A 60.0-kg skier slides down a hill. She starts with a speed of 0.50 m/s. The friction force is only 10 N. The slope is 80 m long and 50 m high. What is the speed of the skier at the bottom of the hill?

1 answer

M*g = 60*9.8 = 588 N. = Wt. of skier.

PE = 0.5M*V^2 + Fki*d = 588h.
30V^2 + 10*80 =588*50,
V = 30.9 m/s.