Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (4th Edition)

Published by Pearson
ISBN 10: 0133942651
ISBN 13: 978-0-13394-265-1

Chapter 6 - Dynamics I: Motion Along a Line - Exercises and Problems: 48

Answer

$v = \sqrt{2gh - 2g~\mu_k~\sqrt{L^2-h^2}}$

Work Step by Step

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; $v^2 = v_0^2+2aL = 0 + 2aL$ $v = \sqrt{2aL}$ $v = \sqrt{(2L)(g~sin(\theta) - g~cos(\theta)~\mu_k)}$ $v = \sqrt{(2L)(g~\frac{h}{L} - g~\mu_k~\frac{\sqrt{L^2-h^2}}{L})}$ $v = \sqrt{2gh - 2g~\mu_k~\sqrt{L^2-h^2}}$
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