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OH-05-solutions

# OH-05-solutions - obaidi(aao476 OH-05 Bradley(191 This...

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obaidi (aao476) – OH-05 – Bradley – (191) 1 This print-out should have 15 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. This is Online Homework No. 05. It is due Tuesday morning at 06:00 AM 16 September Abu Dhabi Time (or 09:00 PM on 15 Septem- ber in Texas), unless otherwise notified. 001 (part 1 of 3) 4.0 points A hockey puck is hit on a frozen lake and starts moving with a speed of 13.0 m/s. Ex- actly 4.80 s later, its speed is 6.70 m/s. The acceleration of gravity is 9 . 81 m / s 2 . a) What is its average acceleration? Correct answer: - 1 . 3125 m / s 2 . Explanation: Basic Concept: v f = v i + a Δ t Given: v i = 13 . 0 m / s Δ t = 4 . 80 s v f = 6 . 70 m / s Solution: a = v f - v i Δ t = 6 . 7 m / s - 13 m / s 4 . 8 s = - 1 . 3125 m / s 2 002 (part 2 of 3) 3.0 points b) What is the coefficient of kinetic friction between the puck and the ice? Correct answer: 0 . 133792. Explanation: Basic Concepts: F net = ma = - F k F n = mg F k = μ k F n Solution: ma = - μ k mg a = - μ k g μ k = - a g = - - 1 . 3125 m / s 2 9 . 81 m / s 2 = 0 . 133792 003 (part 3 of 3) 3.0 points c) How far does the puck travel during this 4.80 s interval? Correct answer: 47 . 28 m. Explanation: Basic Concept: Δ x = v i Δ t + 1 2 a t ) 2 Solution: Δ x = (13 m / s)(4 . 8 s) + 1 2 ( - 1 . 3125 m / s 2 ) (4 . 8 s) 2 = 47 . 28 m 004 (part 1 of 2) 5.0 points A block weighing 9 . 7 N requires a force of 3 . 2 N to push it along at constant velocity. What is the coefficient of friction for the surface? Correct answer: 0 . 329897. Explanation: Constant velocity implies the system is in equilibrium. Thus Σ F right = Σ F left The force necessary to overcome friction is F = μ N . Friction always opposes the mo- tion, and acts in the opposite direction of the motion. Consider the free body diagram: μ W 1 f W 1

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obaidi (aao476) – OH-05 – Bradley – (191) 2 The normal force is W 1 , so f - μ W 1 = 0 μ = f W 1 = 3 . 2 N 9 . 7 N = 0 . 329897 . 005 (part 2 of 2) 5.0 points A weight W is now placed on the block and 5 . 9 N is needed to push them both at constant velocity. What is the weight W of the block? Correct answer: 8 . 18438 N. Explanation: Consider the free body diagram: μ 1 N f 1 W W 1 The normal force is W 1 + W , so F net = f 1 - μ ( W 1 + W ) = 0 f 1 = μ W 1 + μ W W = f 1 - μ W 1 μ = 5 . 9 N - 9 . 7 N0 . 4 0 . 4 = 8 . 18438 N .
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