A 58.0-kg skier starts from rest at the top of a ski slope of height 70.0 m. If

Question

A 58.0-kg skier starts from rest at the top of a ski slope of height 70.0 m. If frictional forces do -1.07 times 10^4 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. Now moving horizontally, the skier crosses a patch of soft snow, where the coefficient of friction is 0.25. If the patch is of width 70.0 m and the average force of air resistance on the skier is 180 N, how fast is A 58.0-kg skier starts from rest at the top of a ski slope of height 70.0 m. Now moving horizontally, the skier crosses a patch of soft snow, where the coefficient of friction is 0.25. If the patch is of width 70.0 m and the average force of air resistance on the skier is 180 N, how fast is she going after crossing the patch? 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?

Explanation / Answer

B)

while crossing the rough patch workdone on the skier,

W = change in kinetic energy

fk*d*cos(180) + F_air*d*cos(180) = (1/2)*m*(vf^2 - vi^2)

-mue_k*m*g*d - F_Air*d = (1/2)*m*(vf^2 - vi^2)

-0.25*58*9.8*70 - 180*70 = (1/2)*58*(vf^2 - 31.7^2)

on solving the above equation we get

vf = 15.1 m/s

C) again use Work-energy theorem

Workdone = change in kinetic energy

F*d*cos(180) = (1/2)*m*(vf^2 - vi^2)

F*3*(-1) = (1/2)*58*(0^2 - 15.1^2)

F = 2204 N

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