Answer
$v_A=7.18ft/s$
Work Step by Step
According to the work and energy principle
$\frac{1}{2}m_Av_{A_{\circ}}^2+\frac{1}{2}m_Bv_{B_{\circ}}^2+W_A\Delta S_A+W_B\Delta S_B=\frac{1}{2}m_Av_A^2+\frac{1}{2}m_Bv_B^2$
This can be rearranged as:
$\frac{1}{2}m_Av_{A_{\circ}}^2+\frac{1}{2}m_B2v_{A_{\circ}}^2+W_A\Delta S_A-2W_BS_A=\frac{1}{2}m_Av_A^2+\frac{1}{2}m_B(2v_A)^2$
We plug in the known values to obtain:
$_0+0+6(5)-10(10)=\frac{1}{2}(\frac{60}{32.2})v_A^2+\frac{1}{2}(\frac{10}{32.2})(2v_A)^2$
This simplifies to:
$v_A=7.18ft/s$
