Physics Help..

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The cable of an elevator of mass M = 1900 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 = 10.6 m above a cushioning spring whose spring constant is k = 8500 N/m. A safety device clamps the elevator against the guide rails so that a constant frictional force of f = 14499 N opposes the motion of the elevator. Find the maximum distance by which the cushioning spring will be compressed.

  • Physics Help.. -

    net force down = m g - 14499
    = 1900 (9.81) - 14499
    = 4140 Newtons

    Work done by downward force = 4140(10.6+x)Joules

    Energy stored in spring = (1/2) k x^2
    so
    4140(10.6+x) = 4250 x^2

    4250 x^2 - 4140 x - 43884 = 0

    x^2 - .974 x - 10.3 = 0

    x = [ .974 +/- sqrt (.949+41.2) /2
    = [ .974+/-6.49 ]/2
    = 3.73

  • Physics Help.. -

    Thank you sooo much!

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