p p - = = = . = = = 14.23. Model: Assume the pendulum to have small-angle oscillations. In this case, the pendulum undergoes simple harmonic motion. Solve: Using the formula g = GM R2 , the periods of the pendulums on the moon and on the earth are 2 2 earth earth moon moon earth moon earth moon T 2 L 2 L R and T 2 L R g GM GM = p = p = p Because earth moon T = T , ( ) 2 2 2 earth earth moon moon moon earth moon earth earth moon earth moon 22 6 2 24 6 2 2 7.36 10 kg 6.37 10 m 2.0 m 33 cm 5.98 10 kg 1.74 10 m L R L R L M R L GM GM M R p p . .. . = . =. .. . . .. . . .. . = . .. . = . .. . 14.24. Model: Assume a small angle of oscillation so that the pendulum has simple harmonic motion. Solve: The time periods of the pendulums on the earth and on Mars are earth Mars earth Mars T 2 L and T 2 L g g = p = p Dividing these two equations, ( ) 2 2 earth Mars earth 2 2 Mars earth Mars earth Mars 9.8 m s 1.50 s 3.67 m s 2.45 s T g g g T T g T . . . . = . = . . = . . = . . . . 14.25. Visualize: Please refer to Figure Ex14.25. Solve: The mass of the wrench can be obtained from the length that it stretches the spring. From Equation 14.41, ( ) 2 360 N/m 0.030 m 1.10 kg 9.8 m/s L mg m k L k g . . = . = = = When swinging on a hook the wrench is a physical pendulum. From Equation 14.52, 2 f mgl 2 mgl I T I p p = . = From the figure, l = 0.14 m. Thus ( )( )( ) 2 2 0.90 s 1.10 kg 9.8 m/s2 0.14 m 3.1 10 2 kg m2 2 2 I T mgl p p = .. .. = .. .. = - . . . . 14.26. Model: The spider is in simple harmonic motion. Solve: Your tapping is a driving frequency. Largest amplitude at fext =1.0 Hz means that this is the resonance frequency, so 0 ext f = f =1.0 Hz. That is, the spiders natural frequency of oscillation 0 f is 1.0 Hz and 0 0 . = 2p f = 2p rad s. We have 2 ( )( )2 0 0k k m 0.0020 kg 2 rad s 0.079 N m m . = . = . = p = 14.27. Model: The motion is a damped oscillation. Solve: The amplitude of the oscillation at time t is given by Equation 14.58: ( ) 2 0 A t = A e-t t , where t = m b is the time constant. Using x = 0.368 A and t =10.0 s, we get ( ) ( ) 0.368 10.0 s 2 ln 0.368 10 s 10.0 s 5.00 s 2 2ln 0.368 A Ae t t t - - = . = . =- = 14.28. Model: The motion is a damped oscillation. Solve: The position of a damped oscillator is ( ) ( 2 ) ( ) 0 x t Ae t cos t . = - t . +f The frequency is 1.0 Hz and the damping time constant t is 4.0 s. Let us assume 0 f = 0 rad and A =1 with arbitrary units. Thus, x(t ) = e-t (8.0 s) cos ..2p (1.0 Hz)t... x(t ) = e-0.125 t cos(2p t ) where t is in s. Values of x(t) at selected values of t are displayed in the following table: t(s) x(t) t(s) x(t) t(s) x(t) 0 0.25 0.50 0.75 1.00 1.25 1.50 1.75 1 0 -0.939 0 0.882 0 -0.829 0 2.00 2.50 3.00 3.50 4.00 4.50 5.00 5.50 0.779 -0.732 0.687 -0.646 0.607 -0.570 0.535 -0.503 6.00 6.50 7.00 7.50 8.00 8.50 9.00 9.50 10.00 0.472 - 0.444 0.417 - 0.392 0.368 - 0.346 0.325 - 0.305 0.286 14.29. Model: The pendulum is a damped oscillator. Solve: The period of the pendulum and the number of oscillations in 4 hours are calculated as follows: ( ) 2 osc 15.0 m 4 3600 s 2 2 7.773 s 1853 9.8 m s 7.773 s T L N g = p = p = . = = The amplitude of the pendulum as a function of time is A(t ) = Ae-bt / 2m. The exponent of this expression can be calculated to be ( )( ) ( ) 0.010 kg s 4 3600 s 0.6545 2 2 110 kg bt m - =- =- We have A(t ) = (1.50 m)e-0.6545 = 0.780 m. 14.30. Model: The vertical oscillations are damped and follow simple harmonic motion. Solve: The position of the ball is given by ( ) ( 2 ) ( ) 0 x t Ae t cos t . = - t . +f The amplitude A(t ) Ae (t 2 ) = - t is a function of time. The angular frequency is (15.0 N m) 2 5.477 rad s 1.147 s 0.500 kg k T m p . . = = = . = = Because the balls amplitude decreases to 3.0 cm from 6.0 cm after 30 oscillations, that is, after 301.147 s = 34.41 s, we have 3.0 cm (6.0 cm) (34.414 s 2 ) 0.50 (34.41 s 2 ) ln(0.50) 34.41 s 25 s 2 e e t t t t - - - = . = . = .= 14.31. Visualize: Please refer to Figure P14.31. Solve: The position and the velocity of a particle in simple harmonic motion are ( ) ( ) ( ) ( ) ( ) 0 0max 0 cos x t = A .t +f and vx t = -A. sin .t +f = -v sin .t +f (a) At t = 0 s, the equation for x yields ( ) ( ) ( ) 1 ( ) 2 0 0 3 -5.0 cm = 10.0 cm cos f .f = cos- -0.5 = p rad Because the particle is moving to the left at t = 0 s, it is in the upper half of the circular motion diagram, and the phase constant is between 0 and p radians. Thus, 2 0 3 f = p rad. (b) The period is 4.0 s. At t = 0 s, ( ) 0 0 sin 10.0 cm 2 sin 2 13.6 cm/s 3 x v A T p p = - . f = - .. .. .. .. = - . . . . (c) The maximum speed is ( ) max 2 10.0 cm 15.7 cm s 4.0 s v A p =. = .. .. = . . Assess: The negative velocity at t = 0 s is consistent with the position-versus-time graph and the positive sign of the phase constant. 14.32. Visualize: Please refer to Figure P14.32. Solve: The position and the velocity of a particle in simple harmonic motion are ( ) ( ) ( ) ( ) ( ) 0 0max 0 cos x t = A .t +f and vx t = -A. sin .t +f = -v sin .t +f From the graph, T =12 s and the angular frequency is 2 2 rad s T 12 s 6 p p p . = = = (a) Because max v = A. = 60 cm s, we have 60 cm s 60 cm s 115 cm 6 rad s A . p = = = (b) At t = 0 s, ( ) ( ) ( ) ( ) 0 0 0 1 1 5 0 6 6 sin 30 cm/s 60 cm s sin 30 cm/s sin 0.5 rad rad 30 or rad 150 x v A. f f f - p p = - = - . - = - . = = Because the velocity at t = 0 s is negative and the particle is slowing down, the particle is in the second quadrant of the circular motion diagram. Thus 5 0 6 f = p rad. (c) At t = 0 s, ( ) ( 5 ) 0 6 x = 115 cm cos p rad = -100 cm. 14.33. Model: The vertical mass/spring systems are in simple harmonic motion. Visualize: Please refer to Figure P14.33. Solve: (a) For system A, the maximum speed while traveling in the upward direction corresponds to the maximum positive slope, which is at t = 3.0 s. The frequency of oscillation is 0.25 Hz. (b) For system B, all the energy is potential energy when the position is at maximum amplitude, which for the first time is at t =1.5 s. The time period of system B is thus 6.0 s. (c) Spring/mass A undergoes three oscillations in 12 s, giving it a period A T = 4.0 s. Spring/mass B undergoes 2 oscillations in 12 s, giving it a period B T = 6.0 s. We have A B A A B A B A B B BA 2 and 2 4.0 s 2 6.0 s 3 T m T m T m k k k T mk p p . .. . = = . = . .. . = = . .. . If A Bm = m , then B A A B 4 9 2.25 9 4 k k k k = . = = 14.34. Solve: The objects position as a function of time is x(t ) = Acos(.t +f0 ). Letting x = 0 m at t = 0 s, gives 1 0 0 2 0 = Acosf .f = p Since the object is traveling to the right, it is in the lower half of the circular motion diagram, giving a phase constant between -p and 0 radians. Thus, 1 0 2 f = - p and ( ) ( 1 ) ( ) ( ) ( 1 ) 2 2 x t = Acos .t - p . x t = Asin.t = 0.10 m sin p t where we have used A = 0.10 m and 2 2 rad rad s T 4.0 s 2 p p p . = = = Let us now find t where x = 0.60 m: 0.060 m (0.10 m)sin 2 sin 1 0.060 m 0.41 s 2 0.10 m t t p p = .. ... = - .. .. = . . . . Assess: The answer is reasonable because it is approximately 18 of the period. 14.35. Model: The block attached to the spring is in simple harmonic motion. Visualize: The position and the velocity of the block are given by the equations ( ) ( ) x t = Acos .t +f0 and vx (t) = -A. sin(.t +f0 ) Solve: To graph x(t) we need to determine ., 0 f , and A. These quantities will be found by using the initial (t = 0 s) conditions on x(t) and ( ). x v t The period is 2 2 1.0 kg 1.405 s 2 2 rad 4.472 rad s 20 N m 1.405 s T m k T p p = p = p = .. = = = At t = 0 s, 0 0 0 0 cos and sin . x x = A f v = -A. f Dividing these equations, ( ) ( )( ) 0 0 0 0 1.0 m s tan 1.1181 0.841 rad 4.472 rad s 0.20 m x v x f f . - = - = - = . = From the initial conditions, ( ) 2 2 2 0 2 0 0.20 m 1.0 m s 0.300 m 4.472 rad s x A x v . . . . - . = + . . = + . . = . . . . The position-versus-time graph can now be plotted using the equation x(t ) = (0.300 m)cos ..(4.472 rad s)t + 0.841 rad.. 14.36. Model: The astronaut attached to the spring is in simple harmonic motion. Visualize: Please refer to Figure P14.36. Solve: (a) From the graph, T = 3.0 s, so we have ( ) 2 2 2 3.0 s 240 N m 55 kg 2 2 T m m T k k p p p = . = . . = . . = . . . . . . . . (b) Oscillations occur about an equilibrium position of 1.0 m. From the graph, 1 ( ) 2 0 A = 0.80 m = 0.40 m, f = 0 rad, and 2 2 2.1 rad s T 3.0 s p p . = = = The equation for the position of the astronaut is ( ) ( ) ( ) ( ) ( ) ( ) cos 1.0 m 0.4 m cos 2.1 rad s 1.0 m 1.2 m= 0.4 m cos 2.1 rad s 1.0 m cos 2.1 rad s 0.5 0.50 s x t A t t t t t = . + = .. .. + . .. .. + . .. .. = . = The equation for the velocity of the astronaut is ( ) ( ) ( )( ) ( )( ) 0.5 s sin 0.4 m 2.1 rad s sin 2.1 rad s 0.50 s 0.73 m s x v t A t v = - . . . = - .. .. = - Thus her speed is 0.73 m/s. 14.37. Model: The particle is in simple harmonic motion. Solve: The equation for the velocity of the particle is ( ) (25 cm)(10 rad s)sin(10 t) x v t = - Substituting into K = 2U gives ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 2 2 2 2 2 2 2 2 2 2 2 2 2 1 1 2 1 1 250 cm s sin 10 t 25 cm cos 10 t 2 2 2 sin 10 t 25 cm 1 2 2 s cos 10 t 250 cm s 100 tan 10 t 2 10 rad s 1 s 2.0 1 tan 2.0 0.096 s 100 10 x mv t kx t m k k m t . - = . .. .- . = . . . . . . . . . . . = . . = . . . . . . . . . . . = . . = . = = . . . . 14.38. Model: The spring undergoes simple harmonic motion. Solve: (a) Total energy is 1 2 2 E = kA . When the displacement is 1 2 x = A, the potential energy is 1 2 1 1 2 1 (1 2 ) 1 3 2 2 2 42 4 4 U = kx = k( A) = kA = E. K = E -U = E One quarter of the energy is potential and three-quarters is kinetic. (b) To have 1 2 U = E requires 1 2 1 1 (1 2 ) 2 2 22 2 U = kx = E = kA . x = A 14.39. Solve: Average speed is vavg = .x/.t. During half a period 1 2 (.t = T), the particle moves from x = -A to x = +A(.x = 2A). Thus avg max max avg 2 4 4 2 ( ) 2 /2 2 / 2 v x A A A A v v v t T T p . p . p p . = = = = = = . = . 14.40. Model: The ball attached to a spring is in simple harmonic motion. Solve: (a) Let t = 0 s be the instant when 0 x = -5.0 cm and 0 v = 20 cm/s. The oscillation frequency is 2.5 N m 5.0 rad / s 0.100 kg k m . = = = Using Equation 14.27, the amplitude of the oscillation is ( ) 2 2 2 0 2 0 5.0 cm 20 cm/ s 6.4 cm 5.0 rad / s A x v . . . . . = + . . = - + . . = . . . . (b) The maximum acceleration is 2 2 max a =. A =160 cm s . (c) For an oscillator, the acceleration is most positive ( ) max a = a when the displacement is most negative ( ) max x = -x = -A . So the acceleration is maximum when x = -6.4 cm. (d) We can use the conservation of energy between 0 x = -5.0 cm and 1 x = 3.0 cm: 1 2 1 2 1 2 1 2 2 ( 2 2 ) 2 0 2 0 2 1 2 1 1 0 0 1 mv kx mv kx v v k x x 0.283 m s m + = + . = + - = The speed is 28 cm/s. Because k is known in SI units of N/m, the energy calculation must be done using SI units of m, m/s, and kg. 14.41. Model: The block on a spring is in simple harmonic motion. Solve: (a) The position of the block is given by x(t ) = Acos(.t +f0 ). Because x(t ) = A at t = 0 s, we have 0 f = 0 rad, and the position equation becomes x(t ) = Acos.t. At t = 0.685 s, 3.00 cm = Acos(0.685. ) and at t = 0.886 s, -3.00 cm = Acos(0.886. ). These two equations give cos(0.685 ) cos(0.886 ) cos( 0.886 ) 0.685 0.886 2.00 rad s . . p . . p . . = - = - . = - . = (b) Substituting into the position equation, 3.00 cm cos((2.00 rad s)(0.685 s)) cos(1.370) 0.20 3.00 cm 15.0 cm 0.20 = A = A = A. A = = 14.42. Model: The oscillator is in simple harmonic motion. Energy is conserved. Solve: The energy conservation equation E1 = E2 is ( )( ) ( ) ( )( ) ( ) 1 2 1 2 1 2 1 2 2 1 2 1 2 2 2 2 1 0.30 kg 0.954 m s 2 1 0.030 m 2 1 0.30 kg 0.714 m s 2 1 0.060 m 2 2 2 2 2 44.48 N m mv kx mv kx k k k + = + + = + . = The total energy of the oscillator is 2 2 ( )( )2 ( )( )2 total 1 1 1 1 1 0.30 kg 0.954 m s 1 44.48 N m 0.030 m 0.1565 J 2 2 2 2 E = mv + kx = + = Because 1 2 total 2 max E = mv , ( ) 2 max max 0.1565 J 1 0.300 kg 1.02 m s 2 = v .v = Assess: A maximum speed of 1.02 m/s is reasonable. 14.43. Model: The transducer undergoes simple harmonic motion. Solve: Newtons second law for the transducer is ( 3 ) 8 2 restoring max max max F = ma .40,000 N = 0.1010- kg a .a = 4.010 m s Because 2 max a =. A, ( ) 8 2 max 5 2 6 2 4.0 10 m s 1.01 10 m 10.1 m 2 1.0 10 Hz A a . p - = = = = .. .. (b) The maximum velocity is ( 6 )( 5 ) max v =. A = 2p 1.010 Hz 1.0110- m = 64 m s 14.44. Model: The block attached to the spring is in simple harmonic motion. Solve: (a) The frequency is 1 1 2000 N m 3.183 Hz 2 2 5.0 kg f k p m p = = = The frequency is 3.2 Hz. (b) From energy conservation, 2 2 2 0 2 0 (0.050 m) 1.0 m/s 0.0707 m 2 3.183 Hz A x v . p = + . . = + . . = . . . . . . . . The amplitude is 7.1 cm. (c) The total mechanical energy is 1 2 1 ( )( )2 2 2E = kA = 2000 N m 0.0707 m = 5.0 J 14.45. Model: The tips of the tuning fork are in simple harmonic oscillation. Solve: (a) The maximum speed is related to the amplitude. ( )( 4 ) max v =. A = 2p fA = 2p 440 Hz 5.010- m =1.38 m/s (b) The acceleration of the flea can not be greater than that allowed by the maximum force with which it can hold on. From Newtons second law, the maximum acceleration that the flea can withstand is 3 2 flea 6 1.0 10 N 100 m/s 10 10 kg a F m - - = = = The maximum acceleration at the tip of the prong is ( ) ( ( )) ( ) 2 2 2 4 2 max a =. A = 2p f A = 2p 440 Hz 5.010- m = 382 m/s The flea will not be able to hold on to the tuning fork. 14.46. Model: The block undergoes simple harmonic motion. Visualize: Solve: (a) The frequency of oscillation is 1 1 10 N/m 1.125 Hz 2 2 0.20 kg f k p m p = = = The frequency is 1.13 Hz. (b) Using conservation of energy, 1 2 1 2 1 2 1 2 2 1 2 1 2 0 2 0 mv + kx = mv + kx , we find 2 2 2 2 2 2 1 0 0 1 ( ) ( 0.20 m) 0.20 kg ((1.00 m/s) (0.50 m/s) ) 10 N/m 0.2345 m or 23 cm x x m v v k = + - = - + - = (c) At time t, the displacement is 0 x = Acos(.t +f ). The angular frequency is . = 2p f = 7.071 rad/s. The amplitude is 2 2 2 0 2 0 ( 0.20 m) 1.00 m/s 0.245 m 7.071 rad/s A x v . = + . . = - + . . = . . . . . . . . The phase constant is 1 0 1 0 cos cos 0.200 m 2.526 rad or 145 0.245 m x A f - - . . . - . = . . = . . = . . . . A negative displacement (below the equilibrium point) and positive velocity (upward motion) indicate that the corresponding circular motion is in the third quadrant, so 0 f = -2.526 rad. Thus at t =1.0 s, x = (0.245 m)cos((7.071 rad/s)(1.0 s) - 2.526 rad) = -0.0409 m = -4.09 cm The block is 4.1 cm below the equilibrium point. 14.47. Model: The mass is in simple harmonic motion. Visualize: The high point of the oscillation is at the point of release. This conclusion is based on energy conservation. Gravitational potential energy is converted to the springs elastic potential energy as the mass falls and stretches the spring, then the elastic potential energy is converted 100% back into gravitational potential energy as the mass rises, bringing the mass back to exactly its starting height. The total displacement of the oscillationhigh point to low pointis 20 cm. Because the oscillations are symmetrical about the equilibrium point, we can deduce that the equilibrium point of the spring is 10 cm below the point where the mass is released. The mass oscillates about this equilibrium point with an amplitude of 10 cm, that is, the mass oscillates between 10 cm above and 10 cm below the equilibrium point. Solve: The equilibrium point is the point where the mass would hang at rest, with Fsp = FG = mg. At the equilibrium point, the spring is stretched by .y =10 cm = 0.10 m. Hookes law is sp F = k.y, so the equilibrium condition is [ ] 2 2 sp G 9.8 m s 98 s 0.10 m F k y F mg k g m y . = . . = = . = = = - . . . The ratio k m is all we need to find the oscillation frequency: 1 1 98 s 2 1.58 Hz 2 2 f k p m p
= = - = 14.48. Model: The spring is ideal, so the apples undergo SHM. Solve: The spring constant of the scale can be found by considering how far the pan goes down when the apples are added. 20 N 222 N/m 0.090 m L mg k mg k L . = . = = = . The frequency of oscillation is 2 1 1 222 N/m 1.66 Hz 2 2 (20 N 9.8 m/s ) f k p m p = = = Assess: An oscillation of fewer than twice per second is reasonable. 14.49. Model: The compact car is in simple harmonic motion. Solve: (a) The mass on each spring is (1200 kg) 4 = 300 kg. The spring constant can be calculated as follows: ( ) ( ) ( ) 2 k k m 2 m 2 f 2 300 kg 2 2.0 Hz 2 4.74 104 N m m . = . = . = p = .. p .. = The spring constant is 4.7104 N/m. (b) The car carrying four persons means that each spring has, on the average, an additional mass of 70 kg. That is, m = 300 kg + 70 kg = 370 kg. Thus, 1 1 4.74 104 N m 1.80 Hz 2 2 2 370 kg f k m . p p p = = = = Assess: A small frequency change from the additional mass is reasonable because frequency is inversely proportional to the square root of the mass. 14.50. Model: Hookes law for the spring. The springs compression and decompression constitutes simple harmonic motion. Visualize: Solve: (a) The springs compression or decompression is one-half of the oscillation cycle. This means the contact time is 1 .t = 2T, where T is the period. The period is calculated as follows: 50 N m 10 rad s 1 2 2 0.628 s 0.500 kg 10 rad s 0.31 s 2 k T m f t T p p . . = = = . = = = = . . = = (b) There is no change in contact time, because period of oscillation is independent of the amplitude or the maximum speed. 14.51. Model: The two blocks are in simple harmonic motion, without the upper block slipping. We will also apply the model of static friction between the two blocks. Visualize: Solve: The net force acting on the upper block m1 is the force of friction due to the lower block m2. The model of static friction gives the maximum force of static friction as ( ) ( ) s max s s 1 1 max max s f = n = m g = m a .a = g Using s = 0.5, ( )( 2 ) 2 max S a = g = 0.5 9.8 m s = 4.9 m s . That is, the two blocks will ride together if the maximum acceleration of the system is equal to or less than max a . We can calculate the maximum value of A as follows: ( ) ( 2 )( ) 2 max 1 2 max max max max 1 2 4.9 m s 1.0 kg 5.0 kg 0.59 m 50 N m k a m m a A A A m m k . + + = = . = = = + 14.52. Model: Assume simple harmonic motion for the two-block system without the upper block slipping. We will also use the model of static friction between the two blocks. Visualize: Solve: The net force on the upper block m1 is the force of static friction due to the lower block m2. The two blocks ride together as long as the static friction doesnt exceed its maximum possible value. The model of static friction gives the maximum force of static friction as ( ) ( ) s max s s 1 1 max max s 2 2 2 max max max s 2 2 2 0.40 m 0.72 1.5 s 9.8 m s f n mg ma a g a A A g g T g . p p = = = . = . . . . . . . . . = = = . . . . = . . . . = . . . . . . . . Assess: Because the period is given, we did not need to use the block masses or the spring constant in our calculation. 14.53. Model: The DNA and cantilever undergo SHM. Visualize: Please refer to figure P14.53. Solve: The cantilever has the same spring constant with and without the DNA molecule. The frequency of oscillation without the DNA is 1 1 3 k M . = With the DNA, the frequency of oscillation is 2 1 3 k M m . = + where m is the mass of the DNA. Divide the two equations, and express .2 =.1 - .., where .. = 2p.f = 2p (50 Hz). 1 1 1 1 1 3 3 1 2 2 1 3 1 3 ( ) ( ) k f f M M m f f f k M M m . . + = = = = - . + Thus ( ) ( ) ( ) ( ( ) ) ( ) ( ) 2 1 2 1 1 3 1 3 2 2 1 1 2 1 1 1 3 2 3 2 1 1 1 f M f f M m f f f f m M M f f f f = -. + - -. . . = = . - . - . . - . . . . Since 1.f .. f , (50 Hz ..12MHz), ( ) 2 2 2 2 1 1 1 1 1 f f f 1 f f 1 2 f . f f - - - - . . . . . . - . = . - . . + . . . . . Thus 1 2 3 3 1 1 m M 1 2 f 1 M f f f . . . . = . + - . = . . The mass of the cantilever M = (2300 kg/m3)(400010-9 m)(10010-9 m) = 3.6810-16 kg Thus the mass of the DNA molecule is ( 16 ) 21 6 2 3.68 10 kg 50 Hz 1.02 10 kg 3 1210 Hz m = - .. .. = - . . Assess: The mass of the DNA molecule is about 6.2105 atomic mass units, which is reasonable for such a large molecule. 14.54. Model: Assume that the swinging lamp makes a small angle with the vertical so that there is simple harmonic motion. Visualize: Solve: (a) Using the formula for the period of a pendulum, ( ) 2 2 2 9.8 m s2 5.5 s 7.5 m 2 2 T L L g T g p p p = . = . . = . . = . . . . . . . . (b) The conservation of mechanical energy equation 0 g0 1 g1 K +U = K +U for the swinging lamp is ( ) ( )( )( ) 1 2 1 2 1 2 2 0 0 2 1 1 2 max max 2 0 J 0 J 2 2 cos3 2 9.8 m s 7.5 m 1 cos3 0.45 m s mv mgy mv mgy mgh mv v gh gLL + = + . + = + . = = - = - = 14.55. Model: Assume that the angle with the vertical that the pendulum makes is small enough so that there is simple harmonic motion. Solve: The angle . made by the string with the vertical as a function of time is . (t ) =. max cos(.t +f0 ) The pendulum starts from maximum displacement, thus 0 f = 0. Thus, ( ) max . t =. cos.t. To find the time t when the pendulum reaches 4.0 on the opposite side: (-4.0) = (8.0)cos.t ..t = cos-1 (-0.5) = 2.094 rad Using the formula for the angular frequency, 9.8 m s2 2.0944 rad 2.094 rad 3.130 rad s 0.669 s 1.0 m 3.130 rad s g t L . . = = = . = = = The time t = 0.67 s. Assess: Because T = 2p . = 2.0 s, a value of 0.67 s for the pendulum to cover a little less than half the oscillation is reasonable. 14.56. Model: Assume a small angle oscillation of the pendulum so that it has simple harmonic motion. Solve: (a) At the equator, the period of the pendulum is equator 2 2 1.000 m 2.009 s 9.78 m s T = p = The time for 100 oscillations is 200.9 s. (b) At the north pole, the period is pole 2 2 1.000 m 2.004 s 9.83 m s T = p = The time for 100 oscillations is 200.4 s. (c) The difference between the two answers is 0.5 s, and this difference is quite measurable with a hand-operated stopwatch. (d) The period on the top of the mountain is 2.010 s. The acceleration due to gravity can be calculated by rearranging the formula for the period: ( ) 2 2 2 mountain mountain 2 1.000 m 2 9.772 m s 2.010 s g L T . p . . p . = . . = . . = . . . . Assess: This last result is reasonable because g decreases with altitude. 14.57. Model: The mass is a particle and the string is massless. Solve: Equation 14.52 is Mgl I . = The moment of inertia of the mass on a string is I = Ml2 , where l is the length of the string. Thus 2 Mgl g Ml l .= = This is Equation 14.49 with L = l. Assess: Equation 14.49 is really a specific case of the more general physical pendulum described by Equation 14.52. 14.58. Model: The rod is thin and uniform with moment of inertia described in Table 12.2. The clay ball is a particle located at the end of the rod. The ball and rod together form a physical pendulum. The oscillations are small. Visualize: Solve: The moment of inertia of the composite pendulum formed by the rod and clay ball is ( )( ) ( )( ) 2 2 rod + ball rod ball rod ball 2 2 3 2 1 3 1 0.200 kg 0.15 m 0.020 kg 0.15 m 1.95 10 kg m 3 I I I mL mL - = + = + = + = The center of mass of the rod and ball is located at a distance from the pivot point of ( ) ( )( ) ( ) 2 cm 0.200 kg 0.15 m 0.020 kg 0.15 m 2 8.18 10 m 0.200 kg 0.020 kg y - . . + . . = . . = + The frequency of oscillation of a physical pendulum is ( )( 2 )( 2 ) 3 2 1 1 0.220 kg 9.8 m/s 8.18 10 m 1.51 Hz 2 2 1.95 10 kg m f Mgl p I p - - = = = The period of oscillation T 1 f = = 0.66 s. 14.59. Model: The circular hoop can be modeled as a cylindrical hoop and its moment of inertia about the point of rotation found with the parallel-axis theorem. Visualize: Please refer to Figure P14.59. Solve: Using the parallel-axis theorem, the moment of inertia of the cylindrical hoop about the rotation point is I = MR2 + MR2 = 2MR2 The frequency of small oscillations is given by Equation 14.52. 1 2 f Mgl p I = The center of mass of the hoop is its center, so l = R. Thus 2 1 1 2 2 2 2 f MgR g p MR p R = = 14.60. Model: The motion is a damped oscillation. Solve: The position of the air-track glider is ( ) ( 2 ) ( ) 0 x t Ae t cos t , = - t . +f where t = m b and 2 4 2 k b m m . = - Using A = 0.20 m, 0 f = 0 rad, and b = 0.015 kg/s, ( ) ( ) 2 4 2 4.0 N m 0.015 kg s 16 9 10 rad s 4.0 rad s 0.250 kg 4 0.250 kg . = - = - - = Thus the period is 2 2 rad 1.57 s 4.0 rad s T p p . = = = The amplitude at t = 0 s is 0 x = A and the amplitude will be equal to e-1A at a time given by 1 A Ae (t 2 ) t 2 2m 33.3 s e b = - t . = t = = The number of oscillations in a time of 33.3 s is (33.3 s)/(1.57 s) = 21. 14.61. Model: A completely inelastic collision between the two gliders resulting in simple harmonic motion. Visualize: Let us denote the 250 g and 500 g masses as m1 and m2 , which have initial velocities i1 v and i2v . After 1 m collides with and sticks to 2 m , the two masses move together with velocity f v . Solve: The momentum conservation equation f i p = p for the completely inelastic collision is ( ) 1 2 f m + m v = 1 i1 2 i2 . mv + m v Substituting the given values, ( ) ( )( ) ( )( ) f f 0.750 kg v = 0.250 kg 1.20 m s + 0.500 kg 0 m s .v = 0.400 m s We now use the conservation of mechanical energy equation: ( ) ( ) ( ) ( ) 1 2 1 2 s compressed s equilibrium 2 2 1 2 f 1 2 f 0 J 0 J 0.750 kg 0.400 m s 0.110 m 10 N m K U K U kA m m v A m m v k + = + . + = + + + . = = = The period is 1 2 2 2 0.750 kg 1.72 s 10 N m T m m k p p + = = = 14.62. Model: The block attached to the spring is oscillating in simple harmonic motion. Solve: (a) Because the frequency of an object in simple harmonic motion is independent of the amplitude and/or the maximum velocity, the new frequency is equal to the old frequency of 2.0 Hz. (b) The speed 0 v of the block just before it is given a blow can be obtained by using the conservation of mechanical energy equation as follows: ( ) ( )( )( ) 1 2 1 2 1 2 2 2 max 2 0 0 2 2 2.0 Hz 0.02 m 0.25 m s kA mv mv v k A A f A m . p p = = . = = = = = The blow to the block provides an impulse that changes the velocity of the block: ( )( ) ( ) ( )( ) f 0 3 f f 20 N 1.0 10 s 0.200 kg 0.200 kg 0.25 m s 0.150 m s x x J F t p mv mv - v v = . = . = - - = - . = Since f v is the new maximum velocity of the block at the equilibrium position, it is equal to A.. Thus, ( ) 0.150 m s 0.150 m s 0.012 m 1.19 cm 2 2.0 Hz A . p = = = = Assess: Because f v is positive, the block continues to move to the right even after the blow. 14.63. Model: The pendulum falls, then undergoes small-amplitude oscillations in simple harmonic motion. Visualize: We placed the origin of the coordinate system at the bottom of the arc. Solve: We need to find the length of the pendulum. The conservation of mechanical energy equation for the pendulums fall is ( g )top ( g )bottom K +U = K +U : ( ) ( ) ( ) 1 2 1 2 1 2 2 0 0 2 1 1 2 2 0 J 2 5.0 m s 0 J 1 5.0 m s 0.6377 m 4 mv mgy mv mgy mg L m L g + = + . + = + . = = Using L = 0.6377 m, we can find the frequency f as 1 1 9.8 m s2 0.62 Hz 2 2 0.6377 m f g p L p = = = 14.64. Model: Assume the small-angle approximation. Visualize: Solve: The tension in the two strings pulls downward at angle . . Thus Newtons second law is 2 sin y y SF = - T . = ma From the geometry of the figure we can see that 2 2 sin y L y . = + If the oscillation is small, then y .. L and we can approximate sin. y / L. Since y/L is tan. , this approximation is equivalent to the small-angle approximation sin. tan. if . ..1 rad. With this approximation, Newtons second law becomes 2 2 2 2 2 sin 2 2 y T T y ma m d y d y T y L dt dt mL - . - = = . = - This is the equation of motion for simple harmonic motion (see Equations 14.33 and 14.47). The constants 2T/mL are equivalent to k/m in the spring equation or g/L in the pendulum equation. Thus the oscillation frequency is 1 2 2 f T p mL = 14.65. Visualize: Please refer to Figure P14.65. Solve: The potential energy curve of a simple harmonic oscillator is described by 1 ( )2 2 U = k .x , where 0 .x = x - x is the displacement from equilibrium. From the graph, we see that the equilibrium bond length is 0 x = 0.13 nm. We can find the bonds spring constant by reading the value of the potential energy U at a displacement .x and using the potential energy formula to calculate k. x (nm) .x (nm) U (J) k (N/m) 0.11 0.10 0.09 0.02 0.03 0.04 0.810-19 J 1.910-19 J 3.410-19 J 400 422 425 The three values of k are all very similar, as they should be, with an average value of 416 N/m. Knowing the spring constant, we can now calculate the oscillation frequency of a hydrogen atom on this spring to be 13 27 1 1 416 N m 7.9 10 Hz 2 2 1.67 10 kg f k p m p - = = = 14.66. Model: Assume that the size of the ice cube is much less than R and that . is a small angle. Visualize: Solve: The ice cube is like an object on an inclined plane. The net force on the ice cube in the tangential direction is ( ) 2 2 G 2 2 F sin ma mR mR d mg sin mR d dt dt . . - . = = a = .- . = where a is the angular acceleration. With the small-angle approximation sin. . , this becomes 2 2 2 d g dt R . = - . = -. . This is the equation of motion of an object in simple harmonic motion with a period of T 2 2 R g p p . = = 14.67. Visualize: Solve: (a) Newtons second law applied to the penny along the y-axis is Fnet = n - mg = may net F .. is upward at the bottom of the cycle (positive ), y a so n > mg. The speed is maximum when passing through equilibrium, but 0 y a = so n = mg. The critical point is the highest point. net F .. points down and y a is negative. If y a becomes sufficiently negative, n drops to zero and the penny is no longer in contact with the surface. (b) When the penny loses contact (n = 0), the equation for Newtons law becomes max a = g. For simple harmonic motion, 2 2 max 9.8 m s 15.65 rad 0.040 m 15.65 rad s 2.5 Hz 2 2 a A g A f . . . p p = . = = = . = = = 14.68. Model: The vertical oscillations constitute simple harmonic motion. Visualize: Solve: At the equilibrium position, the net force on mass m on Planet X is: X Fnet k L mgX 0 N k g m L = . - = . = . For simple harmonic motion k m =. 2 , thus ( ) 2 2 2 X X 2 X 2 2 2 0.312 m 5.86 m s 14.5 s 10 g g g L L L T T p p p . . . . . . = . = = . = . . . = . . = . . . . . . 14.69. Model: The dolls head is in simple harmonic motion and is damped. Solve: (a) The oscillation frequency is 1 (2 )2 (0.015 kg)(2 )2 (4.0 Hz)2 9.475 N m 2 f k k m f m p p p = . = = = The spring constant is 9.5 N/m. (b) The maximum speed is ( ) max 9.475 N m 0.020 m 0.50 m s 0.015 kg v A k A m =. = = = (c) Using ( ) / 2 0 A t = A e-bt m, we get ( ) ( ) ( ) 0.5 cm 2.0 cm (4.0 s)/(2 0.015 kg) 0.25 (133.3 s/kg) 133.33 s kg ln0.25 0.0104 kg / s e b e b b b = - . = - . - = . = 14.70. Model: The oscillator is in simple harmonic motion. Solve: (a) The maximum displacement at time t of a damped oscillator is ( ) 2 max ( ) max ln 2 t t x t x t Ae A t t - . . = .- = . . . . Using max x = 0.98A at t = 0.50 s, we can find the time constant t to be ( ) 0.50 s 12.375 s 2ln 0.98 t = - = 25 oscillations will be completed at t = 25T =12.5 s. At that time, the amplitude will be ( ) 12.5 s (2)(12.375 s) max, 12.5 s x 10 cm e 6.0 cm = - = (b) The energy of a damped oscillator decays more rapidly than the amplitude: 1/ 0 E(t) = E e- t . When the energy is 60% of its initial value, 0 E(t)/E = 0.60. We can find the time this occurs as follows: ( ) ( ) ( ) ( ) 0 0 ln ln 12.375 s ln 0.60 6.3 s t E t E t t E E t t . . . . - = . .. = - . . = - = . . . . 14.71. Model: The oscillator is in simple harmonic motion. Solve: The maximum displacement, or amplitude, of a damped oscillator decreases as 2 max x (t) = Ae-t t , where t is the time constant. We know max x A = 0.60 at t = 50 s, so we can find t as follows: ( ) ( ) ln max 50 s 48.9 s 2 2ln 0.60 t x t A t t . . - = . .. = - = . . Now we can find the time 30 t at which max x A = 0.30 : max ( ) ( ) ( ) 30 2 ln 2 48.9 s ln 0.30 118 s x t t A t . . = - . . = - = . . The undamped oscillator has a frequency f = 2 Hz = 2 oscillations per second. Damping changes the oscillation frequency slightly, but the text notes that the change is negligible for light damping. Damping by air, which allows the oscillations to continue for well over 100 s, is certainly light damping, so we will use f = 2.0 Hz. Then the number of oscillations before the spring decays to 30% of its initial amplitude is ( )( ) 30 N = f t = 2 oscillations s 118 s = 236 oscillations 14.72. Solve: The solution of the equation 2 2 d x b dx k x 0 dt m dt m + + = is ( ) 2 ( ) 0 x t = Ae-bt m cos .t +f . The first and second derivatives of x(t) are ( ) ( ) ( ) ( ) / 2 / 2 0 0 2 2 2 /2 2 2 0 0 cos sin 2 cos sin 4 bt m bt m bt m dx Abe t Ae t dt m d x Ab A t Ab t e dt m m . f . . f . . . f . f - - - = - + - + .. . . = .. - . + + + . .. . . Substituting these expressions into the differential equation, the terms involving 0 sin(.t +f ) cancel and we obtain the simplified result ( ) 2 2 2 0 cos 0 4 b k t m m . . f . - . . - + . + = . . Because ( ) 0 cos .t +f is not equal to zero in general, 2 2 2 2 2 0 4 4 b k k b m m m m . . - - + = . = - 14.73. Model: The two springs obey Hookes law. Visualize: Solve: There are two restoring forces on the block. If the blocks displacement x is positive, both restoring forcesone pushing, the other pullingare directed to the left and have negative values: ( net )x ( sp 1 )x ( sp 2 )x 1 2 ( 1 2 ) eff F = F + F = -k x - k x = - k + k x = -k x where eff 1 2 k = k + k is the effective spring constant. This means the oscillatory motion of the block under the influence of the two springs will be the same as if the block were attached to a single spring with spring constant eff k . The frequency of the blocks, therefore, is eff 1 2 1 2 2 2 2 2 1 2 1 1 2 2 4 4 f k k k k k f f p m p m p m p m + = = = + = + 14.74. Model: The two springs obey Hookes law. Assume massless springs. Visualize: Each spring is shown separately. Note that .x = .x1 + .x2. Solve: Only spring 2 touches the mass, so the net force on the mass is 2 on . m m F = F Newtons third law tells us that 2 on m m on 2 F = F and that 2 on 1 1 on 2F = F . From net F = ma, the net force on a massless spring is zero. Thus w on 1 F = 2 on 1 1 1 F = k .x and on 2 1 on 2 2 2. mF = F = k .x Combining these pieces of information, m 1 1 2 2 F = k .x = k .x The net displacement of the mass is 1 2.x = .x + .x , so 1 2 1 2 1 2 1 2 12 1 1 m m m m x x x F F F k k F k k k k kk . . + . = . + . = + = . + . = . . Turning this around, the net force on the mass is 1 2 1 2 eff eff 1 2 1 2 where m F k k x k x k k k k k k k = . = . = + + eff k , the proportionality constant between the force on the mass and the masss displacement, is the effective spring constant. Thus the masss angular frequency of oscillation is eff 1 2 1 2 k 1 k k m mk k .= = + Using 2 1 1. = k /m and 2 2 2. = k /m for the angular frequencies of either spring acting alone on m, we have 2 2 1 2 1 2 2 2 1 2 1 2 ( / )( / ) ( / ) ( / ) k m k m k m k m . . . . . = = + + Since the actual frequency f is simply a multiple of ., this same relationship holds for f: 2 2 1 2 2 2 1 2 f f f f f = + 14.75. Model: The blocks undergo simple harmonic motion. Visualize: The length of the stretched spring due to a block of mass m is .L1. In the case of the two block system, the spring is further stretched by an amount 2.L . Solve: The equilibrium equations from Newtons second law for the single block and double block systems are ( ) ( ) ( ) 1 1 2 .L k = mg and .L + .L k = 2m g Using 2 .L = 5.0 cm, and subtracting these two equations, gives us ( ) ( ) ( ) 1 2 1 .L + .L k - .L k = 2m g - mg . 0.05 m k = mg 2 9.8 m/s 196 s2 0.05 m k m . = = With both blocks attached, giving total mass 2m, the angular frequency of oscillation is 1 1196 s2 9.90 rad/s 2 2 2 k k m m .= = = = Thus the oscillation frequency is f =./2p =1.58 Hz. 14.76. Model: A completely inelastic collision between the bullet and the block resulting in simple harmonic motion. Visualize: Solve: (a) The equation for conservation of energy after the collision is 2 ( ) 2 ( ) b B f f b B 1 1 2500 N m 0.10 m 5.0 m s 2 2 1.010 kg kA m m v v k A m m = + . = = = + The momentum conservation equation for the perfectly inelastic collision after before p = p is ( ) ( )( ) ( ) ( )( ) b B f bb BB 2 b b 1.010 kg 5.0 m s 0.010 kg 1.00 kg 0 m s 5.0 10 m s m m v mv mv v v + = + = + . = (b) No. The oscillation frequency ( ) b B k m + m depends on the masses but not on the speeds. 14.77. Model: The block undergoes SHM after sticking to the spring. Energy is conserved throughout the motion. Visualize: Its essential to carefully visualize the motion. At the highest point of the oscillation the spring is stretched upward. Solve: Weve placed the origin of the coordinate system at the equilibrium position, where the block would sit on the spring at rest. The spring is compressed by .L at this point. Balancing the forces requires k.L = mg. The angular frequency is w2 = k/m = g/.L, so we can find the oscillation frequency by finding .L. The block hits the spring (1) with kinetic energy. At the lowest point (3), kinetic energy and gravitational potential energy have been transformed into the springs elastic energy. Equate the energies at these points: 1 2 1 2 1 1g 3s 3g 2 1 2K +U =U +U . mv + mg.L = k(.L + A) + mg(-A) Weve used 1y = .L as the block hits and 3y = -A at the bottom. The spring has been compressed by .y = .L + A. Speed 1 v is the speed after falling distance h, which from free-fall kinematics is 2 1 v = 2gh. Substitute this expression for 2 1 v and mg/.L for k, giving ( )2 ( ) 2( ) mgh mg L mg L A mg A L + . = . + + - . The mg term cancels, and the equation can be rearranged into the quadratic equation (.L)2 + 2h(.L) - A2 = 0 The positive solution is .L = h2 + A2 - h = (0.030 m)2 + (0.100 m)2 - 0.030 m = 0.0744 m Now that .L is known, we can find 9.80 m/s2 11.48 rad/s 1.83 Hz 0.0744 m 2 g f L . . p = = = . = = . 