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4 Chapter Homework 9. A box rests on a frozen pond, which serves as a frictionless horizontal surface. If a fisherman applies a horizontal force with magnitude 48.0 N to the box and produces an acceleration of magnitude 3.00 m/s2, what is the mass of the box? Fx=max 48.0 N=m3.00ms2 m=16 kg 10. A dockworker applies a constant horizontal force of 80.0 N to a block of ice on a smooth horizontal floor. The frictional force is negligible. The block starts from rest and moves 11.0 m in 5.00 s. (a) What is the mass of the block of ice? x-x0=v0xt+12 axt2 11.0 m=0ms5.00 s+12 ax5.00 s2 ax=0.880ms2 Fx=max 80.0 N=m0.880ms2 m=90.9 kg (b) If the worker stops pushing at the end of 5.00 s, how far does the block move in the next 5.00 s? vx= v0x+ axt =0ms+ 0.880ms25.00 s =4.40 m/s x-x0=v0xt+12 axt2 =4.40 m/s5.00 s+12 0ms25.00 s2 =22.0 m 12. A crate with mass 32.5 kg initially at rest on a warehouse floor is acted on by a net horizontal force of 140 N. (a) What acceleration is produced? Fx=max 140 N=32.5 kgax ax=4.31ms2 (b) How far does the crate travel in 10.0 s? Page | 1 2.1.11 x-x0=v0xt+12 axt2 =0 m/s10.0 s+12 4.31 ms210.0 s2 =215 m (c) What is its speed at the end of 10.0 s? vx= v0x+ axt =0ms+ 4.31ms210.0 s =43.1 m/s 16. An electron (mass = 9.11 x 10-31 kg) leaves one end of a TV picture tube with zero initial speed and travels in a straight line to the accelerating grid, which is 1.80 cm away. It reaches the grid with a speed of 3.00 x 106 m/s. If the accelerating force is constant, compute (a) the acceleration vx 2= v0x 2+2axx-x0 3.00 106ms2=0ms2+2ax0.0180 m ax=2.50 1014ms2 (b) the time to reach the grid vx= v0x+ axt 3.00 106ms=0ms+ 2.50 1014ms2 t t=1.20 10-8 s (c) the net force, in newtons Fx=max =9.11 10-31 kg2.50 1014ms2 =2.28 10-16 N 17. Superman throws a 2400-N boulder at an adversary. What horizontal force must Superman apply to the boulder to give it a horizontal acceleration of 12.0 m/s2? w=mg 2400 N=m9.80ms2 m=245 kg Fx=max =245 kg12.0ms2 =2940 N 18. A bowling ball weighs 71.2 N (16.0 Ib). The bowler applies a horizontal force of 160 N (36.0 lb) to the ball. What is the magnitude of the horizontal acceleration of the ball? w=mg Page | 2 2.1.11 71.2 N=m9.80ms2 m=7.27 kg Fx=max 160 N =7.27 kgax ax=22.0 m/s2 19. At the surface of Jupiter's moon Io, the acceleration due to gravity is g = 1.81 m/s2. A watermelon weighs 44.0 N at the surface of the earth. (a) What is the watermelon's mass on the earth's surface? w=mg 44.0 N=m9.80ms2 m=4.49 kg (b) What are its mass and weight on the surface of Io? m=4.49 kg w=mg =4.49 kg1.81ms2 =8.13 N 20. An astronaut's pack weighs 17.5 N when she is on earth but only 3.24 N when she is at the surface of an asteroid. (a) What is the acceleration due to gravity on this asteroid? wE=mgE 17.5 N=m9.80ms2 m=1.79 kg wA=mgA 3.24 N=1.79 kggA gA=1.81 m/s2 (b) What is the mass of the pack on the asteroid? m=1.79 kg 22. Imagine that you are holding a book weighing 4 N at rest on the palm of your hand. Complete the following sentences: (a) A downward force of magnitude 4 N is exerted on the book by the earth. Page | 3 2.1.11 (b) An upward force of magnitude 4 N is exerted on the book by your hand. (c) Is the upward force in part (b) the reaction to the downward force in part (a)? No (d) The reaction to the force in part (a) is a force of magnitude 4 N exerted on the earth by the book. Its direction is upward. (e) The reaction to the force in part (b) is a force of magnitude 4 N exerted on the hand by the book. Its direction is downward. (f) The forces in parts (a) and (b) are equal and opposite because of Newton's Second law. The forces are exerted on the same object and this object has zero acceleration. (g) The forces in parts (b) and (e) are equal and opposite because of Newton's Third law. The forces are between a pair of objects. Now suppose that you exert an upward force of magnitude 5 N on the book. (h) Does the book remain in equilibrium? No (i) Is the force exerted on the book by your hand equal and opposite to the force