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: 54

Answer

$d_{min} = \frac{v_0^2}{2g~\mu_s}$

Work Step by Step

While the small box rides on the large box without slipping, the force of static friction provides the force to bring the small box to rest. We can find the maximum rate of deceleration that the force of static friction can provide for the small box as; $F_f = ma$ $mg~\mu_s = ma$ $a = g~\mu_s$ We can find the distance $d_{min}$ it takes to stop when the magnitude of deceleration of the system is $g~\mu_s$; $d_{min} = \frac{v_f^2-v_0^2}{-2g~\mu_s}$ $d_{min} = \frac{0-v_0^2}{-2g~\mu_s}$ $d_{min} = \frac{v_0^2}{2g~\mu_s}$
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