#### Answer

(a) v = 25.5 m/s
(b) d = 370 m

#### Work Step by Step

We can use work and energy to solve this problem.
(a) We first find the height $y$ at the top of the hill:
$y = (85~m)~sin(28^{\circ})$
$y = 39.9~m$
The kinetic energy at the bottom will be equal to the sum of the potential energy at the top plus the work done by friction as the skier descends. So,
$KE = PE + W$
$\frac{1}{2}mv^2 = mgy - F_f~d$
$\frac{1}{2}mv^2 = mgy - mg ~cos(\theta) ~\mu_k~d$
$v^2 = 2gy - 2g ~cos(\theta)~\mu_k~d$
$v^2 = (2g)~(y - ~cos(\theta)~\mu_k~d)$
$v^2 = (2g)~[(39.9~m) - ~cos(28^{\circ})(0.090)(85~m)]$
$v^2 = 324.825~m^2/s^2$
$v = \sqrt{324.825~m^2/s^2}$
$v = 25.5 ~m/s$
(b) The work done by friction will be equal in magnitude to the kinetic energy at the bottom of the hill. So,
$W = KE$
$F_f ~d = \frac{1}{2}mv^2$
$mg ~\mu_k ~d = \frac{1}{2}mv^2$
$d = \frac{v^2}{2g ~\mu_k}$
$d = \frac{(25.5 ~m/s)^2}{(2)(9.80 ~m/s^2)(0.090)}$
$d = 370 ~m$