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We know that $$\oint \boldsymbol{F}\cdot d\boldsymbol{r}= \iint (\nabla \times \boldsymbol{F})\cdot d\boldsymbol{s}.$$ Now if $\boldsymbol{F}$ is a constant vector, then $\nabla \times \boldsymbol{F}=0$, this gives that $\oint \boldsymbol{F}\cdot d\boldsymbol{r}=0$. And $\oint \boldsymbol{F} \cdot d\boldsymbol{r}$ represents the work $$W=\int \boldsymbol{F} \cdot d\boldsymbol{r}$$ done by the vector field $\boldsymbol{F}$ along a curve. So this gives that work done by a constant vector field is $0$. How can it be possible?

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This isn't surprising. Remember, $\oint$ means that you're integrating around a closed path. Without that requirement you can't use Stokes's Theorem to get a curl. All you've shown is that a constant vector does no work if you go in a big loop. This is the situation with, for instance, gravity close to Earth's surface. You throw a ball in the air, and when it returns to where it started it has the same amount of energy (it's just going the opposite direction).

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  • $\begingroup$ I got beaten by 20 seconds! I knew I should have typed faster. $\endgroup$
    – Javier
    Oct 10 '15 at 14:37
  • $\begingroup$ Haha, right down to the wording of "gravity close to Earth's surface" $\endgroup$
    – zeldredge
    Oct 10 '15 at 14:37
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It's perfectly possible, because it's not just any work; it's the work done around a closed curve. If you have a uniform force field (which is a special case of a conservative force field) and you move an object around in a closed curve you won't get any net work out of the force field. A good example is gravity close to the Earth's surface.

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It is possible for conservative forces. Closed integral of a conservative force (say Electric force) is zero.enter image description here

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