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Falling elevator energy conservation

  1. Feb 23, 2013 #1
    The cable of an elevator of mass M = 1640 kg snaps when the elevator is at rest at one of the floors of a skyscraper. At this point the elevator is a distance d = 36.4 m above a cushioning spring whose spring constant is k = 13600 N/m. A safety device clamps the elevator against the guide rails so that a constant frictional force of f = 12724 N opposes the motion of the elevator. Find the maximum distance by which the cushioning spring will be compressed.


    (1/2)kx^2=Spring energy
    K=(1/2)mv^2
    U=mgh
    Vf^2=Vi^2+2a(X-Xi)


    F=ma
    mg-Ff=ma
    a=2.05 m/s^2

    Vf^2=Vi^2+2a(X-Xi)
    Vf^2=0+2(-2.05)(-36.4)
    Vf=12.22 m/s

    (1/2)kx^2=(1/2)mv^2
    (1/2)(13600)x^2=(1/2)(1640)(12.22^2)
    x= 4.24 m (that didn't work)

    (then I tried this)
    (1/2)kx^2=(1/2)mv^2+mgh
    x=10.20 m (still wasn't correct)
     
  2. jcsd
  3. Feb 23, 2013 #2
    What energy changes are taking place when the elevator falls?
     
  4. Feb 23, 2013 #3

    TSny

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    Don't forget that the friction force continues to act while the spring is compressing.
     
  5. Feb 23, 2013 #4
    Potential energy is being converted to kinetic energy. So with that, (1/2)mv^2=mgh. Would the final equation be (1/2)kx^2=mgh?
     
  6. Feb 23, 2013 #5
    What about the friction force?
     
  7. Feb 23, 2013 #6
    The friction force is acting over a distance throughout the process of the elevator falling and compressing the spring. So the friction force is doing work against the falling elevator.
    (1/2)kx^2=mgh-Ffd?
     
  8. Feb 23, 2013 #7
    Ok, looking good.
    Are h and d the same?
    How are they related to the initial height you are given.
    Hint : think what happens to the spring.
     
  9. Feb 23, 2013 #8
    h is the initial height and d is the initial height + the distance the spring compresses when the elevator hits it. Thus Ffd=Ff(h+x), right?
     
  10. Feb 23, 2013 #9
    Yes, good.
     
  11. Feb 23, 2013 #10
    Awesome! Thank you!
     
  12. Feb 23, 2013 #11

    TSny

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    Does the force of gravity also act through the entire distance h+x?
     
  13. Feb 23, 2013 #12
    Yes it does. So will mgh be mg(h+x) instead?
     
  14. Feb 23, 2013 #13

    TSny

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    Yes, I think that's right.
     
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