I have some questions about Hooke's law. I have posted my thoughts, but getting stuck at the end. Can someone please help me? I would appreciate!

**1) An archer pulls her bow string back 0.400m by exerting a force that increases uniformly from 0 to 230N. How much work is done in pulling the bow?**
F=kx, so the equivalent spring constant is 575N/m. The elastic potential energy is (1/2)kx^2=(1/2)(575)(0.400)^2=46J. So is the work done by the archer 46J? I am not sure about that because I was told that WORK DONE and ENERGY are not entirely the same thing.....

2)

Total spring constant=65.0x4=260N/m

Amount of stretch from equilibrium position=0.365m

F=kx

=(260)(0.365)

=94.9N

Now do I have to divide this by 2 to get 47.4N? I think this is getting really TRICKY. Is the man stretching and applying a force on both sides? Or is he applying a force of 94.9N on each side?