exerted on the book by the earth? No (j) Is the force exerted on the book by the earth equal and opposite to the force exerted on the earth by the book? Yes (k) Is the force exerted on the book by your hand equal and opposite to the force exerted on your hand by the book? Yes Finally, suppose you snatch your hand away while the book is moving upward. (l) How many forces then act on the book? 1 (gravity) (m) Is the book in equilibrium? No 23. A bottle is given a push along a table top and slides off the edge of the table. Do not ignore air resistance. (a) What forces are exerted on the bottle while it is falling from the table to the floor? Only gravity (downward) and air resistance (upward) (b) What is the reaction to each force; that is, on which body and by which body is the reaction exerted? The earth is exerting a downward force on the bottle and the bottle exerts an upward force on the earth. The air exerts an upward force on the bottle and the bottle exerts downward force on the air. Page | 4 2.1.11 24. The upward normal force exerted by the floor is 620 N on an elevator passenger who weighs 650 N. What are the reaction forces to these two forces? Is the passenger accelerating? If so, what are the magnitude and direction of the acceleration? The elevator exerts a downward force on the floor and the passenger exerts an upward force on the earth. Fy =may 650N-620N=650N9.8 ms2 ay ay=0.452 ms2 downward 27. Two crates, A and B, sit at rest side by side on a frictionless horizontal surface. The crates have masses mA and mB. A horizontal force F is applied to crate A and the two crates move off to the right. (a) Draw clearly labeled free-body diagrams for crate A and for crate B. Indicate which pairs of forces, if any, are third-law action-reaction pairs. The only action-reaction pair is the force exerted on the mass of crate B by the mass of crate A and the force exerted on the mass of crate A by the mass of crate B. (b) If the magnitude of force F is less than the total weight of the two crates, will it cause the crates to move? Explain. Yes, since the horizontal surface is frictionless, a force of any magnitude will cause the crate to move (there is no horizontal force in the negative direction opposing F). 31. A chair of mass 12.0 kg is sitting on the horizontal floor; the floor is not frictionless. You push on the chair with a force F = 40.0 N that is directed at an angle of 37.0 below the horizontal and the chair slides along the floor. (a) Draw a clearly labeled free-body diagram for the chair. Page | 5 2.1.11 (b) Use your diagram and Newton's laws to calculate the normal force that the floor exert on the chair. Fy=may 0=n-mg-Fsin37 0 =n-12.0 kg9.8 ms2-(4.0 N)sin37 n=142 N 32. A skier of mass 65.0 kg is pulled up a snow-covered slope at constant speed by a tow rope that is parallel to the ground. The ground slopes upward at a constant angle of 26.0 above the horizontal, and you can ignore friction. (a) Draw a clearly labeled free-body diagram for the skier. (b) Calculate the tension in the tow rope. Fx=max 0=T-mgsin T=65.0 kg9.8 ms2sin26.0 =279 N 33. A truck is pulling a car on a horizontal highway using a horizontal rope. The car is in neutral gear, so we can assume that there is no appreciable friction between its tires and the highway. As the truck is accelerating to highway speeds, draw a free-body diagram of Page | 6 2.1.11 (a) the car and (b) the truck (c) What force accelerates this system forward? Explain how this force originates. The friction force f accelerates this system forward. The tires of the truck push backwards on the road and the road pushes forwards on the tires. 