## College Physics (4th Edition)

(a) Let $W_f$ be the work done by friction as the block slides. We can use work and energy to find the compression of the spring: $\frac{1}{2}kx^2 = mgh+W_f$ $\frac{1}{2}kx^2 = mgd~sin~\theta-mgd~cos~\theta~\mu_k$ $x^2 = \frac{2mgd~(sin~\theta-\mu_k~cos~\theta)}{k}$ $x = \sqrt{\frac{2mgd~(sin~\theta-\mu_k~cos~\theta)}{k}}$ $x = \sqrt{\frac{(2)(0.50~kg)(9.80~m/s^2)(0.85~m)~(sin~30.0^{\circ}-(0.25)~cos~30.0^{\circ})}{35~N/m}}$ $x = 0.26~m$ The spring is compressed a distance of 0.26 meters. (b) We can use conservation of energy to find the distance the block slides back up the incline after it leaves the spring: $mgh = \frac{1}{2}kx^2 +W_f$ $mgd~sin~\theta = \frac{1}{2}kx^2 -mgd~cos~\theta~\mu_k$ $mgd~sin~\theta + mgd~cos~\theta~\mu_k = \frac{1}{2}kx^2$ $d = \frac{kx^2}{2mg~(sin~\theta + \mu_k~cos~\theta)}$ $d = \frac{(35~N/m)(0.26~m)^2}{(2)(0.50~kg)(9.80~m/s^2)~(sin~30.0^{\circ} + (0.25)~cos~30.0^{\circ})}$ $d = 0.34~m$ The block slides back up the incline a distance of 0.34 meters.