## Physics: Principles with Applications (7th Edition)

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, $W = KE$ $F_f ~d = \frac{1}{2}mv^2$ $mg ~\mu_k ~d = \frac{1}{2}mv^2$ $d = \frac{v^2}{2g ~\mu_k}$ $d = \frac{(25.5 ~m/s)^2}{(2)(9.80 ~m/s^2)(0.090)}$ $d = 370 ~m$