## Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (3rd Edition)

We can find the distance $d$ that the skier travels down the slope. $\frac{h}{d} = sin(\theta)$ $d = \frac{h}{sin(\theta)}$ $d = \frac{50~m}{sin(20^{\circ})}$ $d = 146~m$ The skier's kinetic energy at the bottom is equal to the sum of the initial potential energy and the work $W_w$ done on the skier by the wind. $KE = PE+W_w$ $\frac{1}{2}mv^2 = mgh+F~d~cos(160^{\circ})$ $v^2 = \frac{2mgh+2F~d~cos(160^{\circ})}{m}$ $v = \sqrt{\frac{2mgh+2F~d~cos(160^{\circ})}{m}}$ $v = \sqrt{\frac{(2)(75~kg)(9.80~m/s^2)(50~m)+(2)(200~N)(146~m)~cos(160^{\circ})}{75~kg}}$ $v = 15.8~m/s$ The skier's speed at the bottom is 15.8 m/s