Chapter 7 - Section 7.3 - Conservation of Mechanical Energy - Example - Page 116: 7.4

$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 $