14.78. Model: Model the bungee cord as a spring. The motion is damped SHM. Visualize: Solve: (a) For light damping, the oscillation period is 2 2 2 2 (75 kg) 2 185 N/m 4.0 s T m k m k T p p = p . = .. .. = .. .. = . . . . (b) The maximum speed is max v =. A = (2p /4.0 s)(11.0 m) =17.3 m/s. (c) Jose oscillates about the equilibrium position at which he would hang at rest. Balancing the forces, .L = mg / k = (75 kg)(9.80 m/s2 )(185 N/m) = 3.97 m. Joses lowest point is 11.0 m below this point, so the bungee cord is stretched by max .y = .L + A =14.97 m. Choose this lowest point as y = 0. Because Jose is instantaneously at rest at this point, his energy is entirely the elastic potential energy of the stretched bungee cord. Initially, his energy was entirely gravitational potential energy. Equating his initial energy to his energy at the lowest point, 1 2 lowest point highest point 2 max 2 2 max 2 ( ) ( ) (185 N/m)(14.97 m) 28.2 m 2 2(75 kg)(9.80 m/s ) U U ky mgh h k y mg = . . = . = = = Jose jumped 28 m above the lowest point. (d) The amplitude decreases due to damping as A(t) = Ae-bt/2m. At the time when the amplitude has decreased from 11.0 m to 2.0 m, 2.0 m / 2 2 ln 2 2(75 kg) ( 1.705) 42.6 s 11.0 m 11 6.0 kg/s e bt m t m b = - . = - .. .. = - - = . . With a period of 4.0 s, the number of oscillations is osc N = (42.6 s)/(4.0 s) =10.7 oscillations. 14.79. Model: The vertical movement of the car is simple harmonic motion. Visualize: The fact that the car has a maximum oscillation amplitude at 5 m/s implies a resonance. The bumps in the road provide a periodic external force to the cars suspension system, and a resonance will occur when the bump frequency fext matches the cars natural oscillation frequency f0. Solve: Now the 5.0 m/s is not a frequency, but we can convert it to a frequency because we know the bumps are spaced every 3.0 meters. The time to drive 3.0 m at 5.0 m/s is the period: 3.0 m 0.60 s 5.0 m s T x v . = = = The external frequency due to the bumps is thus ext f =1 T =1.667 Hz. This matches the cars natural frequency 0 f , which is the frequency the car oscillates up and down if you push the car down and release it. This is enough information to deduce the spring constant of the cars suspension: ( )2 ext 1.667 Hz 1 2 131,600 N 2 m k k m f m p p = . = = where we used total car passenger m = m = m + 2m =1200 kg. When at rest, the car is in static equilibrium with net F = 0 N. The downward weight total m g of the car and passengers is balanced by the upward spring force k.y of the suspension. Thus the compression .y of the suspension is total y m g k . = Initially total car passenger m = m + 2m =1200 kg, causing an initial compression i .y = 0.0894 m = 8.94 cm. When three additional passengers get in, the mass increases to total car passenger m = m + 5m =1500 kg. The final compression is f .y = 0.1117 m =11.17 cm. Thus the three new passengers cause the suspension to sag by 11.17 cm- 8.94 cm = 2.23 cm. 14-1 14.80. Model: The rod is thin and uniform. Visualize: Please refer to Figure CP14.80. Solve: We must derive our own equation for this combination of a pendulum and spring. For small oscillations, s F .. remains horizontal. The net torque around the pivot point is G cos sin 2 net s t = Ia = -F L . - F . L . . . . . . With 2 2 d , dt . a = G F = mg, s F = k.x = kLsin. , and 1 2 , 3 I = mL 2 2 3 sin cos 3 sin 2 d k g dt m L . = - . . - . We can use sin cos 1 sin 2 . 2 . . = . For small angles, sin. . and sin 2. 2. . So 2 2 3 3 2 d k g dt m L . = -.. + ... . . This is the same as Equations 14.33 and 14.47 with 3 3 2 k g m L . = + The frequency of oscillation is thus ( ) ( ) ( ) ( ) 1 3 3.0 N/m 3 9.8 m/s2 1.73 Hz 2 0.200 kg 2 0.20 m f p = + = The period T 1 f = = 0.58 s. Assess: Fewer than two oscillations per second is reasonable. The rods angle from the vertical must be small enough that sin 2. 2. . This is more restrictive than other examples, which only require that sin. . . 15.1. Solve: The density of the liquid is 3 3 3 3 0.240 kg 0.240 kg 960 kg m 250 mL 250 10 10 m m V . - - = = = = Assess: The liquids density is near that of water (1000 kg/m3) and is a reasonable number. 15.2. Solve: The volume of the helium gas in container A is equal to the volume of the liquid in container B. That is, VA =VB. Using the definition of mass density . = m V , the above relationship becomes A B He He ( ) ( )( 3 ) 3 B He A B He B m m m 7000 m . 7000 . 7000 0.18 kg m 1260 kg m . . . . = . = . = = = Referring to Table 15.1, we find that the liquid is glycerine. 15.3. Model: The density of water is 1000 kg/m3. Visualize: Solve: Volume of water in the swimming pool is 1 ( ) 3 2 V = 6 m12 m3 m - 6 m12 m 2 m =144 m The mass of water in the swimming pool is m = .V = (1000 kg m3 )(144 m3 ) =1.44105 kg 15.4. Model: The densities of gasoline and water are given in Table 15.1. Solve: (a) The total mass is mtotal = mgasoline + mwater = 0.050 kg + 0.050 kg = 0.100 kg The total volume is total gasoline water V =V +V gasoline water gasoline water m m . . = + 4 3 3 3 0.050 kg 0.050 kg 1.24 10 m 680 kg m 1000 kg m = + = - total 2 3 avg 4 3 total 0.100 kg 8.1 10 kg m 1.24 10 m m V . - . = = = (b) The average density is calculated as follows: ( )( ) total gasoline water water water gasoline gasoline 3 3 3 water water gasoline gasoline 2 3 avg 3 water gasoline 50 cm 1000 kg/m 680 kg/m 8.4 10 kg/m 100 cm m m m V V V V V V . . . . . = + = + + + . = = = + Assess: The above average densities are between those of gasoline and water, and are reasonable. 15.5. Model: The density of sea water is 1030 kg/m3. Solve: The pressure below sea level can be found from Equation 15.6 as follows: 5 ( 3 )( 2 )( 4 ) 0 5 8 8 3 1.013 10 Pa 1030 kg m 9.80 m s 1.1 10 m 1.013 10 Pa 1.1103 10 Pa 1.1113 10 Pa 1.10 10 atm p = p + . gd = + = + = = where we have used the conversion 1 atm =1.013105 Pa. Assess: The pressure deep in the ocean is very large. 15.6. Model: The density of water is 1000 kg/m3 and the density of ethyl alcohol is 790 kg/m3. Solve: (a) The volume of water that has the same mass as 8.0 m3 of ethyl alcohol is ( ) 3 water alcohol alcohol alcohol 3 3 water 3 water water water 790 kg/m 8.0 m 6.3 m 1000 kg/m V m m V . . . . . . = = = = . . = . . (b) The pressure at the bottom of the cubic tank is 0 water p = p + . gd : p =1.013105 Pa + (1000 kg/m3 )(9.80 m s2 )(6.3)1 3 =1.19105 Pa where we have used the relation ( )1 3 water d = V . 15.7. Visualize: Solve: The pressure at the bottom of the vat is p = p0 + . gd =1.3 atm. Substituting into this equation gives 1.013105 Pa + . (9.8 m s2 )(2.0 m) = (1.3)(1.013105 )Pa.. =1550.5 kg m3 The mass of the liquid in the vat is m = .V = .p (0.50 m)2 d = (1550.5 kg m3 )p (0.50 m)2 (2.0 m) = 2.4103 kg 15.8. Model: 3 3 oil water The density of oil 900 kg . = m and the density of water . =1000 kg m . Visualize: Solve: The pressure at the bottom of the oil layer is 1 0 oil 1 , p = p + . gd and the pressure at the bottom of the water layer is ( ) ( )( )( ) ( )( )( ) 2 1 water 2 0 oil 1 water 2 5 3 2 3 2 5 2 1.013 10 Pa 900 kg m 9.80 m s 0.50 m 1000 kg m 9.80 m s 1.20 m 1.18 10 Pa p p gd p gd gd p = + . = + . + . . = + + = Assess: A pressure of 1.18105 Pa =1.16 atm is reasonable. 15.9. Model: The density of seawater 2 .seawater =1030 kg m . Visualize: Solve: The pressure outside the submarines window is out 0 seawater p = p + . gd, where d is the maximum safe depth for the window to withstand a force F. This force is out in F A = p - p , where A is the area of the window. With in 0 p = p , we simplify the pressure equation to ( )( )( ) 6 out 0 seawater 2 2 2 seawater 1.0 10 N 3.2 km 0.10 m 1030 kg m 9.8 m s p p F gd d F d A A g . . p - = = . = = = Assess: A force of 1.0106 N corresponds to a pressure of ( ) 6 2 1.0 10 N 314 atm 0.10 m F A . p = = = A depth of 3 km is therefore reasonable. 15.10. Visualize: We assume that the seal is at a radius of 5 cm. Outside the seal, atmospheric pressure presses on both sides of the cover and the forces cancel. Thus, only the 10 cm diameter opening inside the seal is relevant, not the 20 cm diameter of the cover. Solve: Within the 10 cm diameter area where the pressures differ, to l F eft = patmosA Fto right = pgasA where A =p r2 = 7.8510-3 m2 is the area of the opening. The difference between the forces is ( ) ( )( 3 2 ) to left to right atmos gas F - F = p - p A = 101,300 Pa - 20,000 Pa 7.8510- m = 0.64 kN Normally, the rubber seal exerts a 0.64 kN force to the right to balance the air pressure force. To pull the cover off, an external force must pull to the right with a force = 0.64 kN. 15.11. Model: The density of water is . =1000 kg m3 . Visualize: Please refer to Figure 15.17. Solve: From the figure and the equation for hydrostatic pressure, we have 0 atmos p + . gh = p Using 0 p = 0 atm, and 5 atmos p =1.01310 Pa, we get 0 Pa + (1000 kg m3 )(9.8 m s2 )h =1.013105 Pa.h =10.3 m Assess: This large value of h is due to water having a much smaller density than mercury. 15.12. Model: Assume that the oil is incompressible. Its density is 900 kg/m3. Visualize: Please refer to Figure 15.19. Because the liquid is incompressible, the volume displaced in the left cylinder of the hydraulic lift is equal to the volume displaced in the right cylinder. Solve: Equating the two volumes, ( ) ( ) ( ) 2 2 2 2 2 1 1 2 2 1 1 2 2 1 2 1 0.04 m 0.20 m 3.2 m 0.01 m Ad A d r d r d d r d r p p . . . . = . = . = . . = . . = . . . . 15.13. Visualize: Solve: By sucking on the top and reducing the pressure, the straw is essentially a barometer. Atmospheric pressure pushes the liquid up the straw. The length of the longest straw can be obtained from the formula p = p0 + . gd. If one could reduce the mouth pressure to zero, the length of the straw would be 1.013105 Pa = 0 Pa + (1000 kg m3 )(9.8 m s2 )d .d =10.3 m 15.14. Model: The vacuum cleaner can create zero pressure. Visualize: Solve: The gravitational force on the dog is balanced by the force resulting from the pressure difference between the atmosphere and the vacuum ( phose = 0) in the hose. The force applied by the hose is ( ) ( )( ) atmos hose atmos 2 4 2 5 10 kg 9.8 m/s 9.7 10 m 1.013 10 Pa F p p A p A mg A - = - = = . = = Since 2 , 2 A =p . d . . . . . the diameter of the hose is d 2 A 0.035 m 3.5 cm. p = = = Assess: The dog will have a terrible skin rash. Its a good thing the miscreant holding the vacuum does not have a younger sibling. 15.15. Model: The buoyant force on the sphere is given by Archimedes principle. Visualize: Solve: The sphere is in static equilibrium because it is neutrally buoyant. That is, SFy = FB - FG = 0 N..lVlg - msg = 0 N The sphere displaces a volume of liquid equal to its own volume, l sV =V , so ( ) s s 2 3 l 4 3 4 3 s 3 s 3 0.0893 kg 7.9 10 kg m 0.030 m m m V r . p p = = = = A density of 790 kg/m3 in Table 15.1 identifies the liquid as ethyl alcohol. Assess: If the density of the fluid and an object are equal, we have neutral buoyancy. 15.16. Model: The buoyant force on the cylinder is given by Archimedes principle. Visualize: cyl V is the volume of the cylinder and w V is the volume of the water displaced by the cylinder. Note that the volume displaced is only from the part of the cylinder that is immersed in water. Solve: The cylinder is in static equilibrium, so FB = FG. The buoyant force is the weight w w . V g of the displaced water. Thus ( ) ( ) ( ) w B w w G cyl cyl w w cyl cyl cyl w cyl 3 2 3 cyl 0.040 m 1000 kg m 6.7 10 kg m 0.060 m F V g F mg V g V V V V A A . . . . . . . = = = = . = . = . = = Assess: cyl w . < . for a cylinder floating in water is an expected result. 15.17. Model: The buoyant force on the sphere is given by Archimedes principle. Visualize: Solve: The sphere is in static equilibrium. The free body diagram on the sphere shows that ( ) B G B G G G G 3 3 sphere sphere sphere sphere 0 N 1 4 3 3 4 331000 kg m 750 kg m 3 44 y w w F F T F F T F F F F . V g . V g . . = - - = . = + = + = . = . = = = S 15.18. Model: The buoyant force on the rock is given by Archimedes principle. Visualize: Solve: Because the rock is in static equilibrium, Newtons first law is ( ) net B G rock rock water rock rock water rock rock water rock rock water rock rock rock 0 N 1 1 1 1 2 2 2 2 F T F F T V g V g V g m g m g . . . . . . . . . = + - = . . . . . .. . . . . = - . . = . - . = . - .. . = . - . . . . . . .. . . . Using 3 rock . = 4800 kg m and rock m = 5.0 kg, we get T = 44 N. 15.19. Model: The buoyant force on the aluminum block is given by Archimedes principle. The density of aluminum and ethyl alcohol are 3 3 Al . = 2700 kg m and .ethyl alcohol = 790 kg m . Visualize: The buoyant force B F and the tension due to the string act vertically up, and the gravitational force on the aluminum block acts vertically down. The block is submerged, so the volume of displaced fluid equals Al V , the volume of the block. Solve: The aluminum block is in static equilibrium, so ( ) ( )( )( ) B G fAl AlAl Al Al f 6 3 2 3 3 0 N 0 N 100 10 m 9.80 m s 2700 kg m 790 kg m 1.87 N y F F T F V g T V g T V g T . . . . - = + - = . + - = . = - = - = S where we have used the conversion ( )100 cm3 =100 10-2 m 3 =10-4 m3. Assess: The gravitational force on the aluminum block is Al Al . V g = 2.65 N. A similar order of magnitude for T is reasonable. 15.20. Model: The buoyant force on the steel cylinder is given by Archimedes principle. Visualize: The length of the cylinder above the surface of mercury is d. Solve: The cylinder is in static equilibrium with FB = FG. Thus ( ) ( ) ( ) ( ) B Hg Hg G cyl cyl Hg Hg cyl cyl Hg cyl 3 cyl 3 Hg 0.20 m 0.20 m 0.20 m 0.20 m 0.20 m 1 7900 kg m 0.084 m 8.4 cm 13,600 kg m F Vg F mg Vg V V A d A d . . . . . . . . = = = = . = . - = . . . = - = . - . = = . . That is, the length of the cylinder above the surface of the mercury is 8.4 cm. 15.21. Model: The buoyant force is determined by Archimedes principle. Ignore any compression the air in the beach ball may undergo as a result of submersion. Solve: The mass of the beach ball is negligible, so the force needed to push it below the water is equal to the buoyant force. 4 3 (1000 kg/m3 ) 4 (0.30 m)3 (9.8 m/s2 ) 1.11 kN 3 3 B w F = p .. p R .. g = .. p .. = . . . . Assess: It would take a 113 kg (250 lb) person to push the ball below the water. Two people together could do it. This seems about right. 15.22. Model: The buoyant force on the sphere is given by Archimedes principle. Visualize: Solve: For the Styrofoam sphere and the mass not to sink, the sphere must be completely submerged and the buoyant force B F must be equal to the sum of the gravitational force on the Styrofoam sphere and the attached mass. The volume of displaced water equals the volume of the sphere, so ( ) ( ) ( ) ( ) ( ) ( ) ( ) 3 4 3 2 B water water 3 3 4 3 2 G Styrofoam Styrofoam Styrofoam 3 1000 kg m 0.25 m 9.80 m s 641.4 N 150 kg m 0.25 m 9.80 m s 96.2 N F V g F V g . p . p = = = = = . . = . . Because ( )G Styrofoam BF + mg = F , ( ) B GStyrofoam 2 641.4 N 96.2 N 55.6 kg 9.80 m s F F m g - - = = = The mass is 56 kg. 15.23. Model: Treat the water as an ideal fluid. The pipe is a flow tube, so the equation of continuity applies. Solve: The volume flow rate is 3 3 300 L 300 10 m 1.0 10 3 m3 s 5.0 min 5.0 60 s Q - - = = = Using the definition Q = vA, we get ( ) 3 3 2 1.0 10 m s 3.2 m s 0.010 m v Q A p - = = = 15.24. Model: Treat the water as an ideal fluid. The pipe itself is a flow tube, so the equation of continuity applies. Visualize: Note that A1, A2 , and A3 and v1, v2 , and v3 are the cross-sectional areas and the speeds in the first, second, and third segments of the pipe. Solve: (a) The equation of continuity is ( )( ) ( ) ( ) ( ) ( ) 2 2 2 2 2 2 1 1 2 2 3 3 1 1 2 2 3 3 1 1 2 2 3 3 2 2 2 2 3 2 2 2 3 0.0050 m 4.0 m s 0.010 m 0.0025 m 0.0050 m 4.0 m s 1.00 m s 0.0050 m 4.0 m s 16.0 m s 0.010 m 0.0025 m Av Av Av r v r v r v r v r v r v v v v v = = .p =p =p . = = . = = . = . . = = . . = . . . . . . . . (b) The volume flow rate through the pipe is ( )2 ( ) 4 3 1 1 Q = Av =p 0.0050 m 4.0 m s = 3.110- m s 15.25. Model: Treat the water as an ideal fluid so that the flow in the tube follows the continuity equation. Visualize: Solve: The equation of continuity is 0 0 1 1 , v A = v A where 2 0 A = L and ( )1 2 1 2 A =p L . The above equation simplifies to 2 2 0 1 1 0 0 4 1.27 2 v L vp L v v v p = . . . = . . = . . . . . . . . 15.26. Model: Treat the oil as an ideal fluid obeying Bernoullis equation. Consider the path connecting point 1 in the lower pipe with point 2 in the upper pipe a streamline. Visualize: Please refer to Figure EX15.26. Solve: Bernoullis equation is 1 2 1 2 1 ( 2 2 ) ( ) 2 2 2 2 1 2 1 1 2 1 2 1 2 1 2 p + .v + . gy = p + .v + . gy . p = p + . v - v + . g y - y Using 5 1 p = 200 kPa = 2.0010 Pa, . = 900 kg m3 , 2 1 y - y =10.0 m, 1 v = 2.0 m s, and 2 v = 3.0 m s, we get 2 p = 1.096105 Pa =110 kPa. 15.27. Model: Turning the tuning screws on a guitar string creates tensile stress in the string. Solve: The tensile stress in the string is given by T/A, where T is the tension in the string and A is the crosssectional area of the string. From the definition of Youngs modulus, / / Y T A L T L LL AY = .. = . . . . . . . Using T = 2000 N, L = 0.80 m, A =p (0.00050 m)2 , and Y = 201010 N/m2 (from Table 15.3), we obtain .L = 0.010 m = 1.02 cm. Assess: 1.02 cm is a large stretch for a length of 80 cm, but 2000 N is a large tension. 15.28. Model: The dangling mountain climber creates tensile stress in the rope. Solve: Youngs modulus for the rope is / stress / strain Y F A L L = = . The tensile stress is ( )( ) ( ) 2 6 2 70 kg 9.8 m/s 8.734 10 Pa p 0.0050 m = and the strain is 0.080 m 50 m = 0.00160. Dividing the two quantities yields Y = 5.5109 N m2 . 15.29. Model: The hanging mass creates tensile stress in the wire. Solve: The force (F) pulling on the wire, which is simply the gravitational force (mg) on the hanging mass, produces tensile stress given by F/A, where A is the cross-sectional area of the wire. From the definition of Youngs modulus, we have ( ) ( )( )( ) ( )( ) 2 4 2 10 2 3 2 / 2.50 10 m 20 10 N/m 1.0 10 m 2.0 kg / 9.80 m/s 2.0 m mg A r Y L Y m L L gL p . p - - = . = = = . 15.30. Model: The load supported by a concrete column creates compressive stress in the concrete column. Solve: The gravitational force on the load produces tensile stress given by F/A, where A is the cross-sectional area of the concrete column and F equals the gravitational force on the load. From the definition of Youngs modulus, ( ) 2 2 10 2 / 200,000 kg 9.8 m/s 3.0 m 1.0 mm / 0.25 m 3 10 N/m Y F A L F L L L A Y p . .. . . .. . = .. = . .. . = . .. . = . . .. . . .. . . . Assess: A compression of 1.0 mm of the concrete column by a load of approximately 200 tons is reasonable. 15.31. Model: Water is almost incompressible and it applies a volume stress. Solve: (a) The pressure at a depth of 5000 m in the ocean is ( ) 5 ( 3 )( 2 )( ) 7 0 sea water p = p + . g 5000 m =1.01310 Pa + 1030 kg m 9.8 m s 5000 m = 5.05710 Pa (b) Using the bulk modulus of water, 7 10 5.057 10 Pa 0.025 0.2 10 Pa V p V B . = - = - = - (c) The volume of a mass of water decreases from V to 0.975V. Thus the waters density increases from . to . /0.975. The new density is 3 3 5000m 1030 kg/m 1056 kg/m 0.975 . = = 15.32. Solve: The pressure p at depth d in a fluid is p = p0 + . gd. Using 1.29 kg/m3 for the density of air, 3 bottom top air bottom top p = p + . gd . p - p = 202 Pa = 1.9910- atm Assuming bottom p =1 atm, 3 bottom top bottom 1.99 10 atm 0.2% 1 atm p p p - - = = 15.33. Model: We assume that there is a perfect vacuum inside the cylinders with p = 0 Pa. We also assume that the atmospheric pressure in the room is 1 atm. Visualize: Please refer to Figure P15.33. Solve: (a) The flat end of each cylinder has an area A =p r2 =p (0.30 m)2 = 0.283 m2. The force on each end is thus ( 5 )( 2 ) 4 atm 0 F = p A = 1.01310 Pa 0.283 m = 2.8610 N The force on each end is 2.9104 N. (b) The net vertical force on the lower cylinder when it is on the verge of being pulled apart is ( ) ( ) ( )( ) 4 atm G players G players atm 4 2 0 N 2.86 10 N number of players 2.86 10 N 29.2 100 kg 9.8 m s y F = F - F = . F = F = . = = S That is, 30 players are needed to pull the two cylinders apart. 15.34. Model: Assume that the oil is incompressible and its density is 900 kg/m3. Visualize: Please refer to Figure P15.34. Solve: (a) The pressure at depth d in a fluid is 0 p = p + . gd. Here, pressure 0 p at the top of the fluid is due both to the atmosphere and to the gravitational force on the piston. That is, 0 atm Gp p = p + (F ) / A. At point A, ( ) ( ) ( )( ) ( ) ( )( )( ) ( ) ( ) G P A atm 2 5 3 2 2 2 A A 1.00 m 0.30 m 10 kg 9.8 m s 1.013 10 Pa 900 kg m 9.8 m s 0.70 m 185,460 Pa 0.02 m 185,460 Pa 0.10 m 5.8 kN F p p g A F p A . p p = + + - = + + = . = = = (b) In the same way, ( G )P ( ) B atm B 1.30 m 190,752 Pa 6.0 kN F p p g F A = + + . = . = Assess: B F is larger than AF , because B p is larger than Ap . 15.35. Model: The tire flattens until the pressure force against the ground balances the upward normal force of the ground on the tire. Solve: The area of the tire in contact with the road is A = (0.15 m)(0.13 m) = 0.0195 m2. The normal force on each tire is ( )( 2 ) G 1500 kg 9.8 m s 3675 N 4 4 n = F = = Thus, the pressure inside each tire is inside 2 3675 N 188,500 Pa 1.86 atm 14.7 psi 27 psi 0.0195 m 1 atm p n A = = = = = 15.36. Visualize: Solve: (a) Because the patients blood pressure is 140/100, the minimum fluid pressure needs to be 100 mm of Hg above atmospheric pressure. Since 760 mm of Hg is equivalent to 1 atm and 1 atm is equivalent to 1.013105 Pa, the minimum pressure is 100 mm =1.333104 Pa. The excess pressure in the fluid is due to force F pushing on the internal 6.0-mm-diameter piston that presses against the liquid. Thus, the minimum force the nurse needs to apply to the syringe is F = fluid pressure area of plunger = (1.333104 Pa)..p (0.0030 m)2 .. = 0.38 N (b) The flow rate is Q = vA, where v is the flow speed of the medicine and A is the cross-sectional area of the needle. Thus, ( ) 6 3 3 2 2.0 10 m 2.0 s 20 m s 0.125 10 m v Q A p - - = = = Assess: Note that the pressure in the fluid is due to F that is not dependent on the size of the plunger pad. Also note that the syringe is not drawn to scale. 15.37. Solve: The fact that atmospheric pressure at sea level is 101.3 kPa =101,300 N/m2 means that the weight of the atmosphere over each square meter of surface is 101,300 N. Thus the mass of air over each square meter is m = (101,300 N)/g = (101,300 N)/(9.80 m/s2 ) =10,340 kg per m2. Multiplying by the earths surface area will give the total mass. Using 6 e R = 6.2710 m for the earths radius, the total mass of the atmosphere is 2 62 2 18 air earth e M = A m = (4p R )m = 4p (6.3710 m) (10,340 kg/m ) = 5.2710 kg 15.38. Visualize: Let d be the atmospheres thickness, p the atmospheric pressure on the earths surface, and 0 p (= 0 atm) the pressure beyond the earths atmosphere. Solve: The pressure at a depth d in a fluid is 0 p = p + . gd. This equation becomes ( )( ) 5 air 3 2 air 1 atm 0 atm 1 atm 1.013 10 Pa 7.95 km 1.3 kg/m 9.8 m/s gd d g . . = + . = = = 15.39. Solve: (a) We can measure the atmospheres pressure by measuring the height of the liquid column in a barometer, because patmos = . gh. In the case of the water barometer, the height of the column at a pressure of 1 atm is ( )( ) 5 atmos 3 2 water 1.013 10 Pa 10.337 m 1000 kg/m 9.8 m/s h p . g = = = Because the pressure of the atmosphere can vary by 5 percent, the height of the barometer must be at least be 1.05 greater than this amount. That is, min h =10.85 m. (b) Using the conversion rate 1 atm = 29.92 inches of Hg, we have 29.55 inches of Hg 29.55 1 atm 0.9876 atm 29.92 = = The height of the water in your barometer will be ( )( ) 5 atmos 3 2 water 0.9876 1.013 10 Pa 10.21 m 1000 kg/m 9.8 m/s h p . g = = = 15.40. Model: Oil is incompressible and has a density 900 kg/m3. Visualize: Please refer to Figure P15.40. Solve: (a) The pressure at point A, which is 0.50 m below the open oil surface, is ( ) ( 3 )( 2 )( ) A 0 oil p = p + . g 1.00 m - 0.50 m =101,300 Pa + 900 kg m 9.8 m s 0.50 m =106 kPa (b) The pressure difference between A and B is 3 2 B A 0 B 0 A B A p - p = ( p + . gd ) - ( p + . gd ) = . g(d - d ) = (900 kg/m )(9.8 m/s )(0.50 m) = 4.4 kPa Pressure depends only on depth, and C is the same depth as B. Thus C A p - p = 4.4 kPa also, even though C isnt directly under A. 15.41. Model: Assume that oil is incompressible and its density is 900 kg/m3. Visualize: Please refer to Figure P15.41. Solve: (a) The hydraulic lift is in equilibrium and the pistons on the left and the right are at the same level. Equation 15.11, therefore, simplifies to ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) left piston right piston G student G elephant 2 2 left piston right piston student elephant G student student elephant G elephant 70 kg 1.0 m 0.2415 m 1200 kg F F F F A A r r F g r r F g p p = . = . . . = . . = = . . . . The diameter of the piston the student is standing on is therefore 2 0.2415 m = 0.48 m. (b) From Equation 15.13, we see that an additional force .F is required to increase the elephants elevation through a distance 2d . That is, ( ) ( )( ) ( )( ) ( ) ( ) left piston right piston 2 2 3 2 2 2 2 2 70 kg 9.8 m s 900 kg m 9.8 m s 0.2415 m 1.0 m 0.0234 m F g A A d d d . p . = + . = . + . . . . = The elephant moves 2.3 cm. 15.42. Model: Assume that the oil is incompressible and its density is 900 kg/m3. Visualize: Solve: The pressures 1 p and 2 p are equal. Thus, 1 2 1 2 0 0 1 2 1 2 p F p F gh F F gh A A A A + = + + . . = + . With 2 2 1 1 2 2 1 1 2 2 F - m g,F = 4m g, A =p r , and A =p r , we have 1 2 1 2 1 2 2 1 2 2 2 2 1 2 1 m g 4m g gh r 4m g m g gh r r r . . p p p p - . . . . = + . = . . . - . . . . . Using 3 1 2 1 m = 55 kg, m =110 kg, r = 0.08 m, . = 900 kg/m , and h =1.0 m, the calculation yields 2 r = 0.276 m. The diameter is 55 cm. Assess: Both pistons are too small to hold the people as shown, but the ideas are correct. 15.43. Model: Assume that the oil is incompressible. Visualize: Solve: When the force 1 F balances the gravitational force mg, the pressure 2 p is related to the pressure 1 p as p1 = p2 + . gh When 1 F is increased to 1F', the weight is raised higher through a distance 2 d and the left piston is lowered through a distance 1d . The pressures 2 p' and 1 p' are now related through ( ) 1 2 1 2 p' = p' + . g h + d + d Subtracting these two equations, ( ) ( ) 1 1 2 2 1 2 p' - p = p' - p + . g d + d Because 1 1 1 1 1 1 2 2 2 p = F A , p' = F' A , and p = p' = mg A , the above equation simplifies to ( ) ( ) ( ) 1 1 1 1 2 1 1 1 2 F'- F A = . g d + d ..F = . g Ad + Ad Since the oil is incompressible, 1 1 2 2 . Ad = A d The equation for .F thus becomes ( ) ( ) 2 2 1 2 1 2 2 .F = . g A d + Ad = . g A + A d 15.44. Model: Water and mercury are incompressible and immiscible liquids. Visualize: The water in the left arm floats on top of the mercury and presses the mercury down from its initial level. Because points 1 and 2 are level with each other and the fluid is in static equilibrium, the pressure at these two points must be equal. If the pressures were not equal, the pressure difference would cause the fluid to flow, violating the assumption of static equilibrium. Solve: The pressure at point 1 is due to water of depth dw =10 cm: 1 atmos w w p = p + . gd Because mercury is incompressible, the mercury in the left arm goes down a distance h while the mercury in the right arm goes up a distance h. Thus, the pressure at point 2 is due to mercury of depth Hg d = 2h : 2 atmos Hg Hg atmos Hg p = p + . gd = p + 2. gh Equating 1 p and 2 p gives ( ) 3 w atmos w w atmos Hg w 3 Hg 2 1 1 1000 kg m 10 cm 3.7 mm 2 2 13,600 kg m p gd p gh h d . . . . + = + . = = = The mercury in the right arm rises 3.7 mm above its initial level. 15.45. Model: Glycerin and ethyl alcohol are incompressible and do not mix. Visualize: Solve: The alcohol in the left arm floats on top of the denser glycerin and presses the glycerin down distance h from its initial level. This causes the glycerin to rise distance h in the right arm. Points 1 and 2 are level with each other and the fluids are in static equilibrium, so the pressures at these two points must be equal: 1 2 0 eth eth 0 gly gly eth gly 3 eth 3 gly (20 cm) (2 ) 1 (20 cm) 1 790 kg/m (20 cm) 6.27 cm 2 2 1260 kg/m p p p gd p gd g g h h . . . . . . = . + = + . = = = = You can see from the figure that the difference between the top surfaces of the fluids is .y = 20 cm - 2h = 20 cm - 2(6.27 cm) = 7.46 cm 7.5 cm 15.46. Model: The water is in hydrostatic equilibrium. Visualize: Please refer to figure P15.46. Solve: (a) Can 2 has moved down with respect to Can 1 since the water level in Can 2 has risen. Since the total volume of water stays constant, the water level in Can 1 has fallen by the same amount. The water level is equalized in the two cans at the middle of the height change, so the change in height of the water is half the relative change in height of the cans. Can 2 has moved relative to Can 1 (6.5 cm - 5.0 cm) 2 = 3.0 cm down. (b) The water level in Can 1 has fallen by the same amount. The new level is 5.0 -1.5 cm = 3.5 cm Assess: The two cans are an inexpensive method of measuring relative changes in height. 15.47. Visualize: The figure shows a dam with water height d. We chose a coordinate system with the origin at the bottom of the dam. The horizontal slice has height dy, width w, and area dA = wdy. The slice is at the depth of d - y. Solve: (a) The water exerts a small force dF = pdA = pwdy on this small piece of the dam, where p = . g(d - y) is the pressure at depth d - y. Altogether, the force on this small horizontal slice at position y is dF = . gw(d - y)dy. Note that a force from atmospheric pressure is not included. This is because atmospheric pressure exerts a force on both sides of the dam. The total force of the water on the dam is found by adding up all the small forces dF for the small slices dy between y = 0 m and y = d. This summation is expressed as the integral ( ) 2 2 total all slices 0 0 0 1 1 2 2 d d d F = dF = dF = . gw d - y dy = . gw.. yd - y .. = . gwd . . . . . (b) The total force is 1 ( 3 )( 2 )( )( )2 9 total 2 F = 1000 kg m 9.8 m s 100 m 60 m =1.7610 N 15.48. Visualize: The figure shows an aquarium tank with water at a height d. We chose a coordinate system with the origin at the bottom of the tank. The slice has a height dy, length l, and area dA = ldy. It is at depth d - y. Solve: (a) Pressure in excess of atmospheric pressure of the water on the bottom is bottom water p = . gd. Atmospheric pressure is ignored because it exerts an equal force on both sides of the bottom. Therefore, the force of the water on the bottom is ( ) ( )( ) ( 3 )( 2 )( )( )( ) bottom bottom water F = p lw = . gd lw = 1000 kg m 9.8 m s 0.40 m 1.0 m 0.35 m =1.37 kN (b) The water exerts a small force dF = pdA = pldy on a small piece of the window, where p = . g(d - y) is the excess pressure at a depth of d - y. The force on this small horizontal slice at position y is dF = . gl (d - y)dy. The total force on the front window is ( ) ( ) ( )( )( )( ) 1 2 2 total water water 2 0 water 0 1 3 2 2 2 1 2 1000 kg m 9.8 m s 1.00 m 0.40 m 0.78 kN d d F = dF = . gL d - y dy = . gl yd - y = . gld = = . . 15.49. Visualize: The figure shows a small column of air of thickness dz, of cross-sectional area A =1 m2 , and of density . (z). The column is at a height z above the surface of the earth. Solve: (a) The atmospheric pressure at sea level is 1.013105 Pa. That is, the weight of the air column with a 1 m2 cross section is 1.013105 N. Consider the weight of a 1 m2 slice of thickness dz at a height z. This slice has volume dV = Adz = (1 m2 )dz, so its weight is 2 / 0 2 0 dw = (. dV)g = . g(1 m )dz = . e-z z g(1 m )dz. The total weight of the 1 m2 column is found by adding all the dw. Integrating from z = 0 to z = 8, ( ) 0 0 2 0 0 2 0 0 0 2 0 0 (1 m ) (1 m ) (1 m ) z z z z w g e dz g z e g z . . . 8 - - 8 = = - .. .. = . Because 2 0 0 w =101,300 N = . g(1 m )z , ( )( ) 3 0 3 2 2 101,300 N 8.08 10 m 1.28 kg/m 9.8 m/s (1.0 m ) z= = (b) Using the density at sea level from Table 15.1, ( ) ( ) . = 1.28 kg m3 e-z /(8.08103 m) = 1.28 kg m3 e-1600 m/(8.08103 m) =1.05 kg m3 This is 82% of 0. . 15.50. Model: The buoyant force on the ceramic statue is given by Archimedes principle. Visualize: Solve: The gravitational force on the statue is the 28.4 N registered on the scale in air. In water, the gravitational force on the statue is balanced by the sum of the buoyant force B F and the springs force on the statue. That is, ( ) ( ) ( ) ( )( ) ( ) statue G statue B spring on statue w statue statue w statue 3 statue 3 3 statue 28.4 N 17.0 N 11.4 N 28.4 N 1000 kg m 2.49 10 kg m 11.4 N 11.4 N w F F F V g V m g m g . . . . . = + . = + . = = . = = = 15.51. Model: The buoyant force on the cylinder is given by Archimedes principle. Visualize: Solve: (a) Initially, as it floats, the cylinder is in static equilibrium, with the buoyant force balancing the gravitational force on the cylinder. The volume of displaced liquid is Ah, so FB = .liq (Ah)g = FG Force F pushes the cylinder down distance x, so the submerged length is h + x and the volume of displaced liquid is A(h + x). The cylinder is again in equilibrium, but now the buoyant force balances both the gravitational force and force F. Thus B liq G F = . (A(h + x))g = F + F Since liq G . (Ah)g = F , were left with liq F = . Agx (b) The amount of work dW done by force F to push the cylinder from x to x + dx is liq dW = Fdx = (. Agx)dx. To push the cylinder from i x = 0 m to f x =10 cm = 0.10 m requires work f f i i 1 2 2 liq 2 liq i f 1 3 2 2 2 2 ( ) (1000 kg/m ) (0.020 m) (9.8 m/s )(0.10 m) 0.62 J x x x x W F dx . Ag x dx . Ag x x p = = = - = = . . 15.52. Model: The buoyant force on the cylinder is given by Archimedes principle. Visualize: Let d1 be the length of the cylinder in the less-dense liquid with density .1, and 2 d be the length of the cylinder in the more-dense liquid with density 2p . Solve: The cylinder is in static equilibrium, so ( ) ( ) ( ) ( ) B1 B2 G 1 1 2 2 1 2 1 1 2 1 2 2 2 0 N y F F F F Ad g Ad g A d d g d d d d . . . . . . . = + - = . + = + . + = + S Since 1 2 1 2 , l = d + d .d = l - d we can simplify the above equation to obtain 1 2 2 1 d l . . . . . - . = . . . - . That is, the fraction of the cylinder in the more dense liquid is ( ) ( ) 1 2 1 . f = . - . . - . Assess: As expected f = 0 when . is equal to 1. , and f =1 when 2. = . . 15.53. Model: The buoyant force on the cylinder is given by Archimedes principle. Visualize: Solve: The tube is in static equilibrium, so ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) B G tube G Pb liquid 2 3 liquid 2 0 N 0.25 m 0.030 kg 0.250 kg 0.280 kg 0.280 kg 8.9 10 kg m 0.25 m 0.020 m 0.25 m y F F F F A g g g A . . p = - - = . = + . = = = S Assess: This is a reasonable value for a liquid. 15.54. Model: Archimedes Principle determines the buoyant force. Visualize: Solve: The plastic hemisphere will hold the most weight when its rim is at the surface of the water. The buoyant force balances the gravitational force on the bowl and rock. ( G )rock ( G )bowl 0 N y B SF = F - F - F = Thus ( ) ( ) ( ) ( )( ) bowl 3 3 2 2 bowl 1000 kg/m 1 4 0.040 m 9.8 m/s 0.021 kg 9.8 m/s 0.155 kg 2 3 w mg = p V g - m g = .. .... p .. - .m = . .. . Assess: Putting a rock as big as 155 g in an 8 cm diameter bowl before it sinks is reasonable. 15.55. Model: The buoyant force is determined by Archimedes principle. The spring is ideal. Visualize: Solve: The spring is stretched by the same amount that the cylinder is submerged. The buoyant force and spring force balance the gravitational force on the cylinder. ( )( ) ( ) ( )( ) 2 3 2 2 0 N 1.0 kg 9.8 m/s 1000 kg/m 0.025 m 9.8 m/s 35 N/m 0.181 m 18.1 cm y B S w w F F F mg p Ayg ky mg y mg p Ag k p = + - = . + = = = + + = = S Assess: This is difficult to assess because we dont know the height h of the cylinder and cant calculate it without the density of the metal material. 15.56. Model: The balloon displaces air, so the air exerts an upward buoyant force on the balloon. The buoyant force on the balloon is given by Archimedes principle. Visualize: FB is the buoyant force on the balloon, (FG )b is the gravitational force on the balloon, G He (F ) is the gravitational force on the helium gas, and G F is the gravitational force on the mass tied to the balloon. Solve: The volume of the balloon 4 ( )3 3 3 balloon 3 V = p 0.10 m = 4.188810- m . We have ( )( )( ) ( ) ( )( )( ) ( ) ( ) 3 2 3 3 B air balloon 3 2 3 3 G He He balloon G b 1.28 kg m 9.8 m s 4.188 10 m 0.05254 N 0.18 kg m 9.8 m s 4.188 10 m 0.007387 N 0.0010 kg 0.0098 N F gV F gV F g . . - - = = = = = = = = For the balloon to be in static equilibrium, ( ) ( ) B G He G b G 0.05254 N=0.01719 N 0.0036 kg 3.6 g F F F F mg m = + + = + . = = 15.57. Model: The buoyant force on the can is given by Archimedes principle. Visualize: The length of the can above the water level is d, the length of the can is L, and the cross-sectional area of the can is A. Solve: The can is in static equilibrium, so SFy = ( ) ( ) ( ) ( ) B G can G water water water F - F - F = 0 N.. A L - d g = 0.020 kg g + m g The mass of the water in the can is ( ) ( ) 6 3 can 3 water water water water 1000 kg m 355 10 m 0.1775 kg 2 2 0.020 kg 0.1775 kg 0.1975 kg 0.1975 kg 0.0654 m m V A L d d L A . . . . . - = . . = = . . . - = + = . - =- = Because ( )2 6 3 can V =p 0.031 m L = 35510- m , L = 0.1176 m. Using this value of L, we get d = 0.0522 m 5.2 cm. Assess: d L = 5.22 cm 11.76 cm = 0.444, thus 44.4% of the length of the can is above the water surface. This is reasonable. 