Answer
$\mu_k = 0.786$
Work Step by Step
$W_f + W_g = K_2-K_1$
$-F_f~d +m_A~g~d = 0-\frac{1}{2}Mv^2$
$m_B~g~\mu_k~d = \frac{1}{2}Mv^2+m_A~g~d$
$\mu_k = \frac{\frac{1}{2}Mv^2+m_A~g~d}{m_B~g~d}$
$\mu_k = \frac{\frac{1}{2}(14.00~kg)(0.900~m/s)^2+(6.00~kg)(9.80~m/s^2)(2.00~m)}{(8.00~kg)(9.80~m/s^2)(2.00~m)}$
$\mu_k = 0.786$
