Essential University Physics: Volume 1 (4th Edition)

Published by Pearson
ISBN 10: 0-134-98855-8
ISBN 13: 978-0-13498-855-9

Chapter 5 - Exercises and Problems - Page 92: 61


a) .098 meters b) The box will not slide back down.

Work Step by Step

We first must find the acceleration on the ramp. We know that the two forces causing the block to slow down are the force of gravity and the force of friction. Thus, we find: $a = \frac{-F_gsin\theta - F_f}{m} \\ a = \frac{-F_gsin\theta - F_n \ mu}{m} \\ a = \frac{-mgsin\theta - mgcos\theta \mu}{m} \\ a = -gsin\theta - gcos\theta \mu \\ a = -9.81sin22-9.81cos22(.7)=-10.08m/s^2$ We now can find the change in distance: $v_f^2 = v_0^2 + 2a\Delta x \\ 0 = 1.4^2 - 2(-10.08) \Delta x \\ \Delta x = \fbox{.098 meters}$ To see if the box will slide back down, we must compare the force of friction to the force of gravity: $F_f = \mu mg cos22 = .65 mg $ $F_g=mgsin22=.37mg$ Thus, we see that the kinetic friction is greater than the force of gravity. Since static friction is always greater than kinetic friction, this means that the box will not slide back down.
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