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Unformatted text preview: sela (ds39632) – HW05 – janow – (11196) 1 This printout should have 11 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 3 . 5 m during the next 1 . 7 s. The acceleration of gravity is 9 . 8 m / s 2 . 8 k g μ k 24 ◦ What is the coefficient of kinetic friction μ k for the incline? Correct answer: 0 . 174681. Explanation: Given : m = 8 kg , ℓ = 3 . 5 m , θ = 24 ◦ , and t = 1 . 7 s . Consider the free body diagram for the block m g s i n θ N = m g c o s θ μ N a m g The acceleration can be obtained through kinematics. Since v = 0, ℓ = v t + 1 2 a t 2 = 1 2 a t 2 a = 2 ℓ t 2 (1) Applying Newton’s Second Law of Motion summationdisplay 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 ℓ t 2 = g parenleftBig sin θ μ k cos θ parenrightBig 2 ℓ = g t 2 parenleftBig sin θ μ k cos θ parenrightBig μ k = g t 2 sin θ 2 ℓ g t 2 cos θ (2) = tan θ 2 ℓ g t 2 cos θ = tan24 ◦ 2 (3 . 5 m) (9 . 8 m / s 2 ) (1 . 7 s) 2 cos 24 ◦ = . 174681 . 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 1 . 96 kg mass has fallen through 0 . 415 m, its downward speed is 1 . 25 m / s. The acceleration of gravity is 9 . 8 m / s 2 . 1 . 96 kg 3 . 53 kg μ a What is the frictional force between the 3 . 53 kg mass and the table? Correct answer: 8 . 87291 N. Explanation: Given : m 1 = 1 . 96 kg , m 2 = 3 . 53 kg , v = 0 m / s , and v = 1 . 25 m / s . Basic Concept: Newton’s Second Law F = M a sela (ds39632) – HW05 – janow – (11196) 2 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 . 25 m / s) 2 (0 m / s) 2 2 (0 . 415 m) = 1 . 88253 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 = (1 . 96 kg) (9 . 8 m / s 2 ) (1 . 96 kg + 3 . 53 kg) × (1 . 88253 m / s 2 ) = 8 . 87291 N ....
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This note was uploaded on 03/09/2012 for the course PHYS 111 taught by Professor Moro during the Spring '08 term at NJIT.
 Spring '08
 moro

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