15.58. Model: The buoyant force on the boat is given by Archimedes principle. Visualize: The minimum height of the boat that will enable the boat to float in perfectly calm water is h. Solve: The boat barely floats if the water comes completely to the top of the sides. In this case, the volume of displaced water is the volume of the boat. Archimedes principle in equation form is FB = .wVboatg. For the boat to float ( ) B Gboat F = F . Let us first calculate the gravitational force on the boat: ( ) ( ) ( ) ( ) ( ) ( )( ) ( ) ( )( ) ( ) ( )( ) G boat G bottom G side 1 G side 2 3 3 G bottom steel bottom 3 3 G side 1 steel side 1 3 3 G side 2 steel side 2 2 2 , where 7900 kg m 5 10 0.02 m 7900 N 7900 kg m 5 0.005 m 197.5 N 7900 kg m 10 0.005 m F F F F F V g g g F Vg h g gh F V g h g . . . = + + = = = = = = = = = ( ) ( ) ( ) ( ) G boat 395 7900 2 197.5 2 395 N 7900+1185 N gh . F = .. g + gh + gh .. = h g Going back to the Archimedes equation and remembering that h is in meters, we obtain ( ) ( )( ( )) w boat 7900 1185 N 1000 10 5 0.020 7900 1185 14.1 cm V g h g h h h . = + . + = + . = 15.59. Model: Treat the water as an ideal fluid obeying Bernoullis equation. There is a streamline connecting point 1 in the wider pipe with point 2 in the narrower pipe. Visualize: Solve: Bernoullis equation, Equation 15.28, relates the pressure, water speed, and heights at points 1 and 2: 1 2 1 2 1 2 1 1 2 2 2 2 p + .v + . gy = p + .v + . gy For no height change 1 2y = y . The flow rate is given as 5.0 L/s, which means ( ) ( ) 3 3 3 3 1 1 2 2 1 2 2 1 2 1 2 5.0 10 m s 5.0 10 m s 0.6366 m s 0.05 m 0.6366 m s 0.050 m 2.546 m s 0.025 m Q vA vA v v v A A p - - = = = . = = . = = . . = . . . . Bernoullis equation now simplifies to ( ) ( )( ) ( ) 1 2 1 2 1 2 1 2 2 2 1 2 2 3 1 3 2 2 1 2 2 2 1 2 50 10 Pa 1000 kg m 2.546 m s 0.6366 m s 53 kPa p v p v p p v v . . . + = + . = + - = + . - . = . . Assess: Reducing the pipe size reduces the pressure because it makes 2 1v > v . 15.60. Model: The two pipes are identical. Visualize: Solve: The water speed is the same in both pipes. The flow rate ( ) ( )( ) ( ) 6 6 3 3 2 3.0 10 L/min 2 3.0 10 10 1 m /s 60 3.5 m/s 2 2 1.5 m Q vA v Q A p - = = . . . . . = = . . = 15.61. Model: Treat the water as an ideal fluid obeying Bernoullis equation. A streamline begins in the bigger size pipe and ends at the exit of the narrower pipe. Visualize: Please see Figure P15.61. Let point 1 be beneath the standing column and point 2 be where the water exits the pipe. Solve: (a) The pressure of the water as it exits into the air is 2 atmos . p = p (b) Bernoullis equation, Equation 15.28, relates the pressure, water speed, and heights at points 1 and 2: 1 2 1 2 1 ( 2 2 ) ( ) 1 2 1 1 2 2 2 2 1 2 2 2 1 2 1 p + .v + . gy = p + .v + . gy . p - p = . v - v + . g y - y From the continuity equation, ( )( 4 2 ) ( 4 2 ) 4 3 1 1 2 2 1 1 v A = v A = 4 m s 510- m .v 1010- m = 2010- m s.v = 2.0 m s Substituting into Bernoullis equation, 1 ( 3 ) ( )2 ( )2 ( 3 )( )( ) 1 2 1 atmos 2 1000 kg m 4.0 m s 2.0 m s 1000 kg m 9.8 m s 4.0 m 6000 Pa 39,200 Pa 45 kPa p - p = p - p = .. - .. + = + = But 1 2 p - p = . gh, where h is the height of the standing water column. Thus ( )( ) 3 3 2 45 10 Pa 4.6 m 1000 kg/m 9.8 m/s h = = 15.62. Model: Treat the water as an ideal fluid obeying Bernoullis equation. A streamline begins at the faucet and continues down the stream. Visualize: The pressure at point 1 is 1 p and the pressure at point 2 is p2. Both 1 p and 2 p are atmospheric pressure. The velocity and the area at point 1 are 1 v and 1 A and they are 2 v and 2 A at point 2. Let the distance of point 2 below point 1 be d. Solve: The flow rate ( ) 4 3 6 3 4 3 1 1 2 2.0 1000 10 m 2.0 10 m 2.0 10 m 2 1.0 m/s 10 s s 0.0080 m Q vA v p - - - = = = . = = Bernoullis equation at points 1 and 2 is 1 2 1 2 1 ( 2 2 ) 1 2 1 1 2 2 2 2 2 2 1 p + .v + . gy = p + .v + . gy .. gd = . v - v From the continuity equation, ( ) ( ) ( ) 3 2 3 2 1 1 2 2 2 2 v A = v A . 1.0 m s p 8.010- m = v p 5.010- m .v = 2.56 m s Going back to Bernoullis equation, we have 1 ( )2 ( )2 2 gd = .. 2.56 m s - 1.0 m s ...d = 0.283 m 28 cm 15.63. Model: Treat the air as an ideal fluid obeying Bernoullis equation. Solve: (a) The pressure above the roof is lower due to the higher velocity of the air. (b) Bernoullis equation, with yinside youtside , is ( ) 2 1 2 2 3 inside outside 2 air air 1 1 1.28 kg m 130 1000 m 835 Pa 2 2 3600 s p p . v p . v . . = + .. = = . . = . . The pressure difference is 0.83 kPa (c) The force on the roof is (.p) A = (835 Pa)(6.0 m15.0 m) = 7.5104 N. The roof will blow up, because pressure inside the house is greater than pressure on the top of the roof. 15.64. Model: The ideal fluid obeys Bernoullis equation. Visualize: Please refer to Figure P15.64. There is a streamline connecting point 1 in the wider pipe on the left with point 2 in the narrower pipe on the right. The air speeds at points 1 and 2 are 1 v and 2 v and the crosssectional area of the pipes at these points are A1 and A2. Points 1 and 2 are at the same height, so 1 2y = y . Solve: The volume flow rate is 6 3 1 1 2 2 Q = Av = A v =120010- m s. Thus 6 3 2 2 1200 10 m s 95.49 m s (0.0020 m) v p - = = ( ) 6 3 1 2 1200 10 m s 3.82 m s 0.010 m v p - = = Now we can use Bernoullis equation to connect points 1 and 2: ( ) ( ) ( )( ) ( ) 1 2 1 2 1 2 1 1 2 2 2 2 1 2 2 1 3 2 2 1 2 2 2 1 2 1 2 1.28 kg m 95.49 m s 3.82 m s 0 Pa 5.83 kPa p v gy p v gy p p v v g y y . . . . . . + + = + + . - = - + - = . - . + = . . Because the pressure above the mercury surface in the right tube is 2 p and in the left tube is 1p , the difference in the pressures 1 p and 2 p is Hg . gh. That is, ( )( ) 3 1 2 Hg 3 2 5.83 kPa 5.83 10 Pa 4.4 cm 13,600 kg m 9.8 m s p p . gh h - = = . = = 15.65. Model: The ideal fluid (that is, air) obeys Bernoullis equation. Visualize: Please refer to Figure P15.65. There is a streamline connecting points 1 and 2. The air speeds at points 1 and 2 are v1 and v2 , and the cross-sectional areas of the pipes at these points are 1 A and 2A . Points 1 and 2 are at the same height, so 1 2y = y . Solve: (a) The height of the mercury is 10 cm. So, the pressure at point 2 is larger than at point 1 by ( ) ( 3 )( 2 )( ) Hg 2 1 . g 0.10 m = 13,600 kg m 9.8 m s 0.10 m =13,328 Pa. p = p +13,328 Pa Using Bernoullis equation, ( ) ( ) ( ) ( ) 1 2 1 2 1 2 2 1 2 air 1 air 1 2 2 air 2 air 2 2 1 2 air 1 2 2 2 2 1 2 2 1 2 3 air 2 2 13,328 Pa 20,825 m s 1.28 kg/m p v gy p v gy p p v v p p v v . . . . . . + + = + + . - = - - . - = = = From the continuity equation, we can obtain another equation connecting 1 v and 2 v : ( ) ( ) 2 2 1 1 2 2 1 2 2 2 2 1 0.005 m 25 0.001 m Av A v v A v v v A p p = . = = = Substituting 1 2 v = 25v in the Bernoulli equation, we get ( )2 2 22 2 2 2 25 v - v = 20,825 m s .v = 5.78 m s Thus 2 v = 5.8 m/s and 1 2 v = 25v =144 m/s. (b) The volume flow rate ( )2 ( ) 3 3 2 2 A v =p 0.0050 m 5.78 m s = 4.510- m s. 15.66. Model: Treat the water as an ideal fluid that obeys Bernoullis equation. There is a streamline from the top of the water to the hole. Visualize: Please refer to Figure P15.66. The top of the water (at y1 = h) and the hole 2 (at y = y) are at atmospheric pressure. The speed of the water at the top is zero because the tank is kept filled. Solve: (a) Bernoullis equation connecting the two points is 1 2 ( ) 2 0 + . gh = .v + . gy.v = 2g h - y (b) For a particle shot horizontally from a height y with speed v, the range can be found using kinematic equations. For the y-motion, using 0 t = 0 s, we have ( ) 1 ( )2 ( ) 1 ( ) 2 1 0 0 1 0 2 1 0 1 2 1 1 0 m 0 m s 2 y y y = y + v t - t + a t - t . = y + t + -g t .t = y g For the x-motion, ( ) 1 ( )2 1 0 0 1 0 2 1 0 1 0 m 0 m 2 x x x = x + v t - t + a t - t . x = + vt + . x = v y g (c) Combining the results of (a) and (b), we obtain x = 2g (h - y) 2y g = 4y(h - y) To find the maximum range relative to the vertical height, ( ) ( ) 1 0 4 4 02 4 dx h y y y h dy y h y = . .. - - .. = . = - 1 2 With y = h, the maximum range is max 4 2 2 x = . h .. h - h . = h . .. . . .. . 15.67. Model: Treat water as an ideal fluid that obeys Bernoullis equation. There is a streamline connecting the top of the tank with the hole. Visualize: Please refer to Figure P15.67. We placed the origin of the coordinate system at the bottom of the tank so that the top of the tank (point 1) is at a height of h +1.0 m and the hole (point 2) is at a height h. Both points 1 and 2 are at atmospheric pressure. Solve: (a) Bernoullis equation connecting points 1 and 2 is ( ) ( ) 1 2 1 2 1 2 1 1 2 2 2 2 1 2 1 2 atmos 2 1 atmos 2 2 2 2 2 2 2 1 1.0 m 2 1.0 m 19.6 m s p v gy p v gy p v gh p v gh v v g . . . . . . . . + + = + + . + + + = + + . - = = Using the continuity equation 1 1 2 2 , Av = A v ( ) ( ) 3 2 2 2 1 2 2 2 1 2.0 10 m 1.0 m 250,000 v A v v v A p p . . - = . . = = . . Because 1 2v ..v , we can simply put 1 v 0 m/s. Bernoullis equation thus simplifies to 2 22 2 2 v =19.6 m s .v = 4.43 m s Therefore, the volume flow rate through the hole is ( )( ) 3 2 5 3 2 2 Q = A v =p 2.010- m 4.43 m s = 5.5610- m s = 3.3 L min (b) The rate at which the water level will drop is 2 2 1 4.43 m s 1.77 10 mm s 1.06 mm min 250,000 250,000 v = v = = - = Assess: Because the hole through which water flows out of the tank has a diameter of only 4.0 mm, a drop in the water level at the rate of 1.06 mm/min is reasonable. 15.68. Model: The aquarium creates tensile stress. Solve: Weight of the aquarium is ( 3 )( 3 )( 2 ) 4 G water F = mg = . Vg = 1000 kg m 10 m 9.8 m s = 9.810 N where we have used the conversion 1 L =10-3 m3. The weight supported by each wood post is 1 4 4 (9.810 N) = 2.45104 N. The cross-sectional area of each post is A = (0.040 m)2 =1.610-3 m2. Youngs modulus for the wood is ( )( ) ( )( ) 10 2 4 3 3 2 10 2 1 10 N m / / 2.45 10 N 0.80 m 1.23 10 m 1.23 mm 1.6 10 m 1 10 N/m Y F A FL L L A L L FL AY - - = = = . . . . = = = = Assess: A compression of 1.23 mm due to a weight of 2.45104 N is reasonable. 15.69. Model: The water pressure applies a volume stress to the sphere. Solve: The volume change is .V = -1.010-3 V. The volume stress is ( ) 3 F p B V 7 1010 N m2 1.0 10 V 7.0 107 Pa A V V . - - = = - = - = Using 0 p = p + . gh, we get 7.0107 Pa =1.013105 Pa + (1030 kg m3 )(9.8 m s2 )d .d = 6.9 km Assess: A pressure of 7.0107 Pa = 690 atm causes only a volume change of 0.1%. This is reasonable because liquids and solids are nearly incompressible. 15.70. Model: Pressure applies a volume stress to water in the cylinder. Solve: The volume strain of water due to the pressure applied is ( )( ) 6 3 10 3 3 3 3 2 10 Pa 1.0 10 0.20 10 Pa 1.0 10 1.30 m 1.30 10 m 1.3 L V p V B V V V - - - . = - = - = - .. = ' - = - = - = - As the safety plug on the top of the cylinder bursts, the water comes back to atmospheric pressure. The volume of water that comes out is 1.30 L. 15.71. Model: Air is an ideal gas and obeys Boyles law. Visualize: Please refer to Figure CP15.71. h is the length of the air column when the mercury fills the cylinder to the top. A is the cross-sectional area of the cylinder. Solve: For the column of air, Boyles law is 0 0 1 1 , p V = pV where 0 p and 0 V are the pressure and volume before any mercury is poured, and 1 p and 1 V are the pressure and volume when mercury fills the cylinder above the air. Using 1 0 p = p + ( ) Hg . g 1.0 m- h , Boyles law becomes ( ) ( ) ( ) ( ) ( ) ( )( ) 0 0 0 Hg 1 0 0 Hg 5 0 0 Hg 3 2 Hg 1.0 1.0 m 1.0 1.0 1.0 1.013 10 Pa 0.76 m 76 cm 13,600 kg/m 9.8 m/s p V p g m h V p A p g m h Ah p m h g m h h h p g . . . . = .. + - .. . = .. + - .. - = - . = = = = 15.72. Model: The buoyant force on the cone is given by Archimedes principle. Visualize: Solve: It may seem like we need the formula for the volume of a cone. You can use that formula if you know it, but it isnt essential. The volume is clearly the area of the base multiplied by the height multiplied by some constant. That is, the cone shown above has V = aAd, where a is some constant. But the radius of the base is r = d tana , where a is the angle of the apex of the cone, and A =p r2 , making A proportional to d2. Thus the volume of a cone of height d is V = cd 3, where c is a constant. Because the cone is floating in static equilibrium, we must have B GF = F . The cones density is 0 . , so the gravitational force on it is 3 G 0 0 F = . Vg = . cl g. The buoyant force is the gravitational force on the displaced fluid. The volume of displaced fluid is the full volume of the cone minus the volume of the cone of height h above the water, or 3 3 disp V = cl = ch . Thus 3 3 B f disp fF = . V g = . c(l - h )g, and the equilibrium condition is 3 1/ 3 3 3 3 0 B G f 0 f 3 0 f F F c(l h )g cl g 1 h h 1 l l . . . . . . . . . . = . - = . . - . = . = . - . . . . . 15.73. Model: The grinding wheel is a uniform disk. We will use the model of kinetic friction and hydrostatics. Visualize: Please refer to figure CP15.73. Solve: This is a three-part problem. First find the desired angular acceleration, then use that to find the force applied by each brake pad, then finally the needed oil pressure. The angular acceleration required to stop the wheel is found using rotational kinematics. f i 2 0 rad/s 900 2 1 rad/s 60 18.85 rad/s t t 5.0 s . . . p a . - - = = = =- . . Each brake pad applies a frictional force k k f = n to the wheel. The normal force is equal to the force applied by the piston by Newtons third law. Rotational dynamics can be used to find the magnitude of the force. k f is applied 12 cm from the rotation axis on both sides of the disk. ( ) ( ) ( )( )( ) ( )( ) net 2 k 2 2 2 k 2 0.12 m 1 2 15 kg 0.13 m 18.85 rad/s 16.6 N 4 0.12 m 4 0.60 0.12 m I f MR n MR t a a a = - = . . . . . . - . = = = - - The oil pressure required to generate this much force at each brake pad is ( )2 16.6 N 53 kPa 0.010 m p F A p = = = relative to atmospheric pressure. Assess: The required oil pressure is about half an atmosphere, which is quite reasonable. 15.74. Model: The buoyant force on the cylinder is given by Archimedes principle. Visualize: A is cross-sectional area of the cylinder. Solve: (a) The gravitational force on the cylinder is FG = .0 ( Al ) g and the weight of the displaced fluid is ( ) B f F = . Ah g. In static equilibrium, B G F = F and we can write ( ) f 0 0 f . Ahg = . Alg.h = . . l (b) Now the volume of the displaced liquid is A(h - y). Applying Newtons second law in the y-direction, ( ) y G B 0 f SF = -F + F = - . Alg + . A h - y g Using f . 0 Ahg = . Alg from part (a), we find net f ( )y F = -. Agy (c) The result in part (b) is F = -ky, where f k = . Ag. This is Hookes law. (d) Since the cylinders equation of motion is determined by Hookes law, the angular frequency for the resulting simple harmonic motion is . = k m, and the period is 0 f f T 2 T 2 m 2 m 2 Al 2 h k Ag Ag g p . p p p p . . . = = = = = = where we have used the expression for h from part (a). (e) The oscillation period for the 100-m-tall iceberg 3 ice ( p = 917 kg/m ) in sea water is ( )( ) ( )( ) 3 0 3 2 f 917 kg/m 100 m 2 2 2 18.9 s 1030 kg/m 9.8 m/s T h l g g . p p p . = = = = 15.75. Model: A streamline connects every point on the surface of the liquid to a point in the drain. The drain diameter is much smaller than the tank diameter (r .. R). Visualize: Solve: The pressures at the surface and drain (points 1 and 2) are equal to one atmosphere. When the liquid is a depth y, Bernoullis principle comparing points 1 and 2 is 2 2 ( ) 1 1 2 2 2 2 2 1 1 1 0 2 2 2 p v gy p v g v v gy + . + . = + . + . . = + The flow rate through the drain is the same as through a horizontal layer in the tank: 2 1 1 2 2 1 2 1 Q v A v A v v A A = = . = Thus the velocity of the liquid through the drain is 2 2 2 2 2 2 2 1 2 2 2 1 v v A gy v gy A r R p p . . = . . + . = . . - Since r .. R, 2 v 2gy, and the flow rate through the drain is Q p r2 2gy The flow rate gives the volume of liquid flowing out per unit time. The inverse gives the time needed for a unit volume of liquid to flow out. For the volume of liquid to decrease by dV requires a time ( 2 ) 2 2 2 2 2 dV R dy R dy dt Q r gy gr y p p - - = = = Note that dV < 0 implies the volume of liquid is decreasing. Integrating both sides from the initial condition (t = 0, y = d) to the final condition (t, y = 0) yields 1 1 2 2 2 0 2 0 2 2 2 2 2 2 2 2 d d t R y dy R y R d gr gr r g = - . - = - = Assess: The time for the tank to drain depends on the ratio of the cross-sectional areas of the tank to drain, which makes sense, as well as on the strength of the free-fall acceleration. Note that if g were larger (say we were on Jupiter) the time to drain would be shorter, while if the tank were taller (larger d) the time would be longer. 15-1 15.76. Visualize: The column of air has a cross-sectional area A. We chose the origin of the coordinate system at z = 0 m and denote the vertical axis as the z-axis. Solve: (a) The pressure at height z is . z p The pressure at a height z + dz is weight of air column of height z dz z z z p p dz p Adzg p gdz A A . . + = - = - = - (b) The change in pressure is . z dz z dp p p . gdz + = - =- The sign is minus because pressure decreases as z increases. (c) From the ideal gas law, 0 0 0 0 p p p p . . . . = . = The result for part (b) thus becomes ( ) 0 0 dp = - p. p gdz. (d) To find the pressure p at a height z, we rewrite the result of part (c) as 0 0 dp g dz p p . . . = -. . . . Integrating both sides 0 0 0 0 0 0 0 0 0 0 0 0 ln p z gz p z z p dp g dz p g z p p e p e p p p p . . -. - . . = - . . . = - . = = . . . . where 0 0 0 . z = p . g (e) The scale height is ( ) ( )( ) 5 0 0 3 2 0 1.013 10 Pa 8076 m. 1.28 kg m 9.8 m s z p . g = = = (f) A table showing values of ( ) 0 8076 m 0 p p e z z e z atm = - = - at selected values of z follows. z (m) p (atm) 0 1000 2000 4000 6000 8000 10,000 12,000 14,000 15,000 1 0.884 0.781 0.609 0.476 0.371 0.290 0.226 0.177 0.156 Fluids and Elasticity 15-2 16.1. Model: Recall the density of water is 1000 kg/m3. Solve: The mass of lead ( 3 )( 3 ) Pb Pb Pb m = . V = 11,300 kg m 2.0 m = 22,600 kg . For water to have the same mass its volume must be water 3 water 3 water 22,600 kg 22.6 m 1000 kg m V m . = = = Assess: Since the lead is 11.3 times as dense we expect the water to take 11.3 times the volume. 16.2. Model: Assume the nucleus is spherical. Solve: The volume of the uranium nucleus is 4 3 4 ( 15 )3 42 3 3 3 V = p R = p 7.510- m =1.76710- m The density of the uranium nucleus is 25 nucleus 17 3 nucleus 42 3 nucleus 4.0 10 kg 2.3 10 kg m 1.767 10 m m V . - - = = = Assess: This density is extremely large compared to the typical density of materials. 16.3. Solve: The volume of the aluminum cube is 10-3 m3 and its mass is mAl = .AlVAl = ( )( ) 2700 kg m3 1.010-3 m3 = 2.7 kg The volume of the copper sphere with this mass is ( ) ( ) 3 Cu 4 3 Cu Cu 3 Cu 1 4 3 3 Cu 4 2.7 kg 3.027 10 m 3 8920 kg m 3 3.027 10 m 0.042 m 4 V r m r p . p - - = = = = . . . = . . = .. .. The diameter of the copper sphere is 0.0833 m = 8.33 cm. Assess: The diameter of the sphere is a little less than the length of the cube, and this is reasonable considering the density of copper is greater than the density of aluminum. 16.4. Model: The volume of a hollow sphere is ( 3 3 ) out in 4 3 V = r - r p Solve: We are given m = 0.690 kg, out r = 0.050 m, and we know that for aluminum r = 2700 kg/m3. Solve the above equation for in r . 3 in 3 out 43 3 3 out 4 3 3 3 3 43 / 0.050 m 0 690 kg/2700 kg/m 0 040 m r r V r m . . = - = - = - = p r p p So the inner diameter is 8.0 cm. Assess: We are happy that the inner diameter is less than the outer diameter, and in a reasonable range. 16.5. Solve: The volume of the aluminum cube V = 8.010-6 m3 and its mass is M = .V = (2700 kg/m3 )(8.010-6 m3) = 0.0216 kg = 21.6 g One mole of aluminum (27Al) has a mass of 27 g. The number of atoms is ( ) 23 6.02 10 atoms 1 mol 21.6 g 4.8 1023 atoms 1 mol 27 g N . .. . = . .. . = . .. . Assess: This is almost one mole of atoms, which is a reasonable value. 16.6. Solve: The volume of the copper cube is 8.0 10-6 m3 and its mass is M = .V = (8920 kg/m3)(8.010-6 m3 ) = 0.07136 kg = 71.36 g Because the atomic mass number of Cu is 64, one mole of Cu has a mass of 64 g. The number of moles in the cube is 1 mol (71.36 g) 1.1 mol 64 g n . . = . . = . . Assess: This answer is in the same ballpark as the previous exercise. 16.7. Solve: (a) The number density is defined as N V, where N is the number of particles occupying a volume V. Because Al has a mass density of 2700 kg/m3, a volume of 1 m3 has a mass of 2700 kg. We also know that the molar mass of Al is 27 g/mol or 0.027 kg/mol. So, the number of moles in a mass of 2700 kg is (2700 kg) 1 mol 1.00 105 mol 0.027 kg n . . = . . = . . The number of Al atoms in 1.00 105 mols is ( 5 )( 23 ) 28 A N = nN = 1.0010 mol 6.0210 atoms mol = 6.0210 atoms Thus, the number density is 28 28 3 3 6.02 10 atoms 6.02 10 atoms m 1 m N V = = (b) Pb has a mass of 11,300 kg in a volume of 1 m3. Since the atomic mass number of Pb is 207, the number of moles in 11,300 kg is (11,300 kg) 1 mole 0.207 kg n . . = . . . . The number of Pb atoms is thus N = nNA, and hence the number density is ( ) A 23 28 3 3 11,300 kg atoms 1 mol atoms 6.02 10 3.28 10 0.207 kg mol 1 m m N nN V V . .. . = = . .. . = . .. . Assess: We expected to get very large numbers like this. 16.8. Solve: The mass density is . = M/V. The mass M of the sample is related to the number of atoms N and the mass m of each atom by M = Nm. Combining these, the atomic mass is 3 26 28 3 1750 kg/m 3.986 10 kg/atom / 4.39 10 atoms/m m M V N N NV . . = = = = = - the atomic mass in m = A u, where A is the atomic mass number. Thus 26 27 3.986 10 kg 24 1 u 1.661 10 kg A m - - = = = The elements atomic mass number is 24. Assess: This is a reasonable answer for an isotope of neon (although neon is a gas at normal temperatures), sodium, or magnesium. 16.9 Model: Assume the gold is shaped into a solid sphere of volume 4 3. 3 V = p r Visualize: We want to know D = 2r. Because the atomic mass number of gold is 197, 1.0 mol of gold has a mass of 197 g or 0.197 kg. Table 16.1 gives . =19,300 kg/m3. Solve: 3 3 3 3 4 / 3 3 0.197 kg 0.0135 m 1.35 cm 3 4 4 19,300 kg/m V p r M . r M p . p = = . = = = = D = 2r = 2(1.35 cm) = 2.70 cm. Assess: This is about the right size for a chunk that contains one mole of material. 16.10. Solve: The mass of mercury is ( )( ) 6 3 3 3 3 13,600 kg m 10 cm 10 m 0.136 kg 136 g 1 cm M .V . - . = = . . = = . . and the number of moles is mol 0.136 g 0.6766 mol 201 g mol n M M = = = The mass of aluminum with 0.6766 mol of Al is ( ) ( ) mol 0.6766 mol 0.6766 mol 27 g 18.27 g 0.01827 kg mol M = M = .. .. = = . . This mass M of aluminum corresponds to a volume of 6 3 3 3 0.01827 kg 6.8 10 m 6.8 cm 2700 kg m V M . = = = - = Assess: We expected an answer in the same order of magnitude. The size of atoms doesnt vary as much as the density of atoms from element to element. 16.11. Solve: The lowest temperature is 9 9 ( ) F 5 C 5 C C k T = T + 32.-127F = T + 32.T = -88C.T = -88.3+ 273 K =185 K In the same way, the highest temperature is 9 5 C C 136F = T + 32.T = 58C = 331 K Assess: On the absolute scale the highest recorded temperature is not quite twice the lowest. 16.12. Solve: Let TF = TC = T : 9 9 F 5 C 5 T = T + 32.T = T + 32.T = -40 That is, the Fahrenheit and the Celsius scales give the same numerical value at -40 . Assess: It is usually unnecessary to specify the scale when the temperature is reported as -40 . 16.13. Model: A temperature scale is a linear scale. Solve: (a) We need a conversion formula for C to Z, analogous to the conversion of C to F. Since temperature scales are linear, TC = aTZ + b, where a and b are constants to be determined. We know the boiling point of liquid nitrogen is 0Z and 196C. Similarly, the melting point of iron is 1000Z and 1538C. Thus 196 0 1538 1000 a b a b - = + = + From the first, b = -196. Then from the second, a = (1538 +196)/1000 =1734/1000. Thus the conversion is C T = Z (1734/1000)T -196. Since the boiling point of water is C T =100C, its temperature in Z is Z 1000 (100 196 ) 171 Z 1734 T = .. .. + = . . (b) A temperature TZ = 500Z is C 1734 500 196 671 C 944 K 1000 T = .. .. - = = . . 16.14. Solve: (a) The triple point of water is T = 0.01C and p = 0.006 atm. Thus 9 9 ( ) F 5 C 5 T = T + 32 = 0.01 + 32 = 32.02F 1.013 105Pa 0.006 atm 608 Pa 1 atm p = = (b) The triple point of carbon dioxide is T = -56C and p = 5 atm. Thus 9 9 ( ) F 5 C 5 T = T + 32 = -56 + 32 = -68.8F ( ) 5 5.0 atm 5.0 atm 1.013 10 Pa 5.06 105 Pa 1 atm p . . = = . . = . . 16.15. Model: Treat the gas in the container as an ideal gas. Solve: From the ideal-gas law pV = nRT, the pressure of the gas is ( )( )( ) ( ) 6 3 3 3.0 mol 8.31 J mol K 273 120 K 1.9 10 Pa 19 atm 2.0 10 m p nRT V - .. - .. = = = = Assess: 19 atm is a high pressure, but not unreasonable. 16.16. Model: Treat the nitrogen gas in the closed cylinder as an ideal gas. Solve: (a) The density before and after the compression are .before = m1 V1 and .after = m2 V2 . Noting that 1 2 m = m and 1 2 2 1 V = V , after 1 after before before 2 m V 2 2 V m . . . . = = . = The mass density has changed by a factor of 2. (b) The number of atoms in the gas is unchanged, implying that the number of moles in the gas remains the same; hence the number density is unchanged. 16.17. Model: Treat the gas in the sealed container as an ideal gas. Solve: (a) From the ideal gas law equation pV = nRT, the volume V of the container is ( )( ) ( ) 3 5 2.0 mol 8.31 J mol K 273 30 K 0.050 m 1.013 10 Pa V nRT p .. + .. = = = Note that pressure must be in Pa in the ideal-gas law. (b) The before-and-after relationship of an ideal gas in a sealed container (constant volume) is ( )( ) ( ) 1 2 2 2 1 1 2 1 273 130 K 1.0 atm 1.3 atm 273 30 K pV p V p p T T T T + = . = = = + Note that gas-law calculations must use T in kelvins. 16.18. Model: Treat the gas as an ideal gas in the sealed container. Solve: (a) For p1 = p0 , the before-and-after relationship of an ideal gas in a sealed container is 1 0 1 1 0 0 0 1 0 0 473 K 1.27 373 K V V V V T V V T T T = . = = = where 0 T = (273 +100) K and 1 T = (273 + 200) K. (b) When the Kelvin temperature is doubled, 1 0 T = 2T = 2(373 K) = 746 K and the above equation becomes 1 1 0 0 0 0 746 K 2 373 K V V T V V T = = = Assess: When we use the Kelvin scale we expect the volume to double when the temperature doubles (if the pressure is kept constant). 16.19. Model: Treat the air in the compressed-air tank as an ideal gas. Solve: (a) From the ideal-gas law pV = nRT, the number of moles n is ( ) ( ) ( ) ( ) ( )( ) 5 2 2 150 atm 1.013 10 Pa 0.075 m 0.50 m 1 atm 55.1 mol 55 mol 8.31 J mol K 273 20 K pV p r h n RT RT p p . . . . . . . . = = = . . = .. + .. (b) At STP, the ideal-gas law yields ( )( )( ) 3 3 5 55.1 mol 8.31 J mol K 273 K 1.234 m 1.2 m 1.013 10 Pa V nRT p = = = Assess: Because the volume of the compressed air tank is (p r2 )h = 8.835710-3 m3, the volume at STP is 140 times the volume of the tank. 16.20. Model: Treat the oxygen gas in the cylinder as an ideal gas. Solve: (a) The number of moles of oxygen is mol 50 g 1.563 mol 1.6 mol 32 g mol n M M = = = (b) The number of molecules is ( )( 23 1 ) 23 23 A N = nN = 1.563 mol 6.0210 mol- = 9.4110 9.410 (c) The volume of the cylinder V =p r2L =p (0.10 m)2 (0.40 m) =1.25710-2 m3. Thus, 23 25 3 2 3 9.41 10 7.5 10 m 1.257 10 m N V - - = = (d) From the ideal-gas law pV = nRT we can calculate the absolute pressure to be ( )( )( ) 2 3 1.563 mol 8.31 J mol K 293 K 303 kPa 1.257 10 m p nRT V - = = = where we used T = 20C = 293 K. But a pressure gauge reads gauge pressure: g p = p -1 atm = 303 kPa -101 kPa = 202 kPa 200 kPa 16.21. Model: Treat the helium gas in the sealed cylinder as an ideal gas. Solve: The volume of the cylinder isV =p r2h =p (0.05 m)2 (0.30 m) = 2.35610-3 m3. The gauge pressure of the gas is 5 120 psi 1 atm 1.013 10 Pa 8.269 105 Pa, 14.7 psi 1 atm = so the absolute pressure of the gas is 8.269105 Pa + 1.013105 Pa = 9.282105 Pa. The temperature of the gas is T = (273 + 20) K = 293 K. The number of moles of the gas in the cylinder is ( )( ) ( )( ) 9.282 105 Pa 2.356 10 3 m3 0.898 mol 8.31 J mol K 293 K n pV RT - = = = (a) The number of atoms is ( )( 23 1 ) 23 23 A N = nN = 0.898 mol 6.0210 mol- = 5.4110 atoms 5.410 atoms (b) The mass of the helium is ( )( ) 3 3 mol M = nM = 0.898 mol 4 g mol = 3.59 g = 3.5910- kg 3.610- kg (c) The number density is 23 26 3 3 3 5.41 10 atoms 2.3 10 atoms m 2.356 10 m N V - = = (d) The mass density is 3 3 3 3 3.59 10 kg 1.5 kg m 2.356 10 m M V . - - = = = 16.22. Model: The gas is assumed to be ideal and it expands isothermally. Solve: (a) Isothermal expansion means the temperature stays unchanged. That is T2 = T1. (b) The before-and-after relationship of an ideal gas under isothermal conditions is 1 1 2 2 1 1 1 2 1 1 1 1 2 1 2 2 pV p V p p V p V p T T V V . . = . = = . . = . . 16.23. Model: The gas is assumed to be ideal. Solve: (a) Isochoric means the volume stays unchanged. That is V2 =V1. (b) The before-and-after relationship of an ideal gas under isochoric conditions is 1 1 1 2 2 2 1 2 1 1 1 1 1 1 1 3 3 pV p V p p T T T T T T p p . . . . = . = = . . = .. .. . . 16.24. Model: The gas is assumed to be ideal, and since the container is rigid V2 =V1. Solve: Convert both temperatures to the Kelvin scale. 1 1 2 2 2 2 1 1 1 1 3 atm 283 K 3.1 atm 275 K pV p V p p T T T T = . = = . . = . . . . Assess: On the absolute scale the temperature only went up a little bit, so we expect the pressure to rise a little bit. 16.25. Model: The rigid spheres volume does not change, so this is an isochoric process. The air is assumed to be an ideal gas. Solve: (a) When the valve is closed, the air inside is at p1 =1 atm and 1 T =100C. The before-and-after relationship of an ideal gas in the closed sphere (constant volume) is ( ) ( ) ( ) 1 2 2 2 1 1 2 1 273 0 K 1.0 atm 0.73 atm 273 100 K pV p V p p T T T T . . + = . = . . = = . . + (b) Dry ice is CO2. From Figure 16.4, we can see that the solid-gas transition line gives a temperature of -78C when p =1 atm. Cooling the sphere to 78C gives ( )( ) 2 2 1 1 273 78 K 1.0 atm 0.52 atm 373 K p p T T . . - = . . = = . . 16.26. Model: Assume the gas is an ideal gas. Visualize: The pressure in the gas must exert exactly enough upward force to counteract the gravitational force on the piston; this is true at both temperatures ( p2 = p1 = p). The initial volume of the cylinder is 2 ( )2 ( ) 3 1 1 V = pr h = p 0.12m 0.84m = 0.038m . The initial temperature is 1 T = 303C + 273 = 576K. Solve: (a) ( )( ) ( ) 2 2 2 mg 20kg 9 80m/s 4333Pa 4300Pa 0 12m F . p A r . = = = = p p < (b) Knowing the initial temperature, pressure, and volume allows us to compute the number of moles of gas (which will stay constant). ( )( ) ( )( ) 3 1 1 4333Pa 0 038m 0 0344mol 0 034mol 8 31J/mol K 576K pV . n . . RT . = = = < Now apply the n just found and the new temperature 2 (T =15C + 273 = 288K) to find 2V . ( )( )( ) 2 3 2 0 0344 mol 8 31 J mol K 288 K 0 019 m 4333 Pa nRT . . / V . p = = = Solve 2 2 2 V = pr h for 2h . ( ) 3 2 2 2 2 0 019m 0 42m 0 12m h V . . r . = = = p p The new height of the piston is 42 cm. Assess: Due to a coincidence in the data we could have used the shortcut that 1 2 2 1 T = T on the absolute scale to deduce that 1 1 2 2 1 2 2 1 V = V 1h = h . 16.27. Model: In an isochoric process, the volume of the container stays unchanged. Argon gas in the container is assumed to be an ideal gas. Solve: (a) The container has only argon inside with 3 6 3 n 0.1 mol, V1 50 cm 50 10 m = = = - , and 1 T = 20C = 293 K. This produces a pressure ( )( )( ) 6 1 6 3 1 0.1 mol 8.31 J mol K 293 K 4.87 10 Pa 4870 kPa 4900 kPa 50 10 m p nRT V - = = = = An ideal gas process has 2 2 2 1 1 1 p V /T = pV /T . Isochoric heating to a final temperature 2 T = 300C = 573 K has 2 1V =V , so the final pressure is 1 2 2 1 2 1 1 573 4870 kPa 9520 kPa 9500 kPa 293 p V T p V T = = = Note that it is essential to express temperatures in kelvins. (b) Assess: All isochoric processes will be a straight vertical line on a pV diagram. 16.28. Model: The isobaric heating means that the pressure of the argon gas stays unchanged. Solve: (a) The container has only argon inside with n = 0.1 mol, 3 6 3 1 V = 50 cm = 5010- m , and 1 T = 20C = 293 K. This produces a pressure ( )( )( ) 1 6 1 6 3 1 0.1 mol 8.31 J mol K 293 K 4.87 10 Pa 4870 kPa 4900 kPa 50 10 m p nRT V - = = = = An ideal gas process has 2 2 2 1 1 1 p V /T = pV /T . Isobaric heating to a final temperature 2 T = 300C = 573 K has 2 1p = p , so the final volume is 1 2 3 3 3 2 1 2 1 1 573 50 cm 97.8 cm 98 cm 293 V p T V p T = = = (b) Assess: All isobaric processes will be a straight horizontal line on a pV diagram. 16.29. Model: In an isothermal expansion, the temperature stays the same. The argon gas in the container is assumed to be an ideal gas. Solve: (a) The container has only argon inside with n = 0.1 mol, 3 6 3 1 V = 50 cm = 5010- m , and 1 T = 20C = 293 K. This produces a pressure 1 6 1 6 1 (0.1 mol)(8.31 J/mol K)(293 K) 4.87 10 Pa = 12.02 atm 12 atm 50 10 Pa p nRT V - = = = An ideal-gas process obeys 2 2 2 1 1 1 p V /T = pV /T . Isothermal expansion to 3 2 V = 200 cm gives a final pressure 2 1 2 1 1 2 1 200 12.02 atm = 48 atm 50 p T V p T V = = (b) 16.30. Model: Assume the gas to be an ideal gas. Solve: (a) Because the volume stays unchanged, the process is isochoric. (b) The ideal-gas law p1V1 = nRT1 gives ( )( ) ( )( ) 5 63 1 1 1 3 1.013 10 Pa 100 10 m 914 K 641 C 0.0040 mol 8.31 J mol K T pV nR - = = = = The final temperature T2 is calculated as follows for an isochoric process: 1 2 1 2 p p T T = 2 ( ) 2 1 1 914 K 1 atm 305 K 32 C 3 atm T T p p . = = . . = = . . . . Assess: All straight vertical lines on a pV diagram represent isochoric processes. 16.31. Model: Assume that the gas is an ideal gas. Solve: (a) The graph shows that the pressure is inversely proportional to the volume. The process is isothermal. (b) From the ideal-gas law, ( )( ) ( )( ) 5 63 1 1 1 3 1.013 10 Pa 100 10 m 914 K 641 C 0.0040 mol 8.31 J mol K T pV nR - = = = = T2 is also 914 K, because the process is isothermal. (c) The before-and-after relationship of an ideal gas under isothermal conditions is 1 1 2 2 pV = p V 1 ( 3 ) 3 2 1 2 100 cm 3 atm 300 cm 1 atm V V p p . = = . . = . . . . 16.32. Model: Assume that the gas is ideal. Solve: (a) Because the process is at a constant pressure, it is isobaric. (b) For an ideal gas at constant pressure, 2 1 2 1 V V T T = ( ) 3 2 2 1 3 1 273 900 K 100 cm 391 K 118 C 300 cm T T V V . = = .. + .. = = (c) Using the ideal-gas law 2 2 2 , p V = nRT ( )( ) ( )( ) 5 63 2 2 3 2 3 1.013 10 Pa 100 10 m 9.35 10 mol 8.31 J mol K 391 K n p V RT - - = = = Assess: All straight horizontal lines on a pV diagram represent isobaric processes. 16.33. Visualize: Solve: Suppose we have a 1 m1 m1 m block of copper of mass M containing N atoms. The atoms are spaced a distance L apart along all three axes of the cube. There are Nx atoms along the x-edge of the cube, Ny atoms along the y-edge, and Nz atoms along the z-edge. The total number of atoms is . x y z N = N N N If L is expressed in meters, then the number of atoms along the x-edge is (1 m)/ . x N = L Thus, 3 3 1 3 3 N 1 m 1 m 1 m 1 m L 1 m L L L L N . . = = . =. . . . This relates the spacing between atoms to the number of atoms in a 1-meter cube. The mass of the large cube of copper is ( 3 )( 3 ) Cu M = . V = 8920 kg m 1 m = 8920 kg But M = mN, where m = 64 u = 64(1.66110-27 kg) is the mass of an individual copper atom. Thus, ( ) 28 27 8920 kg 8.39 10 atoms 64 1.661 10 kg N M m - = = = 3 1 3 10 28 1 m 2.28 10 m 0.228 nm 8.39 10 L - . . . = . . = = . . Assess: This is a reasonable interatomic spacing in a crystal lattice. 16.34. Solve: The volume of the cube associated with each atom is ( 9 )3 29 3 atom V = 0.22710- m =1.169710- m The volume of a mole of atoms is ( 29 3 )( 23 1 ) 6 3 atom A V =V N = 1.169710- m 6.0210 mol- = 7.041610- m mol Thus, the mass of a mole of atoms is ( 3 )( 3 ) mol M =V. = 7.0416 m mol 7950 kg m = 0.056 kg/mol = 56 g/mol The atomic mass number of the element is 56. Assess: This is likely the most common isotope of iron. 16.35. Model: Assume the density of the liquid water is 3 1.0 g . 1.0 cm Visualize: There are 10 protons in each water molecule. 1 mol of water molecules has a mass of 18 g. Solve: 23 3 26 3 1.0 g 1 0L 1000cm 1 mol 6 02 10 molecules 10 protons 3 3 10 protons 1.0 cm 18 g 1 mol 1 molecule . . . . .. .. .. . = . .. .. .. . = . .. .. .. . Assess: This is an unimaginably large number, but reasonable considering the data. 16.36. Model: Air is an ideal gas. The pressure at sea level is 1 atm. Solve: From the ideal gas law pV = NkBT, the number density is 5 25 3 23 B 1.013 10 Pa 2.5 10 molecules /m (1.38 10 J/K)(293 K) N p V kT - = = = Assess: Each cubic meter of air holds an unimaginably large number of molecules. 16.37. Model: Assume the gas in the solar corona is an ideal gas. Solve: The number density of particles in the solar corona is N V . Using the ideal-gas equation, ( ) ( )( ) B B 15 3 23 6 0.03 Pa 1.1 10 particles m 1.38 10 J K 2 10 K pV Nk T N p V kT N V - = . = = = Assess: This answer is a lot smaller than the one in the previous problem. 16.38. Model: Assume the gas in the evacuated volume is an ideal gas. Solve: The number density of particles is N V . Using the ideal-gas equation, B B pV Nk T N p V kT = . = The pressure is ( ) 5 10 8 8 12 3 6 3 23 1 10 mm of Hg 1 atm 1.013 10 Pa 1.33 10 Pa 760 mm of Hg 1 atm 1.33 10 Pa 3.3 10 m =3.3 10 cm (1.38 10 J/K)(293 K) p N V - - - - - - = = . = = Assess: Even the best vacuum we can achieve in the laboratory contains 3 million molecules per cubic centimeter. 16.39. Model: Assume that the gas in the vacuum chamber is an ideal gas. Solve: (a) The fraction is 10 vacuum chamber 13 atmosphere 1.0 10 mm of Hg 1.3 10 760 mm of Hg p p - - = = (b) The volume of the chamber V =p (0.20 m)2 (0.30 m) = 0.03770 m3. From the ideal-gas equation B pV = Nk T, the number of molecules of gas in the chamber is ( )( )( ) ( )( ) 13 5 3 11 23 B 1.32 10 1.013 10 Pa 0.03770 m 1.2 10 molecules 1.38 10 J K 293 K N pV k T - - = = = 16.40 Model: Assume the nebula gas is ideal. Visualize: Use pV = NkBT. We are given N /V =100 atoms/cm3 =1108 atoms/m3. Solve: p NkBT (1.0 108 atoms/m3 )(1.38 10 23 J/K)(7500 K) 1.0 10 11 Pa 1.0 10 16 atm V = = - = - = - Assess: This is much lower pressure than the best vacuum we can achieve in a laboratory, but it is a higher pressure than in non-nebula space. 16.41. Model: Assume the air is pure 2 N , with a molar mass of 28g/mol. Visualize: We will use pV = nRT both before and after. Our intermediate goal is 1 2n - n . We are given 2 1 T = T = T = 20C + 273 = 293 K and 2 2 4 3 2 1 V =V =V = r C = (0.011 m )(2.0 m) = 7.60102 m . p p Solve: We convert the gauge pressures to absolute pressures with g g p = p +1 atm= p +14.7 psi. 1 p =110 psi +14.7 psi = 124.7 psi = 860 kPa. 2 p = 80 psi +14.7 psi = 94.7 psi = 653 kPa. 1 1 2 2 1 2 1 2 1 2 4 3 ( ) 7.60 10 m (860 kPa 653 kPa) (8.31 J/mol K)(293 K) 0.0646 mol n n pV p V RT RT V p p RT - - = - = - = - = Thus 0.0646 mol of N2 was lost; this is 0 0646mol 28g 1 8g 1mol . . . . . . = . . Assess: The result seems to be a reasonable number. 16.42. Model: The carbon dioxide in the cube is an ideal gas. Solve: Using the ideal gas equation and n = M/Mmol , mol pV nRT V nRT MRT p pM = . = = The molar mass of CO2 is 44 g/mol or 0.044 kg/mol. Thus, ( )( )( ) ( )( ) 3 5 10,000kg 8.31 J mol K 273 K 5090 m 1.013 10 Pa 0.044 kg mol V= = The length of the cube is L = (V )1/3 =17.2 m. Assess: This is sobering when you multiply it by the number of people in the industrialized world. Good thing plants take up CO2 in large quantities. 16.43. Model: Assume the gas is an ideal gas. Solve: The before-and-after relationship of an ideal gas is 1 1 2 2 1 2 pV p V T T = 2 2 2 1 1 1 T T p V p V . = ( )1 2 1 1 1 1 298 K 3 p V p V = = 447 K = 174C 16.44. Model: Assume the evaporated water is an ideal gas with a molar mass of 18 g/mol. Assume the pressure is 1 atm =101.3 kPa. Visualize: We are given T = 35C + 273 = 308 K. n =1000 g(1 mol/18 g) = 55.6 mol. Solve: (a) ( )( )( ) 3 55 6 mol 8 31 J/mol K 308 K 1 4 m 101 3 kPa nRT . . pV nRT V . p . = 1 = = = (b) In the liquid state r =1000 kg/m3. 3 3 1 0 kg 0 0010 m 1000 kg/m V = m = . = . r The factor by which the volume of the evaporated water is larger than the liquid water is 3 3 1 4 m 1400 0 0010 m . . = Assess: Gases really do take up a lot more volume than the equivalent mass of a liquid. 16.45. Model: Assume that the steam (as water vapor) is an ideal gas. Solve: The volume of the liquid water is V m . = mol nM . = mol pV M RT . = . . . . . . ( )( )( ) ( )( )( ) 5 63 3 20 1.013 10 Pa 10,000 10 m 0.018 kg mol 8.31 J mol K 473 K 1000 kg m - = = 9.2810-5m3 = 92.8 cm3 93 cm3 Assess: The liquid takes a lot smaller volume than the same number of atoms as a gas. 16.46. Model: Assume that the steam is an ideal gas. Solve: (a) The volume of water is V M . = mol nM . = mol pV M RT . = ( )( )( ) ( )( )( ) 5 3 3 50 1.013 10 Pa 5.0 m 0.018 kg mol 8.31 J mol K 673 K 1000 kg m = = 0.0815 m3 = 81.5 L 82 L (b) Using the before-and-after relationship of an ideal gas, 2 2 1 1 2 1 2 1 2 1 1 2 p V pV V T p V T T Tp = . = ( ) ( 3 ) 3 3 273 150 K 50 atm 5.0 m 78.6 m 79 m 673 K 2.0 atm . + .. . = . .. . = . .. . 16.47. Model: We assume that the volume of the tire and that of the air in the tire is constant. Solve: A gauge pressure of 30 psi corresponds to an absolute pressure of (30 psi) + (14.7 psi) = 44.7 psi. Using the before-and-after relationship of an ideal gas for an isochoric (constant volume) process, 1 2 1 2 p p
