CW6 - cos . Hence the Frictional Force is f = N = m g cos...

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boland (kbb544) – CW6 – mackie – (10611) 1 This print-out should have 1 question. Multiple-choice questions may continue on the next column or page – fnd all choices beFore answering. 001 10.0 points An Olympic skier moving at 37 m / s down a 26 slope encounters a region oF wet snow oF coe±cient oF kinetic Friction μ = 0 . 96. The acceleration oF gravity is 9 . 8 m / s 2 . How Far down the slope does she travel beFore coming to a halt? Correct answer: 164 . 551 m. Explanation: Align the x axis along the incline, and let down the incline be positive. The normal Force is N = m g
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Unformatted text preview: cos . Hence the Frictional Force is f = N = m g cos Applying Newtons second law to the skier along the incline yields s F = m g sin - m g cos = m a , or solving For a , a = g [sin - cos ] = (9 . 8 m / s 2 ) [sin26 -(0 . 96) cos 26 ] =-4 . 15981 m / s 2 . To fnd how Far the slope the skier travels beFore stopping, use the kinematic equation v 2 = v 2 + 2 a d . Setting v = 0 and solving For d , d =-v 2 2 a =-(37 m / s) 2 2 (-4 . 15981 m / s 2 ) = 164 . 551 m ....
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This note was uploaded on 02/20/2011 for the course PHYS 1062 taught by Professor Mackie during the Spring '11 term at Temple.

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