Answer
$v=2.1m/s$
Work Step by Step
Energy can either be kinetic or elastic potential. Since the system is undamped, the amount of energy remains constant. This means that for all velocities v and displacements x, $$\frac{1}{2}kx^2+\frac{1}{2}mv^2=E$$ Solving for $v$ yields $$\frac{1}{2}mv^2=E-\frac{1}{2}kx^2$$ $$v^2=\frac{2E}{m}-\frac{kx^2}{m}$$ $$v=\sqrt{\frac{2E}{m}-\frac{kx^2}{m}}$$ Substituting known values of $E=4.0J$, $x=0.15m$, $k=200N/m$, and $m=0.80kg$ yields a velocity at $x=0.15m$ of $$v=\sqrt{\frac{2(4.0J)}{0.80kg}-\frac{(200N/m)(0.15m)^2}{0.80kg}}$$ $$v=2.1m/s$$