T T = 2 ( ) 2 1 1 273 45 44.7 psi 49.4 psi 273 15 p T p T . + . . = = . . = . + . Your tire gauge will read a gauge pressure g p = 49.4 psi -14.7 psi = 34.7 psi. 35 psi. 16.48. Model: The air is assumed to be an ideal gas. Solve: At 20C and 1 atm pressure, the number of moles in the container is ( )( ) ( )( ) 5 3 3 1 1 1 1 1.013 10 Pa 10 m 0.0416 mol 8.31 J mol K 293 K n pV RT - = = = At 100C and 1 atm pressure, the number of moles is ( )( ) ( )( ) 5 3 3 2 2 2 2 1.013 10 Pa 10 m 0.0327 mol 8.31 J mol K 373 K n p V RT - = = = When heated, the pressure will rise as the number of moles remains n1. When opened, the pressure drops to 1 atm as gas escapes. Thus, the number of moles of air that escape as the container is opened is 1 2 n - n = 0.0416 mol - 0.0327 mol = 0.0089 mol. 16.49. Model: The gass pressure does not change, so this is an isobaric process. Solve: The triple point of water is 0.01C or 273.16 K, so T1 = 273.16 K. Because the pressure is a constant, 1 2 1 2 V V T T = 2 ( ) 2 1 1 273.16 K 1638 mL 447.44 K 174.3 C 1000 mL T T V V . = = . . = = . . . . 16.50. Model: Assume the gas in the manometer is an ideal gas. Solve: In the ice-water mixture the pressure is 1 atoms Hg ( p = p + . g 0.120 m) =1.013105 Pa + (13,600 kg/m3)(9.8 m/s2 )(0.120 m) =1.173105 Pa In the freezer the pressure is 2 atm Hg ( p = p + . g 0.030 m) =1.013105 Pa + (13,600 kg/m3)(9.8 m/s2 )(0.030 m) =1.053105 Pa Assume that a drop in length of 90 mm produces a very small change in gas volume compared with the total volume of the gas cell. This means the volume of the chamber can be considered constant. Hence, ( ) 5 1 2 2 2 1 5 1 2 1 273 K 1.053 10 Pa 245 K= 28 C 1.173 10 Pa p p T T p T T p . . = . = . . = = - . . Assess: This is a reasonable temperature for an industrial freezer. 16.51. Model: The air in the closed section of the U-tube is an ideal gas. Visualize: The length of the tube is l =1.0 m and its cross-sectional area is A. Solve: Initially, the pressure of the air in the tube is 1 atmos p = p and its volume is 1 V = Al. After the mercury is poured in, compressing the air, the air-pressure force supports the weight of the mercury. Thus the compressed pressure equals the pressure at the bottom of the column: 2 atmos p = p + . gL. The volume of the compressed air is 2 V = A(l - L). Because the mercury is poured in slowly, we will assume that the gas remains in thermal equilibrium with the surrounding air, so 2 T = T1. In an isothermal process, pressure and volume are related by 1 1 atmos 2 2 atmos pV = p Al = p V = ( p + . gL)A(l - L) Canceling the A, multiplying through, and solving for L gives atmos 3 2 1.00 m 101,300 Pa 0.240 m = 24.0 cm (13,600 kg/m )(9.8 m/s ) L l p . g = - = - = 16.52. Model: Assume that the air bubble is always in thermal equilibrium with the surrounding water, and the air in the bubble is an ideal gas. Solve: The pressure inside the bubble matches the pressure of the surrounding water. At 50 m deep, the pressure is 5 3 2 5 1 0 water 1.013 p = p + . gd = 10 Pa + (1000 kg/m )(9.8 m/s )(50 m) = 5.91310 Pa At the lakes surface, 5 2 0 p = p =1.01310 Pa. Using the before-and-after relationship of an ideal gas, 2 2 1 1 1 2 2 1 2 1 2 1 p V pV V V p T T T pT = . = ( ) 5 3 3 2 5 4 4 0.005 m 5.913 10 Pa 293 K 3 3 1.013 10 Pa 283 K r p p . .. . . = . .. . . .. . 2 .r = 0.0091 m The diameter of the bubble is 2 2r = 0.0182 m =1.82 cm. Assess: The bubbles diameter just about doubled; this seems reasonable. 16.53. Model: Assume that the compressed air in the cylinder is an ideal gas. The volume of the air in the cylinder is a constant. Solve: Using the before-and-after relationship of an ideal gas, 2 2 1 1 2 1 ( ) 1 2 1 2 1 1 2 1 25 atm 1223 K 104 atm 293 K p V pV p p T V V T T TV V = . = = . . = . . . . where we have converted to the Kelvin temperature scale. Because the pressure does not exceed 110 atm, the compressed air cylinder does not blow. 16.54. Model: Assume that the gas is an ideal gas. Solve: Assess: For the isothermal process, the pressure must be halved as the volume doubles. This is because p1 is proportional to 1/V1 for isothermal processes. 16.55. Model: Assume that the gas is an ideal gas. Solve: Assess: For the isothermal process, the pressure must double as the volume is halved. This is because p is proportional to 1/V for isothermal processes. 16.56. Model: Assume that the helium gas is an ideal gas. Visualize: Process 1.2 is isochoric, process 2.3 is isothermal, and process 3.1 is isobaric. Solve: The number of moles of helium is mol 8.0 g 2.0 mol 4 g mol n M M = = = Using the ideal-gas equation, ( )( )( ) ( ) 1 3 1 5 1 2.0 mol 8.31 J mol K 273 37 K 0.0254 m 2 1.013 10 Pa V nRT p .. + .. = = = 0.025 m3 For the isochoric process 2 1V =V , and 1 2 1 2 p p T T = 2 ( ) 2 1 1 2 atm 657 273 6.0 atm 37 273 p p T T . + . . = = . . = . + . For the isothermal process, the equation 3 3 2 2 p V = p V is 2 ( 3 ) 3 3 3 2 3 0.0254 m 6 atm 0.0762 m 0.076 m 2 atm V V p p = = . . = . . . . For the isothermal process, 3 2 T = T = 657C. 16.57. Model: Assume the nitrogen gas is an ideal gas. Solve: (a) The number of moles of nitrogen is mol 1 g 1 mol 28 g mol 28 n M M = = = . . . . . . Using the ideal-gas equation, ( )( )( ) ( ) 1 5 1 6 3 1 1 28 mol 8.31 J mol K 298 K 8.84 10 Pa 884 kPa 100 10 m p nRT V - = = = = 880 kPa (b) For the process from state 1 to state 3: 1 1 3 3 1 3 pV pV T T = ( ) 3 3 3 1 3 1 3 1 1 1 298 K 1.5 50 cm 223.5 K 49 C 100 cm T T p V p p V p . .. . . = = . .. . = - . .. . For the process from state 3 to state 2: 2 2 3 3 2 3 p V pV T T = ( ) 3 2 2 1 2 3 3 3 3 1 223.5 K 2.0 100 cm 596 K 323 C 1.5 50 cm T T p V p p V p . .. . . .. . . = . .. . = . .. . = = . .. . . .. . For the process from state 1 to state 4: 4 4 1 1 4 1 p V pV T T = 4 4 4 1 1 1 T T p V p V . = ( ) 3 1 3 1 298 K 1.5 150 cm 670.5 K 100 cm p p . .. . = . .. . = . .. . 398C 16.58. Model: The gas is an ideal gas. Solve: (a) Using the ideal-gas equation, ( )( ) ( )( ) 5 3 1 1 1 1.0 10 Pa 2.0 m 301 K 80 mol 8.31 J mol K T pV nR = = = Because points 1 and 2 lie on the isotherm, 2 1 T = T = 301 K. The temperature of the isothermal process is 301 K. (b) The straight-line process 1.2 can be represented by the equation p = (3 -V )105 where V is in m3 and p is in Pa. We can use the ideal gas law to find that the temperature along the line varies as 5 T pV (3V V 2 ) 10 nR nR = = - We can maximize T by setting the derivative dT/dV to zero: 5 2 3 3 max max d (3 2 ) 10 0 3 m 1.50 m d 2 T V V V nR = - = . = = At this volume, the pressure is 5 max p = 1.510 Pa and the temperature is 5 3 max max max (1.50 10 Pa)(1.5 m ) 338 K (80 mol)(8.31 J/mol K) T p V nR = = = 16.59. Model: Assume the gas is an ideal gas. Solve: (a) We can find the temperatures directly from the ideal-gas law after we convert all quantities to SI units: ( )( ) ( )( ) 3 6 3 3 1 1 1 3.0 atm 101,300 Pa atm 1000 cm 10 m cm 366 K 93 C 0.10 mol 8.31 J mol K T pV nR - = = = = ( )( ) ( )( ) 3 6 3 3 2 2 2 1.0 atm 101,300 Pa atm 3000 cm 10 m cm 366 K 93 C 0.10 mol 8.31 J mol K T p V nR - = = = = (b) 2 1T = T , so this is an isothermal process. (c) A constant volume process has 3 2V =V . Because 1 2 p = 3p , restoring the pressure to its original value means that 3 p = 3p2. From the ideal-gas law, 3 3 2 2 3 3 3 2 2 3 2 2 2 p V p V T p V T 3 1 T 3 366 K 1098 K 825 C T T p V . .. . = . = . .. . = = = = . .. . 16.60. Model: Assume the gas is an ideal gas. Solve: (a) Using the ideal-gas law, ( )( ) ( )( ) 5 63 1 1 1 3 1.013 10 Pa 100 10 m 244 K 29 C 5.0 10 mol 8.31 J mol K T pV nR - - = = = =- (b) Using the before and after relationship of an ideal gas, 1 1 2 2 1 2 pV pV T T = 2 1 2 1 1 2 p p T V T V . = ( ) 3 3 1 atm 2926 K 100 cm 4.0 atm 244 K 300 cm . .. . = . .. . = . .. . (c) Using the before and after relationship of an ideal gas, 3 3 2 2 3 2 p V p V T T = 2 3 3 2 3 2 V p T V p T . = 4 atm 2438 K (300 cm3 ) 500 cm3 2 atm 2926 K = . .. . = . .. . . .. . 16.61. Model: We assume the oxygen gas is ideal. Visualize: From the figure we glean p2 = 3p1 and 2 1 V = 3V . We are given 1 T = 20C + 273 = 293K. Solve: Use the before-and-after version of the ideal-gas law. 1 1 2 2 1 2 pV p V T T = 2 2 ( 1 )( 1 ) ( ) 2 1 1 1 1 1 1 1 3 3 9 9 293K 2637K 2364 C p V p V T T T T pV pV = = = = = = Assess: This is a hot temperaturehigher than the melting point of many elements. 16.62. Model: Assume CO2 gas is an ideal gas. Solve: (a) The molar mass for CO2 is Mmol = 44 g/mol, so a 10 g piece of dry ice is 0.2273 mol. This becomes 0.227 mol of gas at 0C. With 3 3 1 V =10,000 cm = 0.010 m and 1 T = 0C = 273 K, the pressure is ( )( )( ) 1 4 1 3 1 0.2273 mol 8.31 J mol K 273 K 5.156 10 Pa 0.509 atm 0.010 m p nRT V = = = = 0.50 atm (b) From the isothermal compression, 1 ( 3 ) 3 3 3 2 2 1 1 2 1 2 0.010 m 0.509 atm 1.70 10 m 1700 cm 3.0 atm p V pV V V p p = . = = .. .. = - = . . From the isobaric compression, ( ) 3 3 3 2 3 2 273 K 1000 cm 161 K 112 C 1700 cm T T V V . . = = . . = = - . . (c) 16.63. Model: The gas in the container is assumed to be an ideal gas. Solve: (a) The gas starts at pressure p1 = 2.0 atm, temperature 1 T =127C = (127 + 273) K = 400 K and volume V1. It is first compressed at a constant temperature 2 1 T = T until 1 2 2 1 V = V and the pressure is p2. It is then further compressed at constant pressure 3 2 p = p until 1 3 2 2V = V . From the ideal-gas law, 2 2 1 1 2 1 pV pV T T = 1 2 ( ) 1 2 1 1 2 1 2 1 p p V T 2.0 atm V 1 4.0 atm V T V . = = = Note that 2 1 T = T = 400 K. Using the ideal-gas law once again, 3 3 2 2 3 2 p V p V T T = ( )1 3 3 2 2 3 2 2 2 2 400 K 1 200 K 73 C T T V p V V p V . = = = = - The final pressure and temperature are 4.0 atm and 73C. (b) 16.64. Model: Assume that the nitrogen gas is an ideal gas. Solve: (a) The molar mass of N2 gas is 28 g/mol. The number of moles is n = (5 g)/(28 g /mol) = 0.1786 mol. The initial conditions are 1 p = 3.0 atm and 1 T = 293 K. We use the ideal gas law to find the initial volume as follows: ( )( )( ) 1 3 3 3 3 1 1 0.1786 mol 8.31 J mol K 293 K 1.430 10 m 1430 cm 1400 cm 3.0 atm 101,300 Pa atm V nRT p = = = - = An isobaric expansion until the volume triples results in 3 2 1 V = 3V = 4290 cm . (b) After the expansion, 2 2 1 1 2 2 2 1 1 1 2 1 1 1 p V pV T p V T 1 3 T 3T 879 K 606 C T T pV = . = = = = = (c) A constant volume decrease at 3 3 2 V =V = 4290 cm back to 1 3 1 3 2 T = T = T results in the following: 3 3 2 2 3 2 3 2 2 3 2 2 3 1 1 1 3.0 atm 1.0 atm 3 3 p V p V p T V p p T T TV = . = = = = (d) An isothermal compression at 4 3 T = T back to the initial volume 1 4 1 3 3 V =V = V results in the following: 4 4 3 3 4 3 4 3 1 3 4 3 3 4 3 p V p V p T V p 1 1 p 3 1.0 atm 3.0 atm T T TV = . = = = = (e) 16.65. Solve: (a) A gas is compressed isothermally from a volume 300 cm3 at 2 atm to a volume of 100 cm3. What is the final pressure? (b) (c) The final pressure is p2 = 6 atm. 16.66. Solve: (a) A gas at 400C and 500 kPa is cooled at a constant volume to a pressure of 200 kPa. What is the final temperature in C? (b) (c) The final temperature is T2 = -3.8C. 16.67. Solve: (a) A gas expands at constant pressure from 200 cm3 at 50C until the temperature is 400C. What is the final volume? (b) (c) The final volume is 3 V2 = 417 cm . 16.68. Solve: (a) 0.12 g of neon gas at 2.0 atm and 100 cm3 expands isobarically to twice its initial volume. What is the final temperature of the gas? (b) (c) The number of moles of neon is mol 0.12 g 0.006 mol 20 g n M M = = = The initial temperature is ( )( ) ( )( ) 5 43 1 1 1 2 1.013 10 Pa 1.0 10 m 406 K 0.006 mol 8.31 J mol K T pV nR - = = = The final temperature is 2 2 1 ( ) 2 1 1 1 1 T p V T p 2V 406 K 812 K p V p V = = = 16.69. Model: Assume that the compressed air is an ideal gas. Solve: (a) Because the piston is floating in equilibrium, Fnet = ( p1 - patoms )A - w = 0 N where the pistons cross-sectional area A =p (r2) =p (0.050 m)2 = 7.85410-3 m2 and the pistons weight w = (50 kg)(9.8) = 490 N. Thus, 5 5 1 atmos 3 2 490 N 1.013 10 Pa=1.637 10 Pa 7.854 10 m p w p A - = + = + Using the ideal-gas equation 1 1 1 , pV = nRT 5 1 (1.63710 Pa)Ah = (0.12 mol)(8.31 J/mol K)[(273 + 30) K] With the value of A given above, this equation yields 1 h = 0.235 m = 23.5 cm. (b) When the temperature is increased from 1 T = 303 K to 2 T = (303 +100) K = 403 K, the volume changes from 1 1 V = Ah to 2 2 V = Ah at a constant pressure 2 1p = p . From the before-and-after relationship of the ideal gas: 1 1 2 2 1 2 p (Ah ) p (Ah ) T T = 1 2 ( ) ( ) 2 1 2 1 1 403 K 0.235 m 0.313 m 31.3 cm 303 K h p T h p T . = = . . = = . . . . Thus, the piston moves 2 1 h - h = 7.8 cm. 16.70. Model: The air in the diving bell is an ideal gas. Visualize: Solve: (a) Initially p1 = p0 (atmospheric pressure), 1 V = AL, and 1 T = 293 K. When the diving bell is submerged to d =100m at the bottom edge, the water comes up height h inside. The volume is 2 V = A(L - h) and the temperature is 2 T = 283 K. Like a barometer, the pressure at points a and b must be the same. Thus 2 0 p + . gh = p + . gd, or 2 0 p = p + . g(d - h). Using the before and after relationship of an ideal gas, 1 1 2 2 0 0 1 2 [ ( )]( ) 293 K 283 K pV pV pAL p gd h AL h T T + . - - = . = Multiplying this out gives the following quadratic equation for h: 2 0 0 [ ( )] 1 283 0 293 . gh - p + . g d + L h + .. - .. p L + . gLd = . . Inserting the known values (using . =1030 kg/m3 for seawater) and dividing by . g gives h2 -113.04h + 301.03 = 0.h =110 m or 2.7 m The first solution is not physically meaningful, so the water rises to height h = 2.7 m. (b) To expel all the water, the air pressure inside the bell must be increased to match the water pressure at the bottom edge of the bell, d =100 m. The necessary pressure is 3 2 0 p = p + . gd =101,300 Pa + (1030 kg/m )(9.8 m/s )(100 m) =1111 kPa =10.96 atm 11 atm 16.71. Model: Assume the trapped air to be an ideal gas. Visualize: Initially, as the pipe is touched to the water surface and the gas inside is thus closed off from the air, the pressure 1 atmos 1 p = p = atm and the volume is 1 1 V = L A, where A is the cross-sectional area of the pipe. By pushing the pipe in slowly, the gas temperature in the pipe remains the same as the water temperature. Thus, this is an isothermal compression of the gas with 2 1T = T . Solve: From the ideal-gas law, ( ) 2 2 1 1 2 2 1 1 2 2 atmos 1 2 atmos 1 2 p V = pV . p L A = p L A. p L = p L . p = p L L As the pipe is pushed down, the increasing water pressure pushes water up into the pipe, compressing the air. In equilibrium, the pressure at points a and b, along a horizontal line, must be equal. (This is like the barometer. If the pressures at a and b werent equal, the pressure difference would cause the liquid level in the pipe to move up or down.) The pressure at point a is just the gas pressure inside the pipe: a 2p = p . The pressure at point b is the pressure at depth L2 in water: b atmos 2p = p + . gL . Equating these gives 2 atmos 2 p = p = . gL Substituting the expression for p2 from the ideal-gas equation above, the pressure equation becomes atmos 1 atmos 2 2 p L p gL L = + . 2 2 atmos 2 atmos 1 0 .. gL + p L - p L = This is a quadratic equation for L2 with solutions ( )2 atmos atmos atmos 1 2 4 2 p p gp L L g . . - + = Length has to be a positive quantity, so the one physically acceptable solution is ( ) ( )( )( )( ) ( )( ) 2 3 2 2 3 2 101,300 Pa 101,300 Pa 4 1000 kg m 9.8 m s 101,300 Pa 3.0 m 2.4 m 2 1000 kg m 9.8 m s L - + + = = 16.72. Model: The gas in the cylinder is assumed to be an ideal gas. Solve: The gas at T1 = 20C = 293 K has a pressure 1 p =1 atm and a volume 1 0V = AL . (The pressure has to be 1 atm to balance the external force of the air on the piston.) At a temperature 2 T =100C = 373 K, its volume becomes 2 0 V = A(L + .x) and its pressure increases to 2 1 1 p p F p k x A A . = + = + where F = k.x is the spring force on the piston. Using the before-and-after relationship of an ideal-gas equation, ( )( ) ( )( ) 1 1 2 2 1 0 1 0 1 2 1 2 2 1 0 1 10 1 0 1 1 pV p V p L A p k x A L x A T T T T T p k x A L x k x x T pL pA L + . + . = . = + . + . . . .. . . . = = . + .. + . . .. . L0 can be obtained from the ideal-gas law as follows: ( )( )( ) ( )( ) 1 1 0 5 42 1 1 0.004 mol 8.31 J mol K 293 K 0.0961 m 9.61 cm 1.013 10 Pa 10 10 m L V nRT A p A - . . = = . . = = = . . Substituting this value of L0 and the values for p1, T1, T2, and k, the above equation can be simplified to 154.02.x2 + 25.210.x - 0.2730 = 0..x = 0.0102 m =1.02 cm The spring is compressed by 1.0 cm. 16.73. Model: The gas in containers A and B is assumed to be an ideal gas. Solve: (a) The number of moles of gas in containers A and B can be expressed as follows: ( ) ( ) 5 A A A A A A 1.0 10 Pa 333.3 300 K p V V V n RT R R = = = ( )( ) ( ) 5 B B A A B B 5.0 10 Pa 4 5000 400 K p V V V n RT R R = = = where nA and nB are expressed in moles. Let A n' and B n' be the number of moles after the valve is opened. Since the total number of moles is the same, A A B A B n n 5333.3V n n R + = = ' + ' The pressure is now the same in both containers: A A B B A B A B p n RT p n RT V V ' ' ' = = ' = B A A A B B A B A B A 400 K 3 300 K 4 n n T V n V n n T V V . ' = ' = ' . . . ' = ' . . . . Solving the above equations, A A A A 5333.3V n 3n 4n R = ' + ' = ' A A n 1333.3V R . ' = A B n 4000V R . ' = The new pressure is A A A 5 A A A A p n RT 1333.3 V RT 4.0 10 Pa V V R ' . . ' = = . . = . . B B A 5 B B B B p n RT 4000 V RT 4.0 10 Pa V V R ' . . ' = = . . = . . Gas flows from B to A until the pressure is 4.0105 Pa. (b) The change is irreversible because opening the valve is like breaking a membrane. It is not a quasi-static process. 16-1 16.74 Model: Assume the gas in the left chamber is ideal. Visualize: For the piston to be in equilibrium (Fx )net = 0. The initial volume is V0 = L0A. Solve: (a) The force to the right on the piston by the gas in the left chamber is simply 0 p A.. The magnitude of the force to the left on the piston by the spring is k(.L). 0 net 0 ( ) ( ) x p A F p A k L Lk = - . .. = (b) When the piston is moved x to the right the spring is compressed a total of .L + x and the pressure in the left chamber changes to 1 p according to the ideal gas law. When the temperature doesnt change the ideal gas equation reduces to 0 0 1 1 p V = pV , where 0 0 V = L A and 1 0 V = (L + x)A . Therefore 0 0 1 0 ( ) p p L L x = + Now, similar to part (a) above, net 1 ( ) ( ) xF = p A- k .L + x Substitute in .L from part (a) and 1 p from above. 0 0 0 net 0 ( ) x ( ) F p L A k p A x L x k = - . + . + . . . . Now it is time to apply the binomial approximation: 1 0 0 0 (L + x)- 1/ L (1- x / L ) for 0x << L . ( ( )) ( ) net 0 0 0 0 0 0 0 0 0 0 0 ( ) ( ) 1/ 1 / / / xF pL L xL A pAkx p A p Ax L p A kx p A L k x = - - - = - - - = - + This is clearly a linear restoring force of the form of Hookes law where the usual k is replaced with 0 0 ( p A/ L + k) . That is, the gas in the left chamber is increasing the k of the system (by 0 0 p A/ L ) and making it act like a stiffer spring. (c) Since we have a Hookes law for which we know the modified k, we simply replace k in the period equation for harmonic oscillators with the modified constant. 0 0 2 / T M p A L k = p + Assess: This is simple harmonic motion when the approximation 0 x << L is valid. 17.1. Model: Assume the gas is ideal. The work done on a gas is the negative of the area under the pV curve. Visualize: The gas is compressing, so we expect the work to be positive. Solve: The work done on the gas is ( ) ( ( 3 )( )) ( 6 3 )( 5 ) area under the curve 200 cm 200 kPa 200 10 m 2.0 10 Pa 40 J W pdV pV - = - = - = - - = = . Assess: The area under the curve is negative because the integration direction is to the left. Thus, the environment does positive work on the gas to compress it. 17.2. Model: Assume the gas is ideal. The work done on a gas is the negative of the area under the pV curve. Visualize: The gas is expanding, so we expect the work to be negative. Solve: The area under the pV curve is the area of the rectangle and triangle. We have ( 6 3 )( 3 ) 1 ( 6 3 )( 3 ) 2 20010- m 20010 Pa + 20010- m 20010 Pa = 60 J Thus, the work done on the gas is W = -60 J. Assess: The environment does negative work on the gas as it expands. 17.3. Model: Assume the gas is ideal. The work done on a gas is the negative of the area under the pV curve. Solve: The work done on gas in an isobaric process is W = - p.V = - p(Vf -Vi ) Substituting into this equation, ( 3 )( ) 1 1 80 J = - 20010 Pa V - 3V 4 3 3 i .V = 2.010- m = 200 cm Assess: The work done to compress a gas is positive. 17.4. Model: Helium is an ideal gas that undergoes isobaric and isothermal processes. Solve: (a) Since the pressure ( pi = pf = p) is constant the work done is ( ) ( )( )( )( ) i on gas f i f i i 3 3 3 ( ) 1000 cm 2000 cm 0.10 mol 8.31 J mol K 573 K 240 J 2000 cm W p V p V V nRT V V V = - . = - - = - - - = - = (b) For compression at a constant temperature, ( ) ( )( )( ) on gas f i 6 3 6 3 ln 0.10 mol 8.31 J mol K 573 K ln 1000 10 m 330 J 2000 10 m W nRT VV - - = - . . = - . . = . . (c) For the isobaric case, i 5 i p nRT 2.38 10 Pa V = = For the isothermal case, 5 i p = 2.3810 Pa and the final pressure is f 5 f f p nRT 4.76 10 Pa V = = 17.5. Visualize: Solve: Because W = -. p dV and this is an isochoric process, W = 0 J. The final point is on a higher isotherm than the initial point, so f iT > T . Heat energy is thus transferred into the gas (Q > 0) and the thermal energy of the gas increases th f th i (E > E ) as the temperature increases. 17.6. Visualize: Solve: Because this is an isobaric process W = -. pdV = - p(Vf -Vi ) . Since Vf is smaller than Vi, W is positive. That is, the gas is compressed. Since the final point is on a lower isotherm than the initial point, f iT < T . In other words, the thermal energy decreases. For this to happen, the heat energy transferred out of the gas must be larger than the work done. 17.7. Visualize: Solve: Because the process is isothermal, .Eth = Eth f - Eth i = 0 J. According to the first law of thermodynamics, th .E =W +Q. This can only be satisfied if W = -Q. W is positive because the gas is compressing, hence Q is negative. That is, heat energy is removed from the gas. 17.8. Visualize: Solve: This is an adiabatic process of gas compression so no heat energy is transferred between the gas and the environment. That is, Q = 0 J. According to the first law of thermodynamics, the work done on a gas in an adiabatic process goes entirely to changing the thermal energy of the gas. The work W is positive because the gas is compressed. 17.9. Solve: The first law of thermodynamics is .Eth =W +Q.-200 J = 500 J +Q = Q.Q = -700 J The negative sign means a transfer of energy from the system to the environment. Assess: Because W > 0 means a transfer of energy into the system, Q must be less than zero and larger in magnitude than W so that th f th i E < E . 17.10. Solve: This is an isobaric process. W > 0 because the gas is compressed. This transfers energy into the system. Also, 100 J of heat energy is transferred out of the gas. The first law of thermodynamics is 5 63 th .E =W +Q = - p.V +Q = -(4.010 Pa)(200 - 600)10- m -100 J = 60 J Thermal energy increases by 60 J. 17.11. Model: The removal of heat from the ice reduces its thermal energy and its temperature. Solve: The heat needed to change an objects temperature is Q = Mc.T. The mass of the ice cube is 3 3 M = .iceV = (920 kg/m )(0.06 0.06 0.06)m = 0.1987 kg The specific heat of ice from Table 17.2 is ice c = 2090 J/kg K, so Q = (0.199 kg)(2090 J/kg K)(243 K - 273 K) = -12,460 J 12,000 J Thus, the energy removed from the ice block is 12,000 J. Assess: The negative sign with Q means loss of energy. 17.12. Model: The spinning paddle wheel does work and changes the waters thermal energy and its temperature. Solve: (a) The temperature change is .T = Tf -Ti = 25C - 21C = 4 K. The mass of the water is M = (20010-6m3)(1000 kg/m3) = 0.20 kg The work done is th water W = .E = Mc .T = (0.20 kg)(4190 J/kg K)(4 K) = 3350 J 3400 J (b) Q = 0. No energy is transferred between the system and the environment because of a difference in temperature. 17.13. Model: Heating the mercury at its boiling point changes its thermal energy without a change in temperature. Solve: The mass of the mercury is M = 20 g = 2.010-2 kg, the specific heat mercury c =140 J/kg K, the boiling point b T = 357C, and the heat of vaporization 5 v L = 2.9610 J/kg. The heat required for the mercury to change to the vapor phase is the sum of two steps. The first step is 2 1 mercury Q = Mc .T = (2.010- kg)(140 J/kg K)(357C - 20C) = 940 J The second step is 2 5 2 V Q = ML = (2.010- kg)(2.9610 J/kg) = 5920 J The total heat needed is 6860 J. 17.14. Model: Heating the mercury changes its thermal energy and its temperature. Solve: (a) The heat needed to change the mercurys temperature is Hg ( )( ) Hg 100 J 35.7 K 35 C 0.020 kg 140 J kg K Q Mc T T Q Mc = . .. = = = (b) The amount of heat required to raise the temperature of the same amount of water by the same number of degrees is water Q = Mc .T = (0.020 kg)(4190 J/kg K)(35.7 K) = 3000 J Assess: Q is directly proportional to cwater and the specific heat for water is much higher than the specific heat for mercury. This explains why Qwater > Qmercury. 17.15. Model: Changing ethyl alcohol at 20C to solid ethyl alcohol at its melting point requires two steps: lowering its temperature from 20C to -114C, then changing the ethyl alcohol to its solid phase at -114C. Solve: The change in temperature is -114C - 20C = -134C = -134 K. The mass is M = .V = (789 kg/m3 )(20010-6m3 ) = 0.1578 kg The heat needed for the two steps is 4 1 alcohol Q = Mc .T = (0.1578 kg)(2400 J/kg K)(-134 K) = -5.0710 J 5 4 2 f Q = -ML = -(0.1578 kg)(1.0910 J/kg) = -1.7210 J The total heat required is 4 4 1 2 Q = Q +Q = -6.7910 J -6.810 J Thus, the minimum amount of energy that must be removed is 6.8104 J. Assess: The negative sign with Q indicates that 6.8 104 J will be removed from the system. 17.16. Model: Changing solid lead at 20C to liquid lead at its melting point (Tm = 328C) requires two steps: raising the temperature to Tm and then melting the solid at Tm to a liquid at Tm. Solve: The equation for the total heat is Q = Q1 +Q2 .1000 J = Mclead (Tf -Ti ) + MLf .1000 J = M(128 J/kg K)(328 - 20) K + M(0.25105 J/kg) ( ) 1000 J 15.5 g 64,424 J/kg .M = = The maximum mass of lead you can melt with 1000 J of heat is 15.5 g. 17.17. Model: We have a thermal interaction between the copper pellets and the water. Solve: The conservation of energy equation Qc +Ww = 0 is c c f w w f M c (T -300C) +M c (T - 20C) = 0 J Solving this equation for the final temperature Tf gives c c w w f c c w w (300 C) (20 C) (0.030 kg)(385 J/kg K)(300 C) (0.10 kg)(4190 J/kg K)(20 C) 28 C (0.030 kg)(385 J/kg K) (0.10 kg)(4190 J/kg K) T M c M c M c M c + = + + = = + The final temperature of the water and the copper is 28C. 17.18. Model: We have a thermal interaction between the copper block and water. Solve: The conservation of energy equation Qcopper +Qwater = 0 J is copper copper f i copper water water f i water M c (T -T ) + M c (T -T ) = 0 J Both the copper and the water reach the common final temperature Tf = 25.5C. Thus 3 3 3 copper M (385 J/kg K)(25.5C - 300C) + (1.0010- m )(1000 kg/m )(4190 J/kg K)(25.5C - 20C) = 0 J copper .M = 0.218 kg 17.19. Model: We have a thermal interaction between the thermometer and the water. Solve: The conservation of energy equation Qthermo +Qwater = 0 J is thermo thermo f i thermo water water f i water M c (T - (T ) ) +M c (T - (T ) ) = 0 J The thermometer slightly cools the water until both have the same final temperature f T = 71.2C. Thus 6 3 3 i water i water i water (0.050 kg)(750 J/kg K)(71.2 C 20.0 C) (200 10 m )(1000 kg/m )(4190 J/kg K)(71.2 C ) 1920 J 838 (J/K)(71.2 C ) 0 J 73.5 C T T T - + - - = + - = . = Assess: The thermometer reads 71.2C for a real temperature of 73.5C. This is reasonable. 17.20. Model: We have a thermal interaction between the aluminum pan and the water. Solve: The conservation of energy equation QAl +Qwater = 0 J is Al Al f i A1 water water f i water M c (T -T ) + M c (T -T ) The pan and water reach a common final temperature Tf = 24.0C 3 3 3 i Al i Al (0.750 kg)(900 J/kg K)(24.0 C ) (10.0 10 m )(1000 kg/m )(4190 J/kg K)(24.0 C 20.0 C ) (675.0 J/K)(24.0 C ) 167,600 J 0 J T T - + - - = - + = i Al .T = 272C = [(272)(9/5) + 32]F = 522F 17.21. Model: We have a thermal interaction between the metal sphere and the mercury. Solve: The conservation of energy equation Qmetal + QHg = 0 J is metal M cmetal (Tf -Ti metal ) + MHgcHg (Tf -Ti Hg ) = 0 J The metal and mercury reach a common final temperature Tf = 99.0C. Thus 6 3 3 metal (0.500 kg)c (99C - 300C) + (30010- m )(13,600 kg/m )(140 J/kg K)(99C - 20C) = 0 J We find that metal c = 449 J kg K. The metal is iron. 17.22. Model: Use the models of isochoric and isobaric heating. Note that the change in temperature on the Kelvin scale is the same as the change in temperature on the Celsius scale. Solve: (a) The atomic mass number of argon is 40. That is, Mmol = 40 g/mol. The number of moles of argon gas in the container is mol 1.0 g 0.025 mol 40 g mol n M M = = = The amount of heat is V Q = nC .T = (0.025 mol)(12.5 J/mol K)(100C) = 31.25 J 31 J (b) For the isobaric process P Q = nC .T becomes 31.25 J = (0.025 mol)(20.8 J/mol K).T ..T = 60C 17.23. Model: The heating processes are isobaric and isochoric. O2 is a diatomic ideal gas. Solve: (a) The number of moles of oxygen is mol 1.0 g 0.03125 mol 32 g mol n M M = = = For the isobaric process, p Q = nC .T = (0.03125 mol)(29.2 J/mol K)(100C) = 91.2 J 91 J (b) For the isochoric process, V Q = nC .T = 91.2 J = (0.03125 mol)(20.9 J/mol K).T ..T =140C 17.24. Model: The heating is an isochoric process. Solve: The number of moles of helium is mol 2.0 g 0.50 mol 4 g mol n M M = = = For the isochoric processes, ( )( ) He V Q = nC .T = 0.50 mol 12.5 J mol K .T ( ) O2 V 20.9 J mol K 32 g mol Q nC T M T . . = . = . . . . . Because He O2 Q = Q , (0.50 mol)(12.5 J mol K) (20.9 J mol K) 32 g mol . M . =. . . . .M = 9.6 g 17.25. Model: The gas is an ideal gas that is subjected to an adiabatic process. Solve: (a) For an adiabatic process, f f i i p V. = pV. f i i f p V p V . . . . =. . . . 2.5 (2.0). . = ln(2.5) 1.32 ln(2.0) .. = = (b) Equation 17.39 for an adiabatic process is 1 1 f f i i T V. - = TV. - ( ) ( ) 1 f i 1.32 1 0.32 i f T V 2.0 2.0 1.25 T V . - . . - . = . . = = = . . 17.26. Model: We assume the gas is an ideal gas and . =1.40 for a diatomic gas. Solve: Using the ideal-gas law, ( )( )( ) ( ) i 3 3 i 5 i 0.10 mol 8.31 J mol K 423 K 1.157 10 m 3 1.013 10 Pa V nRT p = = = - For an adiabatic process, i i f f pV. = p V. ( ) 1 1 1.40 i 3 3 i 3 3 f i f i 1.157 10 m 1.9 10 m 0.5 V V p p p p . - - . . . . . = . . = . . = . . . . To find the final temperature, we use the ideal-gas law once again as follows: ( ) 3 3 f f i f i 3 3 i i i 423 K 0.5 1.90 10 m 346.9 K 74 C 1.157 10 m T T p V p p V p - - . .. . = = . .. . = . .. . 17.27. Model: The O2 gas has . =1.40 and is an ideal gas. Solve: (a) For an adiabatic process, pV. remains a constant. That is, i i f f pV. = p V. ( ) 1.40 i i f i f i 3.0 atm 2 p p V V V V . . . . . . = . . = . . . . . . ( ) 1.40 3.0 atm 1 1.14 atm 1.1 atm 2 = . . = . . . . (b) Using the ideal-gas law, the final temperature of the gas is calculated as follows: i i f f i f pV p V T T = f f ( ) i f i i i i 423 K 1.14 atm 2 321.5 K 3.0 atm T T p V V p V V . .. . . = = . .. . = . .. . 48C 17.28. Visualize: We are asked for the heat-loss rate which is given by Equation 17.48: Q k A T t L = D D We are given A =10 m14 m=140 m2 , L = 0.12 m, and DT = 22C- 5C =17C =17K. We look up the thermal conductivity of concrete in Table 17.5: k = 0.8W/mK. Solve: ( ) ( ) 140 m2 0 8W/m K 17 K 16 kW 0 12 m Q k A T . t L . . . = = . . = . . D D Assess: The answer is in a reasonable range. The heat loss could be reduced with thicker concrete or adding a layer of a different material. 17.29. Visualize: To determine the material we will solve for k in Equation 17.48: Q k A T t L = D D We are given L = 0.20m and DT =100 K. We compute A = r2 = (0.010m)2 = 3.141024m2 p p . We also convert the heat conduction to watts. 4 5 104 J/h 1 h 12 5 W 3600 s . .. .. = . . . Solve: Solve the equation for k. ( ) 4 2 1 12 5 W 0 20m 1 80 W/m K 3 14 10 m 100K k Q L . . t A T . . . . .. . = . . = . .. . = . . . .. . 