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Homework Help: Power done by Elevator

  1. Nov 27, 2007 #1
    First of all, I know I've posted this already, but I didn't really understand the reply. I tried going back to the thread, but I can't find it anymore.

    1. The problem statement, all variables and given/known data
    A 750 kg elevator starts from rest. It moves upward for 3.00 s with constant acceleration until it reaches its cruising speed, 1.75 m/s.
    (a) What is the average power of the elevator motor during this period?
    (b) How does this power compare with the motor power when the elevator moves at its cruising speed?

    2. Relevant equations

    3. The attempt at a solution

    (a), since it was asking for the "average" power, so I thought I'd take the average velocity for my equation
    so Force = ma = 750*(1.75/3)
    and P= 750*(1.75/3)*(1.75/2) = 382.81
    However, that is the wrong answer
    so I tried it again.. may be I used the wrong numbers? This time I did the following:
    Force = ma = 750*(9.8)
    and P= 750*(9.8)*(1.75/2) = 6431.25
    However, this is still wrong
    what am I missing?

    (b) I think I am stuck on this part because I don't fully understand part a... but I was actually thinking that may be the power of done by the elevator at cruising speed would just be
    P=Fv=mgv = 750*9.8*1.75 = 12862.5
    this is definitely wrong because I typed it in to my online hw, and it told me its wrong
  2. jcsd
  3. Nov 27, 2007 #2


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    ok wait...The resultant force acting on the elevator is upwards...the only forces acting to provide this are the upward force of the elevator,[itex]F_E[/tex] and the downward force of the elevator(it's weight,[itex]W[/itex]). So that the resultant force,[itex]F_R[/itex] moves upward for 3.00 s with constant acceleration until it reaches its cruising speed, 1.75ms[itex]^{-1}[/itex]
    Last edited: Nov 27, 2007
  4. Nov 27, 2007 #3
    1.75/2 because it started at rest where Vi=0, and vf=1.75
    so the average would be 1.75/2
  5. Nov 27, 2007 #4


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    It's unfortunate that you couldn't find your first post buried on page 3 a moment ago, because you practically had the correct answer by correctly calculating the work done by the motor, then you forgot to divide the work by the time (3 seconds) to get the average power delivered by the motor. Anyway, now you have chosen wisely to determine the average power using the motor force times the average velocity, however, you are now incorrectly using the net force acting on the elevator, when instead, you should be using the force provided by the motor, [tex]F_m[/tex], where [tex]F_{net} = F_m - mg [/tex]. For part b, your answer appears correct, because at cruise speed, the motor force must equal the weight (per Newton 1), so I don't know why the computer doesn't like your answer.
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