#### Answer

The speed of both blocks will be 2.90 m/s.

#### Work Step by Step

Note that we can calculate the speed of the total mass $M$ since both blocks will move with the same speed.
$W_f + W_g = K_2-K_1$
$-F_f~d +m_B~g~d = \frac{1}{2}Mv^2 - 0$
$v^2 = \frac{-2m_A~g~\mu_k~d + 2m_B~g~d}{M}$
$v = \sqrt{\frac{-2m_A~g~\mu_k~d + 2m_B~g~d}{M}}$
$v = \sqrt{\frac{(-2)(8.00~kg)(9.80~m/s^2)(0.250)(1.50~m) + (2)(6.00~kg)(9.80~m/s^2)(1.50~m)}{14.00~kg}}$
$v = 2.90~m/s$
The speed of both blocks will be 2.90 m/s.