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Conservation of Energy airplane pilot

  1. Oct 7, 2004 #1
    An airplane pilot fell 400 m after jumping without his parachute opening. He landed in a snowbank, creating a crater 2 m deep, but survived with only minor injuries. Assume that the pilot's mass was 50 kg and his terminal velocity was 65 m/s.

    (a) Estimate the work done by the snow in bringing him to rest.

    Is it just work done by gravity, which is mgh, h being 402 m?

    (b) Estimate the average force exerted on him by the snow to stop him.

    No idea...

    (c) Estimate the work done on him by air resistance as he fell.

    No clue...
     
  2. jcsd
  3. Oct 7, 2004 #2

    Chi Meson

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    Work-energy theorem: The net work done on an object is equal to the change in kinetic energy of the object.

    Assume that the snow exerted the only force on the pilot as he came to a stop. Since he was at a terminal velocity, there's no need to use kinematics.

    b)
    Once you find the work done by the snow, use the standard equation for work to find the force

    c)
    at terminal velocity, air resistance equals weight.
     
  4. Oct 7, 2004 #3

    HallsofIvy

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    No, there is also work done by air resistance- because he has a maximum velocity of 65 m/s. That's the speed with which he hits the snow so the work the snow does is equal to his kinetic energy at that speed.

    You know the work done by the snow from (a) and you know that work is "force times distance"...

    NOW use your idea from (a). His potential energy is what you said there- the total work done to bring him to a stop is equal to that. You calculated the work done by the snow in (a). The work done by the air resistance is the difference between his initial potential energy and the work done by the snow.

    (Rember than the work done by the snow was the same as his kinetic energy at his terminal speed. That is, the work done by the air is the initial potential energy less the kinetic energy at terminal speed.)
     
  5. Oct 7, 2004 #4
    Is a) W=mgh+1/2*mv^2 , where h = height of crater and v = terminal velocity?
     
  6. Oct 7, 2004 #5

    Doc Al

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    Yes, if by h you mean the depth of the crater (2 m).
     
  7. Oct 7, 2004 #6
    For part c, is it mgh, where h is 400m, minus what I got for a) ??
     
  8. Oct 7, 2004 #7

    Doc Al

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    Almost. When it just hits the ground it has a KE that is less than its original PE (measured from the ground). The "missing" energy is that lost to air resistance.
     
  9. Oct 7, 2004 #8
    mgh, where h is 402m, minus what I got for a) ??
     
  10. Oct 8, 2004 #9
    wait... is Work done by friction=mgh-1/2*mv^2-mgd, where h is 400m and v is 65m/s, and d=2.0m?
     
    Last edited: Oct 8, 2004
  11. Oct 8, 2004 #10

    Doc Al

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    That would work.
     
  12. Oct 8, 2004 #11

    Doc Al

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    I would drop the mgd term. Assume that air resistance just acts while the object is falling, but not after it hits the snow.
     
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