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Work and Potential Energy

  1. May 24, 2004 #1
    Hi everyone,

    Here is the problem: Two masses m and n are initially infinitely far appart from each other. Calculate the amount of work to get them a distance R from each other.

    Taking mass m as my reference point, the potential energy of mass n a distance r from m is U(r) = -Gmn/r. Now, if only conservative forces are involved -W = U(r') - U(r) where W is the work need to move a mass a distance r' - r as I understand it. So -W = U(R) - 0 => W = -U(R) = Gmn/R. However, the book where I got this problem from has as the solution -Gmn/R. That means either W = U(r') - U(r) or U(r) = Gmn/r, but these would be contradictions.

    Confuzzled here,
    e(ho0n3
     
  2. jcsd
  3. May 24, 2004 #2

    Doc Al

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    Staff: Mentor

    work against gravity

    You are calculating the work done by gravity, but that's not what the question asks. The work done on this system (against gravity) equals the change in the system's energy. Since the applied force opposes gravity as the masses are brought closer, the work done is negative.
     
  4. May 25, 2004 #3
    I seem to be a little confused about the work done by a force on a body and the work done on a body. When I think of work, I think
    [tex]W = \int_{p}^{p'}{\vec{F}\cdot\vec{dr}}[/tex]​
    I guess I went wrong by assuming that gravity will be doing the work of getting m and n a distance R from each other. But as I understand from your explanation, there must be another force which must be applied for this to happen and that this force is opposing gravity.

    I'm still confused though. Gravity is an attractive force, meaning that mass m will attract mass n and vice versa. Why would I apply a force that opposes gravity and impede m and n from getting closer.

    e(ho0n3
     
  5. May 25, 2004 #4

    Doc Al

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    Staff: Mentor

    Correct, of course.
    What if the question asked: What work is done in separating two masses from distance R to infinity? I assume you would not hesitate is stating that positive work is done by some external force, right? Same thing here, only backwards. Somehow the masses are brought together (not just let go): the KE is kept at zero.

    I will admit that it's a tricky question. :smile:

    Sure, it's a bit of a semantic game. You can just let them fall together. In which case no external work is done, and the energy is constant. Or you can lower them together slowly, via an exernal force against gravity: in this case you've done negative work--and lowered the total energy of the two mass system.

    It's slippery stuff.
     
  6. May 25, 2004 #5

    turin

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    Homework Helper

    I'm wondering, and this is to support you, not argue with you:

    If you just let them go, you will eventually have to stop them. To do that, you would have to supply a force, infinite if you stop them dead in there tracks (decelerating for v to 0 instantaneously). In other words, isn't it a nontrivial issue that both initial and final states are stationary and therefore KE is not a factor? Of course, even this is an assumption that must be made in order to solve the problem.
     
  7. May 25, 2004 #6

    Doc Al

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    Good observation, as usual. But note that the amount of work needed to stop the masses is the same amount of work needed to bring them together slowly with no KE. (No need to add in the complication of infinite force--assume some non-zero time for deceleration.)

    Right--to even begin to answer this question you need to make some assumptions about KE. That's why I agree that it's a tricky question.
     
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