What Happens to the Excess Energy in a Compressed Spring?

AI Thread Summary
When a spring is compressed by a constant force, the work done is calculated as Fx, resulting in kx^2. However, the energy stored in the spring is actually 1/2kx^2, indicating that the additional work done is converted into kinetic energy when a mass is attached. As the spring compresses, the force increases from zero to kx, meaning the average force is F/2, leading to the energy equation. When an external electric field is applied, it exerts a constant force, causing the mass to oscillate around a new equilibrium point. The discussion clarifies the relationship between work done, energy storage, and kinetic energy in spring dynamics.
jd12345
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If you apply a force F and compress a spring of spring constant k by x distance what is the work done by us?
I know energy stored is 1/2kx2 but is the work done by us the same?
We are applying a cosntant force F and dispalcement is x so work done by us should be Fx right? And F = kx so work done by us is kx2

Where does the other 1/2kx2 go?
 
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jd12345 said:
Where does the other 1/2kx2 go?
Imagine there was a mass on the end of the spring. If you used the minimum force required to compress the spring, then that force would vary from 0 (at the start) to the maximum value of kX at full compression. The work you'd do would equal 1/2kx2. But if you use a constant force F = kX to compress the spring, then the extra work that you do goes into increasing the kinetic energy of the mass.
 
I agree with DocAl
When you start to compress(or stretch) the spring you start with zero force and increase to final force,F and F = kx. The work done = average force x extension.
If the spring behaves linearly the average force = F/2 therefore work done = kx/2 . x
= 0.5kx^2
 
Okay thank you - but one more doubt
Suppsoe a mass m attached to a spring with charge q. Suddenly electric field is switched on so it will exert a constant force right? So in this case the other 1/kx^2 goes into kinetic energy and the spring oscillates? Have i got it right?
 
That's right. The mass will oscillate about the new equilibrium point (where the net force on the mass is zero).
 

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