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Unformatted text preview: This printout should have 20 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. The due time is Central time. 001 (part 1 of 1) 10 points Two blocksare arrangedattheendsofamass less string as shown in the figure. The system starts from rest. When the 1 . 48 kg mass has fallen through 0 . 308 m, its downward speed is 1 . 29 m / s. The acceleration of gravity is 9 . 8 m / s 2 . 1 . 48 kg 3 . 03 kg μ a What is the frictional force between the 3 . 03 kg mass and the table? Correct answer: 2 . 32041 N. Explanation: Given : m 1 = 1 . 48 kg , m 2 = 3 . 03 kg , v = 0 m / s , and v = 1 . 29 m / s . Basic Concept: Newton’s 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 . 29 m / s) 2 (0 m / s) 2 2(0 . 308 m) = 2 . 70146 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 . 48 kg)(9 . 8 m / s 2 ) (1 . 48 kg + 3 . 03 kg) × (2 . 70146 m / s 2 ) = 2 . 32041 N . 002 (part 1 of 2) 10 points A block is released from rest on an inclined plane and moves 2 m during the next 4 . 3 s. The acceleration of gravity is 9 . 8 m / s 2 . 5 k g μ k 26 ◦ What is the magnitude of the acceleration of the block? Correct answer: 0 . 216333 m / s 2 . Explanation: Given : m = 5 kg , ‘ = 2 m , θ = 26 ◦ , and t = 4 . 3 s . – Homework 4 – Due: Feb 15 2006, midnight – Inst: Fitzpatrick 1 Consider the free body diagram for the block m g s i n θ N = m g c o s θ μ N a mg The acceleration can be obtained through kinematics. Since v = 0, ‘ = v t + 1 2 at 2 = 1 2 at 2 a = 2 ‘ t 2 (1) = 2(2 m) (4 . 3 s) 2 = . 216333 m / s 2 . 003 (part 2 of 2) 10 points What is the coefficient of kinetic friction μ k for the incline? Correct answer: 0 . 463172 . Explanation: Applying Newton’s Second Law of Motion X F i = ma 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 ma = mg sin θ μ k mg 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 = tan26 ◦ 2(2 m) (9 . 8 m / s 2 )(4 . 3 s) 2 cos26 ◦ = . 463172 . 004 (part 1 of 1) 10 points A block sits motionless on an inclined plane, held in place by friction. The plane could be tilted even more upwards and the block would remain motionless....
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This note was uploaded on 09/14/2009 for the course PHY 302K taught by Professor Kaplunovsky during the Fall '08 term at University of Texas.
 Fall '08
 Kaplunovsky
 Work, Heat

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