Problem involving a skier, friction and conservation of energy

  • #1
::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/s2 .

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/s2
H=62.0m




What I've done:
A)
U= mgH= (63.0kg)(9.8m/s2)(62.0m)= 38278.8 N-m

NET FORCE= 38278.8-1000= 37278.8

37278.8= (1/2)mv2
37278.8= (31.5)v2
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?
 
Last edited:

Answers and Replies

  • #2
351
3
Sorry but this is completely wrong, the formula you are using for potential energy gives you energy for an answer, not a force. The equation you want to use is the law of conservation of energy:
[itex] \frac{1}{2}m{v_1}^2 + mgy_1 = \frac{1}{2}m{v_2}^2 + mgy_2 + F_{fr}l [/itex]
 
  • #3
gneill
Mentor
20,875
2,838
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?
U= mgH= (63.0kg)(9.8m/s2)(62.0m)= 38278.8 N-m

NET FORCE= 38278.8-1000= 37278.8
You're calculating energy, not force. Note that from the question statement that the work done by friction has a magnitude of 1 x 104. That's 10,000 not 1000 Joules.
 

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