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Work And Watts

  1. Mar 4, 2008 #1
    1. The problem statement, all variables and given/known data

    A 523 kg elevator starts from rest. It moves upward for 3.79s with a constant acceleration until it reaches its cruising speed of 1.78 m/s.

    The acceleration of gravity is 9.8 m/s^2.

    Find the average power delivered by the elevator motor during the period of this acceleration. Answer in units of kW.

    2. Relevant equations

    w = m(a+g)
    p = w/t

    3. The attempt at a solution

    m = 523
    a = 1.78
    t = 3.79

    w=m(a+g)=523(1.78+9.8)=6056.34

    p=w/t=6056.34/3.79=1597.98W

    1597.98W becomes 1.59798kW

    I was incredibly confident in this answer, but it is wrong. Where did I make my silly error?
     
  2. jcsd
  3. Mar 4, 2008 #2

    Doc Al

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    Staff: Mentor

    m(a+g) is the force exerted, not the work. What's missing?
     
  4. Mar 4, 2008 #3
    Hmm. I guess it is missing d, as in W=Fd. Am I capable of calculating d?
     
  5. Mar 4, 2008 #4

    Doc Al

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    Staff: Mentor

    Right.
    One easy way to find distance is to use average speed X time. What's the average speed?

    Your value for the acceleration is incorrect. That's the final speed after 3.79 seconds. Use the change in speed and the time to calculate the acceleration.
     
  6. Mar 4, 2008 #5
    Ok. So a = 1.78/3.79 = .469657
    avg v = (V + Vo)/2 = (1.78 + 0)/2 = .89

    So, d = avg v * t = .89 * 3.79

    Am I at least on the right side of the highway?
     
  7. Mar 4, 2008 #6

    Doc Al

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    Staff: Mentor

    You are back on track.
     
  8. Mar 4, 2008 #7
    So F = m(a+g) = 523 (.469657 + 9.8) = 5371.03

    d = .89 * 3.79 = 3.3731

    W = Fd = 5371.03 * 3.3731 = 18117W becomes 18.117 kW

    Where am I still going wrong? This was not correct.
     
  9. Mar 4, 2008 #8

    Doc Al

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    Staff: Mentor

    You calculated the work (in Joules). Now find the power (in Watts).
     
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