Why isn't constant pull a conservative force? Consider the following diagram:

The force $\mathbf{F} = 1 \textrm{ N} \hat{\imath}$ is being applied all time as the ball goes from A to B (assume positive $x$ to the right.) Now, there are a few equivalent definitions for 'conservative force', e.g. a force whose work is $0$ for any closed path. Let me use this definition.
It is clear that if the ball went from $A$ to $B$ and then back from $B$ to $A$, the work done by $\mathbf{F}$ is 0. Of course this is only one possible closed path. We could also think of a triangular path $ABC$ with $\mathbf{r}_C = 0.5(\mathbf{r}_A + \mathbf{r}_B) + 1 \textrm{m } \hat{\jmath}$, and the same reasoning will apply, provided that $\mathbf{F}$ stays the same all along this path.
Essentially I am saying that if $\mathbf{F}$ is a constant field, then it is conservative, no big deal. This could be a Coulomb field $\mathbf{E}$ by two infinite charged planes and the ball a charge between them.
The thing is, if $\mathbf{F}$ is actually a pulling force (by someone's hand) on a ball, then we always say this is a non-conservative force, but why? What is the difference between $\mathbf{F}$ and $\mathbf{E}$ with respect to the definition I have used, that makes one conservative and the other non-conservative?
 A: A constant force is an example of a conservative force (weight is also a constant force). When you pull with your hand, what are the odds that you are going to be able to maintain the same level of force? The force applied will have a ramp-up period where it is not constant.
Taking the block-wire as one system, we can exploit conservation of energy to solve this problem:

Here is a relevant excerpt from O'Reilly (Intermediate Dynamics for Engineers):

A: Recall that a force $\bf\vec F$ is conservative iff for any closed path $C$, $\displaystyle\oint_C\mathbf{\vec F}\cdot\mathrm d\mathbf{\vec r}=0$. The key word here is any, and you can use your hand to move an object in a closed path by accelerating its speed, such that $\mathrm{KE}_i\neq\mathrm{KE}_f$. So, $\bf\vec F$ cannot be conservative. There is no associated potential $U$ such that $\mathbf{\vec F}=-\nabla U$ since a pull force does not depend on position, but rather depends on how you choose to pull.
A: You need to make a distinction between work done by the overall system vs. work done on the ball.  When the ball moves from A to B and then back to A (regardless of the path taken) the work done by that system is zero.  That is, the ball is exactly where it started and therefore no work was done.  However, when you look only at the ball, work is done in moving it.  If you were to pull it from A to B and then back to A you would feel as if you had done some work.  However, even in this case the work done by the system (not by you personally as the one pulling the ball) will be zero and by your definition it too will be a conservative force.
