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Question regarding work.

  1. Feb 22, 2007 #1

    ssb

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    My TA gave us a conceptual example today that I dont understand. She was explaining how to derive work if you are given acceleration.

    She wrote on the board that an object accelerates with 3t^2. the object weighs 3kg, starts at t=0. She asked us how much work is done on the object durring the first 2 seconds and then she said the answer was zero.

    Is this right? what am I not getting:confused: :confused:
     
  2. jcsd
  3. Feb 22, 2007 #2

    arildno

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    No, it is not.
    The net work done equals the change in the object's kinetic energy, which certainly is non-zero.
     
  4. Feb 22, 2007 #3

    ssb

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    Thanks I knew she was crazy. That makes me feel better.

    Do you know how to calculate this in joules by the way so I can go back to her and talk to her about this?
     
  5. Feb 22, 2007 #4

    Doc Al

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    You have the acceleration as a function of time. What would you need to do to get the velocity as a function of time?
     
  6. Feb 22, 2007 #5

    ssb

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    take the derivative right? so it would be 6x right? sorry im new to this stuff
     
  7. Feb 22, 2007 #6

    ssb

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    i tried solving for the position function which is just 6. graphicly speaking, a graph of y = 6 from 0 to 2 would cover 12 meters.
    since work is force x distance and force is mass x acceleration:

    work = (3kg * 3x^2) * (12)
    so at time = 2, the work would be 432 joules?

    Im sorry but im thinking i did something wrong.
     
  8. Feb 22, 2007 #7

    Doc Al

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    Just the opposite! Acceleration is the rate of change of velocity--which means the derivitive of velocity. Given the acceleration, you'd need to integrate to get velocity.

    Read this: Acceleration, velocity, and Position
     
  9. Feb 22, 2007 #8

    ssb

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    ahhhh cool it is starting to make sence. Thanks!
     
  10. Feb 22, 2007 #9
    That would work if your force was constant, but it acceleration is changing and so is force. The best way is to figure out the change in kinetic energy
     
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