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Work Done on Object When Net Force is Zero

  1. Jul 7, 2012 #1
    Suppose I have an object moving in a free space at initial velocity of v (no friction). Then two forces of equal magnitude but opposite directions are applied on the moving object of which one of them is in the same direction as the moving object. Is there work done by that force?
     
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  3. Jul 7, 2012 #2
    Think about it.

    If the two applied forces are in exactly opposite directions they are colinear.

    So whether you say the work done by one = minus work done by the other = 0
    (since the displacement against one is positive and the other negative)

    Or

    There is not net force so work done = 0

    It comes out to the same thing.
     
  4. Jul 7, 2012 #3

    Doc Al

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    Short answer: Yes.

    But as Studiot explained, the net work done by both forces is zero.
     
  5. Jul 7, 2012 #4

    phinds

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    I'm clearly missing something here. By your logic, If I push really hard on the side of my house and do not manage to move it at all, I have done work. You seem to be saying that both forces do work but cancel each other out. I content that both forces are just "applying force" and that neither of them do any work at all. What am I missing?
     
  6. Jul 7, 2012 #5

    Doc Al

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    You are missing the fact that the object being pushed on is moving. (Why in the world would you think that pushing on a nonmoving object involves doing work?)
     
  7. Jul 7, 2012 #6

    CWatters

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    Yes.

    Consider an aircraft flying at a constant airspeed. The thrust and drag forces sum to zero. The engines are clearly doing work.

    Work = force * distance
    Power = force * distance/time = force * velocity

    In the example of your house the oposing force is friction but you haven't moved the house so you haven't done any work against it.
     
  8. Jul 7, 2012 #7
    Thanks guys. I bumped on the answer here:
    http://electron9.phys.utk.edu/phys135d/modules/m6/Work.htm [Broken]

    and I did not know that many have asked this similar question before. The thread below seems to have a good discussion on that matter:
    https://www.physicsforums.com/showthread.php?t=482584
     
    Last edited by a moderator: May 6, 2017
  9. Jul 7, 2012 #8

    phinds

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    But the movement was there BEFORE the pushing started, according to the original post, and the forces are applied AFTER it starts moving at a uniform velocity, so I still don't see what you mean. NEITHER of the applied forces is causing any motion, they are just canceling each other out.
     
  10. Jul 7, 2012 #9
    Work is done when a Force moves its point of application.

    With a moving object the point of application is constantly moving. With a brick wall it is not.

    If there is no resistance then no force is required to keep it moving - that is Newtons First Law. Since the force is zero the work = force times distance = 0
     
  11. Jul 7, 2012 #10

    russ_watters

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    Consider what happens to a car if you push down on both the gas and brake when it is already moving.
     
  12. Jul 7, 2012 #11

    phinds

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    Yes, I agree w/ everything you have said but I'm not sure if you are answering my quesion or not, since you have not applied your statements to the OP's situation.

    The OP posited a situation where the item is moving in free space at a constant velocity and THEN opposing forces are applied. DOC AI said that these opposing forces cause work to be performed and I say they do not. There IS resistance to each force, but because they cancel each other out, I say that neither is doing any work. Is this not correct?
     
  13. Jul 7, 2012 #12

    Doc Al

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    So what? Whether work is done by a force has nothing to do with whether that force has 'caused' the motion.
     
  14. Jul 7, 2012 #13

    Doc Al

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    No, it's not correct.

    According to your thinking, if you push a block along the floor against friction at constant speed no further work is done by you once the block has attained uniform speed. Do you think that is reasonable?
     
  15. Jul 7, 2012 #14

    phinds

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    No, that's not what I'm saying. What I'm saying is that if you push the block along the floor with a force F1 and then as it's moving uniformly along, you increase the force you are applying by amount F2 AND someone else applies F2 directly against you, nothing changes and the extra force you apply does no work. That seems to me to be exactly the original question except that in that case F1 was posited as zero.
     
  16. Jul 7, 2012 #15

    Doc Al

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    What do you mean 'nothing changes'? You're pushing harder! That means more work is done (for a given displacement).
     
  17. Jul 7, 2012 #16

    phinds

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    OK, I see what you're saying. Not sure why I was having trouble with this, but I see a way of looking at it that I believe now clarifies it for me.

    Take my example above and turn it vertical. If you have to exert force F1 to lift a rock 1 foot and you add another rock (which I see as the oposing force F2) then you have to exert F2 more force to lift the total the same distance. The fact that your additional force F2 in some sense balances the extra rock, does NOT mean it doesn't do any work, which is what I was somehow getting myself confused about.

    Thanks.
     
  18. Jul 7, 2012 #17
    A good question to ask oneself when faced with a situation like this is

    How is(are) the force(s) applied?

    Its not too bad with a stationary object, but a moving one.

    How would you apply two equal and opposite forces?

    Not so easy - the force must move along with the body or the body must move through a force field.
    If you have a force field (eg electric) you can't just apply the same in the opposite direction or the fields would just cancel.
     
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