Essential University Physics: Volume 1 (4th Edition) Clone

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

Chapter 5 - Exercises and Problems - Page 91: 27

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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