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Unformatted text preview: hamdeh (ah37785) HW 5 opyrchal (111213) 1 This printout should have 10 questions. Multiplechoice questions may continue on the next column or page find all choices before answering. 001 10.0 points A block is released from rest on an inclined plane and moves 2 . 7 m during the next 3 . 8 s. The acceleration of gravity is 9 . 8 m / s 2 . 1 3 k g k 28 What is the coe ffi cient of kinetic friction k for the incline? Correct answer: 0 . 488492. Explanation: Given : m = 13 kg , ` = 2 . 7 m , = 28 , and t = 3 . 8 s . Consider the free body diagram for the block m g s i n N = m g c o s N a m g The acceleration can be obtained through kinematics. Since v = 0, ` = v t + 1 2 a t 2 = 1 2 a t 2 a = 2 ` t 2 (1) Applying Newtons Second Law of Motion X F i = m a and Eq. 1, the sine component of the weight acts down the plane and friction acts up the plane. The block slides down the plane, so m a = m g sin  k m g cos 2 ` t 2 = g sin  k cos 2 ` = g t 2 sin  k cos k = g t 2 sin  2 ` g t 2 cos (2) = tan  2 ` g t 2 cos = tan28  2 (2 . 7 m) (9 . 8 m / s 2 ) (3 . 8 s) 2 cos 28 = . 488492 . 002 10.0 points Two blocks are arranged at the ends of a mass less string as shown in the figure. The system starts from rest. When the 3 . 29 kg mass has fallen through 0 . 405 m, its downward speed is 1 . 27 m / s. The acceleration of gravity is 9 . 8 m / s 2 . 3 . 29 kg 4 . 88 kg a What is the frictional force between the 4 . 88 kg mass and the table? Correct answer: 15 . 9736 N. Explanation: Given : m 1 = 3 . 29 kg , m 2 = 4 . 88 kg , v = 0 m / s , and v = 1 . 27 m / s . hamdeh (ah37785) HW 5 opyrchal (111213) 2 Basic Concept: Newtons Second Law F = M a Solution: The acceleration of m 1 is obtained from the equation v 2 v 2 = 2 a ( s s ) a = v 2 v 2 2 h = (1 . 27 m / s) 2 (0 m / s) 2 2 (0 . 405 m) = 1 . 99123 m / s 2 . Consider free body diagrams for the two masses T m 1 g a T N N a m 2 g Because m 1 and m 2 are tied together with string, they have same the speed and the same acceleration, so the net force exerted on m 2 is F 2 = m 2 a The net force on m 1 is m 1 a = m 1 g T , so that T = m 1 g m 1 a ....
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This note was uploaded on 04/07/2012 for the course PHYSICS 121 taught by Professor James during the Spring '12 term at University of Texas at Austin.
 Spring '12
 James
 Physics

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