Question

A skier traveling 11.0 m/s reaches the foot of a steady upward 19° incline and

glides 15 m up along this slope before coming to rest. What was the average

coefficient of friction? (use conservation of energy method or work energy

theorem)

Answer 1

$$\Delta E=W$$

$$0-\frac{mv^2}{2}=-mgl \sin\theta-F_fl$$

$$F_f=\frac{mv^2}{2l}-mg \sin\theta=\mu mg\cos\theta$$

$$\mu=\frac{v^2}{2gl\cos\theta}-\tan\theta$$

$$\mu=\frac{11.0^2}{2*9.8*15\cos19^\circ}-\tan19^\circ=0.091$$