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: 51

Answer

Please see the work below.

Work Step by Step

We know that the acceleration of the first motion is given as: $a_1=\frac{0.96}{0.42}=2.2857\frac{m}{s^2}$ The frictional force is $f=\mu_1 mg=ma_1$ This can be rearranged as: $\mu_1=\frac{a_1}{g}$ We plug in the known values to obtain: $\mu_1=\frac{2.2857}{9.8}=0.23$ The acceleration of the second motion is given as: $a_1=\frac{0.96}{0.33}=2.909\frac{m}{s^2}$ The frictional force is $f=\mu_2 mg=ma_2$ This can be rearranged as: $\mu_2=\frac{a_2}{g}$ We plug in the known values to obtain: $\mu_2=\frac{2.909}{9.8}=0.30$ Thus, the frictional coefficient is between 0.23 and 0.30.
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