Answer
$v=6.97m/s$
Work Step by Step
We can determine the required speed as follow:
$T_1+V_1=T_2+V_2$
$\implies \frac{1}{2}mv_1^2+mgh_1+2\frac{1}{2}kx_1^2=\frac{1}{2}mv_2^2+mgh_2+2\frac{1}{2}kx_2^2$
We plug in the known values to obtain:
$0+0=\frac{1}{2}(20)v^2+2(\frac{1}{2}(40)\sqrt{(3)^2+(2)^2-2})-(20)(9.81)(3)$
This simplifies to:
$v=6.97m/s$
