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Spring and elevator

  1. Feb 17, 2008 #1
    1. The problem statement, all variables and given/known data
    The cable of an elevator of mass M = 1920 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 = 15.6 m above a cushioning spring whose spring constant is k = 23100 N/m. A safety device clamps the elevator against the guide rails so that a constant frictional force of f = 13217 N opposes the motion of the elevator. Find the maximum distance by which the cushioning spring will be compressed.


    2. Relevant equations
    K1 + U(grav,1) + U(elastic,1) + W(other) = K2 + U(grav, 2) + U(elastic, 2)


    3. The attempt at a solution
    I tried using the above formula for work-energy. I set my origin at the point at which the elevator initially hits the spring. So,
    0 (initially at rest so K1=0) + (1920)(9.8)(15.6) + 0 (spring not yet compressed) -(13217)y2 = 0 (v2=0 so K2=0) + (1920)(9.8)y2 + .5(23100)(y2)^2

    Basically I then used quadratics to solve for negative value of y2. What am I doing wrong?
     
  2. jcsd
  3. Feb 17, 2008 #2

    Doc Al

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    Staff: Mentor

    Recalculate the work done by the friction force. Over what distance does it act?
     
  4. Feb 17, 2008 #3
    oh, should it be y1 - y2, because y2 is negative?
     
  5. Feb 17, 2008 #4

    Doc Al

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    Staff: Mentor

    Sounds right.
     
  6. Feb 17, 2008 #5
    ok now i think my algebra is just screwy because i keep getting an incorrect answer. i have

    .5k(y2)^2 + (mg - f)y2 - mgy1 + fy1 = 0

    solve for negative value of y2..
     
  7. Feb 17, 2008 #6

    Doc Al

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    Staff: Mentor

    Looks good to me. Just plug in the numbers and solve.
     
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