Answer
(a) $v = 14.1~m/s$
(b) $d = 32.1~m$
Work Step by Step
(a) We can use conservation of energy to find the speed:
$K = U_e$
$\frac{1}{2}mv^2 = \frac{1}{2}kx^2$
$v^2 = \frac{kx^2}{m}$
$v = (\sqrt{\frac{80,000~N/m}{100~kg}})~(0.50~m)$
$v = 14.1~m/s$
(b) We can use conservation of energy to find the kinetic energy at the bottom of the incline:
$K+U_{g2} = U_e+U_{g1}$
$K+0 = \frac{1}{2}kx^2+mgh$
$K = \frac{1}{2}(80,000~N/m)(0.50~m)^2+(100~kg)(9.8~m/s^2)(10~m)$
$K = 19,800~J$
We can use work and energy to find the distance traveled along the incline:
$U_{gf} = K+W_{ext}$
$mgh = K-mg~cos~\theta~\mu_k~d$
$mg~d~sin~\theta + mg~cos~\theta~\mu_k~d= K$
$d = \frac{K}{(mg~sin~\theta + mg~cos~\theta~\mu_k)}$
$d = \frac{19,800~J}{(100~kg)(9.8~m/s^2)(sin~30^{\circ}+0.15~cos~30^{\circ})}$
$d = 32.1~m$