Elastic Potential Energy...

[sps27]

I have a really weird doubt about Elastic Potential energy for a spring system, U = (1/2)kx^2. I also know it can never be negative but my doubt is regarding the relationship of Restoring Force and displacement which is given by F = -kx , force exerted by the spring on the mass 'm' is always opposing the displacement 'x' hence we use the negative sign.

But what about work done.

If the spring is compressed, then Work Done (by the spring on the mass m) = Average Force * Displacement, so in case of compression we have W = -kx/2 (-x) = (1/2)kx^2. For compression the displacement could be interpreted as (-x) from equilibrium position.

For extension, we will then have displacement = (+x), from equilibrium position. So what will be work done by spring in case of extension. W = -kx/2 (x) = -(1/2)kx^2. So work done by the spring in case of extension will be negative.

So should not the Elastic Potential Energy of spring (equated to work done) be also -ve in case of extension of spring? This doubt has been eating me alive. I know I am missing some key concept here. Any thoughts on this?????

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[milski]

You are missing a negative sign in one case. When you change from extension to compression both the direction of the force and the direction of the displacement change, thus preserving the same sign for the work.

Somewhat related note - potential energy is always relative to something, the choice of zero is arbitrary. That means that you can have negative potential energy, although that's rarely practical for springs.

 

[sps27]

You are missing a negative sign in one case. When you change from extension to compression both the direction of the force and the direction of the displacement change, thus preserving the same sign for the work.

Somewhat related note - potential energy is always relative to something, the choice of zero is arbitrary. That means that you can have negative potential energy, although that's rarely practical for springs.

Thanks, I appreciate your response. But I thought that Hooke's law F = -kx is the same regardless of compression or extension. When a spring gets compressed the force exerted by the spring on the mass will be opposite to the direction of displacement of mass and when a spring gets extended, once again the force will be opposite to the direction of displacement. Isn't that the case?

 

[sps27]

You are missing a negative sign in one case. When you change from extension to compression both the direction of the force and the direction of the displacement change, thus preserving the same sign for the work.

Somewhat related note - potential energy is always relative to something, the choice of zero is arbitrary. That means that you can have negative potential energy, although that's rarely practical for springs.

I think I understand now. The work done can be +ve or -ve depending on direction of displacement as I mentioned above -- but "energy which is more of a capacity to do work" will always be positive because it drives work, and not the other way round......does that sound right???

 

[milski]

Thanks, I appreciate your response. But I thought that Hooke's law F = -kx is the same regardless of compression or extension. When a spring gets compressed the force exerted by the spring on the mass will be opposite to the direction of displacement of mass and when a spring gets extended, once again the force will be opposite to the direction of displacement. Isn't that the case?

Yes, the displacement and the direction of the force will have opposite signs. Then the work (F*x) will always be negative, which is correct - when the spring is being compressed/expanded it is doing negative work and its potential energy is increasing.

I think I understand now. The work done can be +ve or -ve depending on direction of displacement as I mentioned above -- but "energy which is more of a capacity to do work" will always be positive because it drives work, and not the other way round......does that sound right???

No, the work is -ve in both cases. In one direction it is -kx * x = -kx², in the other it is -k(-x)*(-x)=-kx².

 

[anticarbon]

I think milski answered the question quite well.

If the signs are confusing, I think just using intuition is another good option. Whether you compress or extend the spring, a force is needed, and since the direction of the spring movement will accordingly change, the work done on the spring is positive and by the spring is negative.

 

[anticarbon]

I think milski answered the question quite well.

If the signs are confusing, I think just using intuition is another good option. Whether you compress or extend the spring, a force is needed, and since the direction of the spring movement will accordingly change, the work done on the spring is positive and by the spring is negative.

 

[sps27]

Thank you guys! Much appreciated. I think I got it now. What started this confusion was a paragraph in TBR Physics book 1, page 227 which is quoted below. So there was not much clear distinction between Potential energy and Work done (on the spring, by the spring) here and I got confused. But as you clarified above, the work done by the spring = -(1/2)kx^2 for both compression and extension and work done on the spring = (1/2)kx^2 once again for both cases.

" Because the restoring force of the spring system increases as it gets further compressed, the work required to compress a spring increases the more the spring gets compressed. The work needed to compress a spring is found using the standard W = F(average) x displacement, where F(average) is -kx/2 and the displacement is -x. The average force is -kx/2 because the force goes from 0 to -kx as its deformed, so the average force during deformation is -kx/2. The work done on a spring becomes potential energy, which means that the potential energy of a spring system can be found using Eqn.... U(spring) = (1/2)kx^2 ".