A 57.0-kg skier starts from rest at the top of a ski slope of height 63.0 m . If

Question

A 57.0-kg skier starts from rest at the top of a ski slope of height 63.0 m . If frictional forces do -1.01 times 104 J of work on her as she descends, how fast is she going at the bottom of the slope? Take free fall acceleration to be g = 9.80 m/s^2 . v = m/s Now moving horizontally, the skier crosses a patch of soft snow, where the coefficient of friction is 0.22. If the patch is of width 62.0 m and the average force of air resistance on the skier is 170 N . how fast is she going after crossing the patch? V = m/s After crossing the patch of soft snow, the skier hits a snowdrift and penetrates a distance 3.0 m into it before coming to a stop. What is the average force exerted on her by the snowdrift as it stops her? F = N

Explanation / Answer

by energy conservation

K = U + Wnc

K = mgh - 1.01 x 10^4

K = 25091.8 J

1/2*mv^2 = 25091.8

v = sqrt(2*25091.8/m)

v = 29.67 m/s

part b )

friction and air resistance in opposite direction of skier

Kf = -mu*mg - Wair - Ko

Kf = -0.22 * 57 * 9.8 - 170 * 62 + 25091.8

Kf = 14427.79 J

1/2 * mv^2 = 14427.79

v = 22.5 m/s

part c )

a = v^2/2s

a = 22.5^2/2*3

a = 84.375 m/s^2

t = 22.5/84.375 = 0.267 s

Favg = Impulse/time

Favg = 57 * 22.5 / 0.267

Favg = 4803.37 N

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