Answer
$v_B=5.42m/s$
Work Step by Step
We can determine the required speed as follows:
$\Sigma U_{A\rightarrow B}=F(AC-BC)$
$\implies \Sigma U_{A\rightarrow B}=300(\sqrt{(8)^+(6)^2})=300(3.675)$
Now, according to the principle of work and energy
$\frac{1}{2}mv_A^2+\Sigma U_{\rightarrow B}=\frac{1}{2}mv_B^2$
We plug in the known values to obtain:
$0+(300\times 3.675)=\frac{1}{2}(75)v_B^2$
This simplifies to:
$v_B=5.42m/s$
