Answer
$v_B=3.34m/s$
Work Step by Step
The required velocity can be determined as follows:
We know that
$d=\sqrt{(0.7)^2+(0.4)^2}-\sqrt{(0.2)^2+(0.2)^2}=0.523m$
Now, according to the principle of work and energy
$\frac{1}{2}mv_A^2+\Sigma U_{1\rightarrow 2}=\frac{1}{2}mv_B^2$
$\frac{1}{2}mv_A^2+(-mgd+Fd)=\frac{1}{2}mv_B^2$
We plug in the known values to obtain:
$\frac{1}{2}(15)(0)^2+(-15\times 9.8\times 0.5)+(300\times 0.523)=\frac{1}{2}(15)v_B^2$
This simplifies to:
$v_B=3.34m/s$