34. A .22 rifle bullet, traveling at 350 m/s, strikes a large tree, which it penetrates to a depth of 0.130 m. The mass of the bullet is 1.80 g. Assume a constant retarding force. (a) How much time is required for the bullet to stop? x-x0= v0x+vx 2t t= 2x-x0v0x+vx = 20.130 m350ms + 0ms = 7.43 10-4 s (b) What force, in newtons, does the tree exert on the bullet? vx 2= v0x 2+2axx-x0 0ms2=350 ms2+2ax0.130 m ax=-4.71 105ms2 FTree on Bullet= -FBullet on Tree --4.71 = 105ms2 = 4.71 105ms2 36. You have just landed on Planet X. You take out a 100-g ball, release it from rest from a height of 10.0 m, and measure that it takes 2.2 s to reach the ground. You can ignore any force on the ball from the atmosphere of the planet. How much does the 100-g ball weigh on the surface of Planet X? y- y0=v0yt+12 ayt2 10 m=0ms(2.2 s)+12 ay2.2 s2 ay= 4.13 m/s2 Page | 7 2.1.11 wX=mgX =0.100 kg4.13ms2 =0.41 N 38. An oil tanker's engines have broken down, and the wind is blowing the tanker straight toward a reef at a constant speed of 1.5 m/s. When the tanker is 500 m from the reef, the wind dies down just as the engineer gets the engines going again. The rudder is stuck, so the only choice is to try to accelerate straight backward away from the reef. The mass of the tanker and cargo is 3.6 x 107 kg, and the engines produce a net horizontal force of 8.0 x 104 N on the tanker. Will the ship hit the reef? If it does, will the oil be safe? The hull can withstand an impact at a speed of 0.2 m/ s or less. You can ignore the retarding force of the water on the tanker's hull. Fx=max -8.0 104 N =3.6 107kgax ax=-2.22 10-3 m/s2 vx 2= v0x 2+2axx-x0 x-x0= vx2-v0x22ax = 0ms2-1.5ms22-2.22 10-3ms2 = 506 m vx= v0x 2+ 2axx-x0 = 1.5ms2+ 2-2.22 10-3ms2500 m = 0.167ms The ship will hit the reef; however, the hull can withstand the impact and the oil will be safe. 40. An advertisement claims that a particular automobile can "stop on a dime." What net force would actually be necessary to stop a 850-kg automobile traveling initially at 45.0 km/h in a distance equal to the diameter of a dime, which is 1.8 cm? 45.0kmhr1000 m1 km1 hr3600 s=12.5 m/s Page | 8 2.1.11 vx 2= v0x 2+2axx-x0 0ms2=12.5ms2+2ax0.018 m ax= -4340 m/s2 Fx=max =850 kg-4340ms2 =-3.70 106 N [Negative sign implies a force in the opposite direction of the car] 41. A 4.80-kg bucket of water is accelerated upward by a cord of negligible mass whose breaking strength is 75.0 N. (a) Draw the free-body force diagram for the bucket. In terms of the forces on your diagram, what is the net force on the bucket? (b) Apply Newton's second law to the bucket and find the maximum upward acceleration that can be given to the bucket without breaking the cord. Fy=may T-mg=may 75.0 N-4.80 kg9.8 ms2=4.80 kgay ay=5.83 ms2 Page | 9 2.1.11 42. A parachutist relies on air resistance (mainly on her parachute) to decrease her downward velocity. She and her parachute have a mass of 55.0 kg, and air resistance exerts a total upward force of 620 N on her and her parachute. (a) What is the weight of the parachutist? W=mg =55.0 kg9.8 ms2 =539 N (b) Draw a free-body diagram for the parachutist. Use that diagram to calculate the net on the parachutist. Is the net force upward or downward? Fy=Fair- W =620 N-539 N =81 N force (c) What is the acceleration (magnitude and direction) of the parachutist? Fy=may 81 N=55.0 kgay ay=1.47 ms2 upward Page | 10 2.1.11 43. Two crates, one with mass 4.00 kg and the other with mass 6.00 kg, sit on the frictionless surface of a frozen pond, connected by a light rope. A woman wearing golf shoes (so she can get traction on the ice) pulls horizontally on the 6.00-kg crate with a force F that gives the crate an acceleration of 2.50 m/s2. (a) What is the acceleration of the 4.00-kg crate? ax=2.50 ms2 (b) Draw a free-body diagram for the 4.00-kg crate. Use that diagram and Newton's second law to find the tension T in the rope that connects the two crates. Fx=max T=m1ax =4.00 kg2.50 ms2 =10.0 N (c) Draw a free-body diagram for the 6.00-kg crate. What is the direction of the net force on the 6.00-kg crate? Which is larger in magnitude, force T or force F? The net force is to the right since the crate is accelerating to the right, so F must be larger than T. (d) Use part (c) and Newton's second law to calculate the magnitude of the force F. Fx=max F-T=m2ax F-10.0 N=6.00 kg2.50 ms2 F=25.0 N Page | 11 2.1.11 44. An astronaut is tethered by a strong cable to a spacecraft. The astronaut and her spaceship have a