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

The kinetic energy of the box when it leaves the spring will be equal to the initial energy stored in the spring plus the work done on the box by friction. $KE = U_s+W_f$ $\frac{1}{2}mv^2 = \frac{1}{2}kx^2-mg~\mu_k~d$ $v^2 = \frac{kx^2-2mg~\mu_k~d}{m}$ $v = \sqrt{\frac{kx^2-2mg~\mu_k~d}{m}}$ $v = \sqrt{\frac{(250~N/m)(0.12~m)^2-(2)(0.25~kg)(9.80~m/s^2)(0.23)(0.12~m)}{0.25~kg}}$ $v = 3.7~m/s$ The box's launch speed is 3.7 m/s.