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

  • Thread starter Jim01
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  • #1
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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(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

Fx= 0

[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?
 

Answers and Replies

  • #2
954
117
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
 
  • #3
36
0
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)
 
Last edited:
  • #4
954
117
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 [tex]2\sqrt{y^2 + (\frac{L}{2})^2}[/tex].
 

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