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Classical Mechanics: Work

  1. Jan 22, 2009 #1
    1. The problem statement, all variables and given/known data
    Hi all.

    We have that the work W equals Ui - Uf (respectively the initial and final potential energy). Let us say that a stone has a height 10 m, and falls down. We say that positive x is upwards.

    Now W = [itex]\int F \cdot dl = F(x_f-x_i)=U_f-U_i[/itex], since F and dl both point downwards. But this result does not correspond to what I wrote above. What am I doing wrong here?

    Thanks in advance.

  2. jcsd
  3. Jan 22, 2009 #2


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    If positive x is upwards, the vector F carries a negative sign doesn't it? It's pointing - X.
  4. Jan 22, 2009 #3
    Yes, but so does dl (i.e. dl points from x_initial to x_final, which is the same direction as for F).
  5. Jan 22, 2009 #4
    When considering work, you must be careful whether you are talking about the work done on the system or the work done by the system. Both these meanings are subject to how you wish to define work, both are equally valid but you must stick to the same one. Your first statement describes the work done by the system whereas your second, derived, statement describes the work done on the system.
  6. Jan 22, 2009 #5


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    It's the Force in the direction of the motion.

    When you move it upwards, the work supplied is a force against gravity pointing down. You're going positive x and your work has increased potential energy.

    W = ΔU

    Going down reduces potential energy. When you let it fall, the work by gravity - not you by you - is negative because it returns the work as kinetic energy.

    W = -ΔU
  7. Jan 22, 2009 #6
    In my first post, I was talking about the work done on the system in both cases. The work done on a system equals -ΔU, where U is the internal (potential) energy. This is why I am confused.

    And I can't see why the work done by gravity is negative, when the object falls. Gravity points downwards just like dl does. So where does this extra minus come from?
  8. Jan 23, 2009 #7
    The work done on the system by gravity is W = -ΔU, whether you are moving the stone up or down a gravitational potential, you just get either a positive or negative answer. When you were performing the calculation of the work you should have noted that F.x is the kinetic energy, thus:

    F(xf-xi) = KEf - KEi = Ui - Uf = -ΔU

    by conservation of energy, i haven't been using this site long and don't know how to type equations yet, do you know where I can learn?
  9. Jan 23, 2009 #8
    Just write:


    Write equations here...

    [* /tex].

    But remove the asterix's (i.e. *).

    I will think about your reply.
  10. Jan 23, 2009 #9
    Hmm, I thought mgh ('h' for height) was the potential energy, and not kinetic?
  11. Jan 23, 2009 #10

    Doc Al

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    The change in gravitational PE when you move a mass from one point to another equals the work done against gravity (not by gravity). (Think of it as the work you have to do to lift the mass.) So if F is the gravitational force, -F is the force you must exert against gravity. Thus when a stone drops 10 m, the change in PE is negative since the force opposing gravity is positive but the displacement is negative.
    The work done by gravity is positive; thus the change in PE is negative.
  12. Jan 23, 2009 #11
    Ok, I will take this from the very top, since I am getting confused with the signs. I think the easiest is if you can point out where I am wrong in this analysis, please:

    1) The work done by a conservative force on a system equals the minus of the change in the systems potential energy.

    2) Thus when I lift a book with constant velocity against gravity from point A to point B (B>A), I get:

    W = \int_A^B F\cdot dl = \int_A^B Fdl = F(B-A) = mg(B-A).

    Now this has to equal the minus of the change in potential energy of the system (i.e. book):

    -\delta U = - mg(B-A).

    Where is my reasoning wrong?
  13. Jan 23, 2009 #12


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    Assuming x increases in the upward direction,
    [tex]W_{\vec W}=\int_A^B \vec W\cdot d\vec l=\int_{x_A}^{x_B} (-mg \hat x)\cdot (\hat x dx)=\int_{x_A}^{x_B} (-mg ) dx= (-mg )(x_B-x_A)

    When it "falls" to a lower position, [tex]x_B-x_A<0[/tex]... thus [tex]W_{\vec W} > 0[/tex]. (The work done by gravity is positive in this case.)

    In short, when writing the integrand in terms of chosen coordinate system, [tex]d\vec l[/tex] in terms of the coordinate system... is [tex]\hat x dx[/tex].... that is... upward, in the direction of increasing x. Let the endpoints handle whether the "actual" displacement is "up" or "down".
  14. Jan 23, 2009 #13
    I don't agree with your signs based on the following explanation: Shouldn't [itex]dl[/itex] be tangent to the path from A to B? Thus it must also point in [itex]-\hat x dx[/itex].

    (Using your explanation, everything is OK, but this is how I learned vector calculus, and I would just like to see my math courses coincide with my physics course).
  15. Jan 23, 2009 #14

    Doc Al

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    You are calculating the work done by gravity. mg is just the magnitude of the force--gravity acts down, of course. Using up as positive, F = -mg.

    You had the sign of the gravitational force wrong.

    Read what robphy says; that's the way I think of it when doing these kinds of integrals. I let the coordinate system dictate the sign convention. In this case, +dl is up, so gravity = -mg (down).
  16. Jan 23, 2009 #15
    Ok, so we found my problem: I am letting the sign of [itex]dl[/itex] be determined by the direction of the tangent from A -> B instead of letting the coordinate system determine it

    But why is this wrong? In my math-course, this was how I was told to determine the direction of dl. On this forum, I am being told both things (see post #6: https://www.physicsforums.com/showthread.php?t=283845 for the contrary of what you are saying).

    Using your method, it all works out.
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