Why can't the work done by a non-conservative force be zero? 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?
 A: 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.
A: Well, we can do a simple counter-example. Let
$$
  \vec{F}(\vec{x}) = F_0 \cdot \varrho(\vec x)
$$
where $\varrho$ is the function that rotates vectors by 90° counter-clockwise (in matrix form $(\begin{smallmatrix}0 & -1\\1 & 0\end{smallmatrix})$ if you prefer that). Clearly, for the closed path
$$
  \vec{\gamma}\colon\quad [0, 2\pi]\ \to\ \mathbb{R}^2
   , \qquad t\ \mapsto \begin{pmatrix}\cos(t) \\ \sin(t)\end{pmatrix}
$$
we have always $\dot{\vec\gamma} \cdot \vec{F}(\vec\gamma) = F_0$. So
$$
  \oint_\gamma \mathrm{d}{\vec x}\cdot \vec{F}(\vec{x}) = 2\pi\cdot F_0.
$$
That's basically the answer to your question, although you've worded it wrongly:

Why can't the work... be zero

It can actually be zero. Consider
$$
  \vec{\gamma}_2\colon\quad [0, 2\pi]\ \to\ \mathbb{R}^2
  ,\qquad t \mapsto \begin{cases}\vec\gamma(t) & \text{if $t<\pi$} \\ \vec\gamma(2\pi - t) & \text{otherwise}\end{cases}
$$
This simply takes half the path of $\gamma$, but then turns around and goes back the same way it came, thereby integrating the opposite force scalar product, so the result comes out as $0$ here although the force, as I proved above, is not conservative. Only, in a non-conservative field not all closed paths have zero total work. (More obviously: in any field the closed "path" that simply stays at the same point forever has zero work.)
A: While the total displacement as shown in your figure is zero, this does not mean that the work is zero! Work is force scalar displacement, $W = \vec F\cdot \vec s$. Diving the path $A \to B \to A$ into infinitesimal steps we are led to $$dW = \vec F \cdot d\vec s$$ and for the total work, adding up the contribution $$W = \int \vec F\cdot d\vec s.$$
I think that from here, you want to distribute the integral over the product, $$W \overset{?}{=} \int F \cdot \int d\vec s$$
and since the total displacement is $0$, the product should vanish. But this equality does not hold in general, which is why you can't conclude that total work done vanishes just because total displacement does.
Of course as other answers have remarked it could happen by accident that the total work is still zero, it is just that for a non-conservative force we are not guaranteed that it does.
A: As non conservative forces are those forces whose work done depend upon path followed by a particle. 
So around a closed path it's value will be positive because some path have been followed by particle in moving Around a closed path. 
                         But in case of conservative forces work depend only upon initial and final points not on path. 
    So around a closed path initial and final points coincide so work done is zero. 
A: Be practical minded. If a ball depends on the path like  air force or any other force, work we given is loss on some point but when it returns,  work done by air force is given to the ball.so work done is not zero at any other point in non conservative force
