Fundamentals of Physics Extended (10th Edition)

Published by Wiley
ISBN 10: 1-11823-072-8
ISBN 13: 978-1-11823-072-5

Chapter 6 - Force and Motion-II - Problems - Page 140: 4

Answer

$\mu = 0.43$

Work Step by Step

Let $d$ be the length of the ramp. We can find the acceleration when the slide is frictionless: $F = ma$ $mg~sin~\theta = ma$ $a = \frac{mg~sin~\theta}{m}$ $a = g~sin~\theta$ We can write an expression for the time required to slide down a frictionless ramp: $d = \frac{1}{2}at_1^2$ $t_1 = \sqrt{\frac{2d}{a}}$ $t_1 = \sqrt{\frac{2d}{g~sin~\theta}}$ We can find the acceleration when there is friction: $mg~sin~\theta - mg~\mu~cos~\theta = ma$ $a = g~sin~\theta - g~\mu~cos~\theta$ We can write an expression for the time required to slide down the ramp when there is friction: $d = \frac{1}{2}at_2^2$ $t_2 = \sqrt{\frac{2d}{a}}$ $t_2 = \sqrt{\frac{2d}{g~sin~\theta - g~\mu~cos~\theta}}$ It is given in the question that $t_2 = 2t_1$ We can find $\mu$: $t_2 = 2t_1$ $\sqrt{\frac{2d}{g~sin~\theta - g~\mu~cos~\theta}} = 2\sqrt{\frac{2d}{g~sin~\theta}}$ $\frac{2d}{g~sin~\theta - g~\mu~cos~\theta} = \frac{8d}{g~sin~\theta}$ $4g~sin~\theta - 4g~\mu~cos~\theta = g~sin~\theta$ $4\mu~cos~\theta = 3~sin~\theta$ $\mu = 0.75~tan~\theta$ $\mu = 0.75~tan~35^{\circ}$ $\mu = 0.43$
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