Why are gravity and spring-force conservative forces? I know that the amount of work does not depend on the followed path but on the position when speaking about conservative forces. But could someone please prove/explain why exactly those two are conservative forces?
Has it something to do with the potential energy?
 A: A conservative force has the property that the work done in moving a particle between two points is independent of the path taken. It implies that the force is dependent only on the position of the particle. Now we can use this idea to define a function called the potential energy. 
It is a conservative force that gives rise to the concept of potential energy and not the other way round. If the force were non conservative, the force would not be dependent on position only and thus we could not have defined a potential energy function.
A simpler way to find out whether a force is conservative or not is to find out the closed line integral of force, i.e $\oint dr F$ and convert it into the area integral of the curl of the force by using Green's theorem, i.e
$$ \oint  F dr = \int_A (\nabla \times F) da$$. Thus if the curl of the force is zero, it automatically means that the force is zero. It is now trivial to see that the gravitational as well as the spring force are conservative as the curl of both forces vanish.
A: I know this answer is pretty late, but I'm hoping it'll help at least a little. The idea here is, of course, using Green's Theorem:
$$\oint \mathbf F \cdot \mathrm{d}\mathbf r = \int_A (\mathbf \nabla \times \mathbf F ) \; \mathrm{d}\mathbf a $$
Here, you're asking about the line integral around a closed path. When using this equation, you end up making any arbitrary closed path (in other words, you can choose any path as long as it's closed). To choose a path, imagine an object (like a box, for example), and imagine the box under the influence of the gravitational field. Move the box in space in any direction, along any path you so choose. To make a closed path, move the box back to its starting point at the end of the path.
The nice trick about Green's Theorem is that, instead of calculating the closed line integral (which can be messy since the maths isn't so nice), you can instead calculate 
$$ \int_A (\mathbf \nabla \times \mathbf F ) \; \mathrm{d}\mathbf a $$
which can sometimes be a lot easier. In the case of gravity, electric force, or a spring force, it turns out that $$\mathbf \nabla \times \mathbf F = \mathbf 0 \;.$$ So it follows that $$ \oint \mathbf F \cdot \mathrm{d}\mathbf r = \int_A (\mathbf \nabla \times \mathbf F ) \; \mathrm{d}\mathbf a =\int_A \mathbf 0 \; \mathrm{d}\mathbf a = 0 $$ and $$ \oint \mathbf F \cdot \mathrm{d}\mathbf r = 0 $$ This is nice because it works for any closed path.
Basically, the curl of a vector field, ($\mathbf \nabla \times \mathbf F$), tells you if a vector field is conservative or not.
If $\mathbf \nabla \times \mathbf F =\mathbf  0,$ like for gravitational and spring forces, the force $\bf F$ is conservative. If $\mathbf \nabla \times \mathbf F \neq 0,$ like for magnetic forces, the vector-field is non-conservative. If the field is conservative, then $$ \oint \mathbf F \cdot \;\mathrm{d}\mathbf r = 0\;. $$
This is true for any conservative vector field.
