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Homework Help: Unit test review

  1. Nov 3, 2004 #1
    Hey guys -> this is for a friend and i havent gotten this far in our class so yeah if you could expain the answer that would be appreciated.

    the motor of an elevator can produce 14KW of power. The elevator has a mass of 1100Kg including its contents. At what constant speed will the elevator rise?

  2. jcsd
  3. Nov 3, 2004 #2


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    [tex] P = \frac{dW}{dt} [/tex]

    [tex] P = F \frac{dx}{dt} [/tex]

    [tex] P = Fv [/tex]

    He got the power, and the mass of the body.
    Last edited: Nov 3, 2004
  4. Nov 3, 2004 #3
    [tex]W=F\cdot s[/tex]
    Since the elevator moves with constant velocity, the accleration of the elevator is zero. so, from newton's 2nd law, the cable tension pulling the elevator is mg with m=1100 kg.so F in this case is a constant.
    [tex]P=\frac{d(Fs)}{dt};\ W=F\cdot s=Fs;\ cos\theta=0[/tex] since the angle between the tension and the displacement of the elevator is 0 (ie they are in the same direction).
    Actually we are talking about the power of the motor which produce the pulling force (tension of the cable) which acts on the elevator and the work this force does on the elevator.
  5. Nov 3, 2004 #4
    Alright well we're only in grade 11 so could you please explain it a little more clearly?
  6. Nov 3, 2004 #5
    Thanks Leong that was very discriptive:)
  7. Nov 3, 2004 #6


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    Use Newton's 1st Law

    [tex] \sum_{i=1}^{n} \vec{F}_{i} = 0 \rightarrow \vec{v} = constant [/tex]

    so, the forces acting on the elevator are the tension of the rope and its weight, and because the speed is constant then the sum of the forces must be equal to 0.

    [tex] T - mg = 0 [/tex]

    [tex] T = mg [/tex]


    [tex] P = mgv [/tex]

    I'm not sure if you understood.....
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