## College Physics (4th Edition)

Published by McGraw-Hill Education

# Review & Synthesis: Chapters 6-8 - Review Exercises - Page 323: 5

#### Answer

(a) $935~J$ of energy was dissipated by friction. (b) The coefficient of sliding friction is $0.73$

#### Work Step by Step

(a) The amount of energy dissipated by friction is equal in magnitude to the negative work done by friction, which is equal to the change in energy of the crate. We can use energy and work to find the work done by friction. We can assume that the crate's final energy is zero: $U_0+KE_0+Work = 0$ $Work = -U_0-KE_0$ $Work = -mgh-\frac{1}{2}mv^2$ $Work = -(100~kg)(9.80~m/s^2)(1.50~m)~sin~30.0^{\circ}-\frac{1}{2}(100~kg)(2.00~m/s)^2$ $Work = -935~J$ The work done by friction is $-935~J$. Therefore, $935~J$ of energy was dissipated by friction. (b) We can find the coefficient of sliding friction $\mu_k$: $F_f~d = 935~J$ $mg~cos~\theta~\mu_k~d = 935~J$ $\mu_k = \frac{935~J}{mg~cos~\theta~d}$ $\mu_k = \frac{935~J}{(100~kg)(9.80~m/s^2)(1.5~m)~cos~30.0^{\circ}}$ $\mu_k = 0.73$ The coefficient of sliding friction is $0.73$.

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