Essential University Physics: Volume 1 (3rd Edition)

Published by Pearson
ISBN 10: 0321993721
ISBN 13: 978-0-32199-372-4

Chapter 5 - Exercises and Problems: 58

Answer

$\mu=.46$

Work Step by Step

We must first find an expression for the acceleration, which is equal to the force of gravity causing the boxes to move down the ramp minus the force of friction, which is all divided by the mass of the box. Thus, we find: $a = \frac{mgsin\theta-F_n \mu}{m} \\ a = \frac{mgsin\theta-mgcos\theta \mu}{m}\\ a = gsin\theta-gcos\theta \mu \\ a = gsin30-gcos30 \mu \\ a = 4.905-8.5\mu $ Thus, we find: $\Delta x = v_0t + \frac{1}{2}at^2 \\ 5.4 = 0t+\frac{1}{2}( 4.905-8.5\mu)(3.3^2)\\ \mu=.46$
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