#### Answer

(a) $\mu_k = \frac{v_0^2}{2~g~d}$
(b) $\mu_k = 0.037$

#### Work Step by Step

(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.
$\mu_k = \frac{v_0^2}{2~g~d}$
$\mu_k = \frac{(1.2~m/s)^2}{(2)(9.80~m/s^2)(2.0~m)}$
$\mu_k = 0.037$