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?
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 Fy, 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]Fx = 0
wx = 0
Fx = L/2
[tex]\sum[/tex]Fy = 0
wy = -mg
N = mg
Fy = -Fcos[tex]\theta[/tex]
wy-N-Fy = 0
Fy = wy-N = -2mg
I attempted to use the Fy value as b in the W = -1/2k(b2-a2) formula but this led me to a wrong answer. I then tried to put Fx into the equation, thinking that maybe I needed to get Fy 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?