2 D D Look this up in Table 17.5; the value corresponds to iron. Assess: We are grateful that our answer was one of the entries in the table. 17.30. Model: Assume the lead sphere is an ideal radiator with e =1 . Also assume that the highest temperature the solid lead sphere can have is the melting temperature of lead. Visualize: Use Equation 17.49. First look up the melting temperature of lead in Table 17.3: m T = 328C = 601 K. Then compute the surface area of the sphere: A = 4pR2 = 4p(0.050 m)2 = 0.0314 m2 * Solve: Q e AT 4 (1)(5.67 10 8 W/m2 K4 )(0.0314 m2 )(601 K)4 230 W t = = 2 = s 3 ? D Assess: If the sphere were larger it could radiate more power without melting. 17.31. Model: We will ignore the bottom of the head and model it with just the cylindrical sides and top, all covered in skin. As instructed, assume an emissivity of e = 0.95 . Visualize: The area of the cylinder is 2 2 side top A = A + A = 2prh + pr = 2p(0.10 m)(0.20 m) + p(0.10 m) = 0.157 m2. Solve: Use Equation 17.50. net 4 4 8 2 4 2 4 4 0 Q e A(T T ) (0.95)((5.67 10 W/m K )(0.157 m ))((308 K) (278 K) ) 26 W t = - = 2 - = s 3 ? D Assess: This is a significant amount of the heat lost by the body. Wearing a hat can help a lot. 17.32. Model: There are various steps to the problem. We must (1) raise the the temperature of the ice to 0C, (2) melt the ice to liquid water at 0C, (3) raise the water temperature to 100C, (4) boil the water to produce steam at 100C , and (5) raise the temperature of the steam to 200C. Solve: The heat needed for each step is ( )( )( ) 1 ice ice 0 Q = Mc DT = .0050 kg 2090 J/kg?K 20 K = 209 J<210 J ( )( 5 ) 2 f Q = ML = 0.0050 kg 3.33310 J/kg =1665 J<1670 J ( )( )( ) 3 water water Q = Mc DT = 0.0050 kg 4190 J/kg?K 100 K = 2095 J<2100 J ( )( 5 ) 4 v Q = ML = 0.0050 kg 22.6310 J/kg =11,300 J ( )( )( ) 5 steam steam Q = Mc DT = 0.0050 kg 2009 J/kg?K 100 K =1005 J<1000 J The total heat is 1 2 3 4 5 Q = Q +Q +Q +Q +Q =16,300 J Assess: It is interesting to note which step required the most heat: The boiling to change the liquid to the gas is by far the largest contributor to the total. 17.33. Solve: The area of the garden pond is A =p (2.5 m)2 =19.635 m2 and its volume is V = A(0.30 m) = 5.891 m3. The mass of water in the pond is M = .V = (1000 kg m3 )(19.635 m3 ) = 5891 kg The water absorbs all the solar power which is (400 W m2 )(19.635 m2 ) = 7854 W This power is used to raise the temperature of the water. That is, ( ) ( )( )( ) water Q = 7854 W .t = Mc .T = 5891 kg 4190 J kg K 10 K ..t = 31,425 s 8.7 h 17.34. Model: The potential energy of the bowling ball is transferred into the thermal energy of the mixture. We assume the starting temperature of the bowling ball to be 0C. Solve: The potential energy of the bowling ball is ( )( 2 ) ( 2 ) g ball U = M gh = 11 kg 9.8 m s h = 107.8 kg m s h This energy is transferred into the mixture of ice and water and melts 5 g of ice. That is, ( 2 ) th w f 107.8 kg m s h = .E = M L ( )( ) ( ) 5 2 0.005 kg 3.33 10 J kg 15.4 m 107.8 kg m s h . = = 17.35. Model: Heating the water and the kettle raises the temperature of the water to the boiling point and raises the temperature of the kettle to 100C. Solve: The amount of heat energy from the electric stoves output in 3 minutes is Q = (2000 J s)(3 60 s) = 3.6105 J This heat energy heats the kettle and brings the water to a boil. Thus, water water kettle kettle Q = M c .T + M c .T Substituting the given values into this equation, 5 ( )( ) ( )( )( ) water 3.610 J = M 4190 J kg K 100C - 20C + 0.750 kg 449 J kg K 100C - 20C water .M = 0.994 kg The volume of water in the kettle is water 3 3 3 3 3 water 0.994 kg 0.994 10 m 994 cm 990 cm 1000 kg m V M . = = = - = Assess: 1 L = 103 cm3, so V 1 L. This is a reasonable volume of water. 17.36. Model: Each cars kinetic energy is transformed into thermal energy. Solve: For each car, 1 2 2 th car K = Mv = .E = Mc .T 2 car 2 T v c . . = Assume ccar = ciron. The speed of the car is 80 km hr = 80 1000 m 22.22 m s 3600 s v = = ( ) ( ) 2 22.22 m s 0.55 C 2 449 J kg K ..T = = Assess: Notice the answer is independent of the cars mass. 17.37. Model: There are three interacting systems: aluminum, copper, and ethyl alcohol. Solve: The aluminum, copper, and alcohol form a closed system, so Q = QAl + QCu + Qeth = 0 J. The mass of the alcohol is ( 3 )( 6 3 ) eth M = .V = 790 kg m 5010- m = 0.0395 kg Expressed in terms of specific heats and using the fact that .T = Tf Ti, the Q = 0 J condition is Al Al Al Cu Cu Cu eth eth eth M c .T + M c .T + M c .T = 0 J Substituting into this expression, ( )( )( ) ( )( )( ) ( )( )( ) ( )( ) 0.010 kg 900 J kg K 298 K 473 K 0.020 kg 385 J kg K 298 K 0.0395 kg 2400 J kg K 298 K 288 K 1575 J 7.7 J K 298 948 J 0 J T T - + - + - =- + -+ = .T = 216.6 K = -56.4C -56C 17.38. Model: There are two interacting systems: aluminum and ice. The system comes to thermal equilibrium in four steps: (1) the ice temperature increases from -10C to 0C, (2) the ice becomes water at 0C, (3) the water temperature increases from 0C to 20C, and (4) the cup temperature decreases from 70C to 20C. Solve: The aluminum and ice form a closed system, so Q = Q1 + Q2 + Q3 + Q4 = 0 J. These quantities are ( )( )( ) ( )( ) ( )( )( ) ( )( ) ( ) 1 iceice 5 2 ice f 3 icewater 4 AlAl Al Al 0.100 kg 2090 J kg K 10 K 2090 J 0.100 kg 3.33 10 J kg 33,300 J 0.100 kg 4190 J kg K 20 K 8380 J 900 J kg K 50 K 45,000 J kg Q M c T Q M L Q M c T Q M c T M M = . = = = = = = . = = = . = - =- The Q = 0 J equation now becomes 43,770 J (45,000 J/kg)MAl = 0 J The solution to this is MAl = 0.973 kg. 17.39. Model: There are three interacting systems: metal, aluminum, and water. Solve: The metal, aluminum container, and water form a closed system, so Qm + QAl + Qw = 0 J, where Qm is the heat transferred to the metal sample. This equation can be written: Mmcm.Tm + MAlcAl.TAl + Mwcw.T = 0 J Substituting in the given values, ( ) ( ) ( )( )( ) ( )( )( ) m m 0.512 kg 351 K 288 K 0.100 kg 900 J kg K 351 K 371 K 0.325 kg 4190 J kg K 351 K 371 K 0 J 900 J kg K c c - + - + - =.= From Table 17.2, we see that this is the specific heat of aluminum. 17.40. Solve: For a monatomic gas, the molar specific heat at constant volume is CV =12.5 J mol K . From Equation 17.22, 12.5 J mol K mol 625 J kg K 1000 g kg = M mol .M = 20 g mol The gas is therefore neon. 17.41. Model: Heating the water raises its thermal energy and its temperature. Solve: A 5.0 kW heater has power P = 5000 W. That is, it supplies heat energy at the rate 5000 J/s. The heat supplied in time .t is Q = 5000.t J. The temperature increase is .TC = (5/9).TF = (5/9)(75) = 41.67C. Thus w 5000 Q = .t J = Mc .T = (150 kg)(4190 J/kg K)(41.67C)..t = 5283 s 87 min Assess: A time of 1.5 hours to heat 40 gallons of water is reasonable. 17.42. Model: Heating the material increases its thermal energy. Visualize: Heat raises the temperature of the substance from 40C to 20C, at which temperature a solid to liquid phase change occurs. From 20C, heat raises the liquids temperature up to 40C. Boiling occurs at 40C where all of the liquid is converted into the vapor phase. Solve: (a) In the solid phase, .Q = Mc.T 1 20 kJ 1 2000 J kg K 20 K 0.50 kg c Q T M . . . . .. . . = . . = . .. . = . . . . .. . (b) In the liquid phase, 1 1 80 kJ 2667 J kg K 0.50 kg 60 K c Q M T . . . . .. . = . . = . .. . = . . . . .. . (c) The melting point m T = -20C and the boiling point b T = +40C . (d) The heat of fusion is 4 f 20,000 J 4.0 10 J kg 0.50 kg L Q M = = = The heat of vaporization is 5 v 60,000 J 1.2 10 J kg 0.50 kg L Q M = = = 17.43. Model: The liquefaction of the nitrogen occurs in two steps: lowering nitrogens temperature from 20C to -196C, and then liquefying it at -196C. Assume the cooling occurs at a constant pressure of 1 atm. Solve: The mass of 1.0 L of liquid nitrogen is M = .V = (810 kg m3 )(10-3 m3 ) = 0.810 kg . This mass corresponds to mol 810 g 28.9 mols 28 g mol n M M = = = At constant atmospheric pressure, the heat to be removed from 28.93 mols of nitrogen is ( )( ) ( )( )( ) v P 0.810 kg 1.99 105 J kg 28.9 mols 29.1 J mol K 77 K 293 K 3.4 105 J Q = ML + nC .T = - + - = - 17.44. Model: There are two interacting systems: coffee (i.e., water) and ice. Changing the coffee temperature from 90C to 60C requires four steps: (1) raise the temperature of ice from -20C to 0C, (2) change ice at 0C to water at 0C, (c) raise the water temperature from 0C to 60C, and (4) lower the coffee temperature from 90C to 60C. Solve: For the closed coffee-ice system, Q = Qice +Qcoffee = (Q1 +Q2 +Q3 ) + (Q4 ) = 0 J ( )( ) ( ) 1 ice ice ice ice Q = M c .T = M 2090 J kg K 20 K = M 41,800 J kg ( ) 2 ice f ice Q = M L = M 330,000 J kg ( )( ) ( ) 3 ice water ice ice Q = M c .T = M 4190 J kg K 60 K = M 251,400 J kg ( 6 3 )( 3 )( )( ) 4 coffee coffee Q = M c .T = 30010- m 1000 kg m 4190 J kg K -30 K = -37,000 J The Q = 0 J equation thus becomes ( ) ice M 41,800 + 330,000 + 251,400 J kg - 37,710 J = 0 J ice .M = 0.061 kg = 61 g Assess: 61 g is the mass of approximately 1 ice cube. 17.45. Model: We have three interacting systems: the aluminum, the air, and the firecracker. The energy released by the firecracker raises the temperature of the aluminum and the air. We will assume that air is predominantly N2. Assume that the surrounding insulation is excellent. Solve: For the closed firecracker + air + aluminum system, energy conservation requires that Q = Qfirecracker + QAl + Qair = 0 J QAl = mAlcAl.T = (2.0 kg)(900 J/kg K)(3 K) = 5400 J air V V Q nC T pV C T RT = . = . . . . . . . ( )( ) ( )( ) ( )( ) 1.01 105 Pa 20 10 3m3 20.8 J/mol K 3 K 8.31 J/mol K 298 K - = = 51 J Thus, Qfirecracker = -QAl - Qair = 5450 J That is, 5500 J of energy are released on explosion of the firecracker. Assess: The negative sign with Qfirecracker means that the firecracker has lost energy. 17.46. Model: There are two interacting systems: the nuclear reactor and the water. The heat generated by the nuclear reactor is used to raise the water temperature. Solve: For the closed reactor-water system, energy conservation per second requires Q = Qreactor +Qwater = 0 J The heat from the reactor in .t = 1 s is 9 reactor Q = -2000 MJ = -2.010 J and the heat absorbed by the water is ( )( ) water water water water Q = m c .T = m 4190 J kg K 12 K 9 ( )( ) water .-2.010 J + m 4190 J kg K 12 K = 0 J 4 water .m = 3.9810 kg Each second, 3.98 104 kg of water is needed to remove heat from the nuclear reactor. Thus, the water flow per minute is 3 4 3 3 3.98 10 kg 60 s 1m 1 L s min 1000 kg 10- m = 2.4 106 L/min 17.47. Model: We have two interacting systems: the water and the gas. For the closed system comprised of water and gas to come to equilibrium, heat is transferred from one interacting system to the other. Solve: Energy conservation requires that Qair + Qwater = 0 J . ngasCV(Tf Ti gas) + mwaterc(Tf Ti water) = 0 J Using the ideal-gas law, ( )( ) ( )( ) 5 63 gas gas i gas gas 10 1.013 10 Pa 4000 10 m =1219 K 0.40 mol 8.31 J mol K p V T n R - = = The energy conservation equation with Ti water = 293 K becomes ( )( )( ) ( )( )( ) 3 0.40 mol 12.5 J mol K 1219 K 20 10 kg 4190 J kg K 293 K 0 J T T - - + - =f .T = 345 K We can now use the ideal-gas equation to find the final gas pressure. i i f f i f pV p V T T = f f i i p p T T . = 345 K (10 atm) 2.8 atm 1219 K = . . = . . . . 17.48. Model: These are isothermal and isobaric ideal-gas processes. Solve: (a) The work done at constant temperature is ( ) ( ) ( )( )( ) ( ) f f i i f i f i 1 3 ln ln ln 2.0 mol 8.31 J mol K 303 K ln 5500 J V V V V W pdV nRT dV nRT V V nRT V V V = - = - = - - = - = - = . . (b) The work done at constant pressure is ( ) ( )( )( ) f i i f i i i 2 3 3 2 22.0 mol 8.31 J mol K 303 K 3400 J 3 3 V V W p dV p V V p V V pV nRT = - = - - = - . - . = . . . . = = = . (c) For an isothermal process in which 1 f 3 i V = V , the pressure changes to f i p = 3p = 4.5 atm. 17.49. Model: This is an isothermal process. The work done is positive for a compression. Solve: For an isothermal process, W = -nRT ln(Vf Vi ) For the first process, ( 1 ) 2 W = 500 J = -nRT ln .nRT = 721.35 J For the second process, ( 1 ) ( ) ( 1 ) 10 10 W = -nRT ln .W = - 721.35 J ln = 1660 J 17.50. Visualize: Solve: (a) The gas exerts a force on the piston of magnitude Fgas on piston = pgasA ( ) ( )2 = 3 atm101,300 Pa atm ..p 0.080 m .. = 6100 N This force is directed toward the right. (b) The piston is in static equilibrium, so the environment must exert a force on the piston of equal magnitude environ on piston F = 6100 N but in the opposite direction, toward the left. (c) The work done by the environment is ( )( ) environ environ on piston environ on piston W = F .r = -F .x = - 6110 N 0.10 m = -610 J .. .. .. The work is negative because the force and the displacement are in opposite directions. (d) The work done by the gas is ( )( ) gas gas on piston gas on piston W = F .r = +F .x = 6110 N 0.10 m = 610 J .. .. This work is positive because the force and the displacement are in the same direction. (e) The first law of thermodynamics is W + Q = .Eth where W is Wenviron. So here W = -610 J and we find ( ) ( ) th Q = .E -W = 196 J - -610 J = 806 J Thus, 806 J of heat is added to the gas. 17.51. Model: This is an isobaric process. Visualize: Solve: (a) The initial conditions are p1 = 10 atm = 1.013 106 Pa, T1 = 50C = 323 K, V1 = pr2L1 = p(0.050 m)2 (0.20 m) = 1.57 10-3 m3. The gas is heated at a constant pressure, so heat and temperature change are related by Q = nCP.T. From the ideal gas law, the number of moles of gas is ( )( ) ( )( ) 6 33 1 1 1 1.013 10 Pa 1.57 10 m 0.593 mol 8.31 J mol K 323 K n pV RT - = = = The temperature change due to the addition of Q = 2500 J of heat is thus ( )( ) P 2500 J 203 K 0.593 mol 20.8 J mol K T Q nC . = = = The final temperature is T2 = T1 + .T = 526 K = 253C. (b) Noting that the volume of a cylinder is V = pr2L and that r doesnt change, the ideal gas relationship for an isobaric process is 2 1 2 1 2 ( ) 2 1 2 1 2 1 1 526 K 20 cm 33 cm 323 K V V L L L T L T T T T T = . = . = = = 17.52. Model: The process in part (a) is isochoric and the process in part (b) is isobaric. Solve: (a) Initially V1 = (0.20 m)3 = 0.0080 m3 = 8.0 L and T1 = 293 K. Helium has an atomic mass number A = 4, so 3 g of helium is n = M/Mmol = 0.75 mole of helium. We can find the initial pressure from the ideal-gas law: ( )( )( ) 1 1 3 1 0.75 mol 8.31 J mol K 293 K 228 kPa 2.25 atm 0.0080 m p nRT V = = = = Heating the gas will raise its temperature. A constant volume process has Q = nCV.T, so ( )( ) V 1000 J 107 K 0.75 mol 12.5 J mol K T Q nC . = = = This raises the final temperature to T2 = T1 + .T = 400 K. Because the process is isochoric, 2 1 2 ( ) 2 1 2 1 1 400 K 2.25 atm 3.1 atm 293 K p p p T p T T T = . = = = (b) The initial conditions are the same as part a, but now Q = nCP.T. Thus, ( )( ) P 1000 J 64.1 K 0.75 mol 20.8 J mol K T Q nC . = = = Now the final temperature is T2 = T1 + .T = 357 K. Because the process is isobaric, 2 1 2 ( 3 ) 3 2 1 2 1 1 357 K 0.0080 m 0.0097 m 9.7 L 293 K V V V T V T T T = . = = = = (c) 17.53. Model: Assume the air and the nitrogen are ideal gases. For the piston to be in equilibrium the forces on it must sum to zero. Also assume that the temperature doesnt change in the second part. Solve: The area is A = r2 = (0.040 m)2 = 5.0 1023 m2 p p 3 . ( 3 2 )( ) 3 1 1 V = Ah = 5.0 102 m 0.26 m = 0.0013 m 3 . (a) net in above F = p A - mg - p A = 0 N ( )( 2 ) ( )( 3 2 ) above in 3 2 5 1 kg 9 8 m/s 100 kPa 5 0 10 m 110 kPa 5 0 10 m mg p A . . . p A . + + = = = 2 2 3 3 (b) Since 1 2 T = T , then 1 1 2 2 pV = p V where 1 p is in p from above, and 2 p is the inside pressure after the new weight is added. Compute 2 p exactly as above replacing 5.1 kg with 5.1 kg + 3.5kg = 8.6kg. ( )( 2 ) ( )( 3 2 ) above 2 3 2 8 6 kg 9 8 m/s 100 kPa 5 0 10 m 117 kPa 5 0 10 m mg p A . . . p A . + + = = = 2 2 3 3 ( )( 3 ) 1 1 3 3 2 2 110 kPa 0 0013 m 1 22 10 m 117 kPa pV . V . p = = = 2 3 3 3 2 2 3 2 1 22 10 m 0 2447 m 24 cm 5 0 10 m h V . . A . = = = 2 2 3 3 < Assess: The result seems to be a reasonable number; we expected the piston to be a little lower than before due to the increased mass on it. 17.54. Model: This is an isothermal process. Solve: (a) The final temperature is T2 = T1 because the process is isothermal. (b) The work done on the gas is 2 2 1 1 1 2 1 1 ln V V V V W pdV nRT dV nRT V V V = -. = -. = - = -nRT1ln 2 (c) From the first law of thermodynamics .Eth = W + Q = 0 J because .T = 0 K. Thus, the heat energy transferred to the gas is 1 Q = -W = nRT ln 2 . 17.55. Model: The gas is an ideal gas and it goes through an isobaric and an isochoric process. Solve: (a) The initial conditions are p1 = 3.0 atm = 304,000 Pa and T1 = 293 K. Nitrogen has a molar mass Mmol = 28 g/mol, so 5 g of nitrogen gas has n = M/Mmol = 0.1786 mol. From this, we can find the initial volume: 1 ( )( )( ) 3 3 3 1 1 0.1786 mol 8.31 J mol K 293 K 1.430 10 m 1400 cm 304,000 Pa V nRT p = = = - The volume triples, so V2 = 3V1 = 4300 cm3. The expansion is isobaric (p2 = p1 = 3.0 atm), so 2 1 2 ( ) 2 1 2 1 1 V V T V T 3 293 K 879 K 606 C T T V = . = = = = (b) The process is isobaric, so ( )( )( ) P Q = nC .T = 0.1786 mol 29.1 J mol K 879 K - 293 K = 3000 J (c) The pressure is decreased at constant volume (V3 = V2 = 4290 cm3) until the original temperature is reached (T3 = T1 = 293 K). For an isochoric process, 3 2 3 ( ) 3 2 3 2 2 293 K 3.0 atm 1.0 atm 879 K p p p T p T T T = . = = = (d) The process is isochoric, so ( )( )( ) V Q = nC .T = 0.1786 mol 20.8 J mol K 293 K - 879 K = -2200 J So, 2200 J of heat was removed to decrease the pressure. (e) 17.56. Model: The gas is an ideal gas. Visualize: In the figure, call the upper left corner (on process A) 2 and the lower right corner (on process B) 3. Solve: The change in thermal energy is the same for any gas process that has the same .T. Processes A and B have the same .T, since they start and end at the same points, so (.Eth)A = (.Eth)B. The first law is then (.Eth)A = QA + WA = (.Eth)B = QB + WB . QA QB = WB WA In process B, work W = p.V = pi(2Vi Vi) = piVi is done during the isobaric process i . 3. No work is done during the isochoric process 3 . f. Thus WB = piVi. Similarly, no work is done during the isochoric process i . 2 of process A, but W = p.V = 2pi(2Vi Vi) = 2piVi is done during the isobaric process 2. f. Thus WA = 2piVi. Combining these, QA QB = WB WA = piVi (2piVi) = piVi 17.57. Model: The two processes are isochoric and isobaric. Solve: Process A is isochoric which means ( ) ( ) 1 f i f i f i f i i 3 i T T = p p .T = T p p = T 1 atm 3 atm = T From the ideal-gas equation, i i i T pV nR = ( )( ) ( )( ) 3 1.013 105 Pa 2000 10 6 m3 731.4 K 0.10 mol 8.31 J mol K - = = 1 f 3 i .T = T = 243.8 K f i .T -T = -487.6 K Thus, the heat required for process A is ( )( )( ) A V Q = nC .T = 0.10 mol 20.8 J mol K -487.6 K = -1000 J Process B is isobaric which means ( ) ( 3 3 ) f f i i f i f i i i T V = T V .T = T V V = T 3000 cm 1000 cm = 3T From the ideal-gas equation, i i i T pV nR = ( )( ) ( )( ) 2 1.013 105 Pa 1000 10 6 m3 243.8 K 0.10 mol 8.31 J mol K - = = f i .T = 3T = 731.4 K f i .T -T = 487.6 K Thus, heat required for process B is ( )( )( ) B P Q = nC .T = 0.10 mol 29.1 J mol K 487.6 K = 1400 J Assess: Heat is transferred out of the gas in process A, but transferred into the gas in process B. 17.58. Model: We have an adiabatic and an isothermal process. Solve: For the adiabatic process, no heat is added or removed. That is Q = 0 J. Isothermal processes occur at a fixed temperature, so .T = 0 K. Thus .Eth = 0 J, and the first law of thermodynamics gives Q = -W = ( ) f i nRT ln V V The temperature T can be obtained from the ideal-gas equation as follows: ( )( ) ( )( ) 5 63 i i i i 1.013 10 Pa 3000 10 m 0.10 mol 8.31 J mol K pV nRT T pV nR - = . = = = 366 K Substituting into the equation for Q we get ( )( )( ) 6 3 6 3 0.10 mol 8.31 J mol K 366 K ln 1000 10 m 330 J 3000 10 m Q - - . . = . . = - . . That is, 330 J of heat energy is removed from the gas. 17.59. Model: The monatomic gas is an ideal gas which is subject to isobaric and isochoric processes. Solve: (a) For the isochoric process, V2 = V1 = 800 10-6 m3, p1 = 4.0 atm, p2 = 2.0 atm. The temperature T1 of the gas is obtained from the ideal-gas equation as: 1 1 1 T pV 390 K nR = = where n = 0.10 mol. T2 can be obtained from the ideal-gas equation as follows: 1 1 2 2 ( ) ( ) 2 1 2 1 1 2 390 K 2.0 atm 195 K 4.0 atm pV p V T T p p T T = . = = . . = . . . . The heat required for the process 1 . 2 is ( ) ( )( )( ) V 2 1 Q = nC T -T = 0.10 mol 20.8 J/mol K 195 K - 390 K = -406 J -410 J Because of the negative sign, this is the amount of heat removed from the gas. (b) For this isobaric process, p2 = p3 = 2.0 atm, V2 = 800 10-6 m3, and V3 = 1600 10-6 m3. 6 3 3 3 2 2 6 3 2 2 1600 10 m 2 800 10 m T T V T T V - - . . = = . . = . . = 390 K Thus, the heat required for the process 2 . 3 is ( ) ( )( )( ) P 3 2 Q = nC T -T = 0.10 mol 29.1 J mol K 195 K = 567 J 570 J This is heat transferred to the gas. (c) The change in the thermal energy of the gas is ( ) ( ) th 1 2 2 3 1 2 2 3 2 3 E Q Q W W 406 J 567 J 0 J +W . . . . . . = + + + = - + + =162 J - p.V = 162 J (2.0 1.013 105 Pa)(1600 10-6 m3 800 10-6 m3) = 0 J Assess: This result was expected since T3 = T1. 17.60. Model: Assume that the gas is an ideal gas. Visualize: The volume of container A is a constant. On the other hand, heating container B causes the volume to change, but the pressure remains the same. Solve: (a) For the heating of the gas in container A, .TA = QA / nCV . Similarly, for the gas in container B, B B P .T = Q / nC . Because QA = QB and CP > CV, we see that A B.T > .T . The gases started at the same temperature, so TA > TB. (b) (c) The pressure in container B exerted by the gas is equal to the pressure on the gas by the piston. That is, piston B atmos piston w p p A = + ( )( 2 ) 5 4 2 10 kg 9.8 m s 1.013 10 Pa 1.0 10- m = + = 1.08 106 Pa Container A has the same volume, temperature, and number of moles of gas as container B, so A B P = P = 1.08106 Pa. (d) The heating of container B is isobaric, so f i f f i f i i V V V V T T T T = . = We have Ti = 293 K, and Tf can be obtained from ( ) P f i Q = P.t = nC T -T The number of moles of gas is i i i n = PV / RT = 0.355 mol. Thus ( )( ) ( )( )( ) f 25 W 15 s = 0.355 mol 20.8 J mol K T - 293 K f .T = 344 K ( 4 3 )( ) f .V = 8.010- m 344 K 293 K = 9.39 10-4 m3 = 939 cm3 940 cm3 17.61. Model: Assume that the gas is an ideal gas. A diatomic gas has . = 1.40. Solve: (a) For container A, Ai ( )( )( ) 4 3 iA 5 Ai 0.10 mol 8.31 J mol K 300 K 8.20 10 m 3.0 1.013 10 Pa V nRT p = = = - For an isothermal process pAfVAf = pAiVAi. This means Af Ai T = T = 300 K and ( ) ( 4 3 )( ) 3 3 Af Ai Ai Af V =V p p = 8.2010- m 3.0 atm 1.0 atm = 2.510- m The gas in container B starts with the same initial volume. For an adiabatic process, Bf Bf Bi Bi p V. = p V. ( ) 1 1 1.40 Bi 4 3 3 3 Bf Bi Bf 8.20 10 m 3.0 atm 1.8 10 m 1.0 atm V V p p . - - . . . . . = . . = . . = . . . . The final temperature TBf can now be obtained by using the ideal-gas equation: ( ) 3 3 Bf Bf Bf iB 4 3 Bi Bi 300 K 1.0 atm 1.80 10 m 3.0 atm 8.20 10 m T T p V p V - - . .. . = = . .. . . .. . = 220 K (b) 17.62. Model: The gas is assumed to be an ideal gas that is subjected to isobaric and isochoric processes. Solve: (a) The initial conditions are p1 = 3.0 atm = 304,000 Pa, V1 = 100 cm3 = 1.0 10-4 m3, and T1 = 100C = 373 K. The number of moles of gas is ( )( ) ( ) 4 3 1 1 3 1 304,000 Pa 1.0 10 m 9.81 10 mol 8.31 J mol K (373 K) n pV RT - - = = = At point 2 we have p2 = p1 = 3.0 atm and V2 = 300 cm3 = 3V1. This is an isobaric process, so 2 1 2 2 1 2 1 1 V V T V T 3(373 K) 1119 K T T V = . = = = The gas is heated to raise the temperature from T1 to T2. The amount of heat required is ( 3 )( )( ) P Q = nC .T = 9.8110- mol 20.8 J mol K 1119 K - 373 K =152 J This amount of heat is added during process 1 . 2. (b) Point 3 returns to T3 = 100C = 373 K. This is an isochoric process, so ( 3 )( )( ) V Q = nC .T = 9.8110- mol 12.5 J mol K 373 K -1119 K = -91.5 J This amount of heat is removed during process 2 . 3. 17.63. Assume the gas to be an ideal gas. Solve: (a) The work done on the gas is the negative of the area under the p-versus-V graph, that is W = -area under curve = -50.7 J (b) The change in thermal energy is ( ) th V V f i .E = nC .T = nC T -T Using the ideal-gas law to calculate the initial and final temperatures, i i i T pV nR = ( )( ) ( )( ) 4.0 1.013 105 Pa 100 10 6 m3 325 K 0.015 mol 8.31 J mol K - = = ( )( ) ( )( ) 5 63 f f f 1.013 10 Pa 300 10 m 244 K 0.015 mol 8.31 J mol K T p V nR - = = = ( )( )( ) th ..E = 0.015 mol 12.5 J mol K 244 K - 325 K = -15.2 J -15 J (c) From the first law of thermodynamics, .Eth = Q + W ( ) th .Q = .E -W = -15.2 J - -50.7 J = 35.5 J 36 J That is, 36 J of heat energy is transferred to the gas. 17.64. Model: Assume that the gas is an ideal gas and that the work, heat, and thermal energy are connected by the first law of thermodynamics. Solve: (a) For point 1, V1 = 1000 cm3 = 1.0 10-3 m3, T1 = 133C = 406 K, and the number of moles is 3 mol 120 10 g 0.030 mol 4 g/mol n M M . - . = = . . = . . Thus, the pressure p1 is 1 5 1 1 p nRT 1.012 10 Pa 1.0 atm V = = = The process 1 . 2 is isochoric (V2 = V1) and p2 = 5p1 = 5.0 atm. Thus, ( ) ( )( ) 2 1 2 1 T = T p p = 406 K 5 = 2030 K =1757C The process 2 . 3 is isothermal (T2 = T3), so V3 = V2(p2/p3) = V2(p2/p1) = 5V2 = 5000 cm3 p (atm) T (C) V (cm3) Point 1 1.0 133 1000 Point 2 5.0 1757 1000 Point 3 1.0 1757 5000 (b) The work 1 2 W 0 . = J because it is an isochoric process. The work in process 2 . 3 can be found using Equation 17.16 as follows: ( ) 2 3 2 3 2 W nRT ln V V . = - = -(0.030 mol)(8.31 J mol K)(2030 K)ln(5) = -815 J The work in the isobaric process 3 . 1 is ( ) 3 1 f i W pV V . = - - = -(1.012105 Pa)(1.010-3 m3 - 5.010-3 m3 ) = 405 J (c) The heat transferred in process 1 . 2 is ( )( )( ) 1 2 V Q nC T 0.030 mol 12.5 J mol K 2030 K 406 K . = . = - = 609 J The heat transferred in the isothermal process 2 . 3 is 2 3 2 3 Q W 815 J . . = - = . The heat transferred in the isobaric process 3 . 1 is ( )( )( ) 3 1 P Q nC T 0.030 mol 20.8 J mol K 406 K 2030 K 1013 J . = . = - =- 17.65. Model: The air is assumed to be an ideal diatomic gas that is subjected to an adiabatic process. Solve: The air admitted into the cylinder at T0 = 30C = 303 K and p0 = 1 atm = 1.013 105 Pa has a volume V0 = 600 10-6 m3 and contains 0 0 0 n p V 0.024 mol RT = = Using Equation 17.36 and the fact that Q = 0 J for an adiabatic process, th V .E = Q +W = nC .T . W = nCVT . 400 J = (0.024 mol)(20.8 J/mol K)(Tf 303 K) f .T =1100 K For an adiabatic process Equation 17.40 is 1 1 f f 0 0 T V. - = T V. - ( ) 1 1 1 1.4 1 0 4 3 5 3 3 f 0 f 6.0 10 m 303 K 2.39 10 m 24 cm 1100 K V V T T . - - - - . . . . . = . . = . . = . . . . Assess: Note that W is positive because the environment does work on the gas. 17.66. Model: . is 1.40 for a diatomic gas and 1.67 for a monoatomic gas. Solve: (a) We will assume that air is a diatomic gas. For an adiabatic process, 1 1 f f i i T V. - = TV. - Thus 1 1 1 1.40 1 i f f i 1123 K 26.4 303 K V T V T . . . .. - . . - . . = . . = . . = . . . . . . (b) For argon, a monatomic gas, 1 1.67 1 i f 1123 K 7.07 303 K V V . . = . . - = . . . . . . . . 17.67. Model: Air is assumed to be an ideal diatomic gas that is subjected to an adiabatic process. Solve: (a) Equation 17.40 for an adiabatic process is 1 1 f f i i T V. - = TV. - 1 1 f i i f V T V T . .. - . =. . . . For the temperature to increase from Ti = 20C = 293 K to Tf = 1000C = 1273 K, the compression ratio will be 1 1.4 1 f i 293 K 0.02542 1273 K V V . . - = . . = . . max min 1 39.3 0.02542 V V . = = (b) From the Equation 17.39, f i ( )1.4 f f i i i f p V pV p V 39.3 171 p V . . . . . = . = . . = = . . 17.68. Model: The helium gas is assumed to be an ideal gas that is subjected to an isobaric process. Solve: (a) The number of moles in 2.0 g of helium is mol 2.0 g 0.50 mol 4.0 g mol n M M = = = At Ti = 100C = 373 K and pi = 1.0 atm = 1.013 105 Pa, the gas has a volume i 3 i i V nRT 0.0153 m 15.3 L p = = = For an isobaric process (pf = pi) that doubles the volume Vf = 2Vi, ( ) ( ) f i f i T T = V V = 2 ( ) f i .T = 2T = 2 373 K = 746 K = 473C (b) The work done by the environment on the gas is ( ) ( ) i f i i i W = - p V -V = - pV 2 -1 = -(1.013105 Pa)(0.0153 m3 ) = -1550 J (c) The heat input to the gas is ( ) ( )( )( ) P f i Q = nC T -T = 0.50 mol 20.8 J mol K 746 K - 373 K = 3880 J 3900 J (d) The change in the thermal energy of the gas is th .E = Q +W = 3880 J -1550 J = 2330 J 2300 J (e) Assess: The internal energy can also be calculated as follows: ( )( )( ) th V .E = nC .T = 0.5 mol 12.5 J mol K 746 K - 373 K = 2330 J This is the same result as we got in part (d). 17.69. Model: The helium gas is assumed to be an ideal gas that is subjected to an isothermal process. Solve: (a) The number of moles in 2.0 g of helium gas is mol 2.0 g 0.50 mol 4.0 g mol n M M = = = At Ti = 100C = 373 K and pi = 1.0 atm = 1.013 105 Pa, the gas has a volume i 3 i i V nRT 0.0153 m 15.3 L p = = = For an isothermal process (Tf = Ti) that doubles the volume Vf = 2Vi, ( ) ( )( 1 ) f f i i f i i f 2 p V = pV . p = p V V = 1.0 atm = 0.50 atm (b) The work done by the environment on the gas is ( ) i f i W = -nRT ln V V = -(0.50 mol)(8.31 J mol K)(373 K)ln(2) = -1074 J -1070 J (c) Because .Eth = Q + W = 0 J for an isothermal process, the heat input to the gas is Q = -W = 1074 J 1070 J . (d) The change in internal energy .Eth = 0 J. (e) 17.70. Model: The nitrogen gas is assumed to be an ideal gas that is subjected to an adiabatic process. Solve: (a) The number of moles in 14.0 g of N2 gas is mol 14.0 g 0.50 mol 28 g mol n M M = = = At Ti = 273 K and pi = 1.0 atm = 1.013 105 Pa, the gas has a volume i 3 i i V nRT 0.0112 m 11.2 L p = = = For an adiabatic process that compresses to a pressure pf = 20 atm, we can use Equation 17.39 and Equation 17.40 as follows: 1 1 1 1 f i f i f f i i f f i i i f i f T V TV T V p V pV p V T V p V . . . . . . - - - . . . . . . = .. . = . . = . . . = . . . . . . Combining the above two equations yields ( ) ( ) 1 f i f i T T p p . . - = ( ) ( ) 1.4 1.0 1.4 f i i T T 20 T 2.3535 643 K - . = = = (b) The work done on the gas is ( ) ( )( )( ) th V f i W = .E = nC T -T = 0.50 mol 20.8 J mol K 643 K - 273 K = 3850 J 3800 J (c) The heat input to the gas is Q = 0 J. (d) From the above equation, ( ) 1 1 i f 1.4 max f i min V p 20 8.5 V V p V . .. = . . = = = . . (e) 17.71. Model: The gas is assumed to be an ideal gas that is subjected to an isochoric process. Solve: (a) The number of moles in 14.0 g of N2 gas is mol 14.0 g 0.50 mol 28 g mol n M M = = = At Ti = 273 K and pi = 1.0 atm = 1.013 105 Pa, the gas has a volume i 3 i i V nRT 0.0112 m 11.2 L p = = = For an isochoric process (Vi = Vf), f f i i 20 atm 20 1 atm T p T p = = = ( ) f .T = 20 273 K = 5460 K 5500 K (b) The work done on the gas is W = - p.V = 0 J. (c) The heat input to the gas is ( ) ( )( )( ) 4 V f i Q = nC T -T = 0.50 mol 20.8 J mol K 5460 K - 273 K = 5.410 J (d) The pressure ratio is max f min i 20 atm 20 1 atm p p p p = = = (e) 17.72. Model: The air is assumed to be an ideal gas. Because the air is compressed without time to exchange heat with its surroundings, the compression is an adiabatic process. Solve: The initial pressure of air in the mountains behind Los Angeles is pi = 60 103 Pa at Ti = 273 K. The pressure of this air when it is carried down to the elevation near sea level is pf = 100 103 Pa. The adiabatic compression of a gas leads to an increase in temperature according to Equation 17.39 and Equation 17.40, which are 1 1 1 1 f i f i f f i i f f i i i f i f T V TV T V p V pV p V T V p V . . . . . . - - - . . . . . . = .. . = . . = . . . = . . . . . . Combining these two equations, ( ) ( ) ( ) 1.4 1 1 3 1.4 0.286 f i f i f i 3 100 10 Pa 273 K 5 60 10 Pa 3 T T p p T T . . - - . . . . = . = . . = . . . . . . = 316 K = 43C = 109F 17.73. Model: Assume the collector has emissivity e =1.0 . We will use Equation 17.50 and divide both sides by A to get intensity in W/m2 . At the equilibrium temperature the collector will absorb 800 W/m2 on the sun side and must radiate the same amount from the same side. 0 T = 20C = 293 K . Solve: ( 4 4 ) 0 Q/ t e T T A = - D s Solve this for T . 4 4 0 Q/ t 1 T T A e = - D s ( )( )( ) 2 4 4 4 4 0 8 2 4 1 800W/m 293K 383K 110 C 1 0 5 67 10 W/m K T Q/ t T A e . . = + = + = = 2 D s Assess: 110C is hot enough to boil water, so this seems feasible as a hot water heater, at least during the 5 hours a day the collector gets 800 W/m2 . You can store the hot water in an insulated tank for use at other times of the day. 17.74. Model: This is a problem about conduction of the heat from inside the box to outside. 100 W of heat is generated by the light bulb inside the box, so in equilibrium that is how much must be conducted away through the sides of the box. Visualize: We are given L = 0.012m , Q/Dt =100W and the thermal conductivity of concrete k = 0.8W/mK . We compute A = 6(0.20m 0.20m) = 0.24m2 . C T = 20C = 293K . Solve: Use Equation 17.48 for the rate of heat transfer by conduction. ( ) H C Q k A T T t L = - D Solve for H T . ( ) H C Q L T T t kA . . = - . . . D . ( ) H C ( )( 2 ) 100 W 0 012 m 293 K 299 K 26 C 0 8 W/m K 0 24 m T Q L T . t kA . . = . . + = + = = . . . D . ? Assess: We expected H T to be a few degrees hotter than C T , and so it is. 17.75. Model: Refer to Example 17.11. Assume the surface of the earth is an ideal radiator with e =1. Visualize: Only half the earth faces the sun at a time, so the earth intercepts 1370 W/m2 over a cross section of ( )2 6 37 106m 2 1 275 1014 m2 epR = p . = . . If the earths surface absorbs 70% of the incident power then the total power absorbed is (0.70)(1370 W/m2 )(1.2751014 m2 ) =1.2221017 W . This much power must also be radiated away from the earth in equilibrium. Solve: The power is radiated from the whole spherical surface of the earth: 2 e A = 4pR . Use Equation 17.49 to find the temperature of the earth. ( ) ( )( ) ( ) 1 4 1 4 17 2 8 2 6 2 e 1 222 10 W 255 K 18 C 4 1 5 67 10 W/m K 4 6 37 10 m / / T Q/ t . e R . . . . . . = . . = . . = = .. .. . . . . 2 D 3 2 s p 3 ? p 3 Assess: Without a moderate greenhouse effect the earth would be too cold for mammal life. On the other hand, when the greenhouse effect gets out of hand, such as on Venus, it can be too hot for life. 17.76. Solve: (a) 50 J of work are done on a gas to compress it to one-third of its original volume at a constant temperature of 77C. How many moles of the gas are in the sample? (b) The number of moles is ( )( )( 1 ) 3 50 J 0.0156 mol 8.31 J mol K 350 K ln n= = - 17.77. Solve: (a) A heated 500 g iron slug is dropped into a 200 cm3 pool of mercury at 15C. If the mercury temperature rises to 90C, what was the initial temperature of the iron slug? (b) The initial temperature was 217C . 17.78. Solve: (a) A diatomic gas is adiabatically compressed from 1 atm pressure to 10 atm pressure. What is the compression ratio Vmax/Vmin? (b) The ratio is 1 1.4 Vmax Vmin =10 = 5.18 . 17.79. Model: Assume the helium gas to be an ideal gas. The gas is subjected to isothermal, isochoric, and adiabatic processes. Visualize: Please refer to Figure CP17.79. The gas at point 1 has volume V1 = 1000 cm3 = 1.0 10-3 m3 and pressure p1 = 3.0 atm. At point 2, V2 = 3000 cm3 = 3.0 10-3 m3 and p2 = 1.0 atm. These values mean that T2 = T1, so process 1 . 2 is an isothermal process. The process 2 . 3 occurs at constant volume and is thus an isochoric process. Finally, because temperature T3 is lower than T1 or T2, the process 3 . 1 is an adiabatic process. Solve: (a) The number of moles of gas is 0.120 g 0.030 mols 4 g mol n= = The temperature T1 can be calculated to be ( )( ) ( )( ) 5 33 1 1 1 3.0 1.013 10 Pa 1.0 10 m 1219 K 0.030 mol 8.31 J mol K T pV nR - = = = = 946C For the isothermal process 1 . 2, T2 = 1219 K. For the adiabatic process 3 . 1, ( ) ( ) ( ) ( )( ) 3 1.67 1 1 0.67 3 1 1 3 3 3 1 1 3 3 3.0 atm 1000 cm 0.48 atm 1219 K 583 K=310 C 3000 cm p p V V T T V V . . . . - = = . . = = = = . . Point p (atm) T (C) V (cm3) 1 3.0 946 1000 2 1.0 946 3000 3 0.48 310 3000 Note that the values obtained above are consistent with the isochoric process 2 . 3, for which 2 3 2 3 p p T T = ( ) ( )( ) 2 2 3 3 . p = T T p = 1219 K 583 K 0.48 atm =1 atm (b) From Equation 17.15, 2 1 2 1 1 ln V W nRTV . . . = - . . . . = -(0.030 mol)(8.31 J mol K)(1219 K)ln(3) = -334 J The work done in the ischochoric process is 2 3 W 0 J . = . The work done in the adiabatic process is ( ) ( )( )( ) 3 1 V 1 3 W nC T T 0.030 mol 12.5 J mol K 1219 K 583 K 239 J . = - = - = (c) For the process 1 . 2, th .T = 0 K..E = 0 J.Q = -W = 334 J For the process 2 . 3, W = 0 J, ( ) th V 3 2 .E = Q = nC T -T = -239 J For the process 3 . 