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: 31

Answer

Please see the work below.

Work Step by Step

We know that $a=\frac{v^2}{2d}$ $a=\frac{(14)^2}{(2)(56)}$ $a=1.75\frac{m}{s^2}$ Now we can find the coefficient of the kinetic friction as $\mu_k=\frac{ma}{mg}=\frac{a}{g}$ We plug in the known values to obtain: $\mu_k=\frac{1.75}{9.8}=0.18$
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