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Work against gravity confusion

  1. Jun 19, 2013 #1
    Hi I've just started learning work, energy etc and the first question in the book is in the form:

    An object of mass m is lifted by a tension through a displacement of h. Find the work done on the object.

    I know that the answer is simply mgh J, there is something that I don't quite get.

    Why can't it be that the tension could be greater than mg but still lifting the object the same distance (in less time) so therefore does more work? I hope that makes sense.

    Thanks :)
  2. jcsd
  3. Jun 19, 2013 #2


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    You are correct. If the tension is greater than mg then the object is not only lifted. It is accelerated. It ends with more kinetic energy than it had when it started.

    The extra work done in this case manifests as extra kinetic energy.
  4. Jun 19, 2013 #3
    Ok so is this just a case of a poorly worded question, as it does not mention that the object is moving at constant velocity? (the answer in the book is mgh)

    Would the actual answer be that the tension does at least mgh J of work?
  5. Jun 19, 2013 #4


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    The "constant velocity" bit would be key. That would allow you to deduce (through Newton's second law) that the tension is equal to mg.

    So it does not sound as if the problem was poorly worded.
  6. Jun 19, 2013 #5
    Say there is an object on which a force of T acts for a displacement s and there us a constant frictional force of F.

    Would you say that the force T does Ts J of work or that the force T does Ts J of work and Fs J of work against friction? Or is it something else?

    Any explanation would be much appreciated :)

  7. Jun 19, 2013 #6
    You would say that force T does T*s J work, while work done by friction is -F*s J.
  8. Jun 19, 2013 #7


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    The question said the object was lifted a height h. If I threw a rock so that it passed, say, 10 meters height while still going up to a maximum height of 20 meters, I would NOT say that I "lifted" it 10 m. "Lifted" implies that it was motionless at its original position and motionless at the end. Any acceleration or velocity it had in between is irrelevant.
  9. Jun 20, 2013 #8
    For this question, actually the Ts J of work and the Fs J of work is just one same work. It is because the work done by T is used to cancel out the work done by friction. Therefore, the total work done on the object is actually zero.
  10. Jun 20, 2013 #9
    So for this question, I believe that the question is poorly worded. For the sentence "Find the work done on the object", assuming that the tension is same as mg, if it means the TOTAL work done on the object, then the work done will be zero. It's because the work done by both the tension and gravity will cancel each other out. Meanwhile, if it means the work done by the tension on the object, then only the answer will be mgh J.
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