The force pulling the spring back into equilibrium (due to the tension in the spring) is a conservative force - the very definition of a conservative force is that it's minus the gradient of some potential, i.e. $\overrightarrow{F}(x,y,z)=-\mathrm{\nabla}U(x,y,z)$. In the case of a spring in the xy-plane being stretched along the x axis, the equation reduces to $\overrightarrow{F}=-\frac{\text{d}U}{\text{d}x}\overrightarrow{{e}_{x}}$, which is indeed the case, since the right hand side is $-kx\overrightarrow{{e}_{x}}$ so that $\overrightarrow{F}=-kx\overrightarrow{{e}_{x}}$, Hooke's law.

Another way to define a conservative force is that $\oint \overrightarrow{F}\cdot \text{d}\overrightarrow{s}$ must equal zero. In other words, the work of this force along a random closed path must equal zero. This certainly is the case, since along any path, the work done by the spring force is independent of the path in between the end points.

At any rate, the work done by the spring would indeed be minus the work done by you, since your force is anti-parallel to the spring's force.

###### Did you like this example?

Subscribe for all access