A \(\text{60,0}\) \(\text{kg}\) skier with an initial speed of \(\text{12,0}\) \(\text{m·s$^{-1}$}\) coasts up a \(\text{2,50}\) \(\text{m}\)-high rise as shown in the figure. Find her final speed at the top, given that the coefficient of friction between her skis and the snow is \(\text{0,0800}\). (Hint: Find the distance traveled up the incline assuming a straight-line path as shown in the figure.)

We need to determine the length of the slope as this is the distance over which friction acts as well as the normal force of the skier on the slope to determine the magnitude of the force due to friction. The normal force balances the component of gravity perpendicular to the slope, therefore: \begin{align*} F_{\text{friction}}&= - \mu N \\ &= - \mu F_g\cos\theta \\ &= - \mu mg \cos\theta \end{align*} The length of the slope will be \(\Delta x = \dfrac{h}{\sin\theta}\).

\begin{align*} W_{\text{non-conservative}} + E_{k,i} + E_{p,i} & = E_{k,f} + E_{p,f} \\ -\mu mg\cos\theta \dfrac{h}{\sin\theta} + \frac{1}{2}mv_i^2 +mgh_i &= \frac{1}{2}mv_f^2 + mgh_f \\ -\mu g\cos\theta \dfrac{h}{\sin\theta} + \frac{1}{2}v_i^2 +gh_i &= \frac{1}{2}v_f^2 + gh_f \\ \frac{1}{2}v_f^2 + gh_f &= -\mu g\cos\theta \dfrac{h}{\sin\theta} + \frac{1}{2}v_i^2 +gh_i \\ \frac{1}{2}v_f^2 + &= -\mu g\cos\theta \dfrac{h}{\sin\theta} + \frac{1}{2}v_i^2 +g(h_i -h_f) \\ v_f^2 &= -2\mu g\cos\theta \dfrac{h}{\sin\theta} + v_i^2 +2g(h_i -h_f) \end{align*} \begin{align*} v_f^2 &= -2(0,08)(9,8) \cos(35) \dfrac{2.5}{\sin(35)} + (12)^2 +2(9,8)(-2,5) \\ v_f & = \sqrt{89,4016598} \\ v_f & = \text{9,46}\text{ m·s$^{-1}$} \end{align*} \(\text{9,46}\) \(\text{m·s$^{-1}$}\)