## University Physics with Modern Physics (14th Edition)

The kinetic energy of the block will be equal to the sum of the work done by the spring and friction. $K = W_s+W_f$ $\frac{1}{2}mv^2 = \frac{1}{2}kx_1^2 - \frac{1}{2}kx_2^2 - F_f~d$ $v^2 = \frac{k~(x_1^2 - x_2^2) - 2mg~\mu_k~d}{m}$ $v = \sqrt{\frac{k~(x_1^2 - x_2^2) - 2mg~\mu_k~d}{m}}$ $v = \sqrt{\frac{(840~N/m)((0.0300~m)^2 - (0.0100~m)^2) - (2)(2.50~kg)(9.80~m/s^2)(0.40)(0.0200~m)}{2.50~kg}}$ $v = 0.335~m/s$ The speed of the block is 0.335 m/s.