Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (3rd Edition)

Published by Pearson
ISBN 10: 0321740904
ISBN 13: 978-0-32174-090-8

Chapter 6 - Dynamics I: Motion Along a Line - Exercises and Problems - Page 163: 41


(a) $v = \sqrt{2gh - 2g~\mu_k~\sqrt{L^2-h^2}}$ (b) $v = 9.95~m/s$

Work Step by Step

(a) Let $\theta$ be the angle the slope makes with the horizontal. Then $sin(\theta) = \frac{h}{L}$ and $cos(\theta) = \frac{\sqrt{L^2-h^2}}{L}$ We can use a force equation to find an expression for the acceleration as the object slides down the slope. $\sum F = ma$ $mg~sin(\theta) - F_f = ma$ $mg~sin(\theta) - mg~cos(\theta)~\mu_k = ma$ $a = g~sin(\theta) - g~cos(\theta)~\mu_k$ We can use the expression for the acceleration to find the speed at the bottom. $v^2 = v_0^2+2aL = 0 + 2aL$ $v = \sqrt{2aL}$ $v = \sqrt{(2L)(g~sin(\theta) - g~cos(\theta)~\mu_k)}$ $v = \sqrt{(2L)(g~\frac{h}{L} - g~\mu_k~\frac{\sqrt{L^2-h^2}}{L})}$ $v = \sqrt{2gh - 2g~\mu_k~\sqrt{L^2-h^2}}$ (b) We can use the expression in part (a) to solve this question. $v = \sqrt{2gh - 2g~\mu_k~\sqrt{L^2-h^2}}$ $v = \sqrt{(2)(9.80~m/s^2)(12~m) - (2)(9.80~m/s^2)(0.07)~\sqrt{(100~m)^2-(12~m)^2}}$ $v = 9.95~m/s$
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