# Work Power Energy [duplicate]

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, peterh, heather, ACuriousMind♦, BosoneandoOct 8 '16 at 18:42

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.