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Work vs Power. Still confused

  1. Jan 21, 2013 #1
    This isn't homework. I've been thinking.

    If I have say a 1kg mass and I want to accelerate it at 1m/s^2 for 10m, I would calculate the work to be done as 10J.

    Force = 1kg x 1m/s^2 = 1N
    Work = 1N x 10m = 10J

    I hope that is correct so far.

    But what confuses me is power. Power is work/time, but how could I possibly alter the amount of time it takes for me to accelerate a mass for a certain distance? Wouldn't time be constant?

    Is it possible for me to accelerate the mass at 1m/s^2 and then to reach 10m with different amount of times?

    Thanks a lot.
     
  2. jcsd
  3. Jan 21, 2013 #2

    cjl

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    No, because you're holding the acceleration constant. What you need to do is hold the work done constant, but vary the time taken. For example, you could similarly do 10J of work by pushing it with a force of 0.5N for 20m, which will give you the same final KE, the same work done, but a different time (and thus a different average power).
     
  4. Jan 21, 2013 #3
    In your setup, where you have the acceleration specified, you can't change the time.

    But that is not the main point. The main point is that power, like velocity, can be calculated between two points very close to each other, so you get an instantaneous value.

    In your example, the acceleration (and force) is constant, so the instantaneous power is also constant and is equal to the average power.

    However, one could equally move the mass with 10 m/s^2 for the first one meter, and then let the mass slide by inertia for the rest of the distance. Average power would then be different, and, more importantly, the instantaneous power during the first meter would be (much) greater than the average power, while the instantaneous power in the second segment would be zero.
     
  5. Jan 21, 2013 #4

    jbriggs444

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    Science Advisor

    The force is constant, but velocity is not. Power = force multiplied by velocity.
     
  6. Jan 21, 2013 #5
    Indeed. Thanks for correcting that.
     
  7. Jan 21, 2013 #6
    Well thank you very much. That helped a lot.
     
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