## Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (4th Edition)

The spring brings the box to rest when the spring compresses by 6.5 cm. The work done by the spring on the box is $W = -\frac{1}{2}kx^2$. We can use the work-energy theorem to find the initial speed of the box. $KE_1+W = KE_2$ $\frac{1}{2}mv^2-\frac{1}{2}kx^2 = 0$ $\frac{1}{2}mv^2=\frac{1}{2}kx^2$ $v^2=\frac{kx^2}{m}$ $v=\sqrt{\frac{kx^2}{m}}$ $v=\sqrt{\frac{(85~N/m)(0.065~m)^2}{1.5~kg}}$ $v = 0.49~m/s$ The box was moving with a speed of 0.49 m/s