## University Physics with Modern Physics (14th Edition)

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.