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HS5 - Adams Russell Homework 5 Due midnight Inst Opyrchal H...

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Adams, Russell – Homework 5 – Due: Oct 12 2006, midnight – Inst: Opyrchal, H 1 This print-out should have 14 questions. Multiple-choice 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 A block is released from rest on an inclined plane and moves 3 . 4 m during the next 4 s. The acceleration of gravity is 9 . 8 m / s 2 . 15 kg μ k 34 What is the coefficient of kinetic friction μ k for the incline? Correct answer: 0 . 622198 . Explanation: Given : m = 15 kg , = 3 . 4 m , θ = 34 , and t = 4 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 X 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 34 - 2 (3 . 4 m) (9 . 8 m / s 2 ) (4 s) 2 cos 34 = 0 . 622198 . keywords: 002 (part 1 of 2) 10 points A car is traveling at 47 . 3 mi / h on a horizontal highway. The acceleration of gravity is 9 . 8 m / s 2 . If the coefficient of friction between road and tires on a rainy day is 0 . 082, what is the minimum distance in which the car will stop? (1 mi = 1.609 km) Correct answer: 278 . 074 m. Explanation: Newton’s second law in the direction of motion gives - f k = - μ k mg = ma Solving for a a = - μ k g (1) The acceleration a may be found from the following kinematics equation with v f = 0: v 2 f = v 2 0 + 2 ax a = - v 2 0 2 x (2) Combining equations (1) and (2), we obtain v 2 0 2 x = μ k g
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Adams, Russell – Homework 5 – Due: Oct 12 2006, midnight – Inst: Opyrchal, H 2 or solving for x , x = v 2 0 2 μ k g Plugging in the appropriate values, x wet = (21 . 1405 m / s) 2 2(0 . 082) (9 . 8 m / s 2 ) = 278 . 074 m 003 (part 2 of 2) 10 points What is the stopping distance when the sur- face is dry and μ dry = 0 . 536? Correct answer: 42 . 5411 m. Explanation: x dry = (21 . 1405 m / s) 2 2(0 . 536) (9 . 8 m / s 2 ) = 42 . 5411 m (A considerable difference) keywords: 004 (part 1 of 1) 10 points Masses 19 kg and 11 kg are connected by a light string that passes over a frictionless pulley as shown in the figure. The acceleration of gravity is 9 . 8 m / s 2 . 19 kg 11 kg μ If the 19 kg mass, initially held at rest on the table, is released and moves 1 . 7 m in 1 . 5 s, determine the coefficient of kinetic friction between it and the table. Correct answer: 0 . 335482 . Explanation: Given : m 1 = 19 kg , m 2 = 11 kg , t = 1 . 5 s , and s = 1 . 7 m . m 1 m 2 a T N m 1 g a T m 2 g μ m 1 g For the mass on the table, N = m 1 g , so that F f = μ m 1 g . Applying Newton’s second law to the masses, m 1 a = T - μ m 1 g, and m 2 a = m 2 g - T. Adding the two equations eliminates T , so that ( m 1 + m 2 ) a = ( m 2 - μ m 1 ) g.
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