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Work on a static object?

  1. Sep 28, 2007 #1
    So I'm trying to push an big object and it doesn't move, am I doing work on it? Or is work 0 joules?
  2. jcsd
  3. Sep 28, 2007 #2


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    What is the definition of work?
  4. Sep 28, 2007 #3
    Work is the force I would apply times the distance the object moves.... OHHH, it's that simple. It doesn't move so work would be zero. Thanks cristo.
  5. Sep 28, 2007 #4
    Ok, I have another work question. What if the box is sliding down a frictionless slope. Would the equation for work be W=w*d*cos(theta), with w=weight and d=distance? Or would it be W=w*d*cos(90-theta)? I'm guessing it's one of these because since the box is moving, work cannot be zero.
  6. Sep 28, 2007 #5
    In that case the work would be done by gravity. Calculate the force that gravity is exerting on the box at that angle and relate it to the distance.

    You can think of it this way. If there was no gravity, the box would just sit at the top of the slope because nothing is pushing it down.
  7. Sep 28, 2007 #6
    Ohh yes, I got it. It must be W=w*d*cos(90-theta) because W=w*d*cos(theta) would find something else.
  8. Sep 28, 2007 #7
    What if there was no friction?
  9. Sep 28, 2007 #8
    Even if there is no friction, nothing is pushing the box down. Space is a vacuum, there is no air resistance. If you are in space, there is no gravity, and no friction, and you still float. So does the box.
  10. Sep 28, 2007 #9
    so, if

    weight = mass x gravity

    is gravity a constant here? If so, what varies as what now that it's been removed?
  11. Sep 28, 2007 #10
    Gravity on Earth is approximately equal to 9.81 m/s^2. This changes very slightly at higher altitudes, but the change is so small that it can be left out.

    The removal of gravity was just an example in order to show that it was the active force pushing the box down.
  12. Sep 28, 2007 #11
    I understood that, I was just wondering where the equation went from there
  13. Sep 28, 2007 #12
    Oh, I figured it out myself. Thanks for the reply.
  14. Sep 28, 2007 #13
    I think I did too
    you end up with mass varying as mass or something equally useful :smile:
  15. Sep 28, 2007 #14
    Consider that the object would essentially have to be of infinite mass and perfectly rigid not to perform work on it.
  16. Sep 29, 2007 #15
    You should specify that, if the object does not move, you are doing no MECHANICAL work on it. You are, however, doing some combination of elastic work, thermal work, and biological work; it's just that we don't usually count those in this kind of problem.
  17. Sep 29, 2007 #16
    Wouldnt it be easier to use Sin(Theta) rather then Cos(90-Theta)?

    Edit: Since Cos is just a derivative function of Sin, and is 90 out of phase with Sin anyway, so adding 90 is effectively just superimposing a Cos wave onto a Sin?
    Last edited: Sep 29, 2007
  18. Sep 29, 2007 #17
    In fact scratch that, that would be +90+Theta rather then minus i guess...
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