# Work for a force function of x, v, t

1. Sep 19, 2004

### quasar987

Hey people. This is crazy! I haven't had TIME to check out the forum since class began. I do nothing but study and still am (at least a little) late in every class!

Anyway I do have a question about physics to which I haven't been able to find the answer. I was hoping you guys could help! For the record, I use the book 'Mechanics', third edition by Symon. It is written in there that the work, W, is defined has follow

$$W=\int_{0}^{t} F(x, v, t) v dt$$

Now it is obvious that if F is simply a function of x, we can simplify vdt=dx and change the limits of the integral from 0 and t to 0 and x. And if x(t) = x(0) = 0 then W = 0. But for force function of x, v and t, it is not so easy to show that if the particle ends up at his initial position, then W = 0. In fact, according to what the calculations I've made, I would tend to say that it isn't.

So, is it? And how do you know?

2. Sep 19, 2004

### Cyrus

That makes no sense to me, If force is a function of (x,v,t), then the integral must be in terms of dx,dv,dt unless you are assuming they are held fixed. If not, shouldnt the formula be expressed as, F(x(t),v(t),t)dt ?

3. Sep 19, 2004

### Tide

Quasar,

That's because the integral around a closed loop is zero only for special circumstances - namely the force must be a conservative one! As hard as I might try I will not recover the energy I expended in moving my sandpaper in a circular motion on a board because friction is not a conservative force.

4. Sep 19, 2004

### Cyrus

Hey tide! Did you see my comment about the function? Does that make sense to you? I dont see how a function of three variables can be integrated only along one? where is the dx and dv ? That surface charge thing has not worked out for me at all :rofl:

5. Sep 20, 2004

### Tide

Hi, Cyrus! Yes, I saw that and would agree in a strict mathematical sense. Technically, unless Quasar is talking about a phase space (which I'm quite sure he's not!) then the force is just a function of t alone or x alone. Ordinarliy, x maps on to t from the equations of motion.

Regarding the surface charge problem - you need to realize that the "test charge" and the surface charge interact with each other but we'd best leave that discussion in the appropriate thread!

6. Sep 20, 2004

### quasar987

I think this comes down to "W egals 0 for a closed loop only for forces depending strictly on x, because all conservative forces are function of position.