Exact differential and work done

[putongren]

TL;DR

The work done is independent of path if the infinitesimal work 𝐹⃗ ⋅𝑑𝑟⃗
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 ⋅𝑑𝑟⃗ is an exact differential. It further states that in 2 dimensions the condition for 𝐹⃗ ⋅𝑑𝑟⃗ = Fxdx + Fydy to be an exact differential is:

𝑑𝐹𝑥/𝑑𝑦=𝑑𝐹𝑦/𝑑𝑥.

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?

 

[kuruman]

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.

 

[Juanda]

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.

 

[kuruman]

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.

 

[putongren]

I guess knowing the definition of an exact differential would help in answering my question: https://en.wikipedia.org/wiki/Exact_differential

 

[kuruman]

Sorry, I assumed you already knew what an exact differential is.