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March 29, 2017

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The cable of an elevator of mass = 5320 kg snaps when the elevator is a rest at one of the floors of a skyscraper. At this point the elevator is a distance = 31.6 m above a cushioning spring whose spring constant is = 11700 N/m. A safety device clamps the elevator against the guide rails so that a constant frictional force of = 18158 N opposes the motion of the elevator. Find the maximum distance by which the cushioning spring will be compressed.

  • physics - ,

    Kinetic energy that the elevator has when it hits the spring is:
    KE = m*g*h-work done by friction so far

    the work done by friction so far =18158*h

    where
    m = 5320 kg
    g = 9.81 m/s^2
    h - 31.6 m

    Now we compress the spring a distance x

    Work done by gravity in falling additional distance x = m g x
    So at the bottom we have
    gained m g x additional
    so we have a total available of KE + m g x
    That amount goes into compressing the spring and doing further work against friction
    (1/2) k x^2 + 18158 x

    so in the end

    m*g*h-18158*h + m g x = (1/2) k x^2 + 18158 x
    or
    m g (h+x) - 18158 (h+x) = (1/2) k x^2

    find the two solutions to the quadratic. Probably only one will make sense.

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