Answer
(a) The spring is compressed a distance of 0.26 meters.
(b) The block slides back up the incline a distance of 0.34 meters.
Work Step by Step
(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.
