A 60.0-kg skier starts from rest at thetop of a ski slope 95.0 m high. (a) If fr

Question

A 60.0-kg skier starts from rest at thetop of a ski slope 95.0 m high. (a) If frictional forces do 10.5 kJ ofwork on her as she descends, how fast is she going at the bottom ofthe slope?
1 m/s

(b) Now moving horizontally, the skier crosses a patch of softsnow, where µk = 0.2. Ifthe patch is 83.0 m wide and the averageforce of air resistance on the skier is 160 N, how fast is shegoing after crossing the patch?
2 m/s

(c) The skier hits a snowdrift and penetrates 2.2 m into it before coming to a stop. What is theaverage force exerted on her by the snowdrift as it stops her?
3 N
(a) If frictional forces do 10.5 kJ ofwork on her as she descends, how fast is she going at the bottom ofthe slope?
1 m/s

(b) Now moving horizontally, the skier crosses a patch of softsnow, where µk = 0.2. Ifthe patch is 83.0 m wide and the averageforce of air resistance on the skier is 160 N, how fast is shegoing after crossing the patch?
2 m/s

(c) The skier hits a snowdrift and penetrates 2.2 m into it before coming to a stop. What is theaverage force exerted on her by the snowdrift as it stops her?
3 N

Explanation / Answer

(a) If frictional forces do 10.5 kJ ofwork on her as she descends, how fast is she going at the bottom ofthe slope?
mgh = 10.5 kJ + mv2/2
v = 38.9 m/s

b) Now moving horizontally, the skier crosses a patch of soft snow,where µk = 0.2. If thepatch is 83.0 m wide and the average forceof air resistance on the skier is 160 N, how fast is she goingafter crossing the patch?
d = 83.0 m
-mg - 160 = ma
a = -4.627 m/s2
v'2 - v2 = 2ad
v' = 27.3 m/s

(c) The skier hits a snowdrift and penetrates 2.2 m into it before coming to a stop. What is theaverage force exerted on her by the snowdrift as it stops her?
F*2.2 = mv'2/2
F = 10163 N

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