University Physics with Modern Physics (14th Edition)

Published by Pearson
ISBN 10: 0321973615
ISBN 13: 978-0-32197-361-0

Chapter 7 - Potential Energy and Energy Conservation - Problems - Exercises - Page 233: 7.58

Answer

$d = \frac{\frac{1}{2}v_0^2+gL~sin(\alpha)}{g~sin(\beta) +g~cos(\beta)~\mu_r}$

Work Step by Step

$K_2+U_2 = K_1+U_1+W$ $0+mgd~sin(\beta) = \frac{1}{2}mv_0^2+mgL~sin(\alpha)-mg~cos(\beta)~\mu_r~d$ $gd~sin(\beta) +g~cos(\beta)~\mu_r~d = \frac{1}{2}v_0^2+gL~sin(\alpha)$ $d = \frac{\frac{1}{2}v_0^2+gL~sin(\alpha)}{g~sin(\beta) +g~cos(\beta)~\mu_r}$
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