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Understanding Potential Energy

  1. Apr 15, 2012 #1
    1. The problem statement, all variables and given/known data

    My calculus book gives the following definition of a conservative vector field:

    A vector field F is called conservative if there exists a differentiable function f such that F=[itex]\nabla[/itex]f. The function f is called the potential function for F.


    It then gives the fundamental theorem of line integrals:

    The line integral of a conservative vector field is equal to the change in the potential function f.

    And follows this up by saying that the work done on a particle moving in a force field F is given by the change in the potential function f.


    All this pretty much made sense until my physics book came along and said that W=-ΔU. Why is that negative sign there?
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Apr 15, 2012 #2
    Hi 4570562,

    It looks to me like your calculus and physics textbooks are talking about slightly different things. Your calculus book is giving you a discussion of formal vector fields, whereas from the notation, I take it that your physics text is just considering some simple scalar work and energy ideas.

    The matter of the negative sign in an equation like your W = -ΔU is a choice that represents who is doing work on who. try to find in your physics book the exactly what it means by 'W'; is that the work that the system with potential energy U is doing, or is it the work being performed on that system? Judging by the placement of the negative sign you were wondering about, it looks like that equation is saying "the work W done BY the system is equal to the negative of the change in its potential energy'; qualitatively this makes some sense, since the system has to expend some energy to do some work on something external to it.

    Hope this helps,
    Bill Mills
     
    Last edited by a moderator: Apr 16, 2012
  4. Apr 15, 2012 #3
    Let's take an example. Suppose you drop from rest a ball of mass m and let it fall for 3 seconds. You want to find the amount of work that gravity does on the object. There are three ways you could proceed.

    1) Use W=[itex]\int[/itex][itex]\vec{F}[/itex][itex]\cdot[/itex]d[itex]\vec{r}[/itex] the line integral method. (Note that this method can be made simpler by first finding the potential function f such that gradF = f and then calculating the change in f. In other words, use W=Δf.)
    2) Use W=ΔK the work-energy theorem.
    3) Use W=-ΔU where U=mgy

    I'll spare you the details, but either way the work comes out to 4.5mg2. Note that in the first method I used f=-mgy. So in other words U=-f. Why is U defined this way? It seems to me that things would be a whole lot less confusing if U was just defined to be equal to f.
     
    Last edited: Apr 15, 2012
  5. Apr 15, 2012 #4
    Okay I think I figured it out. Suppose toward a contradiction that U=f=-mgy. Going back to the ball example that would mean

    [itex]U_{a}[/itex]=-mgh
    [itex]K_{a}[/itex]=0

    [itex]U_{b}[/itex]=-mg(-4.5+h)=4.5mg2-mgh
    [itex]K_{b}[/itex]=4.5mg2

    But that would violate conservation of energy since [itex]U_{a}[/itex]+[itex]K_{a}[/itex]≠[itex]U_{b}[/itex]+[itex]K_{b}[/itex].



    However, letting U=-f=mgy implies that

    [itex]U_{a}[/itex]=mgh
    [itex]K_{a}[/itex]=0

    [itex]U_{b}[/itex]=mg(-4.5+h)=-4.5mg2+mgh
    [itex]K_{b}[/itex]=4.5mg2

    which makes [itex]U_{a}[/itex]+[itex]K_{a}[/itex]=[itex]U_{b}[/itex]+[itex]K_{b}[/itex].
     
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