1, Q = 0 J. 17.80. Model: The gas is an ideal gas, and its thermal energy is the total kinetic energy of the moving molecules. Visualize: Please refer to Figure P17.80. Solve: (a) The piston is floating in static equilibrium, so the downward force of gravity on the pistons mass must exactly balance the upward force of the gas, Fgas = pA where A = pr2 is the area of the face of the piston. Since the upper part of the cylinder is evacuated, there is no gas pressure force pushing downward. Thus, piston Cu piston ( 3 )( 2 )( ) piston Cu 8920 kg m 9.80 m s 0.040 m 3500 Pa M g V g M g pA p gh A A . = . = = = . = = (b) The gas volume is 2 2 43 1 V =p r L =p (0.030) (0.20 m) = 5.6510- m . The number of moles is ( )( ) ( )( ) 4 3 1 1 4 1 3500 Pa 5.65 10 m 8.12 10 mol 8.31 J mol K 293 K n pV RT - - = = = The number of molecules is N = nNA = (8.12 10-4 mol)(6.02 1023 mol-1) = 4.9 1020 (c) The pressure in the gas is determined simply by the weight of the piston. That will not change as heat is added, so the heating takes place at constant pressure with Q = nCP.T. The temperature increase is ( 4 )( ) P 2.0 J 85 K 8.12 10 mol 29.1 J mol K T Q nC - . = = = This raises the gas temperature to T2 = T1 + .T = 378 K = 105C. (d) Noting that the volume of a cylinder is V = pr2L and that r doesnt change, the ideal-gas relationship for an isobaric process is 2 1 2 1 2 2 1 2 1 2 1 1 378 K 20 cm 25.8 cm 293 K V V L L L T L T T T T T = . = . = = . . = . . . . (e) The work done by the gas is gas gas W = F .y . The force exerted on the piston by the gas is 2 gas F = pA = pp r = 9.90 N This force is applied through .y = 5.8 cm = 0.058 m, so the work done is Wgas = (9.90 N)(0.058 m) = 0.574 J 0.57 J Thus, 0.57 J is the work done by the gas on the piston. The work done on the gas is 0.57 J. 17.81. Model: There is a thermal interaction between the iron, assumed to be initially at room temperature (20C), and the liquid nitrogen. The boiling point of liquid nitrogen is 196C = 77 K. Solve: The piece of iron has mass Miron = 197 g = 0.197 kg and volume Viron = Miron/.iron = (0.197 kg)/(7870 kg/m3) = 25 10-6 m3 = 25 cm3 = 25 mL. The heat lost by the iron is Qiron = Mironciron.T = (0.197 kg)(449 J/kg K)(77 K 293 K) = 1.911 104 J This heat causes mass M of the liquid nitrogen to boil. Energy conservation requires 4 iron iron N2 iron f 5 f 0 1.911 10 J 0.0960 kg 1.99 10 J/kg Q Q Q ML M Q L + = + = . =- = = The volume of liquid nitrogen boiled away is thus 4 3 boil 3 N2 0.0960 kg 1.19 10 m 119 mL 810 kg/m V M . = = = - = Now the volume of nitrogen gas (at 77 K) is 1500 mL before the iron is dropped in. The volume of the piece of iron excludes 25 mL of gas, so the initial gas volume, when the lid is sealed and the liquid starts to boil, is V1 = 1475 mL. The pressure is p1 = 1.0 atm and the temperature is T1 = 77 K. Thus the number of moles of nitrogen gas is 3 3 1 1 1 1 (101,300 Pa)(1.475 10 m ) 0.234 mol (8.31 J/mol K)(77 K) n pV RT - = = = 119 mL of liquid boils away, so the gas volume increases to V2 = V1 + Vboil = 1475 mL + 119 mL = 1594 mL. The temperature is still T2 = 77 K, but the number of moles of gas has been increased by the liquid that boiled. The number of moles that boiled away is boil 96.0 g 3.429 mol 28 mol/g n = = Thus the number of moles of nitrogen gas increases to n2 = n1 + nboil = 0.234 mol + 3.429 mol = 3.663 mol. Consequently, the gas pressure increases to 2 2 6 2 3 3 2 (3.663 mol)(8.31 J/mol K)(77 K) 1.470 10 Pa 14.5 atm 1.594 10 m p n RT V - = = = = 15 atm Assess: Dont try this! The large pressure increase could cause a flask of liquid nitrogen to explode, leading to serious injuries. 17.82. Model: Ignore any radiation and assume all of the heat is transferred by conduction through the compound rod from the hot end to the cold end. Visualize: Look up the thermal conductivities of the two metals in Table 17.5: kCu = 400 W/m?k and Fe k = 80 W/m?K. H T = 373 K and C T = 273 K. Equation 17.48 will be applied to each rod and all of the heat must go through both rods at the same rate. Cu Fe Q Q t t . . = . . . . . . . D . . D . Solve: Since A and L are the same for the two rods, then Cu Cu Fe Fe k DT = k DT Label the temperature at the joining point in the middle with a subscript M. ( ) ( ) Cu H M Fe M C k T -T = k T -T Solve for M T . Cu H Cu M Fe M Fe C k T - k T = k T - k T Cu H Fe C Fe M Cu M k T + k T = k T + k T ( ) Fe Cu M Cu H Fe C k + k T = k T + k T Cu H Fe C M Cu Fe T k T k T k k + = + M (400 W/m K)(373 K) (80 W/m K)(273 K) 356 K 83 C 400 W/m K 80 W/m K T + = = = + Assess: Had we not arrived at an answer between 100C and 0C we would have been very worried. Furthermore, because the conductivity of copper is greater than the conductivity of iron we expected the answer to be above 50C. 17-1 17.83. Model: Assume the gas is ideal. The process is not adiabatic, despite Equation 17.38, because the exponent given in the problem is not equal to g . The gas is identified as diatomic which means that g =1.40 , but the exponent is explicitly given as 2, so while we are given pV 2 = constant, it is not true that pV g = constant. We will use the fact that the gas is diatomic to deduce 5 V 2 C = R = 20.8J/mol k , although we wont specifically need g =1.40 . Visualize: We are given n = 0.020mol , i T = 293K , 3 i V =1500cm , and 3 f V = 500cm . Solve: (a) If we use the ideal-gas-law expression p = nRT/V in pV 2 = constant we get TV = constant . 3 i i f 3 f (293 K)(1500 cm ) 879 K 606 C 500 cm T TV V = = = = (b) The strategy to find Q will be to use the first law th Q = DE -W . We will use th V E = nC DT and W =2. pdV. ( )( )( ) th V DE = nC DT = 0.020 mol 20.8 J/mol k 879 K - 293 K = 243.8 J To do the W =2. pdV integral we need to know what the constant is in pV 2 = constant . Use the ideal gas law to compute i p . ( )( )( ) i i 3 3 i 0 020 mol 8 31 J/mol K 293 K 32,460 Pa 1 5 10 m nRT . . p V . = = = 2 ( )( )2 3 3 2 6 i i constant = pV = 32,460Pa 1.5102 m = 0.0730 Pa m ( ) ( ) 3 3 f 3 i 3 0.00050m 6 0 00050m 6 constant 2 0 0015m 2 0.0015m d d 0 0730 Pa m d 0 0730 Pa m 1 97 4 J V . V V . W p V V . v . . V V = - = = = . . = .. .. . 2 2 th Q = DE -W = 243.8 J -97.4 J =146.4 J<150 J (c) We also need f p to complete the pV diagram. We could use the ideal-gas law again to compute f p , but lets do it another way. 2 2 i i f f pV = p V 2 3 2 i f i i 3 i f 1500 cm 9 292,200 Pa 500 cm p p V p p V . . . . = . . = . . = = . . . . Assess: Both Q and W are positive because heat was added to the system and work was done on the system. 18.1. Solve: We can use the ideal-gas law in the form pV = NkBT to determine the Loschmidt number (N/V): ( ) ( )( ) 5 25 3 23 B 1.013 10 Pa 2.69 10 m 1.38 10 J K 273 K N p V kT - - = = = 18.2. Solve: The volume of the nitrogen gas is 1.0 m3 and its temperature is 20C or 293 K. The number of gas molecules can be found as ( )( ) ( )( )( ) 5 3 23 1 25 A A 1.013 10 Pa 1.0 m 6.02 10 mol 2.5 10 8.31 J mol K 293 K N nN pV N RT - = = = = According to Figure 18.2, 12% of the molecules have a speed between 700 and 800 m/s, 7% between 800 and 900 m/s, and 3% between 900 and 1000 m/s. Thus, the number of molecules in the cube with a speed between 700 m/s and 1000 m/s is (0.22)(2.51 1025) = 5.5 1024. 18.3. Solve: Nitrogen is a diatomic molecule, so r 1.0 10-10 m. We can use the ideal-gas law in the form pV = NkBT and Equation 18.3 for the mean free path to obtain p: ( ) B 2 2 1 4 2 4 2 k T N V r pr . p p = = ( )( ) ( )( ) 23 B 2 10 2 1.38 10 J k 293 K 4 2 4 2 1.0 m 1.0 10 m p k T p.r p - - . = = = 0.023 Pa Assess: In Example 18.1 . = 225 nm at STP for nitrogen. . = 1.0 m must therefore require a very small pressure. 18.4. Solve: (a) Air is primarily comprised of diatomic molecules, so r 1.0 10-10 m. Using the ideal-gas law in the form pV = NkBT, we get ( )( ) 5 10 12 3 23 B 1.0 10 mm of Hg 1.013 10 Pa 760 mm of Hg 3.30 10 m 1.38 10 J K 293 K N p V kT - - - = = = (b) The mean free path is ( )( ) ( )( ) 6 2 12 3 10 2 1 1 1.71 10 m 4 2 N V r 4 2 3.30 10 m 1.0 10 m . p p - - = = = Assess: The pressure p in the vacuum chamber is 1.33 10-8 Pa = 1.32 10-13 atm. A mean free path of 1.71 106 m is large but not unreasonable. 18.5. Solve: (a) The mean free path of a molecule in a gas at temperature T1, volume V1, and pressure p1 is .1 = 300 nm. We also know that 2 1 4 2 ( / ) V N V r . . p = . . Although T2 = 2T1, constant volume (V2 = V1) means that .2 = .1 = 300 nm. (b) For T2 = 2 T1 and p2 = p1, the ideal gas equation gives ( ) 1 1 2 2 1 2 B 1 B 2 B 1 2 pV p V pV Nk T Nk T Nk T = = .V2 = 2V1 Because V . . , .2 = 2 .1 = 2(300 nm) = 600 nm. 18.6. Solve: Neon is a monatomic gas and has a radius r 5.0 10-11 m. Using the ideal-gas equation, ( )( ) ( )( ) 5 27 3 23 B 150 1.013 10 Pa 3.695 10 m 1.38 10 J/K 298 K N p V kT - - = = = Thus, the mean free path of a neon atom is ( ) ( )( ) 9 2 27 3 11 2 1 1 6.09 10 m 4 2 N V r 4 2 3.695 10 m 5.0 10 m . p p - - - = = = Since the atomic diameter of neon is 2 (5.0 10-11 m) = 1.0 1010 m, 9 10 6.09 10 m 61 atomic diameters 1.0 10 m . - - = = 18.7. Solve: The number density of the Ping-Pong balls inside the box is 3 3 2000 2000 m 1.0 m N V = = - With r = (3.0 cm)/2 = 1.5 cm, the mean free path of the balls is ( )( 2 ) 1 0.125 m 12.5 cm 4 2 N V r . p = = = 18.8. Solve: (a) The average speed is 25 avg 15 1 m/s 220 m/s 20.0 m/s 11 11 n n v n = = . . = . . = = . . S (b) The root-mean-square speed is ( 2 ) rms avg v = v 1 25 2 2 15 1 m/s 11 n n n = = . . =. . . . S 1 4510 2 20.2 m/s 11 = . . = . . . . 18.9. Solve: (a) In tabular form we have Particle vx (m/s) vy (m/s) 2 x v (m/s)2 2 y v (m/s)2 v2 (m/s)2 v (m/s) 1 20 30 400 900 1300 36.06 2 -40 70 1600 4900 6500 80.62 3 -80 -10 6400 100 6500 80.62 4 60 -20 3600 400 4000 63.25 5 0 -50 0 2500 2500 50.00 6 40 -20 1600 400 2000 44.72 Average 0 0 3800 59.20 The average velocity is vavg = 0 i + 0 j .. .. .. . (b) The average speed is avg v = 59 m/s . (c) The root-mean-square speed is ( 2 ) 2 2 rms avg v = v = 3800 m / s = 62 m/s . 18.10. Solve: (a) The most probable speed is 4.0 m/s. (b) The average speed is avg 2 2 m/s 4 4 m/s 3 6 m/s 1 8 m/s 4.6 m/s 2 4 3 1 v + + + = = + + + (c) The root-mean-square speed is ( )2 ( )2 ( )2 ( )2 rms 2 2 m/s 4 4 m/s 3 6 m/s 1 8 m/s 4.9 m/s 2 4 3 1 v + + + = = + + + 18.11. Solve: (a) The atomic mass number of argon is 40. This means the mass of an argon atom is m = 40 u = 40(1.66110-27 kg) = 6.6410-26 kg The pressure of the gas is 1 2 3 rms p N mv V = . . . . . . 1 ( 25 3 )( 26 )( )2 3 = 2.0010 m- 6.6410- kg 455 m/s = 9.16 104 Pa (b) The temperature of the gas in the container can be obtained from the ideal-gas equation in the form pV = NkBT: ( )( ) 4 25 3 23 B 9.16 10 Pa 332 K 2.00 10 m 1.38 10 J/K T pV Nk - - = = = 18.12. Model: Pressure is due to random collisions of gas molecules with the walls. Solve: According to Equation 18.8, the collision rate with one wall is coll net coll rate of collisions 2 2 x x N F pA t mv mv = = = . where Fnet = pA is the force exerted on area A by the gas pressure. However, this equation assumed that all molecules are moving in the x-direction with constant speed. The rms speed vrms is for motion in three dimensions at varying speeds. Consequently, we need to replace vx not with (vx)avg, which is zero, but with 2 2 rms rms avg ( ) 3 3 x x v . v = v = v With this change, 25 1 27 rms rate of collisions 3 3(2 101,300 Pa)(0.10 m 0.10 m) 6.5 10 s 2 2(28 1.661 10 kg)(576 m/s) pA mv - - = = = This collision rate can also be found by using the expression in Eq. 18.10, making the same change in vx, and using the ideal-gas law to determine N/V. 18.13. Visualize: We will use Equation 18.18 to find m of the atoms. 2 rms 1 3 p N mv V = We are given p = 2.0 atm = 202,650 Pa , N/V = 4.21025 m-3, and rms v = 660 m/s. Solve: Solve for m. 26 2 2 253 rms 3 3(202,650 Pa) 1 3 32 10 kg 20 u (660 m/s) 4 2 10 m m p V v N - - = = = . = . With an atomic mass of 20 the gas is likely neon. Assess: Neon is gaseous at room temperature, so is a likely choice. As a noble gas, it also doesnt form many molecules. 18.14. Visualize: Equation 18.26 will give us rms v from the temperature and mass of the particles. B rms v 3k T m = We are given T =1100C =1373 K . The atomic mass of neon atoms is 20 u = 3.32110-26 kg; the molecular mass of oxygen molecules is 32 u = 5.3110-26 kg. Solve: (a) 23 rms 26 3(1 38 10 J/K)(1373 K) 1310 m/s 3 321 10 kg v - - . = = . (b) 23 rms 26 3(1 38 10 J/K)(1373 K) 1030 m/s 5 31 10 kg v - - . = = . Assess: These are fast speeds, but the temperature is high. 18.15. Visualize: Use Equation 18.26. B rms v 3k T m = We are given rms v =1.5 m/s and m = 28 u = 4.65010-26 kg. Solve: Solve the equation for T . 2 2 26 rms 23 B (1 5 m/s) (4 650 10 kg) 2 5 mK 3 3(13810 J/K) T v m k - - . . = = = . . Assess: 2.5 mK is close to absolute zero, but thats how cold it would have to be for rms v to be 1.5 m/s. 18.16. Solve: Because the neon and argon atoms in the mixture are in thermal equilibrium, the temperature of each gas in the mixture must be the same. That is, using Equation 18.26, 2 2 mArvrms Ar = mNevrms Ne Ne rms Ar rms Ne Ar v v m m = (400 m/s) 20 u 40 u = = 283 m/s 18.17. Solve: The average translational kinetic energy per molecule is 2 avg rms B 1 3 2 2 P = mv = k T B rms v 3k T m . = Since we want the vrms for H2 and N2 to be equal, 2 2 2 B H B H N 3k T 3k T m m = 2 ( ) 2 2 2 H H N N 2 u 373 K 27 K 246 C 28 u m T T m . = = . . = = - . . . . 18.18. Solve: The formula for the root-means-square speed as a function of temperature is ( ) B rms T v 3k T m = (a) For ( ) 1 ( ) rms T 2 rms STP v = v , ( ) B 1 B 2 3k T 3k 273 K m m = 1 (273 K) 68 K= 205 C 4 .T = = - (b) For( ) ( ) rms T rms STP v = 2 v , ( ) B B 3 3 273 K 2 k T k m m = .T = 4(273 K) =1090 K = 817C 18.19. Solve: Solve Equation 18.18 for vrms to see that if both p and V are doubled then vrms is also doubled. rms v 3pV Nm = So the new rms speed will be 800 m/s. Assess: Think microscopically; for the pressure to double when the volume is doubled the particles will have to be going a lot faster and hit the walls more often. 18.20. Solve: The formula for the root-means-square speed as a function of temperature is ( ) B rms T v 3k T m = The ratio at 20C and at 100C is ( ) ( ) rms 100 rms 20 373 K 1.13 293 K v v = = 18.21. Solve: Assuming ideal-gas behavior and ignoring relativistic effects, the root-mean-square speed of a molecule is B rms v 3k T m = The temperature where vrms is the speed of light (c) for a hydrogen molecule ( ) ( )( ) ( ) 27 8 2 2 2 rms 12 23 B B 2 u 2 1.66 10 kg 3.0 10 m/s 7.22 10 K 3 3 3 1.38 10 J/K mv c T k k - - = = = = 18.22. Solve: (a) The average translational kinetic energy per molecule is 2 avg rms B 1 3 2 2 P = mv = k T This means Pavg doubles if the temperature T doubles. (b) The root-mean-square speed vrms increases by a factor of 2 as the temperature doubles. (c) The mean free path is ( ) 2 1 4 2 N V r . p = Because N/V and r do not depend on T, doubling temperature has no effect on .. 18.23. Solve: (a) The total translational kinetic energy of a gas is 3 3 micro 2 A B 2 K = N k T = nRT. For H2 gas at STP, ( )( )( ) micro 3 1.0 mol 8.31 J/mol K 273 K 3400 J 2 K= = (b) For He gas at STP, ( )( )( ) micro 3 1.0 mol 8.31 J/mol K 273 K 3400 J 2 K= = (c) For O2 gas at STP, micro K = 3400 J. Assess: The translational kinetic energy of a gas depends on the temperature and the number of molecules but not on the molecules mass. 18.24. Solve: (a) The mean free path is ( ) 2 1 4 2 N V r . p = where r 0.5 10-10 m is the atomic radius for helium and N/V is the gas number density. From the ideal-gas law, ( )( ) 25 3 23 0.10 atm 101,300 Pa/atm 7.34 10 m 1.38 10 J/K 10 K N p V kT - - = = = ( )( ) 7 25 3 10 2 1 3.1 10 m 310 nm 4 2 7.34 10 m 0.5 10 m . p - - - . = = = (b) The root-mean-square speed is ( )( ) ( ) 23 B rms 27 3 3 1.38 10 J/K 10 K 250 m/s 4 1.661 10 kg v k T m - - = = = where we used A = 4 u as the atomic mass of helium. (c) The average energy per atom is 3 3 ( 23 )( ) 22 avg 2 B 2 e = k T = 1.3810- J/K 10 K = 2.110- J . 18.25. Solve: (a) The average kinetic energy of a proton at the center of the sun is avg B 3 2 P = k T 3 ( 23 )( 7 ) 16 2 1.3810- J/K 2.010 K = 4.110- J (b) The root-mean-square speed of the proton is ( 23 )( 7 ) B 5 rms 27 3 3 1.38 10 J/K 2.0 10 K 7.0 10 m/s 1.67 10 kg v k T m - - = = 18.26. Solve: (a) Since the hydrogen in the suns atmosphere is monatomic, the average translational kinetic energy per atom is avg B 3 2 P = k T 3 ( 23 )( ) 19 2 = 1.3810- J/K 6000 K =1.2410- J (b) The root-mean-square speed is ( 23 )( ) B 4 rms 27 H 3 3 1.38 10 J/K 6000 K 1.22 10 m/s 1.67 10 kg v k T m - - = = = 18.27. Model: For a gas, the thermal energy is the total kinetic energy of the moving molecules. That is, Eth = Kmicro. Solve: The number of atoms is 23 27 0.0020 kg 3.01 10 6.64 10 kg N M m - = = = Because helium atoms have an atomic mass number A = 4, the mass of each helium atom is m = 4 u = 4(1.66110-27 kg) = 6.6410-27 kg The average kinetic energy of each atom is 1 2 1 ( 27 )( )2 avg 2 avg 2 K = mv = 6.6410- kg 700 m s = 1.63 10-21 J Thus the thermal energy of the gas is ( 23 )( 21 ) th micro avg E = K = NK = 3.0110 1.6310- J = 490 J 18.28. Model: For a gas, the thermal energy is the total kinetic energy of the moving molecules. Solve: Neon atoms have an atomic mass number A = 14, so the mass of each molecule is m = 28 u = 28(1.661 10-27 kg) = 4.51 10-26 kg The number of molecules in the gas is 26 0.010 kg 4.51 10 kg N M m - = = = 2.218 1023 The thermal energy is (1 2 ) th avg 2 avg E = NK = N mv ( ) ( )( ) th avg 23 26 2 2 1700 J 2.218 10 4.51 10 kg v E Nm - . = = = 580 m/s 18.29. Solve: The volume of the air is V = 6.0 m 8.0 m 3.0 m = 144.0 m3, the pressure p = 1 atm = 1.013 105 Pa, and the temperature T = 20C = 293 K. The number of moles of the gas is n pV 5991 mol RT = = This means the number of molecules is ( )( 23 1 ) 27 A N = nN = 5991 mols 6.02210 mol- = 3.6110 molecules Since air is a diatomic gas, the rooms thermal energy is ( 5 ) 7 th avg 2 B E = NP = N k T = 3.610 J Assess: The rooms thermal energy can also be obtained as follows: th V E = nC T = (5991 mol)(20.8 J/mol K)(293 K) = 3.6 107 J 18.30. Solve: The thermal energy of a solid is Eth = 3NkBT = 3nRT The volume of lead V = 100 cm3 = 10-4 m3, which means the mass is M = .V = (11,300 kg/m3)(10-4 m3) = 1.13 kg Because the atomic mass number of Pb is 207, the number of moles is mol 1.13 kg 5.459 mol 0.207 kg/mol n M M = = = ( )( )( ) 4 th . E = 3 5.459 mols 8.31 J/mol K 293 K = 3.9910 J 18.31. Solve: (a) For a monatomic gas, .Eth = nCV.T =1.0 J = (1.0 mol)(12.5 J/mol K).T ..T = 0.080C or 0.080 K (b) For a diatomic gas, 1.0 J = (1.0 mol)(20.8 J/mol K).T . .T = 0.048C or K (c) For a solid, 1.0 J = (1.0 mol)(25.0 J/mol K) .T . .T = 0.040C or K 18.32. Solve: The conservation of energy equation (.Eth)gas + (.Eth)solid = 0 J is gas ( V )gas ( f i )gas solid ( V )solid ( f i )solid n C T -T + n C T -T = 0 J . (1.0 mol)(12.5 J/mol K)(-50 K) + (1.0 mol)(25.0 J/mol K)(.T)solid = 0 . (.T)solid = 25C The temperature of the solid increases by 25C. 18.33. Visualize: Refer to Figure 18.13. At low temperatures, 3 CV = 2 R =12.5 J/mol K. At room temperature and modestly hot temperatures, 5 V 2 C = R = 20.8 J/mol K. At very hot temperatures, 7 V 2 C = R = 29.1 J/mol K. Solve: (a) The number of moles of diatomic hydrogen gas in the rigid container is 0.20 g 0.10 mol 2 g/mol = The heat needed to change the temperature of the gas from 50 K to 100 K at constant volume is th V Q = .E = nC .T = (0.10 mol)(12.5 J/mol K)(100 K 50 K) = 62 J (b) To raise the temperature from 250 K to 300 K, Q = .Eth = (0.10 mol)(20.8 J/mol K)(300 K 250 K) = 100 J (c) To raise the temperature from 550 K to 600 K, Q = 100 J. (d) To raise the temperature from 2250 K to 2300 K, Q = .Eth = nCV.T = (0.10 mol)(29.1 J/mol K)(50 K) = 150 J. 18.34. Model: The potential energy of the bowling ball is transferred into the thermal energy of the mixture. We assume the starting temperature of the bowling ball to be 0C. Solve: The potential energy of the bowling ball is 2 2 g ball U = M gh = (11 kg)(9.8 m/s )h = (107.8 kg m/s )h This energy is transferred into the mixture of ice and water and melts 5 g of ice. That is, 5 2 th w f 2 (107.8 kg m/s ) (0.005 kg)(3.33 10 J/kg) 15.4 m (107.8 kg m/s ) h E M L h = . = . = = 18.35. Model: Heating the water and the kettle raises the temperature of the water to the boiling point and raises the temperature of the kettle to 100C. Solve: The amount of heat energy from the electric stoves output in 3 minutes is Q = (2000 J s)(3 60 s) = 3.6105 J This heat energy heats the kettle and brings the water to a boil. Thus, water water kettle kettle Q = M c .T + M c .T Substituting the given values into this equation, 5 ( )( ) ( )( )( ) water 3.610 J = M 4190 J kg K 100C - 20C + 0.750 kg 449 J kg K 100C - 20C water .M = 0.994 kg The volume of water in the kettle is water 3 3 3 3 3 water 0.994 kg 0.994 10 m 994 cm 990 cm 1000 kg m V M . = = = - = Assess: 1 L = 103 cm3, so V 1 L. This is a reasonable volume of water. 18.36. Solve: The equilibrium condition is (eA)avg = (eB)avg = (etot)avg Af Bf tot A B A B E E E n n n n . = = + Thus the final thermal energies are A ( ) Af tot A B 4.0 mols 9000 J 5000 J 4.0 mols 3.0 mols E n E n n . . . . = . . = . . + . + . . + . = 8000 J B ( ) Bf tot A B 3.0 mols 14,000 J 6000 J 7.0 mols E n E n n . . . . = . . = . . = . + . . . Because EAi = 9000 J and EAf = 8000 J, 1000 J of heat energy is transferred from gas A to gas B. 18.37. Solve: The mean free path for a monatomic gas is ( ) 2 1 4 2 N V r . p = V 4 2 r2 N . = p . For . = 20r, meaning that the mean free path equals the atomic diameter, 4 2 (20 )( 2 ) 4 3 60 2 3 V r r r N p = p = .. .. . . 4 3 3 V / N 60 2 84.8 p r . = = 18.38. Visualize: Solve: The average energy of an oxygen molecule at 300 K is th 5 5 ( 23 )( ) 20 avg 2 B 2 E k T 1.38 10 J/K 300 K 1.035 10 J N e = = = - = - The energy conservation equation Ugf + Kf = Ugi + Ki with f avg K =e is f avg i mgy +e = mgy + 0 J With yi = h and yf = 0 m, we have mgh =1.03510-20 J ( )( ) 20 4 27 2 1.035 10 J 1.99 10 m 32 1.66 10 kg 9.8 m/s h - - . = = 18.39. Solve: (a) To identify the gas, we need to determine its atomic mass number A or, equivalently, the mass m of each atom or molecule. The mass density . and the number density (N/V) are related by . = m(N/V) , so the mass is m = . (V/N) . From the ideal-gas law, the number density is ( )( ) 25 3 23 50,000 Pa 1.208 10 m 1.38 10 J/K 300 K N p V kT - - = = = Thus, the mass of an atom is 2 3 27 25 3 8.02 10 kg/m 6.64 10 kg 1.208 10 m m V N . - - - = = = Converting to atomic mass units, 27 27 6.64 10 kg 1 u 4.00 u 1.661 10 kg A - - = = This is the atomic mass of helium. (b) Knowing the mass, we find vrms to be ( 23 )( ) B rms 27 3 3 1.38 10 J/K 300 K 1370 m/s 6.64 10 kg v k T m - - = = = (c) A typical atomic radius is r 0.5 10-10 m. The mean free path is thus ( ) ( )( ) 6 2 25 3 10 2 1 1 1.86 10 m 1.86 m 4 2 N V r 4 2 1.208 10 m 0.5 10 m . p p - - - = = = = 18.40. Solve: (a) The number density is N /V =1 cm-3 =106 m-3. Using the ideal-gas equation, ( 6 3 )( 23 )( ) B 17 22 5 1 10 m 1.38 10 J/K 3 K 4 10 Pa 1 atm 4 10 atm 1.013 10 Pa p N k T V - - - - = = = (b) For a monatomic gas, ( 23 )( ) B rms 27 3 3 1.38 10 J/K 3 K 1.67 10 kg v k T m - - = = = 270 m/s (c) The thermal energy is 3 th 2 B E = Nk T , where N = (106 m-3)V . Thus 3 ( 6 3 ) ( 23 )( ) th 2 E =1.0 J = 10 m- V 1.3810- J/K 3 K .V =1.61016 m3 = L3 . L = 2.5105 m 18.41. Solve: Mass m of a dust particle is ( ) 3 43 m = .V = . p r ( ) ( )3 3 6 12 43 = 2500 kg/m ... p 510- m ... =1.310- kg The root-mean-square speed of the dust particles at 20C is ( 23 )( ) B 5 rms 12 3 3 1.38 10 J/K 293 K 9.6 10 m/s 1.3 10 kg v k T m - - - = = = 18.42. Solve: Fluorine has atomic mass number A = 19. Thus the root-mean-square speed of 238UF6 is ( 238 ) B B rms 6 UF 3 3 238 u 6 19 u v k T k T m = = + The ratio of the root-mean-square speed for the molecules of this isotope and the 235UF6 molecules is ( ) ( ) ( ) ( ) 235 rms 6 238 rms 6 UF 238 6 19 u 352 1.0043 UF 235 6 19 u 349 v v + = = = + 18.43. Solve: (a) If the electron can be thought of as a point particle with zero radius, then it will collide with any gas particle that is within r of its path. Hence, the number of collisions Ncoll is equal to the number of gas particles in a cylindrical volume of length L. The volume of a cylinder is ( 2 ) cyl V = AL = p r L . If the number density of the gas is (N/V) particles per m3, then the number of collisions along a trajectory of length L is ( ) ( ) 2 coll cyl electron 2 coll 1 / N N V N r L L V V N NVr p . p = = . = = Introducing a factor of 2 to account for the motion of all particles, electron ( ) 2 coll 1 2 / L N N V r . p = = (b) Assuming that most of the molecules in the accelerator are diatomic, ( )( ) 4 14 3 10 2 5.0 10 m 1 / 4.50 10 m 2 / 1.0 10 m N V p N V - - = . = From the ideal-gas equation, ( 14 3 )( 23 )( ) 6 11 B p N k T 4.50 10 m 1.38 10 J/K 293 K 1.82 10 Pa 1.80 10 atm V = = - - = - = - 18.44. Solve: The pressure on the wall with area A = 10 cm2 = 10 10-4 m2 is p F (mv)N A At . = = . where N .t is the number of N2 molecules colliding with the wall every second and .(mv) is the change in momentum for one collision. The mass of the nitrogen molecule is m = 28 u = 28 (1.66 10-27 kg) = 4.648 10-26 kg and .v = 400 m/s - (-400 m/s) = 800 m/s . Thus, ( 26 )( )( 23 1 ) 4 3 2 4.648 10 kg 800 m/s 5.0 10 s 1.9 10 Pa 1.0 10 m p - - - = = 18.45. Solve: (a) The cylinder volume is V = pr2L = 1.571 103 m3. Thus the number density is 22 25 3 25 3 3 3 2.0 10 1.273 10 m 1.3 10 m 1.571 10 m N V - - - = = (b) The mass of an argon atom is m = 40 u = 40(1.661 10-27 kg) = 6.64 10-26 kg ( 23 )( ) B rms 26 3 3 1.38 10 J/K 323 K 449 m/s 450 m/s 6.64 10 kg v k T m - - . = = = (c) vrms is the square root of the average of v2. That is, 2 ( 2 ) ( 2 ) ( 2 ) ( 2 ) rms avg x avg y avg z avg v = v = v + v + v An atom is equally likely to move in the x, y, or z direction, so on average ( 2 ) ( 2 ) ( 2 ) x avg y avg z avg v = v = v . Hence, 2 ( 2 ) ( ) ( 2 ) rms rms avg rms avg 3 259 m/s 260 m/s 3 x x x v v = v . v = v = = (d) When we considered all the atoms to have the same velocity, we found the collision rate to be 1 2( / ) x NV Av (see Equation 18.10). Because the atoms move with different speeds, we need to replace vx with (vx)rms. The end of the cylinder has area A = pr2 = 7.85 103 m2. Therefore, the number of collisions per second is 1 ( ) ( ) 1 ( 25 3 )( 3 2 )( ) 2 rms 2 / 1.273 10 m 7.85 10 m 259 m/s x N V A v = - - = 1.3 1025 s-1 (e) From kinetic theory, the pressure is 1 ( 2 ) 1 2 3 avg 3 rms p N m v N mv V V = . . = . . . . . . . . . . 1 ( 25 3 )( 26 )( )2 3 = 1.27310 m- 6.6410- kg 449 m/s = 56,800 Pa 57,000 Pa (f) From the ideal-gas law, the pressure is ( 22 )( 23 )( ) B 3 3 2.0 10 1.38 10 J/K 323 K 56,700 Pa 57,000 Pa 1.571 10 m p Nk T V - - = = = Assess: The very slight difference between parts (e) and (f) is due to rounding errors; to two significant figures they are the same. 18.46. Model: Pressure is due to random collisions of gas molecules with the walls. Solve: According to Equation 18.9, the collision rate with one wall is coll coll rate of collisions 1 2 x N NAv t V = = . However, this equation assumed that all molecules are moving in the x-direction with constant speed. The rms speed vrms is for motion in three dimensions at varying speeds. Consequently, we need to replace vx not with (vx)avg, which is zero, but with 2 2 rms rms avg ( ) 3 3 x x v . v = v = v With this change, rms rate of collisions 1 2 3 N Av V = The molecular mass of nitrogen is A = 28 u, thus the rms speed of the molecules at 20C is 23 B rms 27 3 3(1.38 10 J/K)(293 K) 510 m/s 28(1.661 10 kg) v k T m - = = = With N = 0.10NA = 6.02 1022 molecules, the number density is 22 6.02 10 6.02 1025 m 3 0.10 m 0.10 m 0.10 m N V - = = Thus rate of collisions 1 (6.02 1025 m 3 )(0.10 m 0.10 m)(510 m/s)=8.9 1025 s 1 2 3 = - - 18.47. Model: Assume the gas is ideal so that Equation 18.30 will apply. .Eth = nCV.T Visualize: We see from the graph that .T = 200 K and th .E = 800 J. We are also given n = 0.14 mol. Solve: Solve for VC . th V 800 J 29 J/mol K (0 14 mol)(200 K) C E n T . = = = . . Assess: This is about 7 2 R. 18.48. Solve: (a) The number of molecules of helium is ( )( ) ( )( ) 5 63 21 helium 23 B 2.0 1.013 10 Pa 100 10 m 3.936 10 1.38 10 J/K 373 K N pV k T - - = = = 21 3 helium 23 1 3.936 10 6.536 10 mol 6.022 10 mol n - - . = = The initial internal energy of helium is 3 helium i 2 helium B E = N k T = 30.4 J 30 J The number of molecules of argon is ( )( ) ( )( ) 5 63 21 argon 23 B 4.0 1.013 10 Pa 200 10 m 8.726 10 1.38 10 J/K 673 K N pV k T - - = = = 21 2 argon 23 1 8.726 10 1.449 10 mol 6.022 10 mol n - - . = = The initial thermal energy of argon is 3 argon i 2 argon B E = N k T =121.6 J 122 J (b) The equilibrium condition for monatomic gases is (ehelium f)avg = (eargon f)avg = (etotal)avg ( ) ( ) helium f argon f tot 3 2 helium argon tot 30.4 121.6 J 7228 J/mol 6.54 10 1.449 10 mol E E E n n n - - + . = = = = + ( ) ( )( 3 ) helium f helium . E = 7228 J/mol n = 7228 J/mol 6.5410- mol = 47.3 J 47 J ( ) ( )( 2 ) argon f argon E = 7228 J/mol n = 7228 J/mol 1.44910- mol =104.7 J 105 J (c) The amount of heat transferred is helium f helium i E - E = 47.3 J - 30.4 J =16.9 J argon f argon i E - E =104.7 J -121.6 J = -16.9 J The helium gains 16.9 J of heat energy and the argon loses 16.9 J. Thus approximately 17 J are transferred from the argon to the helium. (d) The equilibrium condition for monatomic gases is ( ) ( ) helium avg argon avg e = e helium f argon f 3 2 B f helium argon E E k T N N . = = Substituting the above values, 3 ( 23 ) 21 21 2 f 47.3 J 104.7 J 1.38 10 J/K 3.936 10 8.726 10 = = - T 580 K 307 C F .T = = (e) The final pressure of the helium and argon are ( 21 )( 23 )( ) helium B helium f 6 3 helium 3.936 10 1.38 10 J/K 580 K 100 10 m p N k T V - - = = = 3.15 105 Pa 3.1 atm ( 21 )( 23 )( ) argon B argon f 6 3 argon 8.726 10 1.38 10 J/K 580 K 200 10 m N kT p V - - = = = 3.49 105 Pa 3.4 atm 18.49. Solve: (a) The number of moles of helium and oxygen are helium 2.0 g 0.50 mol 4.0 g/mol n = = oxygen 8.0 g 0.25 mol 32.0 g/mol n = = Since helium is a monoatomic gas, the initial thermal energy is ( 3 ) helium i helium 2 helium E = n RT ( )( 3 )( )( ) 2 = 0.50 mol 8.31 J/mol K 300 K =1870 J 1900 J Since oxygen is a diatomic gas, the initial thermal energy is ( 5 ) oxygen i oxygen 2 oxygen E = n RT ( )( 5 )( )( ) 2 = 0.25 mol 8.31 J/mol K 600 K = 3116 J 3100 J (b) The total initial thermal energy is Etot = Ehelium i + Eoxygen i = 4986 J As the gases interact, they come to equilibrium at a common temperature Tf. This means ( 3 ) ( 5 ) helium 2 f oxygen 2 f 4986 J = n RT + n RT f ( 1 )( ) 1 ( )( ) 2 helium oxygen 2 4986 J 4986 J 3 5 8.31 J/mol K 3 0.50 mol 5 0.25 mol T R n n . = = = + + 436.4 K = 436 K The thermal energies at the final temperature Tf are ( 3 ) helium f helium 2 f E = n RT ( 3 )( )( )( ) 2 = 0.50 mol 8.31 J/mol K 436.4 K = 2700 J ( 5 ) oxygen f oxygen 2 f E = n RT ( 5 )( )( )( ) 2 = 0.25 mol 8.31 J/mol K 436.4 K = 2300 J (c) The change in the thermal energies are helium f helium i E - E = 2720 J -1870 J = 850 J oxygen f oxygen i E - E = 2266 J - 3116 J = -850 J The helium gains energy and the oxygen loses energy. (d) The final temperature can also be calculated as follows:
( )3 helium f helium 2 f E = n RT . 2720 J = (0.50 mol)(1.5)(8.31 J/mol K)Tf f .T = 436.4 K 436 K 18.50. Solve: The thermal energy of any ideal gas is related to the molar specific heat at constant volume by Eth = nCVT Since CP = CV + R , V V 20.8 J/mol K = C + R.C =12.5 J/mol K The number of moles of the gas is 20 4 23 1 1.0 10 1.66 10 mol 6.02 10 mol n - - = = Thus ( ) ( 4 )( ) 1.0 J 482 K 1.66 10 mols 12.5 J/mol K T - = = 18.51. Solve: For a system with n degrees of freedom, the molar specific heat is CV = nR/2. The specific heat ratio is P V V V V C C R 1 R C C C . + = = = + V R 1 1.29 1 0.29 C . =. - = - = 7 V 2 3.5 0.29 .C = R = R = R Thus, the system has 7 degrees of freedom. 18.52. Solve: As the volume V1 of a gas increases to V2 = 2V1 at a constant pressure p1 = p2, the temperature of the gas changes from T1 to T2 as follows: 2 2 1 1 2 1 p V pV T T = 2 2 2 1 1 1 1 T T V p 2T V p . = = Since the process occurs at constant pressure the heat transferred is ( ) ( ) P P 2 1 P 1 1 P1 Q = nC .T = nC T -T = nC 2T -T = nC T For a monatomic gas, 3 5 P V 2 2 C = C + R = R + R = R For the diatomic gas, 5 7 P V 2 2 C = C + R = R + R = R Thus ( ) ( ) 7 diatomic 2 1 5 monatomic 2 1 1.40 Q n R T Q nRT = = 18.53. Solve: Assuming that the systems are thermally isolated except from each other, the total energy for the two thermally interacting systems must remain the same. That is, E1i + E2i = E1f + E2f ( ) 1f 1i 2i 2f 2f 1f .E - E = E - E = - E - E 1 2 ..E = -.E No work is done on either system, so from the first law .E = Q. Thus 1 2 Q = -Q That is, the heat lost by one system is gained by the other system. 18.54. Solve: (a) From Equation 18.26 vrms = 3kBT m. For an adiabatic process 1 f f TV. - = 1 i i TV. - f .T 1 i i f T V V . - . . = . . . . ( )5 3 1 f i i T T 8 4T - . = = The root-mean-square speed increases by a factor of 2 with an increase in temperature. (b) From Equation 18.3 . = [4 2p (N /V )r2 ]-1. A decrease in volume decreases the mean free path by a factor of 1/8. (c) For an adiabatic process, 1 1 f f i i T V. - = TV. - ( )5 3 1 i 1 f i i i f T T V T 8 4T V . - . . - . = . . = = . . Because the decrease in volume increases Tf , the thermal energy increases by a factor of 4. (d) The molar specific heat at constant volume is 3 V 2 C = R, a constant. It does not change. 18.55. Model: Assume the gas is monatomic. Visualize: From the equipartition theorem there is 1 2 nRT of energy for each degree of freedom. For a twodimensional monatomic gas there are only two degrees of freedom. Solve: (a) V 2 2 C = R = R (b) Equation 17.34 gives p V C = C + R , so p C = R + R = 2R Assess: It would be nice to measure the values and compare with these predictions. 18.56. Solve: (a) The thermal energy of a monatomic gas of N molecules is Eth = NPavg , where 3 avg 2 B P = k T. A monatomic gas molecule has 3 degrees of freedom. However, a two-dimensional monatomic gas molecule has only 2 degrees of freedom. Thus, th B B 2 2 E = N .. k T .. = Nk T = nRT . . If the temperature changes by .T, then the thermal energy changes by th .E = nR.T . Comparing this form with th V .E = nC .T, we have CV = R = 8.31 J/mol K. (b) A two-dimensional solid has 2 degrees of freedom associated with the kinetic energy and 2 degrees of freedom associated with the potential energy (or 2 spring directions), giving a total of 4 degrees of freedom. Thus, Eth = avg NP and 4 avg 2 B P = k T. Or Eth = 2NkBT = 2nRT. For a temperature change of .T, .Eth = 2nR.T = nCV.T V .C = 2R =16.6 J/mol K. 18.57. Solve: (a) The rms speed is B rms v 3k T m = rms hydrogen rms oxygen 32 u 4 2 u v v . = = (b) The average translational energy is 3 2 B P = k T. Thus avg hydrogen hydrogen avg oxygen oxygen 1 T T P = = P (c) The thermal energy is 5 th 2 E = nRT th hydrogen hydrogen hydrogen th oxygen oxygen oxygen 32.0 g/mol 16 2.0 g/mol E n m E n m . = = = 18.58. Visualize: We are given vrms =1800 m/s. For 1.0 g of molecular hydrogen gas 1.0 g 1 mol 0.5 mol 6.02 1023 molecules 3.01 1023 molecules 2.0 g mol N . . . . = . . = . . = . . . . The mass of one molecule is 2(1.66110-27 kg) = 3.32210-27 kg. Solve: (a) 2 ( 23 ) ( 27 )( )2 total rms 1 3.01 10 1 3.322 10 kg 1800 m/s 1.6 kJ 2 2 . = N .. ..mv = .. .. - = . . . . (b) The temperature is given by Equation 18.25. ( )( ) 27 2 2 ave rms 23 B B 2 2 1 3.322 10 kg 1800 m/s 260 K 3 3 2 3 1.38 10 J/K T mv k k - - . . = . = . . = = . . For diatomic gases Equation 18.37 gives the thermal energy. ( )( )( ) th 5 50.50 mol 8.31 J/mol K 260 K 2.7 kJ 2 2 E = nRT = = (c) There is a net loss of 700 J of energy from the system, so the new th E = 2000 J. Solve Equation 18.37 for T. ( )( )( ) th 2 2 2000 J 192.5 K 5 5 0.50 mol 8.31 J/mol K T E nR = = = Now Equation 18.26 gives the new rms speed. ( 23 )( ) B rms 27 3 3 1.38 10 J/K 192.5 K 1549 m/s 1500 m/s 3.322 10 kg v k T m - - = = = Assess: With the net loss of energy we expected rms v to decrease as the temperature decreased. 18.59. Solve: (a) The escape speed is the speed with which a mass m can leave the earths surface and escape to infinity (rf = 8) with no left over speed (vf = 0). The conservation of energy equation Kf + Uf = Ki + Ui is 1 2 e e 2 esc esc e e 0 0 mv GM m v 2GM R R + = - . = The rms speed of a gas molecule is vrms = (3kBT/m)1/2. Equating vesc and vrms, and squaring both sides, the temperature at which the rms speed equals the escape speed is B e e e Be 3 2 2 3 k T GM T m GM m R kR . . = . = . . . . For a nitrogen molecule, with m = 28 u, the temperature is 11 2 2 24 27 23 6 (28 1.661 10 kg) 2(6.67 10 N m / kg )(5.98 10 kg) 141,000 K 3(1.38 10 J/K)(6.37 10 m) T - - - . . = . . = . . (b) For a hydrogen molecule, with m = 2 u, the temperature is less by a factor of 14, or T = 10,100 K. (c) The average translational kinetic energy of a molecule is 3 avg 2 B e = k T = 6.1 1021 J at a typical atmosphere temperature of 20C. The kinetic energy needed to escape is 1 2 esc 2 esc K = mv . For nitrogen molecules, Kesc = 2.9 1018 J. Thus eavg/Kesc = 0.002 = 0.2%. Earth will retain nitrogen in its atmosphere because the molecules are moving too slowly to escape. But for hydrogen molecules, with Kesc = 2.1 1019 J, the ratio is eavg/Kesc = 0.03 = 3%. Thus a large enough fraction of hydrogen molecules are moving at escape speed, or faster, to allow hydrogen to leak out of the atmosphere into space. Consequently, earths atmosphere does not contain hydrogen. 