total mass of 105 kg, while the mass of the cable is negligible. The mass of the spacecraft is 9.05 x 104 kg. The spacecraft is far from any large astronomical bodies, so we can ignore the gravitational forces on it and the astronaut. We also assume that both the spacecraft and the astronaut are initially at rest in an inertial reference frame. The astronaut then pulls on the cable with a force of 80.0 N. (a) What force does the cable exert on the astronaut? The cable exerts a force of 80.0 N on the astronaut. (b) Since Fx=max, how can a "massless" (m = 0) cable exert a force? The cable is experiencing tension from the pull of the astronaut. (c) What is the astronaut's acceleration? F=ma 80.0 N=105 kga a=0.762ms2 (d) What force does the cable exert on the spacecraft? Since there is zero net force on the cable, the cable must exert a force of 80.0 N on the spacecraft. (e) What is the acceleration of the spacecraft? F=ma 80.0 N=9.05 104 kga a=0.000884 ms2 50. A loaded elevator with very worn cables has a total mass of 2200 kg, and the cables can withstand a maximum tension of 28,000 N. (a) Draw the free-body force diagram for the elevator. In terms of the forces on your diagram, what is the net force on the elevator? Apply Newton's second law to the elevator and find the maximum upward acceleration for the elevator if the cables are not to break. Fy=may T-mg=may Page | 12 2.1.11 28000 N-2200 kg9.8 ms2=2200 kgay ay=2.93 ms2 (b) What would be the answer to part (a) if the elevator were on the moon, where g = 1.62 m/s2? T-mg=may 28000 N-2200 kg1.62 ms2=2200 kgay ay=11.1 ms2 51. A 75.0-kg man steps off a platform 3.10 m above the ground. He keeps his legs straight as he falls, but at the moment his feet touch the ground his knees begin to bend, and, treated as a particle, he moves an additional 0.60 m before coming to rest. (a) What is his speed at the instant his feet touch the ground? vf2=vi2+ 2ayy vf2=0 ms2+ 29.8 ms2(3.10 m) vf=7.79 ms downward (b) Treating him as a particle, what is his acceleration (magnitude and direction) as he slows down, if the acceleration is assumed to be constant? vf2=vi2+ 2ayy 0 ms2=7.79 ms2+ 2ay(0.60 m) ay= -50.6 ms2 (upward) (c) Draw his free-body diagram. In terms of the forces on the diagram, what is the net force on him? Use Newton's laws and the results of part (b) to calculate the average force his feet exert on the ground while he slows down. Express this force in newtons and also as a multiple Page | 13 2.1.11 of his weight. Fy=may F-mg=ma F-75.0 kg9.8 ms2= 75.0 kg50.6 ms2 F=4530 N W=mg =75.0 kg9.8 ms2 =735 N F=4530 N w735 N=6.16w 54. Two blocks are connected by a heavy uniform rope with a mass of 4.00 kg. An upward force of 200 N is applied as shown. (a) Draw three free-body diagrams, one for the 6.00-kg block, one for the 4.00-kg rope, and another one for the 5.00-kg block. For each force, indicate what body exerts that force. Page | 14 2.1.11 (b) What is the acceleration of the system? Fy=may F-mg=may 200 N-6 kg+5 kg+4 kg9.8 ms2=6 kg+5 kg+4 kgay ay=3.53 ms2 (c) What is the tension at the top of the heavy rope? Fy=may F-mg-Tt=may 200 N-6.00 kg9.8 ms2- Tt=6.00 kg3.53 ms2 Tt=120 N (d) What is the tension at the midpoint of the rope? Fy=may Tt Tb- mg=may 120 N- Tb- 124.00 kg9.8 ms2=124.00 kg3.53 ms2 Tb=93.3 N Page | 15 2.1.11 55. An athlete whose mass is 90.0 kg is performing weight-lifting exercises. Starting from the rest position, he lifts, with constant acceleration, a barbell that weighs 490 N. He lifts the barbell a distance of 0.60 m in 1.6 s. (a) Draw a clearly labeled free-body force diagram for the barbell and for the athlete. (b) Use the diagrams in part (a) and Newton's laws to find the total force that his feet exert on the ground as he lifts the barbell. Wathlete=mg Wbarbell=mg =90.0 kg9.8 ms2 490 N= m9.8 ms2 =882 N mbarbell= 50 kg y=vit+12 ayt2 0.60 m=0 ms1.6 s+12 ay1.6 s2 Page | 16 2.1.11 ay=0.469 ms2 Fy=may Flift- wbarbell=may Flift=490 N+50 kg0.469 ms2 =513 N Fy=may Ffloor- Flift- W=0 Ffloor=513 N+882 N =1395 N Page | 17 2.1.11 ... View Full Document