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Why is work done by non-conservative force path dependent? If friction does work then it is given by Force * displacement (no matter what path i take).


marked as duplicate by user36790, user259412, heather, ACuriousMind, Bosoneando Oct 8 '16 at 18:42

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From your comment:

work done = force* displacement

That's not accurate, even in the case of a constant-magnitude force. $$W = \int \vec{F}\cdot d\vec{x}$$

In differential form, during a tiny displacement $d\vec{x}$, the tiny amount of work done by friction resisting the motion is: $$dW = \vec{F}_{fr}\cdot d\vec{x} = F_{fr}dx\cos\theta = -F_{fr}dx$$ If the sliding friction force is always opposing the motion, then we'd add up all of these tiny works along the path, getting: $$W = -F_{fr}d$$ This is notably different from $-F_{fr}\Delta x$, and the difference is about the displacement $\Delta x$ vs the distance traveled $d$. That is, the work done by the friction force is dependent upon the path.

Commenters are correct that there's much reference material available on this (and proving that for conservative forces, like gravity, the integral is indeed about the displacement, not the distance, so they're path independent).


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