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Work done by elevator

  1. Nov 14, 2007 #1
    A 4500kg elevator moves upward 10m in a times of 3.8s
    a) The minimum work done by the motor is
    b) The minimum motor power rewuired is

    a W=F* X
    So F= m*a , that 4500kg * 3.8s = 171,000

    therfore P= 171,000/10 = 17,100

    Is this right?
     
  2. jcsd
  3. Nov 14, 2007 #2
    You have the right equation for work however since the elevator is moving at a constant velocity and its direction doesn't change, there actually is no acceleration present.

    Something to consider is that because the elevator is moving in the y-direction, gravity also has an affect on the forces that are acting on it. Therefore, to be able to move vertically, the elevator has to have a force equal to or greater than its weight (not mass). Weight is equal to its mass times gravity. Therefore, your equation would look like this: W = (mass)(gravity)(distance travel)

    Hope this sets you on the right track.
     
  4. Nov 14, 2007 #3
    So it will be W= (4500kg)(9.8)(10) = 441,000
     
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