## College Physics (4th Edition)

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