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Work and Potential Energy

  1. Jul 1, 2006 #1
    Hi, I haven't taken physics yet, but Im reading a Sparknotes Physics book for fun, and there's something that i dont understand

    In one part of the book, when it was talking about basic energy stuff it said that - ΔU = W


    but then a few chapters later it shows a problem with potential gravitational energy, and

    it said "W= U2 - U1" ...... what happened to the negative sign?

    (no source, since the website on this chapter is not current)
  2. jcsd
  3. Jul 1, 2006 #2


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    Probably the later reference has some outside agent (a person?)
    doing Work against the gravitational Force.
    If the motion is always slow, then the person's Force is opposite gravity's
    (that is, F by person = NEGATIVE F by gravity).

    So that Energy is *transferred* as the Force application point is moved.
    Last edited: Jul 1, 2006
  4. Jul 1, 2006 #3
    Hmm...im not sure

    but here is the actual problem

    A satellite of mass m is launched from the surface of the Earth into an orbit of radius 2r, where r is the radius of the Earth. How much work is done to get it into orbit?
  5. Jul 1, 2006 #4
    W = U2-U1 = -G(m1m2)/2r - -Gm1m2/r

    in the book it said potential grav. energy is -G(m1m2)/r where r is displacement and m1 is object mass, m2 is earth mass
  6. Jul 1, 2006 #5


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    It takes an "outside agent" (a rocket booster!) that does Work
    to get the satellite into orbit ...
    gravity's Force removes Energy from the satellite as the satellite rises.

    By the way, Grav.P.E. is negative, which reminds us that we are trapped
    down here ... we need to add Energy to something just to get it far away,
    and additional Energy to make it go fast there.
  7. Jul 1, 2006 #6
    hmmm..ok i sort of get it
  8. Jul 1, 2006 #7


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    I guess I should have said (in post #2) that

    The Work done by gravity's Force = - Delta U = - Ufinal - Uinitial .

    To lift something, your Force apllied to that thing
    is (approx) the negative of gravity's Force applied to it.
  9. Jul 2, 2006 #8
    you mean -DeltaU = -(Ufinal - U).

    but what is the difference between a hand picking up an object against gravity, and a rocketbooster boosting a rocket against gravity?
    shouldn't they be using the same formula?
    but what im seeing is the first problem uses W=-DeltaU and the latter uses W=DeltaU...

    ugh, I can't wait to take physics next year
  10. Jul 2, 2006 #9

    Doc Al

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    Staff: Mentor

    In both cases the work done by the outside force--hand or rocket--will equal the change in total mechanical energy (PE + KE).
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