How much work to push a spring up/down a distance of y?

[Jim01]

Homework Statement

The ends of a relaxed spring of length L and force constant k are attached to two points on two walls separated by a distance L. How much work must you do to push the midpoint of the spring up or down a distance of y?

Homework Equations

W = -1/2k(b²-a²) (after integration), moving from y=a to y=b

The Attempt at a Solution

I'm completely lost on this. The first thing that I notice is that when you push down on the midpoint of the spring, you form two right triangles. This leads me to believe that y will go from a value of zero to the value of F_y, where F is the downward force on the spring. So I made a force diagram and came up with summation of forces.

$$\sum$$F_x = 0

w_x = 0
F_x = L/2

F_x= 0

$$\sum$$F_y = 0

w_y = -mg
N = mg
F_y = -Fcos$$\theta$$

w_y-N-F_y = 0

F_y = w_y-N = -2mg

I attempted to use the F_y value as b in the W = -1/2k(b²-a²) formula but this led me to a wrong answer. I then tried to put F_x into the equation, thinking that maybe I needed to get F_y from the net forces. This led me nowhere either (I was grasping at straws here).


I am obviously going about this the wrong way. Since it involves the work on a spring, I know that I must use the W = -1/2k(b²-a²) formula. I just have no idea how to get the value of b. What concept(s) am I missing here?

 

[Fightfish]

There is no need to resolve forces. The crux of the question lies within this: by what length has the spring been stretched? It then results in a clear application of that formula you are brandishing around.
Hint: It involves the right angle triangles you've been talking about and Pythagora's theorem of course

 

[Jim01]

There is no need to resolve forces. The crux of the question lies within this: by what length has the spring been stretched? It then results in a clear application of that formula you are brandishing around.
Hint: It involves the right angle triangles you've been talking about and Pythagora's theorem of course

I know I'm thickheaded because I'm still not seeing it. If I set the opposite to y, the adjacent to L/2, this gives me a hypotenus of the square root of y²+(L/2)².

So at rest, the spring length is L. So a=L and b= the square root of y²+(L/2)². However when I plug this into my formula, it still doesn't work. I end up with

W = -1/2 k [(square root of y²+(L/2)² - L²]

= - 1/2k [y²+(L/2)² - L²]

= - 1/2k [y² - 3/4 L²]

= k(-1/2y² + 3/8 L²)

 

[Fightfish]

This is because you forgot to consider the fact that the spring is stretched on both ends. So the length of the extended spring is $$2\sqrt{y^2 + (\frac{L}{2})^2}$$.