18.60. Solve: (a) The thermal energy of a monatomic gas of n1 moles is 3 1 2 1 E = n RT. The thermal energy of a diatomic gas of n2 moles is 5 2 2 2 E = n RT . The total thermal energy of the mixture is 1 ( ) th 2 1 2 E = 3n + 5n RT 1 ( ) th 2 1 2 ..E = 3n + 5n R.T Comparing this expression with ( ) th total V 1 2 V .E = n C .T = n + n C .T we get ( ) ( ) 1 2 V 1 2 3 5 2 n n C R n n + = + (b) For a diatomic gas, n1 . 0, and 5 V 2 C = R . For a monotomic gas, n2 . 0, and 3 V 2 C = R . 18.61. Solve: (a) The thermal energy is ( ) ( ) 2 2 2 2 5 5 5 th th N th O 2 N B 2 O B 2 total B E = E + E = N k T + N k T = N k T where Ntotal is the total number of molecules. The identity of the molecules makes no difference since both are diatomic. The number of molecules in the room is ( )( ) ( )( ) 26 total 23 B 101,300 Pa 2 m 2 m 2 m 2.15 10 1.38 10 J/K 273 K N pV k T - = = = The thermal energy is 5 ( 26 )( 23 )( ) 6 6 th 2 E = 2.1510 1.3810- J/K 273 K = 2.0310 J 2.010 J (b) A 1 kg ball at height y = 1 m has a potential energy U = mgy = 9.8 J. The ball would need 9.8 J of initial kinetic energy to reach this height. The fraction of thermal energy that would have to be conveyed to the ball is 6 6 9.8 J 4.8 10 2.03 10 J = - (c) A temperature change .T corresponds to a thermal energy change 5 th 2 total B .E = N k .T . But 5 2 total B th N k = E T . Using this, we can write th th th 6 th 9.8 J 273 K 0.0013 K 2.03 10 J E E T T E T T E . - . = . .. = = =- The room temperature would decrease by 0.0013 K or 0.0013C. (d) The situation with the ball at rest on the floor and in thermal equilibrium with the air is a very probable distribution of energy and thus a state with high entropy. Although energy would be conserved by removing energy from the air and transferring it to the ball, this would be a very improbable distribution of energy and thus a state of low entropy. The ball will not be spontaneously launched from the ground because this would require a decrease in entropy, in violation of the second law of thermodynamics. As another way of thinking about the situation, the ball and the air are initially at the same temperature. Once even the slightest amount of energy is transferred from the air to the ball, the airs temperature will be less than that of the ball. Any further flow of energy from the air to the ball would be a situation in which heat energy is flowing from a colder object to a hotter object. This cannot happen because it would violate the second law of thermodynamics. 18.62. Solve: This is not a wise investment. Although an invention to move energy from the hot brakes to the forward motion of the car would not violate energy conservation, it would violate the second law of thermodynamics. The forward motion of the car is a very improbable distribution of energy. It happens only when a force accelerates the car and then sustains the motion against the retarding forces of friction and air resistance. The moving car is a state of low entropy. By contrast, the random motion of many atoms in the hot brakes is a state of high probability and high entropy. To convert the random motion of the atoms in the brakes back into the forward motion of the car would require a decrease of entropy and thus would violate the second law of thermodynamics. In other words, the increasing temperature of the brakes as the car stops is an irreversible process that cannot be undone. 18.63. Solve: (a) We are given that ( ) B i rms i v 3k T m = ( ) B f ( ) rms f rms i v 3k T 2 v m = = This means that Tf = 4Ti. Using the ideal-gas law, it also means that pfVf = 4piVi. Since the pressure is directly proportional to the volume during the process, we have pi /Vi = pf /Vf. Combining these two equations gives pf = 2pi and Vf = 2Vi. (b) The change in thermal energy for any ideal gas process is related to the molar specific heat at constant volume by ( ) th V f i .E = nC T -T . The work done on the gas is W = -. pdV = - (area under the p-versus-V graph) 3 2 i i = - pV The first law of thermodynamics .Eth = Q + W can be written ( ) ( ) 3 3 th V f i 2 i i V i 2 i i 5 3 15 3 2 i 2 i i 2 i i 2 i i i i 3 3 9 Q E W nC T T pV nCT pV n R T pV pV pV pV = . - = - + = + = + = + = 18.64. Solve: The thermal energy of a monatomic gas of n1 moles is 3 1 2 1 E = n RT . The thermal energy of a diatomic gas of n2 moles is 5 2 2 2 E = n RT . The total thermal energy of the mixture is 1 ( ) th 1 2 2 1 2 E = E + E = 3n + 5n RT 1 ( ) th 2 1 2 ..E = 3n + 5n R.T Comparing this expression with ( ) th total V 1 2 V .E = n C .T = n + n C .T we get ( ) ( ) 1 2 1 2 V 3 5 2 n n n n C R + + = The requirement that the ratio of specific heats is 1.50 means P V V V V V 1.50 C C R 1 R C 2R C C C . + = = = = + . = The above equation is then ( )( ) ( ) 1 2 1 2 1 2 1 2 1 2 3 5 2 4 4 3 5 2 n n n n R R n n n n n n + + = . + = + . = Thus, monatomic and diatomic molecules need to be mixed in the ratio 1:1. Or the fraction of the molecules that are monatomic needs to be 1 2 . 18.65. Solve: (a) The thermal energy of a monatomic gas of n1 moles at an initial temperature T1i is 1i E = 3 2 1 1i n RT . The thermal energy of a diatomic gas of n2 moles at an initial temperature T2i is 5 2i 2 2 2i E = n RT . Consequently, the total initial energy is ( ) 1 1i 2 2i 1 1i 2 2i tot 1i 2i 3 5 3 5 2 2 n RT n RT nT n T R E E E + + = + = = After the gases interact, they come to equilibrium with T1f = T2f = Tf. Then their total energy is ( ) 1 2 f tot 3 5 2 n n RT E + = No energy is lost, so these two expressions for Etot must be equal. Thus, ( ) ( ) 1 1i 2 2i 1 2 f 3 5 3 5 2 2 nT + n T R n + n RT = ( ) ( ) ( ) 1 1i 2 2i tot 1i 2i f 1 2 1 2 1 2 3 5 2 2 3 5 3 5 3 5 n T n T E E E T n n R n n R n n + + . = = = + + + The thermal energies at the final temperature Tf are 3 1f 2 1 f E = n RT ( ) 3 1i 2i 2 1 1 2 2 3 5 n R E E R n n + = + 1 ( ) 1i 2i 1 2 3 3 5 n E E n n . . = . . + . + . 5 5 1i 2i 2f 2 2 f 2 2 1 2 2 3 5 E E E n RT n RR n n + = = + 2 ( ) 1i 2i 1 2 5 3 5 n E E n n . . = . . + . + . (b) In part (a) we found that 1 1i 2 2i f 1 2 3 5 3 5 T nT n T n n + = + (c) 2 g of He at T1i = 300 K are n1 = (2 g)/(4 g/mol) = 0.50 mol. Oxygen has an atomic mass of 16, so the molecular mass of oxygen gas (O2) is A = 32 g/mol. 8 g of O2 at T2i = 600 K are n2 = (8 g)/(32 g/mol) = 0.25 mol. The final temperature is ( )( ) ( )( ) f ( ) ( ) 3 0.50 mol 300 K 5 0.25 mol 600 K 436 K 3 0.50 mol 5 0.25 mol T + = = + The heat flows to the two gases are found from Q = nCV.T. For helium, ( )( )( ) 1 V Q = nC .T = 0.50 mol 12.5 J/mol K 436 K - 300 K = 850 J For O2, ( )( )( ) 2 V Q = n C .T = 0.25 mol 20.8 J/mol K 436 K - 600 K = -850 J So 850 J of heat is transferred from the oxygen to the helium. 19.1. Solve: (a) The engine has a thermal efficiency of . = 40% = 0.40 and a work output of 100 J per cycle. The heat input is calculated as follows: out H H H W 0.40 100 J Q 250 J Q Q .= . = . = (b) Because out H C W = Q -Q , the heat exhausted is C H out Q = Q -W = 250 J -100 J =150 J 19.2. Solve: During each cycle, the work done by the engine is Wout = 20 J and the engine exhausts C Q = 30 J of heat energy. Because out H C W = Q -Q , H out C Q =W +Q = 20 J + 30 J = 50 J Thus, the efficiency of the engine is C H 1 1 30 J 0.40 50 J Q Q . = - = - = 19.3. Solve: (a) During each cycle, the heat transferred into the engine is QH = 55 kJ , and the heat exhausted is C Q = 40 kJ . The thermal efficiency of the heat engine is C H 1 1 40 kJ 0.27 27% 55 kJ Q Q . = - = - = = (b) The work done by the engine per cycle is out H C W = Q -Q = 55 kJ - 40 kJ = 15 kJ 19.4. Solve: The coefficient of performance of the refrigerator is C H in in in 50 J 20 J 1.5 20 J K Q Q W W W - - = = = = 19.5. Solve: (a) The heat extracted from the cold reservoir is calculated as follows: C C C in 4.0 200 J 50 J K Q Q Q W = . = . = (b) The heat exhausted to the hot reservoir is H C in 2 Q = Q +W = 00 J + 50 J = 250 J 19.6. Model: Assume that the car engine follows a closed cycle. Solve: (a) Since 2400 rpm is 40 cycles per second, the work output of the car engine per cycle is out 500 kJ 1 s 12.5 kJ s 40 cycles cycle W = = (b) The heat input per cycle is calculated as follows: out H H 12.5 kJ 62.5 kJ 0.20 W Q Q .= . = = The heat exhausted per cycle is C H in Q = Q -W = 62.5 kJ -12.5 kJ = 50 kJ 19.7. Solve: The amount of heat discharged per second is calculated as follows: out out ( ) 9 C out H C out 1 1 900 MW 1 1 1.913 10 W 0.32 W W Q W Q Q W . . . . . . = = . = . - . = . - . = + . . . . That is, each second the electric power plant discharges 1.913109 J of energy into the ocean. Since a typical American house needs 2.0104 J of energy per second for heating, the number of houses that could be heated with the waste heat is (1.913109 J) (2.0104 J) = 96,000 . 19.8. Solve: The amount of heat removed from the water in cooling it down in 1 hour is QC = mwatercwater.T. The mass of the water is ( 3 )( ) ( 3 )( 3 3 ) water water water m = . V = 1000 kg/m 1 L = 100 kg/m 10- m =1.0 kg ( )( )( ) 4 C .Q = 1.0 kg 4190 J/kg K 20C - 5C = 6.28510 J The rate of heat removal from the refrigerator is 4 C 6.285 10 J 17.46 J/s 3600 s Q = = The refrigerator does work W = 8.0. W = 8.0 J/s to remove this heat. Thus the performance coefficient of the refrigerator is 17.46 J/s 2.2 8.0 J/s K= = 19.9. Model: Process A is isochoric, process B is isothermal, process C is adiabatic, and process D is isobaric. Solve: Process A is isochoric, so the increase in pressure increases the temperature and hence the thermal energy. Because .Eth = Q -Ws and Ws = 0 J , Q increases for process A. Process B is isothermal, so T is constant and hence th .E = 0 J. The work done Ws is positive because the gas expands. Because s th Q =W + .E , Q is positive for process B. Process C is adiabatic, so Q = 0 J. Ws is positive because of the increase in volume. Since s th Q = 0 J =W + .E , th .E is negative for process C. Process D is isobaric, so the decrease in volume leads to a decrease in temperature and hence a decrease in the thermal energy. Due to the decrease in volume, Ws is negative. Because s th Q =W + .E , Q also decreases for process D. .Eth Ws Q A + 0 + B 0 + + C - + 0 D - - - 19.10. Model: Process A is adiabatic, process B is isothermal, and process C is isochoric. Solve: Process A is adiabatic, so Q = 0 J. Work Ws is positive as the gas expands. Since Q = Ws + .Eth = 0 J, .Eth must be negative. The temperature falls during an adiabatic expansion. Process B is isothermal so .T = 0 and .Eth = 0 J. The gas is compressed, so Ws is negative. Q = Ws for an isothermal process, so Q is negative. Heat energy is withdrawn during the compression to keep the temperature constant. Process C is isochoric. No work is done (Ws = 0 J), and Q is positive as heat energy is added to raise the temperature (.Eth positive). .Eth Ws Q A + 0 B 0 C + 0 + 19.11. Solve: The work done by the gas per cycle is the area inside the closed p-versus-V curve. The area inside the triangle is ( )( ) ( ) 1 6 3 6 3 out 2 5 1 6 3 2 3 atm 1 atm 600 10 m 200 10 m 2 atm 1.013 10 Pa 400 10 m 40.5 J 1 atm W - - - = - - . . = . . = . . 19.12. Solve: The work done by the gas per cycle is the area enclosed within the pV curve. We have 1 ( )( 3 3 ) 2 max 60 J = p -100 kPa 800 cm - 200 cm ( ) 5 6 3 max 2 60 J 1.0 10 Pa 600 10 m p - . = - 5 max . p = 3.010 Pa = 300 kPa 19.13. Model: The heat engine follows a closed cycle, starting and ending in the original state. The cycle consists of three individual processes. Solve: (a) The work done by the heat engine per cycle is the area enclosed by the p-versus-V graph. We get 1 ( )( 6 3 ) out 2 W = 200 kPa 10010- m =10 J The heat energy transferred into the engine is QH = 120 J. Because out H C W = Q -Q , the heat energy exhausted is C H out Q = Q -W =120 J -10 J =110 J (b) The thermal efficiency of the engine is out H 10 J 0.0833 120 J W Q . = = = Assess: Practical engines have thermal efficiencies in the range . 0.1- 0.4 . 19.14. Model: The heat engine follows a closed cycle, which consists of four individual processes. Solve: (a) The work done by the heat engine per cycle is the area enclosed by the p-versus-V graph. We get ( )( 6 3 ) out W = 400 kPa -100 kPa 10010- m = 30 J The heat energy leaving the engine is C Q = 90 J + 25 J =115 J . The heat input is calculated as follows: out H C H C out W = Q -Q .Q = Q +W =115 J + 30 J =145 J (b) The thermal efficiency of the engine is out H 30 J 0.207 145 J W Q . = = = Assess: Practical engines have thermal efficiencies in the range . 0.1- 0.4 . 19.15. Model: The heat engine follows a closed cycle. Solve: The work done by the gas per cycle is the area inside the closed p-versus-V curve. We get 1 ( )( 3 3 ) out 2 W = 300 kPa -100 kPa 600 cm - 300 cm 1 ( 3 )( 6 3 ) 2 = 20010 Pa 30010- m = 30 J Because out H C W = Q -Q , the heat exhausted is C H out Q = Q -W = (225 J + 90 J) - 30 J = 315 J - 30 J = 285 J 19.16. Model: The heat engine follows a closed cycle. Solve: (a) The work done by the gas per cycle is the area inside the closed p-versus-V curve. We get 1 ( )( 3 3 ) out 2 W = 300 kPa -100 kPa 600 cm - 200 cm 1 ( 3 )( 6 3 ) 2 = 20010 Pa 40010- m = 40 J The heat exhausted is C Q =180 J +100 J = 280 J . The thermal efficiency of the engine is out out H C out 40 J 0.125 280 J 40 J W W Q Q W . = = = = + + (b) The heat extracted from the hot reservoir is QH = C out Q +W = 320 J. 19.17. Model: The Brayton cycle involves two adiabatic processes and two isobaric processes. Solve: From Equation 19.21, the efficiency of a Brayton cycle is (1 ) B p 1 r . = - -. . , where rp is the pressure ratio pmax/pmin. The specific heat ratio for a diatomic gas is 7 P 2 5 V 2 R 1.4 R C C . = = = Solving the above equation for rp, ( ) (1 )/ B p 1 r -. = -. . ( ) ( ) 1.4 1 0.4 p B r 1 1 0.60 25 . . . - . = - - = - = 19.18. Model: The Brayton cycle involves two adiabatic processes and two isobaric processes. The adiabatic processes involve compression and expansion through the turbine. Solve: The thermal efficiency for the Brayton cycle is (1 )/ B p 1 r . = - -. . , where P V . = C /C and rp is the pressure ratio. For a diatomic gas . = 1.4. For an adiabatic process, ( ) 1 1 2 2 2 1 1 2 pV p V p / p V /V . = . . = . Because the volume is halved, 1 2 2 1 V = V and hence ( ) 1.4 p 2 1 r p / p 2 2 2.639 . = = = = The efficiency is ( ) 0.4 1.4 B . 1 2.639 0.242 - = - = 19.19. Model: The efficiency of a Carnot engine (.Carnot) depends only on the temperatures of the hot and cold reservoirs. On the other hand, the thermal efficiency (.) of a heat engine depends on the heats QH and QC. Solve: (a) According to the first law of thermodynamics, QH =Wout +QC . For engine (a), QH = 50 J, QC = 20 J and Wout = 30 J, so the first law of thermodynamics is obeyed. For engine (b), QH = 10 J, QC = 7 J and Wout = 4 J, so the first law is violated. For engine (c) the first law of thermodynamics is obeyed. (b) For the three heat engines, the maximum or Carnot efficiency is C Carnot H 1 1 300 K 0.50 600 K T T . = - = - = Engine (a) has C out H H 1 30 J 0.60 50 J Q W Q Q . = - = = = This is larger than .Carnot, thus violating the second law of thermodynamics. For engine (b), out Carnot H 4 J 0.40 10 J W Q . = = = <. so the second law is obeyed. Engine (c) has a thermal efficiency that is Carnot 10 J 0.333 30 J . = = <. so the second law of thermodynamics is obeyed. 19.20. Model: For a refrigerator QH = QC + Win, and the coefficient of performance and the Carnot coefficient of performance are C in K Q W = C Carnot H C K T T T = - Solve: (a) For refrigerator (a) H C in(60 J 40 J 20 J) Q Q W = + = + , so the first law of thermodynamics is obeyed. For refrigerator (b) 50 J = 40 J +10 J , so the first law of thermodynamics is obeyed. For the refrigerator (c) 40 J . 30 J + 20 J , so the first law of thermodynamics is violated. (b) For the three refrigerators, the maximum coefficient of performance is C Carnot H C 300 K 3 400 K 300 K K T T T = = = - - For refrigerator (a), C Carnot in 40 J 2 20 J K Q K W = = = < so the second law of thermodynamics is obeyed. For refrigerator (b), C Carnot in 40 J 4 10 J K Q K W = = = > so the second law of thermodynamics is violated. For refrigerator (c), Carnot 30 J 1.5 20 J K = = < K so the second law is obeyed. 19.21. Model: The efficiency of a Carnot engine depends only on the absolute temperatures of the hot and cold reservoirs. Solve: The efficiency of a Carnot engine is C Carnot H 1 T T . = - ( ) 0.60 1 C 427 273 K . = - T + C .T = 280 K = 7C Assess: A real engine would need a lower temperature than 7oC to provide 60% efficiency because no real engine can match the Carnot efficiency. 19.22. Model: Assume that the heat engine follows a closed cycle. Solve: (a) The engines efficiency is out out H C out 200 J 0.25 25% 600 J 200 J W W Q Q W . = = = = = + + (b) The thermal efficiency of a Carnot engine is .Carnot =1-TC TH . For this to be 25%, ( ) C C 0.25 1 504.8 K 232 C 400 273 K = - T .T = = + 19.23. Model: The efficiency of an ideal engine (or Carnot engine) depends only on the temperatures of the hot and cold reservoirs. Solve: (a) The engines thermal efficiency is out out H C out 10 J 0.40 40% 15 J 10 J W W Q Q W . = = = = = + + (b) The efficiency of a Carnot engine is .Carnot =1-TC TH . The minimum temperature in the hot reservoir is found as follows: H H 0.40 1 293 K T 488 K 215 C T = - . = = This is the minimum possible temperature. In a real engine, the hot-reservoir temperature would be higher than 215C because no real engine can match the Carnot efficiency. 19.24. Solve: (a) The efficiency of the Carnot engine is C Carnot H 1 1 300 K 0.40 40% 500 K T T . = - = - = = (b) An engine with power output of 1000 W does Wout = 1000 J of work during each .t = 1 s. A Carnot engine has a heat input that is out in Carnot =1000 J 2500 J 0.40 Q W . = = during each .t = 1 s. The rate of heat input is 2500 J/s = 2500 W. (c) out in out W = Q - Q , so the heat output during .t = 1 s is out in out Q = Q -W =1500 J . The rate of heat output is thus 1500 J/s = 1500 W. 19.25. Visualize: We will use Equation 19.29 for the efficiency of a Carnot engine. C Carnot H 1 T T . = - We are given TH = 673K and the original efficiency Carnot . = 0.40 . Solve: First solve for CT . C H Carnot T = T (1-. ) = (673K)(1- 0.40) = 404K Solve for C T' again with Carnot .' = 0.60. C H Carnot T' = T (1-.' ) = (673K)(1- 0.60) = 269K The difference of these C T s is 135 K, so the temperature of the cold reservoir should be decreased by 135 degrees to raise the efficiency from 40% to 60%. Assess: We expected to have to lower C T by quite a bit to get the better efficiency. 19.26. Model: The maximum possible efficiency for a heat engine is provided by the Carnot engine. Solve: The maximum efficiency is ( ) ( ) C max Carnot H 273 20 K 1 1 0.6644 273 600 K T T . . + = = - = - = + Because the heat engine is running at only 30% of the maximum efficiency, . = (0.30).max = 0.1993 . The amount of heat that must be extracted is out H 1000 J 5.0 kJ 0.1993 Q W . = = = 19.27. Model: The coefficient of performance of a Carnot refrigerator depends only on the temperatures of the cold and hot reservoirs. Solve: (a) The Carnot performance coefficient of a refrigerator is ( ) ( ) ( ) C Carnot H C 20 273 K 6.33 20 273 K 20 273 K K T T T - + = = = - + - - + (b) The rate at which work is done on the refrigerator is found as follows: C C in in 200 J/s 32 J/s = 32 W 6.325 K Q W Q W K = . = = = (c) The heat exhausted to the hot side per second is H C in Q = Q +W = 200 J/s + 32 J/s = 232 J/s = 232 W 19.28. Visualize: We are given TH = 773K and C T = 273K , therefore (by Equation 19.29) the Carnot efficiency is 273 Carnot 773 1 K 0 647 K n = - = . We are also given Carnot . = 0.60(. ) . Solve: Rearrange Equation 19.7: H out Q =W (1-.) . out W is the same for both engines, so it cancels. H out Carnot H Carnot out Carnot Carnot (1 ) 1 0 60( ) 1 0 60(0 647) 1 73 ( ) (1 ) 1 1 0 647 Q W Q W . . . . - - . - . . = = = =. - - -. Assess: This engine requires 1.73 times as much heat energy in during each cycle as a Carnot engine to do the same amount of work. 19.29. Model: The minimum possible value of TC occurs with a Carnot refrigerator. Solve: (a) For the refrigerator, the coefficient of performance is C ( )( ) C in in K Q Q KW 5.0 10 J 50 J W = . = = = The heat energy exhausted per cycle is H C in Q = Q +W = 50 J +10 J = 60 J (b) If the hot-reservoir temperature is 27C = 300 K, the lowest possible temperature of the cold reservoir can be obtained as follows: C C Carnot C H C C 5.0 250 K 23 C 300 K K T T T T T T = . = . = =- - - 19.30. Model: The coefficient of performance of a Carnot refrigerator depends only on the temperatures of the cold and hot reservoirs. Solve: Using the definition of the coefficient of performance for a Carnot refrigerator, C Carnot H C K T T T = - ( ) ( ) H H 13 273 K 260 K 5.0 T 13 273 K T 260 K - + . = = - - + - H .T = 312 K To increase the coefficient of performance to 10.0, the hot-reservoir temperature changes to H T' . Thus, Carnot H 10.0 260 K 260 K K T = = ' - H .T' = 286 K Since H T' is less than TH, the hot-reservoir temperature must be decreased by 26 K or 26 C. 19.31. Solve: The work done by the engine is equal to the change in the gravitational potential energy. Thus, Wout = .Ugrav = mgh = (2000 kg)(9.8 m/s2)(30 m) = 588,000 J The efficiency of this engine is ( ) ( ) C Carnot H 273 20 K 0.40 0.40 1 0.40 1 273 2000 K T T . . . . . + . = = . - . = .. - .. . . . + . = 0.3484 The amount of heat energy transferred is calculated as follows: out out 6 H H 588,000 J 1.7 10 J 0.3484 W Q W Q . . = . = = = 19.32. Solve: The mass of the water is 3 3 3 3 100 10 L 10 m 1000 kg 0.100 kg 1 L m - - = The heat energy is removed from the water in three steps: (1) cooling from +15C to 0C, (2) freezing at 0C, and (3) cooling from 0C to -15C . The three heat energies are ( )( )( ) ( )( ) ( )( )( ) 1 5 2 f 3 0.100 kg 4186 J/kg K 15 K 6279 J 0.100 kg 3.33 10 J/kg 33,300 J 0.100 kg 2090 J/kg K 15 K 3135 J Q mc T Q mL Q mc T = . = = = = = = . = = C 1 2 3 .Q = Q +Q +Q = 42,714 J Using the performance coefficient, C in in K Q 4.0 42,714 J W W = . = . in 42,714 J 10,679 J 4.0 W = = The heat exhausted into the room is thus 4 H C in 4 Q = Q +W = 2,714 J +10,679 J = 5.310 J 19.33. Solve: An adiabatic process has Q = 0 and thus, from the first law, Ws = .Eth. For any ideal-gas process, .Eth = nCV.T, so Ws = nCV.T. We can use the ideal-gas law to find T pV T ( pV ) ( pV )f ( pV )i pfVf piVi nR nR nR nR . - - = .. = = = Consequently, the work is f f i i V s V V ff ii W nC T nC p V pV C ( p V pV ) nR R . - . = - . = - . . = - - . . Because CP = CV + R, we can use the specific heat ratio . to find P V V V V V V / 1 1 / 1 C C R C R C C C C R R . . + + = = = . =- With this, the work done in an adiabatic process is V s f f i i f f i i f f i i ( ) 1 ( ) ( )/(1 ) 1 W C p V pV p V pV p V pV R . . = - - = - - = - - - 19.34. Visualize: We are given 5 6 QH = msteamLv = (10kg)(22.610 J/kg) = 22.610 J and C ice f Q = m L = (55kg) (3.33105 J/kg) =18.3106 J . Solve: Equation 19.6 gives out H C W = Q -Q . 6 6 out H C 22 6 10 J 18 3 10 J 1h 1200W 1h 3600s P W Q Q t t - . - . . . = = = . . = . . . . Assess: This is a reasonable answer. 19.35. Solve: For any heat engine, . = 1 QC/QH. For a Carnot heat engine, .Carnot = 1 TC/TH. Thus a property of the Carnot cycle is that QC/QH = TC/TH. Consequently, the coefficient of performance of a Carnot refrigerator is C C C H CH C Carnot in H C C H C H H C / / 1 / 1 / K Q Q Q Q T T T W Q Q Q Q T T T T = = = = = - - - - 19.36. Visualize: Equation 19.29 gives C Carnot H 1 T T . = - . We are given Carnot . =1/3. Solve: C C Carnot H C H H 1 1 2 3 3 3 2 T T T T T T . = - = . = . = Equation 19.30 gives the coefficient of performance for the Carnot refrigerator. C C Carnot 3 3 H C 2 C C 2 1 2 1 K T T T T T T = = = = - - - Assess: This result is in the ballpark for coefficients of performance. 19.37. Visualize: We are given TH = 298 K and C T = 273 K . See Figure 19.11. Solve: 5 6 C f Q = mL = (10kg)(3.3310 J/kg) = 3.3310 J . (a) For a Carnot cycle C Carnot H 1 T T . = - but that must also equal C H 1 Q Q . = - , so C C H H Q T Q T = . H 6 6 H C C (3 33 10 J) 298K 3 63 10 J 273K Q Q T T . . = = . . . = . . . (b) 6 6 6 5 in H C W = Q -Q = 3.6310 J - 3.3310 J = 0.3010 J = 3.010 J Assess: This is a reasonable amount of work to freeze 10 kg of water. 19.38. Model: We will use the Carnot engine to find the maximum possible efficiency of a floating power plant. Solve: The efficiency of a Carnot engine is ( ) ( ) C max Carnot H 273 5 K 1 1 0.0825 273 30 K T T . . + = = - = - = + 8% 19.39. Model: The ideal gas in the Carnot engine follows a closed cycle in four steps. During the isothermal expansion at temperature TH, heat QH is transferred from the hot reservoir into the gas. During the isothermal compression at TC, heat QC is removed from the gas. No heat is transferred during the remaining two adiabatic steps. Solve: The thermal efficiency of the Carnot engine is C out Carnot H H 1 T W T Q . = - = 1 323 K out 573 K 1000 J . - = W out .W = 436 J Using H C out Q = Q +W , we obtain isothermal C H out Q = Q = Q -W =1000 J - 436 J 560 J 19.40. Solve: Substituting into the formula for the efficiency of a Carnot engine, C Carnot H 1 T T . = - C C 0.25 1 80 K T T . = -+ .TC = 240 K = -33C The hot-reservoir temperature is H C T = T + 80 K = 320 K = 47 C. 19.41. Model: Assume the gas is monatomic. Visualize: We are given TC = 273 K and C Q =15 J per cycle. For 98 cycles 2 out W = mgh = (10kg)(9.8kg/m )(10m) = 9800 J ; for one cycle this would be out W =100 J. Solve: Do the computation for one cycle. H out C Q =W +Q =100J +15J =115J For a Carnot cycle C Carnot H 1 T T . = - but that must also equal C H 1 Q Q . = - , so C C H H Q T Q T = . H H C C (273K) 115J 2093K 2100K 15J T T Q Q . . = = . . = . . Assess: H T must be this high to give the 87% efficiency. 19.42. Solve: If QC = 23 QH, then Wout = QH QC = 1 3 QC. Thus the efficiency is 1 out 3 H H H 1 3 W Q Q Q . = = = The efficiency of a Carnot engine is . = 1 TC/TH. Thus C C H H 1 1 2 3 3 T T T T - = . = 19.43. Solve: (a) Q1 is given as 1000 J. Using the energy transfer equation for the heat engine, QH = QC +Wout .Q1 = Q2 +Wout .Q2 = Q1 -Wout The thermal efficiency of a Carnot engine is C out H 1 1 1 300 K 0.50 600 K T W T Q . = - = - = = ( ) ( )( ) 2 1 1 1 1 .Q = Q -.Q = Q -. = 1000 J 1- 0.50 = 500 J To determine Q3 and Q4, we turn our attention to the Carnot refrigerator, which is driven by the output of the heat engine with Win = Wout. The coefficient of performance is ( )( )( ) C C44 H C in out 1 4 1 400 K 4.0 500 K 400 K 4.0 0.50 1000 J 2000 J K T Q Q Q T T W W Q Q K Q . . = = = = = = - - . = = = Using now the energy transfer equation in 4 3 W +Q = Q , we have ( )( ) 3 out 4 1 4 Q =W +Q =.Q +Q = 0.50 1000 J + 2000 J = 2500 J (b) From part (a) 3 Q = 2500 J and 1 Q =1000 J , so 3 1 Q > Q . (c) Although Q1 = 1000 J and Q3 = 2500 J, the two devices together do not violate the second law of thermodynamics. This is because the hot and cold reservoirs are different for the heat engine and the refrigerator. 19.44. Model: A heat pump is a refrigerator that is cooling the already cold outdoors and warming the indoors with its exhaust heat. Solve: (a) The coefficient of performance for this heat pump is K = 5.0 = QC Win , where QC is the amount of heat removed from the cold reservoir. QH is the amount of heat exhausted into the hot reservoir. QH = QC + Win, where Win is the amount of work done on the heat pump. We have C in H in in in Q = 5.0W .Q = 5.0W +W = 6.0W If the heat pump is to deliver 15 kJ of heat per second to the house, then H in in 15 kJ 6.0 15 kJ 2.5 kJ 6.0 Q = = W .W = = In other words, 2.5 kW of electric power is used by the heat pump to deliver 15 kJ/s of heat energy to the house. (b) The monthly heating cost in the house using an electric heater is 15 kJ 200 h 3600 s 1$ $270 s 1 h 40 MJ = The monthly heating cost in the house using a heat pump is 2.5 kJ 200 h 3600 s 1$ $45 s 1 h 40 MJ = 19.45. Visualize: We are given TC = 275 K , H T = 295 K. We are also given that in one second in W =100 J and C Q = (1 s)(100 kJ/min)(1 min/60 s) =1667 J. Solve: The coefficient of performance of a refrigerator is given in Equation 19.10. C in 1667 J 16 67 100 J K Q W = = = . However the coefficient of performance of a Carnot refrigerator is given in Equation 19.30. C Carnot H C 275 K 13 75 20 K K T T T = = = . - However, informal statement #8 of the second law says that the coefficient of performance cannot exceed the Carnot coefficient of performance, so the salesman is making false claims. You should not buy the DreamFridge. Assess: The second law imposes real-world restrictions. 19.46. Solve: The maximum possible efficiency of the heat engine is C max H 1 1 300 K 0.40 500 K T T . = - = - = The efficiency of the engine designed by the first student is out 1 H 110 J 0.44 250 J W Q . = = = Because .1 > .max, the first student has proposed an engine that would violate the second law of thermodynamics. His or her design will not work. The efficiency of the engine designed by the second student is 2 . = max 90 J 250 J = 0.36 <. in agreement with the second law of thermodynamics. Applying the first law of thermodynamics, H C out Q = Q +W .250 J =170 J + 90 J we see the first law is violated. This design will not work as claimed. The design by the third student satisfies the first law of thermodynamics because H C out Q = Q +W = 250 J . The thermal efficiency of this engine is 3 . = max 0.36 <. , which satisfies the second law of thermodynamics. The data presented by students 1 and 2 are faulty. Only student 3 has an acceptable design. 19.47. Model: The power plant is to be treated as a heat engine. Solve: (a) Every hour 300 metric tons or 3 105 kg of coal is burnt. The volume of coal is 5 3 3 10 kg m 24 h 4800 m3 1 h 1500 kg = The height of the room will be 48 m. (b) The thermal efficiency of the power plant is 8 8 out 5 6 9 H 7.50 10 J/s 7.50 10 J 0.32 32% 3 10 kg 28 10 J 1 h 2.333 10 J 1 h kg 3600 s W Q . = = = = = Assess: An efficiency of 32% is typical of power plants. 19.48. Model: The power plant is treated as a heat engine. Solve: (a) The maximum possible thermal efficiency of the power plant is C max H 1 1 303 K 0.47 47% 573 K T T . = - = - = = (b) The plants actual efficiency is 6 out 6 H 700 10 J/s 0.35 35% 2000 10 J/s W Q . = = = (c) Because QH = QC + Wout, 9 9 9 C H out C Q = Q -W .Q = 2.010 J/s - 0.710 J/s =1.310 J/s The mass of water that flows per second through the condenser is 3 3 8 4 3 1.2 10 L 1 hr 10 m 1000 kg 3.333 10 kg h 3600 s 1 L m m - = = The change in the temperature as 9 C Q =1.310 J of heat is transferred to m = 3.333104 kg of water is calculated as follows: C Q = mc.T .1.3109 J = (3.333104 kg)(4186 J/kg K).T ..T = 9C The exit temperature is18C + 9C = 27C . 19.49. Model: The power plant is treated as a heat engine. Solve: The mass of water per second that flows through the plant every second is 3 3 8 4 3 1.0 10 L 1 hr 10 m 1000 kg 2.778 10 kg/s h 3600 s 1 L m m - = = The amount of heat transferred per second to the cooling water is thus ( 4 )( )( ) 9 C Q = mc.T = 2.77810 kg/s 4186 J/kg K 27C -16C =1.27910 J/s The amount of heat per second into the power plant is 9 9 9 H out C Q =W +Q = 0.75010 J/s +1.27910 J/s = 2.02910 J/s Finally, the power plants thermal efficiency is 9 out 9 H 0.750 10 J/s 0.37 37% 2.029 10 J/s W Q . = = = = 19.50. Solve: (a) The energy supplied in one day is 9 13 out 1.0 10 J 3600 s 24 h 8.64 10 J s 1 h 1 d W = = (b) The volume of water is V =1 km3 =109 m3. The amount of energy is ( 9 3 ) ( )( ) H 3 10 m 1000 kg 4190 J/kg K 1.0 K m Q = mc.T = .. .. . . = 4.191015 J (c) While its true that the ocean contains vast amounts of thermal energy, that energy can be extracted to do useful work only if there is a cold reservoir at a lower temperature. That is, the ocean has to be the hot reservoir of a heat engine. But theres no readily available cold reservoir, so the oceans energy cannot readily be tapped. There have been proposals for using the colder water near the bottom of the ocean as a cold reservoir, pumping it up to the surface where the heat engine is. Although possible, the very small temperature difference between the surface and the ocean depths implies that the maximum possible efficiency (the Carnot efficiency) is only a few percent, and the efficiency of any real ocean-driven heat engine would likely be less than 1%perhaps much less. Thus the second law of thermodynamics prevents us from using the thermal energy of the ocean. Save your money. Dont invest. 19.51. Visualize: If we do this problem on a per second basis then in one second 5 QC = (1 s)(5.010 J/min) (1 min/60 s) = 8.33103 J. 5 H Q = (1 s)(8.010 J/min)(1 min / 60 s) = 13.33103 J. Solve: (a) Again, in one second 3 3 3 in H C W = Q -Q =13.3310 J - 8.3310 J = 5.010 J Since this is per second then the power required by the compressor is P = 5.0 kW. (b) 3 C 3 in 8 33 10 J 1 7 5 0 10 J K Q W . = = =. . Assess: The result is typical for air conditioners. 19.52. Model: The heat engine follows a closed cycle with process 1 . 2 and process 3 . 4 being isochoric and process 2 . 3 and process 4 . 1 being isobaric. For a monatomic gas, 3 CV = 2 R and 5 P 2 C = R. Visualize: Please refer to Figure P19.52. Solve: (a) The first law of thermodynamics is th S Q = .E +W . For the isochoric process 1 . 2, WS 1 . 2 = 0 J. Thus, ( )( ) ( )( )( ) 1 2 th V 3 3 V 2 2 2 1 2 3750 J 3750 J 3750 J 3750 J 301 K 1.0 mol 1.0 mol 8.31 J/mol K 300.8 K 300.8 K 300 K 601 K Q EnCT T nC R T T T . = = . = . . . = = = = . - = . = + = To find volume V2, ( )( )( ) 1 3 3 2 1 5 1 1.0 mol 8.31 J/mol K 300 K 8.31 10 m 3.0 10 Pa V V nRT p = = = = - The pressure p2 can be obtained from the isochoric condition as follows: 2 1 2 ( 5 ) 5 2 1 2 1 1 601 K 3.00 10 Pa 6.01 10 Pa 300 K p p p T p T T T = . = = . . = . . . . With the above values of p2, V2 and T2, we can now obtain p3, V3 and T3. We have 2 3 5 3 2 3 2 3 2 3 3 2 2 3 2 2 2 1.662 10 m 6.01 10 Pa 2 1202 K V V p p T T T V T T V V V = = - = = = . = = = For the isobaric process 2 . 3, ( )( )( ) ( )( )( )( ) ( ) ( )( ) 5 5 2 3 P 2 3 2 2 5 33 S 2 3 3 3 2 th 2 3 S 2 3 1.0 mol 1.0 mol 8.31 J/mol K 601 K 12,480 J 6.01 10 Pa 8.31 10 m 4990 J 12,480 J 4990 J 7490 J Q nC T R T T W pV V E Q W . - . . . = . = - = = = - = = . = - = - = We are now able to obtain p4, V4 and T4. We have ( ) 2 3 5 4 3 4 1 5 4 3 4 4 3 5 4 3 3 1.662 10 m 3.00 10 Pa 3.00 10 Pa 1202 K 600 K 6.01 10 Pa V V p p T T T p T p p p = = - = = . . = . = = . . = . . For isochoric process 3 . 4, ( )( 3 )( ) ( )( 3 )( )( ) 3 4 V 2 4 3 2 S 3 4 th 3 4 S 3 4 1.0 mol 1.0 mol 8.31 J/mol K 602 7500 J 0 J 7500 J Q nC T R T T W EQ W . . . . = . = - = - =- = .. = - =- For isobaric process 4 . 1, ( ) ( )( ) ( ) ( ) ( ) ( ) 5 4 1 P 2 5 33 23 S 4 1 4 1 4 th 4 1 S 4 1 1.0 mol 8.31 J/mol K 300 K 600 K 6230 J 3.00 10 Pa 8.31 10 m 1.662 10 m 2490 J 6230 J 2490 J 3740 J Q nC T W pV V E Q W . - - . . . = . = - =- = - = - =- . = - =- - - =- WS (J) Q (J) .Eth (J) 1 . 2 0 3750 3750 2 . 3 4990 12,480 7490 3 . 4 0 7500 7500 4 . 1 2490 6230 3740 Net 2500 2500 0 (b) The thermal efficiency of this heat engine is out out H 1 2 2 3 2500 J 0.15 15% 3750 J 12,480 J W W Q Q Q . . . = = = = = + + Assess: For a closed cycle, as expected, (Ws)net = Qnet and (.Eth)net = 0 J 19.53. Model: The heat engine follows a closed cycle. For a diatomic gas, 5 CV = 2 R and 7 P 2 C = R. Visualize: Please refer to Figure P19.53. Solve: (a) Since T1 = 293 K, the number of moles of the gas is ( )( ) ( )( ) 5 63 1 1 4 1 0.5 1.013 10 Pa 10 10 m 2.08 10 mol 8.31 J/mol K 293 K n pV RT - - = = = At point 2, 2 1 2 1 V = 4V and p = 3p . The temperature is calculated as follows: 1 1 2 2 2 2 ( )( )( ) 2 1 1 2 1 1 pV p V T p V T 3 4 293 K 3516 K T T pV = . = = = At point 3, 3 2 1 3 1 V =V = 4V and p = p . The temperature is calculated as before: 3 3 ( )( )( ) 3 1 1 1 T p V T 1 4 293 K 1172 K p V = = = For process 1 . 2, the work done is the area under the p-versus-V curve. That is, ( )( ) ( )( ) ( )( ) 3 3 1 3 3 s 2 5 6 3 0.5 atm 40 cm 10 cm 1.5 atm 0.5 atm 40 cm 10 cm 30 10 m 1 atm 1.013 10 Pa 3.04 J 1 atm W - = - + - - . . = . . = . . The change in the thermal energy is ( 4 ) 5 ( )( ) th V 2 .E = nC .T = 2.0810- mol 8.31 J/mol K 3516 K - 293 K =13.93 J The heat is s th Q =W + .E =16.97 J . For process 2 . 3, the work done is Ws = 0 J and ( )( ) ( )( )( ) 5 th V 2 3 2 4 5 2 2.08 10 mol 8.31 J/mol K 1172 K 3516 K 10.13 J Q E nC T n R T T - = . = . = - = - =- For process 3 . 1, ( )( ) ( )( ) ( )( )( ) 3 3 5 6 3 s 4 5 th V 2 0.5 atm 10 cm 40 cm 0.5 1.013 10 Pa 30 10 m 1.52 J 2.08 10 mol 8.31 J/mol K 293 K 1172 K 3.80 J W E nC T - - = - = - =- . = . = - =- The heat is th s Q = .E +W = -5.32 J . Ws (J) Q (J) .Eth 1 . 2 3.04 16.97 13.93 2 . 3 0 -10.13 -10.13 3 . 1 -1.52 -5.32 -3.80 Net 1.52 1.52 0 (b) The efficiency of the engine is net H 1.52 J 0.090 9.0% 16.97 J W Q . = = = = (c) The power output of the engine is net 500 revolutions 1 min 500 1.52 J/s 13 W min 60 s revolution 60 W = = Assess: For a closed cycle, as expected, (Ws)net = Qnet and (.Eth)net = 0 J. 19.54. Model: For the closed cycle, process 1 . 2 is isothermal, process 2 . 3 is isobaric, and process 3 . 1 is isochoric. Solve: (a) We first need to find the conditions at points 1, 2, and 3. We can then use that information to find WS and Q for each of the three processes that make up this cycle. Using the ideal-gas equation the number of moles of the gas is ( )( ) ( )( ) 5 63 1 1 1 1.013 10 Pa 600 10 m 0.0244 mol 8.31 J/mol K 300 K n pV RT - = = = We are given that . = 1.25, which means this is not a monatomic or a diatomic gas. The specific heats are V PV 4 5 1 C R R C C R R . = = = + = - At point 2, process 1 . 