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Unformatted text preview: Thevalingam, Donald – Homework 10 – Due: Dec 5 2006, midnight – Inst: Eslami 1 This printout should have 15 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 2) 10 points A block is released from rest on an inclined plane and moves 2 . 1 m during the next 4 . 9 s. The acceleration of gravity is 9 . 8 m / s 2 . 1 1 k g μ k 27 ◦ What is the magnitude of the acceleration of the block? Correct answer: 0 . 174927 m / s 2 . Explanation: Given : m = 11 kg , ‘ = 2 . 1 m , θ = 27 ◦ , and t = 4 . 9 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 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 . 1 m) (4 . 9 s) 2 = . 174927 m / s 2 . 002 (part 2 of 2) 10 points What is the coefficient of kinetic friction μ k for the incline? Correct answer: 0 . 489493 . 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 = tan27 ◦ 2(2 . 1 m) (9 . 8 m / s 2 )(4 . 9 s) 2 cos27 ◦ = . 489493 . keywords: 003 (part 1 of 1) 10 points There is friction between the block and the table. The suspended 2 kg mass on the left is moving up, the 3 kg mass slides to the right on the table, and the suspended mass 5 kg on the right is moving down. The acceleration of gravity is 9 . 8 m / s 2 . 2 kg 3 kg 5 kg μ = 0 . 15 Thevalingam, Donald – Homework 10 – Due: Dec 5 2006, midnight – Inst: Eslami 2 What is the magnitude of the acceleration of the system? Correct answer: 2 . 499 m / s 2 . Explanation: m 1 m 2 m 3 μ a Let : m 1 = 2 kg , m 2 = 3 kg , m 3 = 5 kg , and μ = 0 . 15 . Basic Concepts: The acceleration a of each mass is the same, but the tensions in the two strings will be different. F net = ma 6 = 0 Solution: Let T 1 be the tension in the left stringand T 2 bethetensionintherightstring. Consider the free body diagrams for each mass T 1 m 1 g a T 2 m 3 g a T 1 T 2 N μ N a m 2 g For the mass m 1 , T 1 acts up and the weight m 1 g acts down, with the acceleration a di rected upward, so F net 1 = m 1 a = T 1 m 1 g . (1) For the mass on the table, a is directed to the right, T 2 acts to the right, T 1 acts to the left, and the motion is to the right so that the frictional force μm 2 g acts to the left and F net 2 = m 2 a = T 2 T 1 μm 2 g . (2) For the mass m 3 , T 2 acts up and the weight m 3 g acts down, with the acceleration a di rected downward, so F net 3 = m 3 a = m 3 g T 2 . (3) Adding these equations yields ( m 1 + m 2 + m 3 ) a = m 3 g μm 2 g m 1 g a = m 3 μm 2 m 1 m 1 + m 2 + m 3 g = 5 kg (0 .....
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This note was uploaded on 03/04/2010 for the course PHY 303K taught by Professor Turner during the Spring '08 term at University of Texas.
 Spring '08
 Turner
 Physics, Work

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