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Work elevator problem

  1. Mar 18, 2009 #1
    Hello, i'm currently studying for the mcat and ran across this problem in one of my practice tests.

    An elevator used in a building has a mass of 1000 kg when empty
    the elevator hangs on a single cable

    step 1. the elevator rises for 2m/s^2 for 3 seconds,
    if there is no friction on the elevator or cable, what is the work done on the elevator?

    the correct answer is 108 kJ

    I understand how to get the answer
    T - mg = ma
    T = ma + mg = 12000

    W = Fd = 12000 X 9 m = 108 kJ

    but why in this case do we only use the force in the Tension, I thought we always use NET force, or in this case Fnet = T - mg = 1000 X 2 = 2000
    w = fd = 2000 X 9 = 18 kJ

    so why just tension and not Net force?

    thank you!
     
  2. jcsd
  3. Mar 18, 2009 #2

    PhanthomJay

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    Gold Member

    You are quite right, the net or total work done on the elevator is 18 kJ; the work done by the cable is 108 kJ and the work done by gravity is -90 kJ. I guess the problem was not properly worded.
     
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