Answer
(a) The skier's speed at the bottom is $22.8~m/s$
(b) $\mu_k = 0.19$
Work Step by Step
(a) We can find the skier's acceleration down the slope:
$\sum F = ma$
$mg~sin~\theta = ma$
$a = g~sin~\theta$
$a = (9.80~m/s^2)~sin~32^{\circ}$
$a = 5.19~m/s^2$
We can find the speed at the bottom:
$v_f^2 = v_0^2+2ad$
$v_f = \sqrt{v_0^2+2ad}$
$v_f = \sqrt{0+(2)(5.19~m/s^2)(50~m)}$
$v_f = 22.8~m/s$
The skier's speed at the bottom is $22.8~m/s$
(b) We can find the deceleration on the horizontal surface:
$v_f^2 = v_0^2+2ad$
$a = \frac{v_f^2-v_0^2}{2d}$
$a = \frac{0-(22.8~m/s)^2}{(2)(140~m)}$
$a = -1.86~m/s^2$
We can use the magnitude of acceleration to find the coefficient of kinetic friction:
$F_f = ma$
$mg~\mu_k = ma$
$\mu_k = \frac{a}{g}$
$\mu_k = \frac{1.86~m/s^2}{9.80~m/s^2}$
$\mu_k = 0.19$
