Answer
$\mu_k = 0.037$
Work Step by Step
The work done by friction is equal in magnitude to the initial kinetic energy of the suitcase;
$W_f = KE$
$F_f~d = \frac{1}{2}mv^2$
$mg~\mu_k~d = \frac{1}{2}mv^2$
$\mu_k = \frac{v^2}{2~g~d}$
$\mu_k = \frac{(1.2~m/s)^2}{(2)(9.80~m/s^2)(2.0~m)}$
$\mu_k = 0.037$
