Why Is There a Negative Sign in the Work-Energy Equation?

[4570562]

Homework Statement

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=$\nabla$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?

 

[billVancouver]

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

 

[4570562]

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=$\int$$\vec{F}$$\cdot$d$\vec{r}$ 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.

 

[4570562]

Okay I think I figured it out. Suppose toward a contradiction that U=f=-mgy. Going back to the ball example that would mean

$U_{a}$=-mgh
$K_{a}$=0

$U_{b}$=-mg(-4.5+h)=4.5mg²-mgh
$K_{b}$=4.5mg²

But that would violate conservation of energy since $U_{a}$+$K_{a}$≠$U_{b}$+$K_{b}$.



However, letting U=-f=mgy implies that

$U_{a}$=mgh
$K_{a}$=0

$U_{b}$=mg(-4.5+h)=-4.5mg²+mgh
$K_{b}$=4.5mg²

which makes $U_{a}$+$K_{a}$=$U_{b}$+$K_{b}$.