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?
Best Answer
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):