#### Answer

$v_B=15.5m/s$

#### Work Step by Step

The required speed can be determined as follows:
We know that
$x_A=\sqrt{(1.5)^2+(2)^2}-0.5=2m$
and $x_B=1.5-0.5=1m$
Now according to the equation of conservation of energy
$\frac{1}{2}mv_A^2+mgh_A+\frac{1}{2}kx_A^2=\frac{1}{2}mv_B^2+mgh_B+\frac{1}{2}kx_B^2$
We plug in the known values to obtain:
$0+(3)(9.81)(2)+\frac{1}{2}(200)(2)^2=\frac{1}{2}(3)v_B^2+(3)(9.81)(0)+\frac{1}{2}(200)(1)^2$
This simplifies to:
$v_B=15.5m/s$