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Elevator and Spring question

  1. Feb 21, 2007 #1
    An elevator cable breaks when a 920kg elevator is 28 m above a huge spring
    (k=2.2 x10^5 N/m) at the bottom of the shaft.
    Calculate a) the work done by gravity on the elevator before it hits
    the spring b) the speed of the elevator just before striking the spring
    and c) the amount of spring compresses
    ( note that work is done by both the spring and the gravity in this part )

    I've calculated part a) and b)
    answer for a) is 2.5x10^5 J b) 23m/s

    For part C) what i think is

    Einitial = Efinal

    E(ball touches spring) = E(spring compresses)
    1/2mv^2 +mgy + 1/2ky^2 = 1/2mv2^2 + mgy2 + 1/2ky2^2

    Apparantly that's the equation in the book, but i have no idea how to do the question

  2. jcsd
  3. Feb 21, 2007 #2
    sc: spring compression
    ev: elevator speed
    ea: elevator accelleration
    sf: total force on elevator

    What you need to do is:
    Start with sc1 = 0 m and ev1 = 23 m/s,
    Turn travel to force, sf2 = 220000 * sc1
    Turn force to accelleration, ea2 = 9.81 + sf2 / 920
    Turn accelleration to speed, ev2 = ev1 + ea2 * dt
    Turn speed to travel, sc2 = sc1 + ev2 * dt
    Advance time a bit,
    And repeat until ev is 0.
    Last edited: Feb 21, 2007
  4. Feb 22, 2007 #3
    Its very easy. take initial to be at the moment the cable snaps, where we only would have potential energy or mgh which u calclated to be 2.5x10^5 + mgy that would be Einitial. Efinal is after the spring compresses as much as it can, so that means the velocity2 is zero, and potential enery 2 is zero so we are left with the term for compression of the spring 1/2ky2^2. the term mgy is taking into account that the potential difference from the point 1 to point 2 and that difference includes the compression of the string not only the distance from initial pt to the spring. sometimes this value can be neglected.
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