Why can't the work done by a non-conservative force be zero? The displacement along a closed path is always zero. So, whatever be the type of force, variable or constant, the work has to be zero. Why do we need to calculate the work done for individual paths?
This is a non-conservative force that starts from $A$ moves via Path 1 to $B$ and then back to $A$ via Path 2. Since the displacement is anyways going to be zero, why can't work done be zero?
Best Answer
For forces that change along the way, displacement is not the thing to calculate work with. Let $\gamma : [0,1] \rightarrow \mathbb{R}^3$ be the (closed or open) path that the particle the force is exerted on follows. Then, the work done along that path is
$$ W[\gamma,F] = \oint_\gamma \vec{F}(\vec{x})\cdot \mathrm{d}\vec{x}$$
which is a line integral. If $\vec{F}$ is conservative, there is a function $V(\vec{x})$ such that $\nabla V(\vec{x}) = \vec{F}(\vec{x})$, then we can apply Stokes' theorem (or, less fancy, the fundamental theorem of calculus) to calculate the work by
$$\oint_\gamma \vec{F}(\vec{x})\cdot \mathrm{d}\vec{x} = \int_{\partial\gamma} V(\vec{x}) = V(\gamma(1)) - V(\gamma(0))$$
For closed paths, $\gamma(1) = \gamma(0)$, so this is zero. If there is no potential with $\nabla V = \vec{F}$, we cannot apply this argument and have to actually calculate the line integral, which may be anything, especially not zero.