Work for a force function of x, v, t

AI Thread Summary
The discussion revolves around the calculation of work, W, defined as W = ∫ F(x, v, t) v dt, and the complexities involved when force is a function of position, velocity, and time. Participants debate whether work can be zero when a particle returns to its initial position, emphasizing that this holds true only for conservative forces. It is noted that integrating a force function of three variables along one dimension raises mathematical concerns, as it overlooks the contributions of other variables. The consensus indicates that work equals zero for closed loops only under specific conditions related to conservative forces. Overall, the intricacies of integrating force functions in physics are highlighted, particularly in non-conservative scenarios.
quasar987
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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?

Thanks for your imputs!
 
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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, shouldn't the formula be expressed as, F(x(t),v(t),t)dt ?
 
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.
 
Hey tide! :smile: Did you see my comment about the function? Does that make sense to you? I don't 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 :mad: :smile:
 
cyrusabdollahi said:
Hey tide! :smile: Did you see my comment about the function? Does that make sense to you? I don't 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 :mad: :smile:

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!
 
Tide said:
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.


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.
 
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