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Unformatted text preview: Version 188 Quiz 2 coker (58245) 1 This printout should have 24 questions. Multiplechoice questions may continue on the next column or page find all choices before answering. Quiz 2 covers only topics covered in class lectures over Chs. 6, 7, 8 and 9 in Oha nian/Markert. 001 10.0 points A 93 . 9 g bullet is fired from a rifle having a barrel 0 . 256 m long. Assuming the origin is placed where the bullet begins to move, the force exerted on the bullet by the expanding gas is F = a + bx cx 2 , where a = 18600 N, b = 10800 N / m, c = 26300 N / m 2 , with x in meters. Determine the work done by the gas on the bullet as the bullet travels the length of the barrel. 1. 8257.95 2. 4431.12 3. 6960.28 4. 10731.7 5. 3206.1 6. 15372.2 7. 16468.4 8. 6738.68 9. 4968.41 10. 11392.7 Correct answer: 4968 . 41 J. Explanation: The work is found by integrating the force over the distance. W = integraldisplay x f x i F ds For the force in this problem we have, W = integraldisplay ( a + bx cx 2 ) dx = ax + 1 2 bx 2 1 3 cx 3 vextendsingle vextendsingle vextendsingle = (18600 N)(0 . 256 m) + 1 2 (10800 N / m)(0 . 256 m) 2 1 3 (26300 N / m 2 )(0 . 256 m) 3 = 4968 . 41 J . 002 (part 1 of 2) 10.0 points Consider a force of 66 . 3 N, pulling 3 blocks of identical masses, where each mass is 0 . 5 kg, along a rough horizontal surface. The coef ficient of kinetic friction between the blocks and the surface is given by 0 . 8. The accelera tion of the blocks is a . The acceleration of gravity is 9 . 8 m / s 2 . m 3 m 2 m 1 T 2 T 1 F Find the acceleration a . 1. 13.58 2. 26.0583 3. 16.9382 4. 36.36 5. 12.2715 6. 27.3378 7. 11.14 8. 11.857 9. 15.1333 10. 42.46 Correct answer: 36 . 36 m / s 2 . Explanation: Given : m 1 = m = 0 . 5 kg , m 2 = m = 0 . 5 kg , m 3 = m = 0 . 5 kg , and = 0 . 8 . From Newtons second law, considering the system with mass m 1 + m 2 + m 3 = 3 m, F f = F 3 mg = 3 ma. So a = F 3 mg 3 m = 66 . 3 N 3 (0 . 8) (0 . 5 kg) (9 . 8 m / s 2 ) 3 (0 . 5 kg) = 36 . 36 m / s 2 . Version 188 Quiz 2 coker (58245) 2 003 (part 2 of 2) 10.0 points The tension of the strings are T 1 and T 2 (see sketch). The equation of motion of m 2 is given by 1. T 1 m 2 g = m 2 a. 2. T 1 + T 2 + m 2 g m 1 g = m 2 a. 3. T 1 T 2 + m 2 g = m 2 a. 4. T 1 + T 2 m 2 g = m 2 a. 5. T 1 + m 2 g = m 2 a. 6. T 1 T 2 m 2 g m 1 g = m 2 a. 7. T 1 + T 2 m 2 g m 1 g = m 2 a. 8. T 1 T 2 m 2 g = m 2 a. correct 9. T 1 + T 2 + m 2 g = m 2 a. 10. T 1 T 2 + m 2 g m 1 g = m 2 a. Explanation: From inspection, the net force acting on m 2 is T 1 T 2 m 2 g , so the equation of motion is given by T 1 T 2 m 2 g = m 2 a....
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This note was uploaded on 10/29/2009 for the course PHY 303K taught by Professor Turner during the Spring '08 term at University of Texas at Austin.
 Spring '08
 Turner
 Physics

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