HW19 - homework 19 PAPAGEORGE MATT Due Mar 4 2008 4:00 am...

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homework 19 – PAPAGEORGE, MATT – Due: Mar 4 2008, 4:00 am 1 Question 1, chap 8, sect 5. part 1 of 2 10 points A 1200 kg car starts from rest and acceler- ates uniformly to 16 . 2 m / s in 16 . 6 s . Assume that air resistance remains con- stant at 309 N during this time. Find the average power developed by the engine. Correct answer: 16 . 0706 hp (tolerance ± 1 %). Explanation: m = 1200 kg , v i = 0 m / s , v f = 16 . 2 m / s , and Δ t = 16 . 6 s . The acceleration of the car is a = v f - v i Δ t = v f Δ t since v i = 0, so a = 16 . 2 m / s 16 . 6 s = 0 . 975904 m / s 2 . Thus the constant forward force due to the engine is found from summationdisplay F = F engine - F air = m a F engine = F air + m a = 309 N + (1200 kg) ( 0 . 975904 m / s 2 ) = 1480 . 08 N . The average velocity of the car during this interval is v av = v f + v i 2 , so the average power output is P = F engine v av = F engine parenleftBig v f 2 parenrightBig = (1480 . 08 N) parenleftbigg 16 . 2 m / s 2 parenrightbigg parenleftbigg 1 hp 764 W parenrightbigg = 16 . 0706 hp . Question 2, chap 8, sect 5. part 2 of 2 10 points Find the instantaneous power output of the engine at t = 16 . 6 s just before the car stops accelerating. Correct answer: 32 . 1412 hp (tolerance ± 1 %). Explanation: The instantaneous velocity is 16 . 2 m / s and the instantaneous power output of the engine is P = F engine v f = (309 N)(16 . 2 m / s) parenleftbigg 1 hp 764 W parenrightbigg = 32 . 1412 hp . Question 3, chap 9, sect 99. part 1 of 2 10 points The universal gravitationan constant is 6 . 6726 × 10 11 N m 2 / kg 2 . Three 2 kg masses are located at points in the xy plane as shown in the figure. 36 cm 52 cm What is the magnitude of the resultant force (caused by the other two masses) on the mass at the origin? Correct answer: 2 . 28377 × 10 9 N (tolerance ± 1 %). Explanation: Let : m = 2 kg x = 36 cm , and y = 52 cm .

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homework 19 – PAPAGEORGE, MATT – Due: Mar 4 2008, 4:00 am 2 Basic Concepts: Newton’s Law of Grav- itation: F g = G m 1 m 2 r 2 We calculate the forces one by one, and then add them using the superposition prin- ciple. The force from the mass on the right is pointing in the x direction, and has magni- tude f 1 = G m m x 2 = G m 2 x 2 = (6 . 6726 × 10 11 N m 2 / kg 2 )(2 kg) 2 (0 . 36 m) 2 = 2 . 05944 × 10 9 N . The other force is pointing in the y direction and has magnitude f 2 = G m m y 2 = G m 2 y 2 = (6 . 6726 × 10 11 N m 2 / kg 2 )(2 kg) 2 (0 . 52 m) 2 = 9 . 87071 × 10 10 N . f 2 f 1 F θ Now we simply add the two forces, using vector addition. Since they are at right angles to each other, however, we can use Pythago- ras’ theorem as well: F = radicalBig f 2 1 + f 2 2 = bracketleftBig (2 . 05944 × 10 9 N) 2 + (9 . 87071 × 10 10 N) 2 bracketrightBig 1 2 = 2 . 28377 × 10 9 N . Note: As you see, this force is very small. If the masses in the picture are standing on a ta- ble (the view being from above), typically the force of static friction will not be overcome. In agreement with common sense, then, the masses will not move. Question 4, chap 9, sect 99.
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