# SOL10 - Thevalingam Donald Homework 10 Due Dec 5 2006...

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Thevalingam, Donald – Homework 10 – Due: Dec 5 2006, midnight – Inst: Eslami 1 This print-out should have 15 questions. Multiple-choice questions may continue on the next column or page – fnd 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 . 11 kg 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 sin θ N = cos a mg The acceleration can be obtained through kinematics. Since v 0 = 0, = v 0 t + 1 2 at 2 = 1 2 2 a = 2 t 2 (1) = 2(2 . 1 m) (4 . 9 s) 2 = 0 . 174927 m / s 2 . 002 (part 2 oF 2) 10 points What is the coe±cient 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 = sin θ - μ k 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 - . 1 m) (9 . 8 m / s 2 )(4 . 9 s) 2 cos27 = 0 . 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

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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 diFerent. F net = ma 6 = 0 Solution: Let T 1 be the tension in the left string and T 2 be the tension in the right string. 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 ±or 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) ±or 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 - 2 g . (2) ±or 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 - 2 g - m 1 g a = m 3 - 2 - m 1 m 1 + m 2 + m 3 g = 5 kg - (0 . 15)(3 kg) - 2 kg 2 kg + 3 kg + 5 kg × (9 . 8 m / s 2 ) = 2 . 499 m / s 2 .
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SOL10 - Thevalingam Donald Homework 10 Due Dec 5 2006...

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