It takes $80\,\textrm{J}$ of work to stretch a spring $0.5\,\textrm{m}$ from its equilibrium position. How much work is needed to stretch it an additional $0.5\,\textrm{m}$?
Here's what I have: $F(x)=kx$ and $F(0.5) = k(80)$.
Solving for $k$, I get $k = 160$.
Then, I find that $\int_{0.5}^1 160x dx = 60$, but that's not an answer..
Best Answer
Integrating $F(x)=kx$ we get the formula for the elastic energy stored in a spring (=the work to stretch the spring from equilibrium):
$$W=\frac 12kx^2$$
Use this to find the spring constant $k$, then the energy needed to stretch to the new position, then subtract the two energies to find the needed work.
This is how I teach this kind of problem in my $12$th grade physics class.