2 is isothermal, so we can find the pressure p2 as follows: 4 3 1 5 1 1 2 2 2 1 4 3 1 1 2 6.00 10 m 3 3 atm 3.039 10 Pa 2.00 10 m pV p V p V p p p V - - = . = = = = = At point 3, process 2 . 3 is isobaric, so we can find the temperature T3 as follows: 4 3 2 3 3 3 2 4 3 2 2 2 3 2 6.00 10 m 3 900 K 2.00 10 m V V T V T T T T T V - - = . = = = = Point P (Pa) V (m3) T (K) 1 1.0 atm = 1.013 105 6.00 10-4 300 2 3.0 atm = 3.039 105 2.00 10-4 300 3 3.0 atm = 3.039 105 6.00 10-4 900 Process 1 . 2 is isothermal: ( ) ( ) S 12 1 1 2 1 W = pV ln V V = -66.8 J ( ) 12 S 12 Q = W = -66.8 J Process 2 . 3 is isobaric: ( ) ( ) S 23 2 2 3 2 W = p .V = p V -V =121.6 J ( ) 23 P P 3 2 Q = nC .T = nC T -T = 608.3 J Process 3 . 1 is isochoric: ( ) S 31 W = 0 J ( ) 31 V V 1 3 Q = nC .T = nC T -T = -486.7 J We find that ( ) S cycle W = -66.8 J +121.6 J + 0 J = 54.8 J cycle Q = -66.8 J + 608.3 J - 486.7 J = 54.8 J These are equal, as they should be. Knowing that the work done is Wout = (WS)cycle = 54.8 J/cycle, an engine operating at 20 cycles/s has power output out 54.8 J 20 cycle 1096 J 1096 W 1.10 kW cycle s s P = = = (b) Only Q23 is positive, so Qin = Q23 = 608 J. Thus, the thermal efficiency is out in 54.8 J 0.090 9.0% 608.3 J W Q . = = = = 19.55. Model: For the closed cycle of the heat engine, process 1 . 2 is isochoric, process 2 . 3 is adiabatic, and process 3 . 1 is isobaric. 3 CV = 2 R and 5 P 2 C = R for a monatomic gas. Visualize: Please refer to Figure P19.55. Solve: (a) Using the ideal-gas equation, the number of moles is ( 5 ) ( 6 3 ) 1 1 1 1.0 10 Pa 100 10 m 0.00401 mol (8.31 J/mol K)(300 K) n pV RT - = = = We can use the adiabat 2 . 3 to calculate p2 as follows: ( ) 3 5 / 3 3 3 5 6 3 3 2 2 2 3 1 3 2 2 1.0 10 Pa 600 cm 1.981 10 Pa 100 cm p V p V p p V p V V V . . . . . . . . . . = . = . . = . . = . . = . . . . . . T2 can be determined from the ideal-gas equation and is ( 6 )( 6 3 ) 2 2 2 1.981 10 Pa 100 10 m 5943 K (8.31 J/mol K)(0.00401 mol) T p V nR - = = = At point 3, p3 = p1 and ( ) 6 3 3 1 3 3 1 6 3 3 1 1 600 10 m 300 K 1800 K 100 10 m T T T V T V V V - - = . = = = Now we can calculate Ws, Q, and .Eth for the three processes involved in the cycle. For process 1 . 2, Ws 1.2 = 0 J and ( ) ( 3 )( ) th 1 2 V 2 1 2 2 1 E Q nC T T n R T T . . = = - = - = 282.2 J For process 2 . 3, Q2.3 = 0 J and ( ) ( 3 )( ) th V 3 2 2 3 2 .E = nC T -T = n R T -T = 207.2 J Because, .Eth = W + Q, W = -Ws, so Ws 2.3 = -.Eth = +207.2 J for process 2 . 3. For process 3 . 1, ( ) ( 3 )( ) th V 1 3 2 1 3 .E = nC T -T = n R T -T = -75.0 J ( ) ( 5 )( ) 3 1 P 1 3 2 1 3 Q nC T T n R T T 125.0 J . = - = - =- The work done Ws 3.1 is the area under the p-versus-V graph. We have ( 3 )( 6 3 6 3 ) s 3 1 W 100 10 Pa 100 10- m 600 10- m 50.0 J . = - =- Ws (J) Q (J) .Eth (J) 1 . 2 0 282.2 282.2 2 . 3 207.2 0 -207.2 3 . 1 -50.0 -125.0 -75.0 Net 157.2 157.2 0 (b) The thermal efficiency of the engine is out H 157.2 J 0.52 52% 282.2 J W Q . = = = = Assess: As expected for a closed cycle, (Ws)net = Qnet and (.Eth)net = 0 J. 19.56. Model: For the closed cycle of the heat engine, process 1 . 2 is isochoric, process 2 . 3 is adiabatic, and process 3 . 1 is isothermal. For a diatomic gas 5 CV = 2 R and 75 . = . Solve: (a) From the graph 3 2 V =1000 cm . The pressure p2 lies on the adiabat from 2 . 3. We can find the pressure as follows: ( ) 3 7 /5 3 5 5 2 2 3 3 2 3 3 2 1.0 10 Pa 4000 cm 6.964 10 Pa 700 kPa 1000 cm p V p V p p V V . . . . . . . = . = . . = . . = . . . . The temperature T2 can be obtained from the ideal-gas equation relating points 1 and 2: ( ) ( ) 5 1 1 2 2 2 2 2 1 5 1 2 1 1 300 K 6.964 10 Pa 1 522.3 K 522 K 4.0 10 Pa pV p V T T p V T T pV . . = . = = . . = . . (b) The number of moles of the gas is ( 5 )( 3 3 ) 1 1 1 4.0 10 Pa 1.0 10 m 0.1604 mol (8.31 J/mol K)(300 K) R pV RT - = = = For isochoric process 1 . 2 , Ws = 0 J and ( 5 ) th V 2 Q = .E = nC .T = n R .T = 741.1 J For adiabatic process 2 . 3, Q = 0 J and ( 5 )( ) th V 2 3 2 .E = nC .T = n R T -T = -741.1 J Using the first law of thermodynamics, .Eth = .Ws + Q, which means Ws = -.Eth = +741.1 J. Ws can also be determined from ( ) ( )( ) ( ) 4 3 3 2 2 3 2 3 s 2 5 J/K 300 K 522.3 K 741.1 J 1 1 p V p V nR T T W . . - - - = = = = - - - For isothermal process 3 . 1, th .E = 0 J and 1 s 1 3 W nRT ln V 554.5 J V = =- Using the first law of thermodynamics, th s .E = -W +Q , s Q =W = -554.5 J . .Eth (J) Ws (J) Q (J) 1 . 2 741.1 0 741.1 2 . 3 -741.1 741.1 0 3 . 1 0 -554.5 -554.5 Net 0 186.6 186.6 (c) The work per cycle is 186.6 J and the thermal efficiency is s H 186.6 J 0.25 25% 741.1 J W Q . = = = = 19.57. Model: For the closed cycle of the heat engine, process 1 . 2 is adiabatic, process 2 . 3 is isothermal, and process 3 . 1 is isochoric. For a diatomic gas 5 CV = 2 R and 75 . = . Solve: (a) From the graph 3 1 V = 4000 cm . The number of moles of gas is ( 5 )( 6 3 ) 2 2 2 4.0 10 Pa 1000 10 m 0.1203 mol (8.31 J/mol K)(400 K) n p V RT - = = = Using 1 1 2 2 pV. = p V. , and reading 1 V from the graph, ( )( )2 5 1 7 / 5 4 1 2 4 1 p p V 4.0 10 Pa 5.743 10 Pa 5.7 kPa V . . . = . . = = . . With p1, V1, and nR having been determined, we can find T1 using the ideal-gas equation: ( 4 )( 6 3 ) 1 1 1 5.743 10 4000 10 m 229.7 K 230 K (8.31 J/mol K)(0.1203 mol) T pV nR - = = = (b) For adiabatic process 1 . 2, Q = 0 J and ( 5 )( 3 3 ) ( 4 )( 3 3 ) 2 2 1 1 s 4.0 10 Pa 1.00 10 m 5.743 10 Pa 4.00 10 m 425.7 K 1 11.4 W p V pV . - - - - = = =- - - Because th s .E = -W +Q , th s .E = -W = 425.7 K . For isothermal process 2 . 3, th .E = 0 J . From Equation 17.15, 3 s 2 2 W nRT lnV 554.5 J V = = From the first law of thermodynamics, s Q =W = 554.5 J for process 2 . 3. For isochoric process 3 . 1, Ws = 0 J and ( 5 )( ) th V 2 1 3 Q = .E = nC .T = n R T -T = -425.7 J .Eth (J) Ws (J) Q (J) 1 . 2 425.7 -425.7 0 2 . 3 0 554.5 554.5 3 . 1 -425.7 0 -425.7 Net 0 128.8 128.8 (c) The work per cycle is 128.8 J and the thermal efficiency of the engine is s H 128.8 J 0.23 23% 554.5 J W Q . = = = = Assess: As expected, for a closed cycle ( )s net net W = Q and ( ) th net .E = 0 J. 19.58. Model: In the Brayton cycle, process 1 . 2 and process 3 . 4 are adiabatic, and process 2 . 3 and process 4 . 2 are isobaric. We will assume the gas to be diatomic, with 7 CP = 2 R and . =1.40. Visualize: Please refer to Figure P19.58. Solve: The number of moles of gas is ( 5 )( 3 ) 1 1 1 1.013 10 Pa 2.0 m 81.27 mol (8.31 J/mol K)(300 K) n pV RT = = = We can find the volume at 2 from the adiabatic equation 1 1 2 2 pV. = p V. , ( ) 1 1 1.4 1 3 3 2 1 2 2.0 m 1 0.3861 m 10 V V p p . .. . . = . . = . . = . . . . The temperature T2 is determined from the ideal-gas equation as follows: ( ) 2 1 1 2 2 2 2 2 1 3 1 2 1 1 300 K 10 atm 0.3861 m 579.2 K 1 atm 2.0 m pV p V T T p V T T pV . .. . = . = = . .. . = . .. . To find T3 we use the heat input: 6 7 7 ( ) H P 2 2 3 2 2.0 10 J 675.33 J/K 846.1 K 579.2 K 846.1 K 1425.3 K Q nCTnRT T T TT T = = . = . = . . . = . = + . = + = We are now able to find V3 using the ideal-gas law. We have 3 2 3 ( 3 ) 3 3 2 3 2 2 0.3861 m 1425.3 K 0.9501 m 579.2 K V V V V T T T T = . = = . . = . . . . The volume at 4 is found by again using the adiabatic equation 3 3 4 4 p V. = p V. , ( )( ) 1 1 3 3 1.4 3 4 3 4 V V p 0.9501 m 10 4.921 m p . .. = . . = = . . T4 is now obtained from ( 5 )( 3 ) 4 4 4 1.013 10 Pa 4.921 m 738.2 K (81.27 mol)(8.31 J/mol K) T p V nR = = = To find the engines efficiency, we must first find the total work done per cycle which is net 1 2 2 3 3 4 4 1 W W W W W . . . . = + + + The four contributions are ( ) ( )( )( ) ( )( ) ( ) ( )( )( ) 2 2 1 1 2 1 5 1 2 6 3 3 5 2 3 2 4 3 3 4 81.27 mol 8.31 J/mol K 579.2 K 300 K 4.714 10 J 1 1 11.4 1.013 10 Pa 0.9501 m 0.3861 m 5.713 10 J 81.27 mol 8.31 J/mol K 738.2 K 1425.3 K 1.160 1 11.4 p V pV nR T T W W p V nR T T W . . . . . . - - - = = = =- - - - = . = - = - - = = =+ - - ( )( ) 6 5 3 3 5 4 1 4 10 J W p V 1.013 10 Pa 2.0 m 4.921 m 2.959 10 J . = . = - = - Thus, 5 net W = 9.6410 J . The thermal efficiency is 5 net 6 H 9.64 10 J 0.482 48.2% 2.0 10 J W Q . . = . = = = As a comparison, we can calculate . from Equation 19.21 which is ( )( 1) ( )0.4 1.4 1 1 1 1 48.2% 10 p r . . . - = - = - = where p max min r = p p =10 . 19.59. Model: Process 1 . 2 of the cycle is isochoric, process 2 . 3 is isothermal, and process 3 . 1 is isobaric. For a monatomic gas, 3 CV = 2 R and 5 P 2 C = R. Visualize: Please refer to Figure P19.59. Solve: (a) At point 1: The pressure 5 1 p =1.0 atm = 1.01310 Pa and the volume 6 3 1 V =100010- m =1.0 10-3 m3 . The number of moles is 0.120 g 0.030 mol 4 g/mol n= = Using the ideal-gas law, ( )( ) ( )( ) 5 33 1 1 1 1.013 10 Pa 1.0 10 m 406 K 0.030 mol 8.31 J/mol K T pV nR - = = = At point 2: The pressure 5 2 p = 5.0 atm = 5.0610 Pa and 3 3 2 V =1.010- m . The temperature is ( )( ) ( )( ) 5 33 2 2 2 5.06 10 Pa 1.0 10 m 2030 K 0.030 mol 8.31 J/mol K T p V nR - = = = At point 3: The pressure is 5 3 p =1.0 atm =1.01310 Pa and the temperature is 3 2 T = T = 2030 K . The volume is 2 ( 3 3 ) 3 3 3 2 3 1.0 10 m 5 atm 5.0 10 m 1 atm V V p p = = - .. .. = - . . (b) For isochoric process 1 . 2, W1.2 = 0 J and ( )( 3 )( ) 1 2 V 2 Q nC T 0.030 mol R 2030 K 406 K . = . = - = 607 J For isothermal process 2 . 3, th 2 3 E 0 J . . = and ( )( )( ) 3 3 3 2 3 2 3 2 3 3 2 ln 0.030 mol 8.31 J/mol K 2030 K ln 5.0 10 m 815 J 1.0 10 m V Q W nRTV - . . - . . = = = . . = . . For isobaric process 3 . 1, ( 5 )( 3 3 3 3 ) 3 1 3 W p V 1.013 10 Pa 1.0 10- m 5.0 10- m 405 J . = . = - = - ( )( 5 )( )( ) 3 1 P 2 Q nC T 0.030 mol 8.31 J/mol K 406 K 2030 K . = . = - = -1012 J The total work done is net 1 2 2 3 3 1 W W W W . . . = + + = 410 J. The total heat input is H 1 2 2 3 Q Q Q 1422 J . . = + = . The efficiency of the engine is net H 410 J 20% 1422 J W Q . = = = (c) The maximum possible efficiency of a heat engine that operates between Tmax and Tmin is min max max 1 1 406 K 80% 2030 K T T . = - = - = Assess: The actual efficiency of an engine is less than the maximum possible efficiency. 19.60. Model: The process 2 . 3 of the heat engine cycle is isochoric and the process 3 . 1 is isobaric. For a monatomic gas 3 CV = 2 R and 5 P 2 C = R . Solve: (a) The three temperatures are ( )( ) ( )( ) 5 3 1 1 1 4.0 10 Pa 0.025 m 601.7 K 602 K 2.0 mol 8.31 J/mol K T pV nR = = = ( )( ) ( )( ) 5 3 2 2 2 6.0 10 Pa 0.050 m 1805.1 K 1805 K 2.0 mol 8.31 J/mol K T p V nR = = = ( )( ) ( )( ) 5 3 3 3 3 4.0 10 Pa 0.050 m 1203.4 K 1203 K 2.0 mol 8.31 J/mol K T p V nR = = = (b) For process 1 . 2, the work done is the area under the p-versus-V graph. The work and the change in internal energy are ( )( ) ( )( ) ( )( )( ) ( )( )( )( ) 1 5 5 3 3 5 3 3 s 2 4 3 th V 2 2 1 3 4 2 6.0 10 Pa 4.0 10 Pa 0.050 m 0.025 m 4.0 10 Pa 0.050 m 0.025 m 1.25 10 J 2.0 mol 2.0 mol 8.31 J/mol K 1805.1 K 601.7 K 3.00 10 J W E nC T R T T = - - + - = . = . = - = - = The heat input is 4 s th Q =W + .E = 4.2510 J . For isochoric process 2 . 3, Ws = 0 J and ( ) 3 ( )( ) 4 th V 2 Q = .E = nC .T = 2.0 mol 8.31 J/mol K 1203.4 K -1805.1 K = -1.5010 J For isobaric process 3 . 1, the work done is the area under the p-versus-V curve. Hence, ( 5 )( 3 3 ) 4 s W = 4.010 Pa 0.025 m - 0.050 m = -1.010 J ( 3 )( ) th V 2 1 3 E nC T n R T T . = . = - ( ) ( )( ) 4 32 = 2.0 mol 8.31 J/mol K 601.7 K -1203.4 K = -1.510 J The heat input is 4 S th Q =W + .E = -2.5010 J . .Eth (J) Ws (J) Q (J) 1 . 2 3.0 104 1.25 104 4.25 104 2 . 3 -1.5 104 0 -1.50 104 3 . 1 -1.5 104 -1.0 104 -2.50 104 Net 0 2.5 103 2.5 103 (c) The thermal efficiency is 3 net 4 H 2.5 10 J 5.9% 4.25 10 J W Q . = = = 19.61. Model: The closed cycle in this heat engine includes adiabatic process 1 . 2, isobaric process 2 . 3, and isochoric process 3 . 1. For a diatomic gas, 5 7 7 V 2 P 2 5 C = R, C = R, and . = =1.4. Visualize: Please refer to Figure P19.61. Solve: (a) We can find the temperature T2 from the ideal-gas equation as follows: ( )( ) ( )( ) 5 33 2 2 2 4.0 10 Pa 1.0 10 m 2407 K 0.020 mol 8.31 J/mol K T p V nR - = = = We can use the equation 2 2 1 1 p V. = pV. to find V1, ( ) 1/ 5 1/1.4 2 3 3 3 3 1 2 5 1 1.0 10 m 4.0 10 Pa 2.692 10 m 1.0 10 Pa V V p p . - - . . . . = . . = . . = . . . . The ideal-gas equation can now be used to find T1, ( )( ) ( )( ) 5 33 1 1 1 1.0 10 Pa 2.692 10 m 1620 K 0.020 mol 8.31 J/mol K T pV nR - = = = At point 3, 3 1 V =V so we have ( )( ) ( )( ) 5 33 3 3 3 4 10 Pa 2.692 10 m 6479 K 0.020 mol 8.31 J/mol K T p V nR - = = = (b) For adiabatic process 1 . 2, Q = 0 J, th s .E = -W , and ( ) ( )( )( ) ( ) 2 2 1 1 2 1 s 0.020 mol 8.31 J/mol K 2407 K 1620 K 327.0 J 1 1 11.4 p V pV nR T T W . . - - - = = = =- - - - For isobaric process 2 . 3, ( 7 )( ) P 2 Q = nC .T = n R .T ( ) 7 ( )( ) 2 = 0.020 mol 8.31 J/mol K 6479 K - 2407 K = 2369 J ( 5 ) th V 2 .E = nC .T = n R .T =1692 J The work done is the area under the p-versus-V graph. Hence, ( 5 )( 3 3 3 3 ) s W = 4.010 Pa 2.69210- m -1.010- m = 677 J For isochoric process 3 . 1, Ws = 0 J and ( )( 5 )( )( ) th V 2 .E = Q = nC .T = 0.020 mol 8.31 J/mol K 1620 K - 6479 K = -2019 J .Eth (J) Ws (J) Q (J) 1 . 2 327 -327 0 2 . 3 1692 677 2369 3 . 1 -2019 0 -2019 Net 0 350 350 (c) The engines thermal efficiency is 350 J 0.15 15% 2369 J .= = = 19.62. Model: The closed cycle of the heat engine involves the following processes: one isothermal compression, one isobaric expansion, and one isochoric cooling. Visualize: Solve: The number of moles of helium is 2.0 g 0.50 mol 4.0 g/mol n= = The total work done by the engine per cycle is net 1 2 2 3 3 1 W W W W . . . = + + . For the isothermal process, 2 1 2 1 1 lnV W nRTV . = = (0.50 mol)(8.31 J/mol K)(273 K)ln(0.5) = -786.2 J For the isobaric process, ( ) ( ) 1 2 3 2 1 1 1 1 2 2 W p V p V pV nRT . = . = . . = = . . . . = (0.50 mol)(8.31 J/mol K)(273 K) =1134.3 J For the isochoric process, W3.1 = 0 J. Thus, Wnet = 348.1 J 350 J . For calculating heat transfers, note that 2 1 T = T = 273 K and, since process 3.1 is isochoric, 3 T = 1 (2 atm/1 atm) T = 546 K. For the isothermal process, 1 2 1 2 Q W 786.2 J . . = =- because th 1 2 E 0 . . = J. For the isobaric process, ( 5 )( ) 5 ( )( )( ) 2 3 P 2 3 2 2 Q nC T n R T T 0.50 mol 8.31 J/mol K 273 K 2835.8 J . = . = - = = For the isochoric process, ( 3 )( ) ( 3 )( )( )( ) 3 1 v 2 1 3 2 Q nC T n R T T 0.50 mol 8.31 J/mol K 273 K 1701.5 J . = . = - = =- From Q1.2, Q2.3, and Q3.1, we see that QH = 2835.8 J. Thus, the thermal efficiency is 348.1 J 0.12 12% 2835.8 J .= = = 19.63. Model: The closed cycle of the heat engine involves the following four processes: isothermal expansion, isochoric cooling, isothermal compression, and isochoric heating. For a monatomic gas 3 CV = 2 R . Visualize: Solve: Using the ideal-gas law, ( )( )( ) 1 5 1 3 3 1 0.20 mol 8.31 J/mol K 600 K 4.986 10 Pa 2.0 10 m p nRT V - = = = At point 2, because of the isothermal conditions, 2 1 T = T = 600 K and ( ) 3 3 1 5 5 2 1 3 3 2 4.986 10 Pa 2.0 10 m 2.493 10 Pa 4.0 10 m p p V V - - . . = = . . = . . At point 3, because it is an isochoric process, 3 3 2 V =V = 4000 cm and 3 ( 5 ) 5 3 2 2 2.493 10 Pa 300 K 1.247 10 Pa 600 K p p T T = = . . = . . . . Likewise at point 4, 4 3 T = T = 300 K and ( ) 3 3 3 5 5 4 3 3 3 4 1.247 10 Pa 4.0 10 m 2.493 10 Pa 2.0 10 m p p V V - - . . = = . . = . . Let us now calculate Wnet = W1.2 + W2.3 + W3.4 + W4.1. For the isothermal processes, 2 ( )( )( ) ( ) 1 2 1 1 ln 0.20 mol 8.31 J/mol K 600 K ln 2 691.2 J V W nRTV . = = = 4 ( )( )( ) ( 1 ) 3 4 3 2 3 ln 0.20 mol 8.31 J/mol K 300 K ln 345.6 J V W nRTV . = = =- For the isochoric processes, W2.3 = W4.1 = 0 J. Thus, Wnet = 345.6 J 350 J . Because Q = Ws + .Eth, ( ) 1 2 1 2 th 1 2 Q W E . . . = + . = 691.2 J + 0 J = 691.2 J For the first isochoric process, ( )( )( ) ( )( )( ) 3 2 3 V 2 3 2 3 2 0.20 mol 0.20 mol 8.31 J/mol K 300 K 600 K 747.9 K Q nC T R T T . = . = - = - =- For the second isothermal process ( ) 3 4 3 4 th 3 4 Q W E . . . = + . = -345.6 J + 0 J = -345.6 J For the second isochoric process, ( )( ) ( )( )( )( ) 3 4 1 V 2 1 4 32 0.20 mol 8.31 J/mol K 600 K 300 K 747.9 K Q nC T n R T T . = . = - = - = Thus, H 1 2 4 1 1 Q Q Q 439.1 J . . = + = . The thermal efficiency of the engine is net H 345.6 J 0.24 24% 1439.1 J W Q . = = = = 19.64. Model: Processes 2 . 1 and 4 . 3 are isobaric. Processes 3 . 2 and 1 . 4 are isochoric. Visualize: Solve: (a) Except in an adiabatic process, heat must be transferred into the gas to raise its temperature. Thus heat is transferred in during processes 4 . 3 and 3 . 2. This is the reverse of the heat engine in Example 19.2. (b) Heat flows from hot to cold. Since heat energy is transferred into the gas during processes 4 . 3 and 3 . 2, which end with the gas at temperature 2700 K, the reservoir temperature must be T > 2700 K. This is the hot reservoir, so the heat transferred is QH. Similarly, heat energy is transferred out of the gas during processes 2 . 1 and 1 . 4. This requires that the reservoir temperature be T < 300 K. This is the cold reservoir, and the energy transferred during these two processes is QC. (c) The heat energies were calculated in Example 19.2, but now they have the opposite signs. QH = Q43 + Q32 = 7.09 105 J + 15.19 105 J = 22.28 105 J QC = Q21 + Q14 = 21.27 105 J + 5.06 105 J = 26.33 105 J (d) For a counterclockwise cycle in the pV diagram, the work is Win. Its value is the area inside the curve, which is Win = (.p)(.V) = (2 101,300 Pa)(2 m3) = 4.05 105 J. Note that Win = QC QH, as expected from energy conservation. (e) Since QC > QH, more heat is exhausted to the cold reservoir than is extracted from the hot reservoir. In this device, work is used to transfer energy downhill, from hot to cold. The exhaust energy is QC = QH + Win > QH. This is the energy-transfer diagram of Figure 19.19. (f) No. A refrigerator uses work input to transfer heat energy from the cold reservoir to the hot reservoir. This device uses work input to transfer heat energy from the hot reservoir to the cold reservoir. 19.65. Solve: (a) If you wish to build a Carnot engine that is 80% efficient and exhausts heat into a cold reservoir at 0C, what temperature (in C) must the hot reservoir be? (b) ( ) ( ) H H H 0 C 273 0.80 1 273 0.20 1092 C 273 273 T T T + = - . = . = + + 19.66. Solve: (a) A refrigerator with a coefficient of performance of 4.0 exhausts 100 J of heat in each cycle. What work is required each cycle and how much heat is removed each cycle from the cold reservoir? (b) We have 4.0 = QC Win .QC = 4Win . This means QH = QC + Win = 4Win + Win = 5Win H in 100 J 20 J 5 5 .W = Q = = Hence, C H in Q = Q -W =100 J - 20 J = 80 J. 19.67. Solve: (a) A heat engine operates at 20% efficiency and produces 20 J of work in each cycle. What is the net heat extracted from the hot reservoir and the net heat exhausted in each cycle? (b) We have 0.20 =1-QC QH . Using the first law of thermodynamics, out H C C H W = Q -Q = 20 J . Q = Q - 20 J Substituting into the definition of efficiency, H H H H 0.20 1 Q 20 J 1 1 20 J 20 J Q Q Q - = - = - + = H 20 J 100 J 0.20 .Q = = The heat exhausted is C H Q = Q - 20 J =100 J - 20 J = 80 J . 19.68. Solve: (a) In this heat engine, 400 kJ of work is done each cycle. What is the maximum pressure? (b)
( 5 )( 3 ) 5 max 1 1.0 10 Pa 2.0 m 4.0 10 J 2 p - = 5 max . p = 5.010 Pa = 500 kPa 19.69. Model: The heat engine follows a closed cycle, starting and ending in the original state. Visualize: The figure indicates the following seven steps. First, the pin is inserted when the heat engine has the initial conditions. Second, heat is turned on and the pressure increases at constant volume from 1 atm to 3 atm. Third, the pin is removed. The flame continues to heat the gas and the volume increases at constant pressure from 50 cm3 to 100 cm3. Fourth, the pin is inserted and some of the weights are removed. Fifth, the container is placed on ice and the gas cools at constant volume to a pressure of 1 atm. Sixth, with the container still on ice, the pin is removed. The gas continues to cool at constant pressure to a volume of 50 cm3. Seventh, with no ice or flame, the pin is inserted back in and the weights returned bringing the engine back to the initial conditions and ready to start over. Solve: (a) (b) The work done per cycle is the area inside the curve: Wout = (.p)(.V) = (2 101,300 Pa)(50 106 m3) = 10.13 J (c) Heat energy is input during processes 1 . 2 and 2 . 3, so QH = Q12 + Q23. This is a diatomic gas, with CV = 3 2 R and CP = 5 2 R. The number of moles of gas is 6 3 1 1 1 (101,300 Pa)(50 10 m ) 0.00208 mol (8.31 J/mol K)(293 K) n pV RT - = = = Process 1 . 2 is isochoric, so T2 = (p2/p1)T1 = 3T1 = 879 K. Process 2 . 3 is isobaric, so T3 = (V3/V2)T2 = 2T2 = 1758 K. Thus 5 5 12 V 2 2 1 2 Q = nC .T = nR(T -T ) = (0.00208 mol)(8.31 J/mol K)(586 K) = 25.32 J Similarly, 7 7 23 P 2 3 2 2 Q = nC .T = nR(T -T ) = (0.00208 mol)(8.31 J/mol K)(879 K) = 53.18 J Thus QH = 25.32 J + 53.18 J = 78.50 J and the engines efficiency is out H 10.13 J 0.13 13% 78.50 J W Q . = = = = 19.70. Model: System 1 undergoes an isochoric process and system 2 undergoes an isobaric process. Solve: (a) Heat will flow from system 1 to system 2 because system 1 is hotter. Because there is no heat input from (or loss to) the outside world, we have Q1 + Q2 = 0 J. Heat Q1, which is negative, will change the temperature of system 1. Heat Q2 will both change the temperature of system 2 and do work by lifting the piston. But these consequences of heat flow dont change the fact that Q1 + Q2 = 0 J. System 1 undergoes constant volume cooling from T1i = 600 K to Tf. System 2, whose pressure is controlled by the weight of the piston, undergoes constant pressure heating from T2i = 300 K to Tf. Thus, Q1 +Q2 = 0 J = n1CV (Tf -T1i ) + n2CP (Tf -T2i ) ( 3 )( ) ( 5 )( ) 1 2 f 1i 2 2 f 2i = n R T -T + n R T -T Solving this equation for Tf gives ( )( ) ( )( ) ( ) ( ) 1 1i 2 2i f 1 2 3 5 3 0.060 mol 600 K 5 0.030 mol 300 K 464 K 3 5 3 0.060 mol 5 0.030 mol T nT n T n n + + = = = + + (b) Knowing Tf , we can compute the heat transferred from system 1 to system 2: ( ) ( 5 )( ) 2 2 P f 2i 2 2 f 2i Q = n C T -T = n R T -T =102 J (c) The change of thermal energy in system 2 is ( 3 )( ) 3 th 2 V 2 2 f 2i 5 2 .E = n C .T = n R T -T = Q = 61.2 J According to the first law of thermodynamics, Q2 = Ws + .Eth. Thus, the work done by system 2 is Ws = Q - .Eth = 102.0 J - 61.2 J = 40.8 J. The work is done to lift the weight of the cylinder and the air above it by a height .y. The weight of the air is 2 2 2 air w = pA = pp r = (101.310 N/m )p (0.050 m) = 795.6 N. So, s cyl air W = (w + w ).y ( )( ) s 2 cyl air 40.8 J 0.050 m ( ) 2.0 kg 9.8 m/s 795.6 N y W w w . . = = = + + (d) The fraction of heat converted to work is s 2 40.8 J 0.40 40% 102.0 J W Q = = = 19.71. Model: Process 1 . 2 and process 3 . 4 are adiabatic, and process 2 . 3 and process 4 . 1 are isochoric. Visualize: Please refer to Figure CP19.71. Solve: (a) For adiabatic process 1 . 2, Q12 = 0 J and 2 2 1 1 ( 2 1 ) 12 1 1 p V pV nR T T W . . - - = = - - For isochoric process 2 . 3, W23 = 0 J and ( ) 23 V 3 2 Q = nC T -T . For adiabatic process 3 . 4, Q34 = 0 J and ( ) 4 4 3 3 4 3 34 1 1 p V p V nR T T W . . - - = = - - For isochoric process 4 . 1, W41 = 0 J and ( ) 41 V 1 4 Q = nC T -T . The work done per cycle is ( ) ( ) 2 1 4 3 net 12 23 34 41 0 J 0 J 1 1 nR T T nR T T W W W W W . . - - = + + + = + + + - - ( ) 2 1 4 3 1 nR T T T T . = - + - - (b) The thermal efficiency of the heat engine is ( ) ( ) out C 41 V 4 1 H H 23 V 3 2 1 1 1 W Q Q nC T T Q Q Q nC T T . - = = - = - = - - The last step follows from the fact that T3 > T2 and T4 > T1. We will now simplify this expression further as follows: 1 1 1 1 1 1 1 1 1 2 2 2 1 1 2 pV pVV nRTV nRTV nRT V T T V T r V . . . . . . . - - - - - - . . = = . = . = . . = . . Similarly, 1 3 4 T = T r. - . The equation for thermal efficiency now becomes 4 1 1 1 4 1 1 T T T r. T r. . - - - = - - 1 1 1 r. - = - (c) 19.72. Model: For the Diesel cycle, process 1 . 2 is an adiabatic compression, process 2 . 3 is an isobaric expansion, process 3 . 4 is an adiabatic expansion, and process 4 . 1 is isochoric. Visualize: Please refer to CP19.72. Solve: (a) It will be useful to do some calculations using the compression ratio, which is 3 max 1 3 min 2 1050 cm 21 50 cm r V V V V = = = = The number of moles of gas is ( )( ) ( )( ) 5 63 1 1 1 1.013 10 Pa 1050 10 m =0.0430 mol 8.31 J/mol K 25+273 K n pV RT - = = .. .. For an adiabatic process, 1 1 2 2 pV. = p V. 1 1.40 6 2 1 1 2 p V p r p 21 1 atm 71.0 atm 7.19 10 Pa V . . . . . = . . = = = = . . 1 1 1 1 1 0.41 1 1 2 2 2 1 1 2 TV T V T V T r T 21 298 K 1007 K V . . . . - - - - . . = . = . . = = = . . Process 2 . 3 is an isobaric heating with Q = 1000 J. Constant pressure heating obeys P P Q nC T T Q nC = . .. = The gas has a specific heat ratio . = 1.40 5 7 V 2 P 2 = 7 /5, thus C = R and C = R. Knowing CP, we can calculate first .T = 800 K and then T3 = T2 + .T = 1807 K. Finally, for an isobaric process we have 2 3 3 ( 6 3 ) 6 3 3 2 2 3 2 1807 K 50 10 m 89.7 10 m 1007 K V V V T V T T T = . = = - = - Process 3 . 4 is an adiabatic expansion to V4 = V1. Thus, ( ) ( ) 6 3 1.4 3 6 5 3 3 4 4 4 3 6 3 4 1 6 3 0.4 1 1 3 3 3 4 4 4 3 6 3 4 89.7 10 m 7.19 10 Pa 2.30 10 Pa 2.27 atm 1050 10 m 89.7 10 m 1807 K 675 K 1050 10 m p V p V p V p V TV T V T V T V . . . . . . - - - - - - - . . . . = . = . . = . . = = . . . . . . . . = . = . . = . . = . . . . Point p V T 1 1.00 atm = 1.013 105 Pa 1050 10-6 m3 298 K = 25C 2 71.0 atm = 7.19 106 Pa 50.0 10-6 m3 1007 K = 734C 3 71.0 atm = 7.19 106 Pa 89.7 10-6 m3 1807 K = 1534C 4 2.27 atm = 2.30 105 Pa 1050 10-6 m3 675 K = 402C (b) For adiabatic process 1 . 2, ( ) 2 2 1 1 s 12 633 J 1 W p V pV . - = =- - For isobaric process 2 . 3, ( ) ( ) s 23 2 2 3 2 W = p .V = p V -V = 285 J For adiabatic process 3 . 4, ( ) 4 4 3 3 s 34 1009 J 1 W p V p V . - = = - For isochoric process 4 . 1, ( ) s 41 W = 0 J. Thus, ( ) ( ) ( ) ( ) ( ) s cycle out s 12 s 23 s 34 s 41 W =W = W + W + W + W = 661 J (c) The efficiency is out H 661 J 66.1% 66% 1000 J W Q . = = = (d) The power output of one cylinder is 661 J 2400 cycle 1 min 26,440 J 26.4 kW cycle min 60 sec s = = For an 8-cylinder engine the power will be 211 kW or 283 horsepower. 20.1. Model: The wave is a traveling wave on a stretched string. Solve: The wave speed on a stretched string with linear density is vstring = TS / . The wave speed if the tension is doubled will be S ( ) string string v 2T 2v 2 200 m/s 283 m/s ' = = = = 20.2. Model: The wave is a traveling wave on a stretched string. Solve: The wave speed on a stretched string with linear density is S string v T 150 m/s 75 N = . = . = 3.33310-3 kg/m For a wave speed of 180 m/s, the required tension will be 2 ( 3 )( )2 S string T = v = 3.33310- kg/m 180 m/s =110 N 20.3. Model: The wave pulse is a traveling wave on a stretched string. Solve: The wave speed on a stretched string with linear density is ( )( ) S S S string 3 2.0 m 2.0 m 20 N 0.025 kg 25 g 50 10 s v T T LT m m/L m - m = = = . = . = = 20.4. Model: This is a wave traveling at constant speed. The pulse moves 1 m to the right every second. Visualize: The snapshot graph shows the wave at all points on the x-axis at t = 0 s. The wave is just reaching x = 5.0. The first part of the wave causes an upward displacement of the medium. The rising portion of the wave is 2 m wide, so it will take 2 s to pass the x = 5.0 m point. The constant part of the wave, whose width is 2 m, will take 2 seconds to pass x = 5.0 m and during this time the displacement of the medium will be a constant (.y = 1 cm). The trailing edge of the pulse arrives at t = 4 s at x = 5.0 m. The displacement now becomes zero and stays zero for all later times. 20.5. Model: This is a wave traveling at constant speed. The pulse moves 1 m to the left every second. Visualize: This snapshot graph shows the wave at all points on the x-axis at t = 2 s. You can see that the leading edge of the wave at t = 2 s is precisely at x = 0 m. That is, in the first 2 seconds, the displacement is zero at x = 0 m. The first part of the wave causes a downward displacement of the medium, so immediately after t = 2 s the displacement at x = 0 m will be negative. The negative portion of the wave pulse is 3 m wide and takes 3 s to pass x = 0 m. The positive portion begins to pass through x = 0 m at t = 5 s and until t = 8 s the displacement of the medium is positive. The displacement at x = 0 m returns to zero at t = 8 s and remains zero for all later times. 20.6. Model: This is a wave traveling at constant speed to the right at 1 m/s. Visualize: This is the history graph of a wave at x = 0 m. The graph shows that the x = 0 m point of the medium first sees the negative portion of the pulse wave at t = 1.0 s. Thus, the snapshot graph of this wave at t = 1.0 s must have the leading negative portion of the wave at x = 0 m. 20.7. Model: This is a wave traveling at constant speed to the left at 1 m/s. Visualize: This is the history graph of a wave at x = 2 m. Because the wave is moving to the left at 1 m/s, the wave passes the x = 2 m position a distance of 1 m in 1 s. Because the flat part of the history graph takes 2 s to pass the x = 2 m position, its width is 2 m. Similarly, the width of the linearly increasing part of the history graph is 2 m. The center of the flat part of the history graph corresponds to both t = 0 s and x = 2 m. 20.8. Visualize: Figure EX20.8 shows a snapshot graph at t = 0 s of a longitudinal wave. This diagram shows a row of particles with an inter-particle separation of 1.0 cm at equilibrium. Because the longitudinal wave has a positive amplitude of 0.5 cm between x = 3 cm and x = 8 cm, the particles at x = 3, 4, 5, 6, 7 and 8 cm are displaced to the right by 0.5 cm. 20.9. Visualize: We first draw the particles of the medium in the equilibrium positions, with an inter-particle spacing of 1.0 cm. Just underneath, the positions of the particles as a longitudinal wave is passing through are shown at time t = 0 s. It is clear that relative to the equilibrium the particle positions are displaced negatively on the left side and positively on the right side. For example, the particles at x = 0 cm and x = 1 cm are at equilibrium, the particle at x = 2 cm is displaced left by 0.5 cm, the particle at x = 3 cm is displaced left by 1.0 cm, the particle at x = 4 cm is displaced left by 0.5 cm, and the particle at x = 5 cm is undisplaced. The behavior of particles for x > 5 cm is opposite of that for x < 5 cm. 20.10. Solve: (a) The wave number is 2 2 3.1 rad/m 2.0 m k p p . = = = (b) The wave speed is (2.0 m) 30 rad/s 9.5 m/s 2 2 v f . . . p p = = . . = . . = . . . . . . . . 20.11. Solve: (a) The wavelength is 2 2 4.2 m k 1.5 rad/m p p . = = = (b) The frequency is 200 m/s 48 Hz 4.19 m f v . = = = 20.12. Model: The wave is a traveling wave. Solve: (a) A comparison of the wave equation with Equation 20.14 yields: A = 3.5 cm, k = 2.7 rad/m, . = 124 rad/s, and f0 = 0 rad. The frequency is 124 rad/s 19.7 Hz 20 Hz 2 2 f . p p = = = (b) The wavelength is 2 2 2.33 m 2.3 m k 2.7 rad/m p p . = = = (c) The wave speed v = . f = 46 m/s . 20.13. Model: The wave is a traveling wave. Solve: (a) A comparison of the wave equation with Equation 20.14 yields: A = 5.2 cm, k = 5.5 rad/m, . = 72 rad/s, and f0 = 0 rad. The frequency is 72 rad/s 11.5 Hz 11 Hz 2 2 f . p p = = = (b) The wavelength is 2 2 1.14 m 1.1 m k 5.5 rad/m p p . = = = (c) The wave speed v = . f =13 m/s. 20.14. Solve: The amplitude of the wave is the maximum displacement, which is 6.0 cm. The period of the wave is 0.60 s, so the frequency f =1 T =1 0.60 s =1.67 Hz . The wavelength is 2 m/s 1.2 m 1.667 Hz v f . = = = 20.15. Solve: According to Equation 20.28, the phase difference between two points on a wave is ( ) 2 1 2 1 2 r 2 r r p f f f p . . . . = - = = - (3 rad 0 rad) 2 (80 cm 20 cm) p p . . - = - .. = 40 cm 20.16. Solve: According to Equation 20.28, the phase difference between two points on a wave is ( ) 2 1 2 1 2 r r p f f f . . = - = - If f1 =p rad at r1 = 4.0 m, we can determine 2 f at any r value at the same instant using this equation. At r2 = 3.5 m, ( ) ( ) 2 1 2 1 2 rad 2 3.5 m 4.0 m rad 2.0 m 2 r r p p p f f p . = + - = + - = At r2 = 4.5 m, 32 f = p rad. 20.17. Visualize: Solve: For a sinusoidal wave, the phase difference between two points on the wave is given by Equation 20.28: ( ) ( ) ( ) 2 1 2 1 2 r r 2 40 m 30 m 2 10 m p p p f f f . . . f . = - = - = - . = . .f = 2p for two points on adjacent wavefronts and .f = 4p for two points separated by 2.. Thus, . = 10 m when .f = 2p , and . = 5 m when .f = 4p . The crests corresponding to these two wavelengths are shown in the figure. One can see that a crest of the wave passes the 40 mlistener and the 30 mlistener simultaneously. The lowest two possible frequencies will occur for the largest two possible wavelengths, which are 10 m and 5 m. Thus, the lowest frequency is 1 340 m/s 34 Hz 10 m f v . = = = The next highest frequency is 2 f = 68 Hz. 20.18. Visualize: Solve: (a) Because the same wavefront simultaneously reaches listeners at x = -7.0 m and x = + 3.0 m, ( ) 2 1 2 1 0 rad 2 r r r r p f . . = = - . = Thus, the source is at x = -2.0 m, so that it is equidistant from the two listeners. (b) The third person is also 5.0 m away from the source. Her y-coordinate is thus y = (5 m)2 - (2 m)2 = 4.6 m. 20.19. Solve: Two pulses of sound are detected because one pulse travels through the metal to the microphone while the other travels through the air to the microphone. The time interval for the sound pulse traveling through the air is air air 4.0 m 0.01166 s 11.66 ms 343 m/s t x v . . = = = = Sound travels faster through solids than gases, so the pulse traveling through the metal will reach the microphone before the pulse traveling through the air. Because the pulses are separated in time by 11.0 ms, the pulse traveling through the metal takes .tmetal = 0.66 ms to travel the 4.0 m to the microphone. Thus, the speed of sound in the metal is metal metal 4.0 m 6060 m/s 6100 m/s 0.00066 s v x t . = = = . 20.20. Solve: (a) In aluminum, the speed of sound is 6420 m/s. The wavelength is thus equal to 3 6 6420 m/s 3.21 10 m 3.21 mm 3.2 mm 2.0 10 Hz v f . = = = - = (b) The speed of an electromagnetic wave is c. The frequency would be 8 10 3 3.0 10 m/s 9.3 10 Hz 3.21 10 m f c . - = = = 20.21. Solve: (a) The frequency is air 343 m/s 1715 Hz 1700 Hz 0.20 m f v . == = (b) The frequency is 8 3.0 10 m/s 1.5 109 Hz 1.5 GHz 0.20 m f c . = = = = (c) The speed of a sound wave in water is vwater = 1480 m/s. The wavelength of the sound wave would be water 7 9 1480 m/s 9.87 10 m 990 nm 1.50 10 Hz v f . = = = - 20.22. Model: Light is an electromagnetic wave that travels with a speed of 3 108 m/s. Solve: (a) The frequency of the blue light is 8 14 blue 9 3.0 10 m/s 6.67 10 Hz 450 10 m f c . - = = = (b) The frequency of the red light is 8 14 red 9 3.0 10 m/s 4.62 10 Hz 650 10 m f - = = (c) Using Equation 20.30 to calculate the index of refraction, vacuum material n . . = vacuum material 650 nm 1.44 450 nm n . . . = = = 20.23. Model: Microwaves are electromagnetic waves that travel with a speed of 3 108 m/s. Solve: (a) The frequency of the microwave is 8 10 microwaves 2 3.0 10 m/s 1.0 10 Hz 10 GHz 3.0 10 m f c . - = = = = (b) The refractive index of air is 1.0003, so the speed of microwaves in air is air v = c /1.00 c. The time for the microwave signal to travel is ( ) 3 8 air 50 km 50 10 m 0.167 ms 0.17 ms 3.0 10 m 1.00 t v = = = Assess: A small time of 0.17 ms for the microwaves to cover a distance of 50 km shows that the electromagnetic waves travel very fast. 20.24. Model: Radio waves are electromagnetic waves that travel with speed c. Solve: (a) The wavelength is 3.0 108 m/s 2.96 m 101.3 MHz c f . = = = (b) The speed of sound in air at 20C is 343 m/s. The frequency is sound 343 m/s 116 Hz 2.96 m f v . = = = 20.25. Model: Light is an electromagnetic wave. Solve: (a) The time light takes is ( ) 3 3 11 8 glass 3.0 mm 3.0 10 m 3.0 10 m 1.5 10 s 3.0 10 m/s 1.50 t v cn - - - = = = = (b) The thickness of water is ( ) 8 11 water water 3.0 10 m/s 1.5 10 s 3.4 mm 1.33 d v t c t n - = = = = 20.26. Solve: (a) The speed of light in a material is given by Equation 20.29: mat mat n c v c v n = . = The refractive index is vac solid ( 8 ) 8 solid mat vac 3.0 10 m/s 420 nm 1.88 10 m/s 670 nm n v c . . . . = . = = = (b) The frequency is 8 solid 14 solid 1.88 10 m/s 4.48 10 Hz 420 nm f v . = = = 20.27. Model: Assume that the glass has index of refraction n =1.5. This means that 8 glass v = c/n = 210 m/s. Visualize: We apply v =. f twice, once in air and then in the glass. The frequency will be the same in both cases. Solve: (a) In the air 8 air 8 8 air air 3 0 10 m/s 8 57 10 Hz 8 6 10 Hz 0 35 m f v . . = = = . . . The frequency is the same in both media, so 8 glass f = 8.610 Hz. (b) Now that we know glass f and glass v , we can find glass. . 8 glass glass 8 glass 2 0 10 m/s 23 cm 8 57 10 Hz v f . . = = = . Assess: We get the same answer from glass air glass . =. /n = 35 cm/1.5 = 23 cm . 20.28. Solve: The energy delivered to the eardrum in time t is E = Pt, where P is the power of the wave. The intensity of the wave is I = P/ a where a is the area of the ear drum. Puttin