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Homework Help: Power done by elevator

  1. Nov 26, 2007 #1
    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?
    Pcruising =

    2. Relevant equations

    3. The attempt at a solution

    [tex]\sum[/tex]F = T-mg=ma
    Vavg=1.75/2 == > a=1.75/2/3
    => T- 750(9.8) = (750)(1.75/2/3)


    Am I doing this correctly?

    (b) how do I do part b?

    I don't think I did part a correctly because I got a much different value... and the internet assignment said I am very close
    Last edited: Nov 26, 2007
  2. jcsd
  3. Nov 26, 2007 #2


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    Part a you calculated the work correctly. But the question asks for the average power, not the work done. You must note the formula for power as a function of work and time.
    For part b, it is simplest to use the formula P=Fv, waht is F in this case??
  4. Nov 26, 2007 #3
    I guess F is just the tension? therefore
    F = 7568.75N
    will v be the final velocity or the average velocity?
  5. Nov 26, 2007 #4


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    F is the tension, but since the elevator is cruising at constant speed, and not accelerating, the F will not be 7568N. What is F at constant velocity?
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