University Physics with Modern Physics (14th Edition)

$v = 3.07~m/s$
The kinetic energy of the two blocks and the pulley will be equal in magnitude to the change in potential energy of the system. Let $m_A$ be 4.00 kg and let $m_B$ be 2.00 kg. $KE = PE$ $\frac{1}{2}m_Av^2+\frac{1}{2}m_Bv^2+\frac{1}{2}I\omega^2 = m_Agd - m_Bgd$ $\frac{1}{2}m_Av^2+\frac{1}{2}m_Bv^2+\frac{1}{2}I(\frac{v}{R})^2 = (m_A - m_B)~gd$ $v^2 = \frac{(2)(m_A - m_B)~gd}{m_A+m_B+\frac{I}{R^2}}$ $v = \sqrt{\frac{(2)(m_A - m_B)~gd}{m_A+m_B+\frac{I}{R^2}}}$ $v = \sqrt{\frac{(2)(2.00~kg)(9.80~m/s^2)(5.00~m)}{4.00~kg+2.00~kg+\frac{0.380~kg~m^2}{(0.160~m)^2}}}$ $v = 3.07~m/s$