## College Physics (4th Edition)

(a) $F_f = 158.9~N$ (b) $\mu_k = 0.19$
(a) Since the skier moves at a constant velocity, the force of kinetic friction is equal in magnitude to the component of the skier's weight that is directed down the slope. We can find the force of kinetic friction: $F_f = mg~sin~\theta$ $F_f = (85~kg)(9.80~m/s^2)~sin~11^{\circ}$ $F_f = 158.9~N$ (b) We can find the coefficient of kinetic friction: $F_N~\mu_k = F_f$ $\mu_k = \frac{F_f}{F_N}$ $\mu_k = \frac{mg~sin~\theta}{mg~cos~\theta}$ $\mu_k = tan~\theta$ $\mu_k = tan~11^{\circ}$ $\mu_k = 0.19$