B Exact differential and work done

AI Thread Summary
The discussion focuses on the relationship between conservative forces and exact differentials in the context of work done. It explains that if the infinitesimal work 𝐹⃗ β‹…π‘‘π‘Ÿβƒ— is an exact differential, the work is independent of the path taken, indicating that the force is conservative. An example involving gravity illustrates that the total work done is the same regardless of the path, confirming gravity as a conservative force. In contrast, friction is shown to be non-conservative as the total work varies with the path. Understanding exact differentials is essential for grasping why conservative forces exhibit this property.
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The work done is independent of path if the infinitesimal work 𝐹⃗ β‹…π‘‘π‘Ÿβƒ—
is an exact differential.
I was researching about conservative and non-conservative forces, and there is some information in a website that sates that the work done is independent of path if the infinitesimal work 𝐹⃗ β‹…π‘‘π‘Ÿβƒ— is an exact differential. It further states that in 2 dimensions the condition for 𝐹⃗ β‹…π‘‘π‘Ÿβƒ— = Fxdx + Fydy to be an exact differential is:

𝑑𝐹π‘₯/𝑑𝑦=𝑑𝐹𝑦/𝑑π‘₯.

My question is this: why is a force conservative if the work is an exact differential? How can we deduce from the definition of a conservative force that this force is conservative if the work done to it is an exact differential?
 
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putongren said:
My question is this: why is a force conservative if the work is an exact differential?
Because if you integrate an exact differential, the result is the difference of some function evaluated at the two limits. This is what "work is independent of the path" means.

Here is an example. Suppose you use an eraser to erase a chalkboard. To do that you push the eraser against the board and, starting at the bottom B, you go to the middle M, then to the top T and then back to M. Let ##h=~##the height of the board.

The work done by the force of gravity ##mg## is
From B to M: ##~W_{BM}=-mg*\dfrac{h}{2}.##
From M to T: ##~W_{MT}=-mg*\dfrac{h}{2}.##
From T to back to M: ##~W_{TM}=+mg*\dfrac{h}{2}.##
Note that ##W_{BM}+W_{MT}+W_{TM}=W_{BM}.## This says that the total work is independent of the path because it is the same when one stops at M or goes past M to T and then back to M. We conclude that the force of gravity is conservative.

The work done by the force of friction ##f## is
From B to M: ##~W_{BM}=-f*\dfrac{h}{2}.##
From M to T: ##~W_{MT}=-f*\dfrac{h}{2}.##
From T to back to M: ##~W_{TM}=-f*\dfrac{h}{2}.##
Note that ##W_{BM}+W_{MT}+W_{TM}=3W_{BM}.## This says that the total work is not independent of the path because it is not the same when one stops at M or goes past M to T and then back to M. We conclude that the force of gravity friction is not conservative.
 
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kuruman said:
The work done by the force of friction ##f## is
From B to M: ##~W_{BM}=-f*\dfrac{h}{2}.##
From M to T: ##~W_{MT}=-f*\dfrac{h}{2}.##
From T to back to M: ##~W_{TM}=-f*\dfrac{h}{2}.##
Note that ##W_{BM}+W_{MT}+W_{TM}=3W_{BM}.## This says that the total work is not independent of the path because it is not the same when one stops at M or goes past M to T and then back to M. We conclude that the force of *gravity* is not conservative.

I believe you meant friction. Besides that, great explanation.
 
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Juanda said:
I believe you meant friction. Besides that, great explanation.
Good catch, thanks. I cut and pasted from above without changing the force. I edited to fix the typo.
 
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Sorry, I assumed you already knew what an exact differential is.
 
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