Kieran takes off down a 50 m high,10° slope on his jet-powered skis.The skis have a thrust of 280 N.The combined mass of skis and Kieran is 50kg (the fuel mass is negligible).Kieran's speed at the bottom is 40 m/s.What is the coefficient of kinetic friction of his skis on snow?
A)0.47
B)0.58
C)0.23
D)0.29

Question

Quizplus · Accepted Answer

The work done by the thrust is equal to the change in kinetic energy plus the work done by friction. The work done by friction is equal to the coefficient of kinetic friction times the normal force times the distance traveled. The normal force is equal to the weight of the skier and skis minus the thrust. The distance traveled is equal to the height of the slope divided by the sine of the slope angle.$$W_f=\mu_k n d$$$$W_f=\mu_k (mg-T) \frac{h}{\sin	heta}$$$$W_f=\mu_k (50	imes9.8-280) \frac{50}{\sin10°}$$$$W_f=11760\mu_k$$$$W_f=\frac{1}{2}mv_f^2-\frac{1}{2}mv_i^2$$$$11760\mu_k=\frac{1}{2}(50)(40)^2-0$$$$11760\mu_k=40000$$$$\mu_k=0.47$$

Kieran Takes Off Down a 50 M High,10° Slope on His