## College Physics (4th Edition)

(a) The work done by gravity is $6087~J$ The work done by the normal force is zero. (b) The work done by gravity is $6087~J$ The work done by the normal force is zero. The work done by friction is $-2337~J$ The force of friction is $73.0~N$ The coefficient of kinetic friction is 0.10
(a) The work done by gravity is the negative of the change in gravitational potential energy: $\Delta U_g = mg~\Delta h$ $\Delta U_g = (75.0~kg)(9.80~m/s^2)(-32.0~m)~sin~15.0^{\circ}$ $\Delta U_g = -6087~J$ The work done by gravity is $6087~J$ The work done by the normal force is zero since the normal force is perpendicular to the direction of motion. (b) The change in kinetic energy is equal to the sum of the work done by gravity $W_g$ and friction $W_f$: $\sum Work = KE$ $W_g+W_f = \frac{1}{2}mv^2$ $W_f = \frac{1}{2}mv^2 - W_g$ $W_f = \frac{1}{2}(75.0~kg)(10.0~m/s)^2 - 6087~J$ $W_f = -2337~J$ The work done by gravity is $6087~J$ The work done by the normal force is zero. The work done by friction is $-2337~J$ We can use the magnitude of the work done by friction to find the force of friction: $F_f ~d = 2337~J$ $F_f = \frac{2337~J}{d}$ $F_f = \frac{2337~J}{32.0~m}$ $F_f = 73.0~N$ The force of friction is $73.0~N$ We can find the coefficient of kinetic friction: $F_N~\mu_k = F_f$ $mg~cos~\theta~\mu_k = F_f$ $\mu_k = \frac{F_f}{mg~cos~\theta}$ $\mu_k = \frac{73.0~N}{(75.0~kg)(9.80~m/s^2)~cos~15.0^{\circ}}$ $\mu_k = 0.10$ The coefficient of kinetic friction is 0.10