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Power of the lift system

  1. Sep 30, 2015 #1
    1. The problem statement, all variables and given/known data
    image.jpg

    2. Relevant equations
    P=Fv
    3. The attempt at a solution
    Since the elevator is rising at a steady speed, a=0, thus, the tension on the string equals to the weight of the elevator. I chose C as I thought just for the elevator case, it would be m1gv, but how does m2 affects the power of the system?
     
  2. jcsd
  3. Sep 30, 2015 #2
    Answer C is not correct.

    Imagine the movement within a finite time Δt (and therefore a change of position of Δh for both m1 and m2). What is the energy difference of the whole system during Δt?
     
  4. Sep 30, 2015 #3

    haruspex

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    Yes, the motor is working at that rate in raising the elevator, but what is happening on the other side? How much power is being transferred, and which way, between the counterweight and the motor?
     
  5. Sep 30, 2015 #4
    Energy difference would be (m1-m2)gΔh ?
     
  6. Sep 30, 2015 #5
    m2gv, and anti-clockwise? But I thought power is scalar? Does the minus sign comes from the velocity?
     
  7. Sep 30, 2015 #6
    Correct. And instead of distance and energy you need velocity and power - so what to do?
     
  8. Sep 30, 2015 #7
    Divide it by Δt and you will get velocity and (m1-m2)g is the force, so force times velocity gives power. I got it! Thanks!
     
  9. Sep 30, 2015 #8

    haruspex

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    Even scalar values can have sign. If A does positive work on B then B does negative work on A.
    If you take up as positive and gravity as -g then the motor does work at rate -m1(-g)v on the elevator and at rate -m2(-g)(-v) on m2.
     
  10. Sep 30, 2015 #9
    Oh I see, thanks. But why is there a negative sign before the mass?
     
  11. Oct 1, 2015 #10

    haruspex

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    Because m(-g) is the force from gravity. The motor has to supply an equal and opposite force, -m(-g).
     
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