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Conservation of Energy ski-jump ramp

  1. Mar 24, 2009 #1
    1. The problem statement, all variables and given/known data

    A 79 kg skier leaves the end of a ski-jump ramp with a velocity of 26 m/s directed 25° above the horizontal. Suppose that as a result of air drag the skier returns to the ground with a speed of 24 m/s, landing 18 m vertically below the end of the ramp. From the launch to the return to the ground, by how much is the mechanical energy of the skier-Earth system reduced because of air drag?

    2. Relevant equations

    Total Mechanical Energy initial = Total Mechanical Energy final where Total mechanical energry is all energies added.

    3. The attempt at a solution

    I took gravitational potential energy initial to be mgh and added that with kinetic energy initial, which is .5*m*velocity initial ^2. I set all this equal to the same thing, but as the finals of each of these energies. Then I saw that the difference between the two is 403.16. This seems like an awful lot. The answer should be in joules. Is my answer correct?
  2. jcsd
  3. Mar 24, 2009 #2


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    It will be more Joules than that right?

    How many Joules in gravitational potential alone? And it ends with less velocity?
  4. Mar 24, 2009 #3
    I got 691.16 J for the total mechanical energy initial. For the total mechanical energy final I came out with 288 J. So it was reduced by 288 J?
  5. Mar 24, 2009 #4


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    1/2*mv2 initial = 691 J ?

    Can you show your calculation?

    Isn't it 1/2 * 79 * 262 ?
  6. Mar 24, 2009 #5
    I added the Kinetic energy plus the gravitational potential energy initial. I then set that equal to total mechanical energy final which only included kinetic energy. since I had mass on both sides I cancelled out the mass. I was then left with gh + (.5*26^2) = (.5*24^2).
  7. Mar 24, 2009 #6


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    Ahhh. That explains it then.

    You can't cancel out the mass.

    You haven't accounted for the unknown work due to air resistance in your equation. You can't divide the mass out of that.
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