Does work done on a spring = elastic potential energy? So the UK exam board specifications (AQA GCSE) clearly state
"...the work done on the spring and the elastic potential energy stored are equal"
Here's my problem,
So work done = Force x displacement
Force = Spring constant x extension
Elastic potential energy = 0.5 x spring constant x extension squared
Extension = displacement for a strecthed spring
However if I take some sample values and calculate the work done on the spring, and then the elastic potential energy stored. The Elastic potential is always exactly half the work done. This contradicts the statement in the specifications.
I have looked all over the place but can't find a satisfactory answer to this question. What am I missing? I get that elastic potential energy is equal to area under the Fx graph but why does that not equal the work done?
 A: Work is not "force times displacement". Work is an integral $$W=\int\mathbf F\cdot \text d\mathbf x$$
which becomes $W=Fx$ under certain conditions.
The work done by a conservative force is always equal to the negative change in potential energy associated with that force:
$$W_\text{cons}=\int\mathbf F\cdot\text d\mathbf x=\int-\nabla U\cdot\text d\mathbf x=-\Delta U$$
The area under the Fx graph should be the work done by the spring, which is the negative change in potential energy.
Your mistake is most likely thinking that $W=Fx$ holds here, but it does not because the spring force varies with displacement. You are probably getting a discrepancy of a factor of $2$.
A: The work is $Fx$ if the force is constant during the displacement from 0 to $x$.
In the case of an elastic spring, the force is a function  of the displacement: $F = kx$
So the work for a small displacement when the spring is streched in a given $x$ position: $\Delta W = kx\Delta x$
Integrating from 0 to x: $W = \frac{1}{2}kx^2$, which is the stored elastic energy.
