A 61.0-kg skier starts from rest at the top of a ski slope of height 67.0m a)The

Question

A 61.0-kg skier starts from rest at the top of a ski slope of height 67.0m
a)The skier crosses a patch of soft snow, where the coefficient of friction is 0.19. If the patch is of width 64.0 and the average force of air resistance on the skier is 180 , how fast is she going after crossing the patch?
b)After crossing the patch of soft snow, the skier hits a snowdrift and penetrates a distance 2.6 into it before coming to a stop. What is the average force exerted on her by the snowdrift as it stops her?

Explanation / Answer

the idea is that... initial PE - work by friction - work by air resistance = final KE 61 * 9.8 * 67 - 0.19*61*9.8*64 - 180*64 = (1/2) * 61 * v^2 v = 18.3 For part b... work done by snowdrift = skiers KE force * distance = (1/2) m v^2 force * 2.6 = (1/2) * 61 * 18.3^2 force = 3929 Newtons

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