Phys111-5sol

# Phys111-5sol - Anderson Matthew – Homework 5 – Due...

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Unformatted text preview: Anderson, Matthew – Homework 5 – Due: Oct 10 2007, 11:00 pm – Inst: Opyrchal, H 1 This print-out should have 12 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 hockey puck is given an initial speed of 20 m / s on a frozen pond. The puck remains on the ice and slides 131 m before coming to rest. The acceleration of gravity is 9 . 8 m / s 2 . What is the coefficient of friction between the puck and the ice? Correct answer: 0 . 155788 . Explanation: Basic Concept: Newton’s Second Law: F net = ma Solution: From kinematics, 0 = v 2 + 2 ax a =- v 2 2 x since the puck comes to a rest. Consider the free body diagram for the situation: μ N N mg Vertically, the puck is in equilibrium, so F net = N - mg = 0 ⇒ N = mg Horizontally, after the puck is given its initial velocity, the friction force f k = μ N = μmg is the only unbalanced force acting on it. Thus, f net = ma = 0- f k =- μmg μ =- a g = v 2 2 gx = (20 m / s) 2 2(9 . 8 m / s 2 )(131 m) = 0 . 155788 keywords: 002 (part 1 of 1) 10 points A 3 . 7 kg block slides down an inclined plane that makes an angle of 33 ◦ with the horizontal. Starting from rest, the block slides a distance of 1 . 7 m in 6 . 5 s. The acceleration of gravity is 9 . 81 m / s 2 . Find the coefficient of kinetic friction be- tween the block and plane. Correct answer: 0 . 639626 . Explanation: Let : m = 3 . 7 kg , θ = 33 ◦ , Δ x = 1 . 7 m , and Δ t = 6 . 5 s . 3 . 7 k g μ 33 ◦ Consider the forces acting on the block as it slides down the incline: ~ N ~ f k m~g From kinematics, since v = 0, Δ x = v Δ t + 1 2 a (Δ t ) 2 = 1 2 a (Δ t ) 2 a = 2Δ x (Δ t ) 2 = 2(1 . 7 m) (6 . 5 s) 2 = 0 . 0804734 m / s 2 . Anderson, Matthew – Homework 5 – Due: Oct 10 2007, 11:00 pm – Inst: Opyrchal, H 2 Applying X ~ F = m~a to the block, X F y = N - mg cos θ = 0 mg cos θ = N and X F x = mg sin θ- f k = ma mg sin θ- μ k N = ma mg sin θ- μ k mg cos θ = ma. Thus μ k = g sin θ- a g cos θ = (9 . 81 m / s 2 ) sin33 ◦- (0 . 0804734 m / s 2 ) (9 . 81 m / s 2 ) cos33 ◦ = . 639626 . keywords: 003 (part 1 of 2) 10 points A 9 kg block rests on a horizontal table, at- tached to a 8 kg block by a light string as shown in the figure....
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## This note was uploaded on 10/21/2010 for the course PHYS 003 taught by Professor Janow during the Fall '10 term at NJIT.

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Phys111-5sol - Anderson Matthew – Homework 5 – Due...

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