Answer
$v_A=1.54m/s$, $v_B=4.62m/s$
Work Step by Step
We can determine the required speed as follows:
We know that
$3v_A=-v_B~~~~$eq(1)
According to the equation of energy conservation
$T_{A_{\circ}}+T_{B_{\circ}}+V_{A_{\circ}}+V_{B_{\circ}}=T_A+T_B+V_A+V_B$
$\implies 0+0+0+0=\frac{1}{2}m_Av_A^2+\frac{1}{2}mBv_B^2+W_A\Delta S_A+W_B\Delta S_B$
We plug in the known values to obtain:
$0=\frac{1}{2}(20)v_A^2+\frac{1}{2}(30)(-3v_A)^2+(20)(9.81)(\frac{1.5}{3})-(30)(9.81)(1.5)$
This simplifies to:
$v_A=1.54m/s$
We plug in this value in eq(1) to obtain:
$3(1.54)=-v_B$
This simplifies to:
$v_B=4.62m/s$
