Answer
$v_f=2.03\frac{m}{s}$
Work Step by Step
$E_{PBi}=m_Bgh=(38.0kg)(9.8\frac{m}{s^2})(2.5m)=931J$
$E_{PBi}=E_{PAi}+E_{KAf}+E_{KBf}+\frac{I\omega^2}{2}$
$=m_Agh+\frac{m_Av_f^2}{2}+\frac{m_Bv_f^2}{2}+\frac{Mv_f^2}{4}$
$(32.0kg)(9.8\frac{m}{s^2})(2.5m)+\frac{(32.0kg)v_f^2}{2}+\frac{(38.0kg)v_f^2}{2}+\frac{(3.1kg)v_f^2}{4}=931J$
$v_f=\sqrt{\frac{2(931J-784J)}{32.0kg+38.0kg+1.55kg}}=2.03\frac{m}{s}$
