## Essential University Physics: Volume 1 (3rd Edition)

$39 \ m/s$
We know that $K=\frac{1}{2}mv^2$ and that $U= \frac{1}{2}kx^2$. Since energy is conserved, it follows: $\frac{1}{2}mv^2 = \frac{1}{2}kx_0^2$ $v = \sqrt{\frac{k}{m}} \times x_0$ $v = \sqrt{\frac{940}{.038}} \times .25 = 39 \ m/s$