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::A skier of mass 63.0kg starts from rest at the top of a ski slope of height 62.0m .A skier of mass 63.0kg starts from rest at the top of a ski slope of height 62.0m .

A) If frictional forces do −1.00×104J 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 = 9.80m/s

B) Now moving horizontally, the skier crosses a patch of soft snow, where the coefficient of friction is μk = 0.210. If the patch is of width 64.0m and the average force of air resistance on the skier is 180N , how fast is she going after crossing the patch?

C) After crossing the patch of soft snow, the skier hits a snowdrift and penetrates a distance 2.90m into it before coming to a stop. What is the average force exerted on her by the snowdrift as it stops her?

m=63.0kg

g= 9.8m/s

H=62.0m

What I've done:

A)

U= mgH= (63.0kg)(9.8m/s

NET FORCE= 38278.8-1000= 37278.8

37278.8= (1/2)mv

37278.8= (31.5)v

so v= 34.401

Mastering physics says that is WRONG. I've checked my math over and over again. Where did I go wrong?

tips?

A) If frictional forces do −1.00×104J 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 = 9.80m/s

^{2}.B) Now moving horizontally, the skier crosses a patch of soft snow, where the coefficient of friction is μk = 0.210. If the patch is of width 64.0m and the average force of air resistance on the skier is 180N , how fast is she going after crossing the patch?

C) After crossing the patch of soft snow, the skier hits a snowdrift and penetrates a distance 2.90m into it before coming to a stop. What is the average force exerted on her by the snowdrift as it stops her?

m=63.0kg

g= 9.8m/s

^{2}H=62.0m

What I've done:

A)

U= mgH= (63.0kg)(9.8m/s

^{2})(62.0m)= 38278.8 N-mNET FORCE= 38278.8-1000= 37278.8

37278.8= (1/2)mv

^{2}37278.8= (31.5)v

^{2}so v= 34.401

Mastering physics says that is WRONG. I've checked my math over and over again. Where did I go wrong?

tips?

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