## Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (3rd Edition)

(a) $\mu_k = \frac{v_0^2}{2~g~d}$ (b) $\mu_k = 0.037$
(a) The work done by friction is equal in magnitude to the initial kinetic energy of the box. We can find an expression for the coefficient of kinetic friction; $W_f = KE$ $F_f~d = \frac{1}{2}mv_0^2$ $mg~\mu_k~d = \frac{1}{2}mv_0^2$ $\mu_k = \frac{v_0^2}{2~g~d}$ (b) We can use the expression from part (a) to solve this question. $\mu_k = \frac{v_0^2}{2~g~d}$ $\mu_k = \frac{(1.2~m/s)^2}{(2)(9.80~m/s^2)(2.0~m)}$ $\mu_k = 0.037$