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Skier's gravitational energy

  1. Nov 10, 2007 #1
    1. The problem statement, all variables and given/known data

    A 70.0 kg skier rides a 2700 m long lift to the top of a mountain. The lift makes an angle of 14.3° with the horizontal. What is the change in the skier's gravitational potential energy?

    2. Relevant equations



    3. The attempt at a solution

    mg cos(?) * (h0 - hf)


    again, not sure why this formula doesnot work.
     
  2. jcsd
  3. Nov 10, 2007 #2
    The GPE near the Earth's surface is mgh where h is the height above some reference height.

    Thus a change in GPE is mgΔh where Δh is the change in height.

    In your equation (h0 - hf) looks like a change in height -- which is what you want, so why should it be multipled by cos(?)?
     
  4. Nov 10, 2007 #3

    rl.bhat

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    Using trig. you can find that (hf - ho ) = 2700*sin14.3. Using this value you can find gravitational potential energy.
     
  5. Nov 10, 2007 #4
    okay so gpe = m * g * (chng in height)

    so this is what i get 70 * 9.8 * (667.)

    and get something like: 4.6E-5

    okay, thats correct. thanks alot.
     
  6. Nov 10, 2007 #5

    rl.bhat

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    A 70.0 kg skier rides a 2700 m long lift to the top of a mountain. Can you explain this sentence? Which is his starting point? "2700m long lift" Is it the distasance between the starting point and the end point?
     
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