Essential University Physics: Volume 1 (4th Edition)

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

Chapter 7 - Exercises and Problems - Page 130: 23

Answer

Please see the work below.

Work Step by Step

We know that $\Delta E_{int}=f_kd$ $\Delta E_{int}=\mu_k mg$ This can be rearranged as: $\implies \mu_k=\frac{\Delta E_{int}}{mgd}$ We plug in the known values to obtain: $\mu_k=\frac{\Delta E_{int}}{mgd}$ We plug in the known values to obtain: $\mu_k=\frac{1500}{33\times 9.8\times 6.2}$ $\mu_k=0.75$
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