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Basic power problem

  1. Oct 19, 2008 #1
    1. The problem statement, all variables and given/known data

    A 1.0 x 10^3 kg elevator carries a maximum load of 800.0 kg. A constant frictional force of
    4.0 x 10^3 N retards the elevator's motion upward. What minimum power, in kilowatts, must the motor deliver to lift the fully loaded elevator at a constant speed of 3.00 m/s?

    2. Relevant equations
    the final answer turns out to be 66 kW

    P=W/time

    P=Fv

    3. The attempt at a solution
    Im really confused here and do not know where to start. all i know is the equations for power
     
  2. jcsd
  3. Oct 19, 2008 #2

    cepheid

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    You know the force. You know the velocity. What is the issue?
     
  4. Oct 19, 2008 #3
    But the force they give is frictional force. I just need some help on how to start out
     
  5. Oct 19, 2008 #4
    I just did this:
    The power of the frictional force part is:
    (4000 N)(3 m/s)=12000 W

    then you find the power of the normal force of the elevator so its:
    1000 kg + 800 kg =1800 kg

    (1800 kg)(9.8)(3.0 m/s)=52920

    then i added both powers:
    1800W + 52920W =64920 W

    then I converted to kW and got 64.92 W. How did they get 66? sig figs would only bring it up to 65.
     
  6. Oct 19, 2008 #5

    cepheid

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    Your method is correct! You do indeed have to calculate the power based on the SUM of the two forces, because the motor has to provide enough force to exactly counteract both friction AND gravity (i.e. the elevator's weight), both of which are working against it.

    When I first read your problem, I did it in my head, using g = 10 m/s^2 in order to simplify the mental math and get a ballpark answer. However, I got exactly 66 kW. So maybe they were expecting you to use g = 10??? I dunno what else to suggest...
     
  7. Oct 19, 2008 #6
    Oh well, 65 is probably acceptable. thanks for the help
     
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