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## Homework Statement

The ends of a relaxed spring of length L and force constant k are attached to two points on two walls seperated 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(b

^{2}-a

^{2}) (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.

[tex]\sum[/tex]F

_{x}= 0

w

_{x}= 0

F

_{x}= L/2

F

_{x}= 0

[tex]\sum[/tex]F

_{y}= 0

w

_{y}= -mg

N = mg

F

_{y}= -Fcos[tex]\theta[/tex]

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(b

^{2}-a

^{2}) 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(b

^{2}-a

^{2}) formula. I just have no idea how to get the value of b. What concept(s) am I missing here?