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HW 5-solutions-1 - hamdeh(ah37785 HW 5 opyrchal(111213 This...

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hamdeh (ah37785) – HW 5 – opyrchal – (111213) 1 This print-out should have 10 questions. Multiple-choice 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 . 13 kg μ k 28 What is the coe ffi cient of kinetic friction μ k for the incline? Correct answer: 0 . 488492. Explanation: Given : m = 13 kg , = 2 . 7 m , θ = 28 , and t = 3 . 8 s . Consider the free body diagram for the block m g sin θ N = m g cos θ μ N a m g The acceleration can be obtained through kinematics. Since v 0 = 0, = v 0 t + 1 2 a t 2 = 1 2 a t 2 a = 2 t 2 (1) Applying Newton’s Second Law of Motion 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 θ - μ k m g cos θ 2 t 2 = g sin θ - μ k cos θ 2 = g t 2 sin θ - μ k cos θ μ k = g t 2 sin θ - 2 g t 2 cos θ (2) = tan θ - 2 g t 2 cos θ = tan 28 - 2 (2 . 7 m) (9 . 8 m / s 2 ) (3 . 8 s) 2 cos 28 = 0 . 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 μ 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 = 0 m / s , and v = 1 . 27 m / s .
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hamdeh (ah37785) – HW 5 – opyrchal – (111213) 2 Basic Concept: Newton’s Second Law F = M a Solution: The acceleration of m 1 is obtained from the equation v 2 - v 2 0 = 2 a ( s - s 0 ) a = v 2 - v 2 0 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 μ 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 . Thus F 2 = T - f k , f k = T - F 2 = m 1 g - ( m 1 + m 2 ) a = (3 . 29 kg) (9 . 8 m / s 2 ) - (3 . 29 kg + 4 . 88 kg) × (1 . 99123 m / s 2 ) = 15 . 9736 N .
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