A downhill skier must dig the edges of her skis into the snow in order to make a

Question

A downhill skier must dig the edges of her skis into the snow in order to make a turn, so that the snow "pushes" the skier sideways. (3 pts) Suppose a 50-kg skier wants to take a turn at a speed of 10 m/s. What centripetal force exerted by the snow is necessary to turn with a radius of curvature of 20 meters? (Assume a flat snow surface.) (3 pts) If the centripetal force came entirely from kinetic friction, what would the coefficient of kinetic friction have to be in order for this skier to make this size turn at this speed?

Explanation / Answer

Fc = m a_c = m v^2 / R

Fc = (50) (10^2) / 20

Fc = 250 N ........Ans
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f = Fc = 250 N  

and N = m g = 50 x 10 = 500 N  

us = f / N = 0.50 ......Ans

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