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    tnhoots's Avatar
    tnhoots Posts: 28, Reputation: 1
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    #1

    Feb 20, 2007, 09:18 AM
    Elevator Power.
    A 550 kg elevator starts from rest. It moves upward for 5.00 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?
    W
    (b) How does this power compare with the motor power when the elevator moves at its cruising speed?
    Pcruising = W


    The formula for average power is avg. power = work/change in time
    However, I am not sure how to find the work in the problem since no force or angle is given... Is gravity the only force doing work and is the angle 90?
    Capuchin's Avatar
    Capuchin Posts: 5,255, Reputation: 656
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    #2

    Feb 20, 2007, 10:35 AM
    The angle is 0 (you don't need to worry about it, because it's a 1d problem), the motor is doing work against the force of gravity.
    tnhoots's Avatar
    tnhoots Posts: 28, Reputation: 1
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    #3

    Feb 20, 2007, 03:12 PM
    Ok so the F in the problem is 9.8m/s^2. I would use cos0. However, the change in distance is unknown. Using the formula d=time * speed, I found it to be 8.75m. So, now would I use the work formula with the numbers (9.8m/s^2)(8.75m)(cos0). Which equals 85.57J. After I find that I would divide work by time (5secs) to find average power which is 17.15 W... just my thinking?
    Capuchin's Avatar
    Capuchin Posts: 5,255, Reputation: 656
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    #4

    Feb 20, 2007, 03:33 PM
    F = ma :)

    your distance is wrong because it's accelerating, you need to use suvat equations
    tnhoots's Avatar
    tnhoots Posts: 28, Reputation: 1
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    #5

    Feb 20, 2007, 04:01 PM
    The force in this problem would be the mass of the elevator time the acceleration which is 1.75m/s correct. So f in the problem would be 962.5N. For distance I mult. The speed times the time. That is the only way I know to find distance.
    Capuchin's Avatar
    Capuchin Posts: 5,255, Reputation: 656
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    #6

    Feb 20, 2007, 04:13 PM
    acceleration is the acceleration of gravity!!

    you need to use s = ut + 1/2at^2 to find distance traveled, because it is under constant acceleration.

    I have to hit the sack now, but I will carry on tomorrow.
    tnhoots's Avatar
    tnhoots Posts: 28, Reputation: 1
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    #7

    Feb 20, 2007, 05:22 PM
    One more try:
    f=mass*acceleration
    f=550*9.8=5390N
    change in distance=122.5m
    cosign of angle 0degrees

    w=f*change in d*cosign0
    w=5390*122.5*cosign0
    ?? Correct?
    Capuchin's Avatar
    Capuchin Posts: 5,255, Reputation: 656
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    #8

    Feb 21, 2007, 12:31 AM
    Oh no, your acceleration for s = ut + 1/2 at^2 is 1.75/5

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