Does Compressing a Spring Decrease Its Potential Energy?

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Compressing a spring increases its potential energy, as the work done on the spring by an external force, like a thumb, adds energy to the system. The equation Fx(x) = -dU(x)/dx applies to conservative forces, indicating that potential energy decreases when a conservative force acts in the direction of displacement. However, the thumb's force is not conservative, which complicates the analysis. The potential energy stored in the spring is independent of any mass attached to it; if the mass is removed, the energy remains in the spring's mechanism. Understanding the distinction between conservative and non-conservative forces clarifies the energy dynamics in this scenario.
Elias Waranoi
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I have just read chapter 7.4 FORCE AND POTENTIAL ENERGY in Sears and Zemansky's university physics 14th edition. There they show that a conservative force always acts to push the system toward lower potential energy in a one-dimensional motion with the equation Fx(x) = - dU(x)/dx. As I understand this it is saying that when a force is pushing in the same direction as its displacement then the potential energy is lowered. I felt like it made sense but after thinking of different scenarios I stumbled upon something that I couldn't understand.

If positive x is to the right and I have a metal spring with it's right side attached to a wall. If I compress the spring with my thumb, my thumb experiences a force to the left from the metal spring while it is moving/pushing to the right. Since the force is in the other direction than the displacement Fx(x) = - dU(x)/dx gives an increase in potential energy.

But what happens if we look at the spring? The spring is feeling the force from my thumb, a force to the right while it being compressed/displaced to the right. The force and displacement on the spring is in the same direction so according to my understanding of Fx(x) = - dU(x)/dx the potential energy should decrease? That doesn't make sense.

According to all this my thumbs potential energy increases but the metal spring's decreases, what have I understood wrong? Writing this I'm thinking that it might be that my thumb is not a conservative force in the system but I'm not sure.

(Was going to post this in the homework board but then I'd have to use the homework template and this doesn't feel like a homework template question, I'm guessing this is the right place to post?)
 
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Elias Waranoi said:
Since the force is in the other direction than the displacement Fx(x) = - dU(x)/dx gives an increase in potential energy.
Yes, the potential energy of the spring-mass system ! Work is being done ON the system.

It's like pushing a ball up the hill: you perform work to do that and your work can be converted to kinetic energy of the ball when you let it roll down again.
 
Elias Waranoi said:
According to all this my thumbs potential energy increases but the metal spring's decreases, what have I understood wrong? Writing this I'm thinking that it might be that my thumb is not a conservative force in the system but I'm not sure.

The equation you are using here is for the energy of a particle under a conservative force. The more the spring is compressed, the greater the potential energy of the particle.

The potential energy in the spring is the potential energy of the particle.

If you try to turn the system round and treat the spring as a particle under the force of your thumb, you will get a mess.
 
PeroK said:
The equation you are using here is for the energy of a particle under a conservative force.
So the equation is describing the change in potential energy for a mass that is affected by a conservative force? And since me pressing my thumb on a spring is not a conservative force I just get trash when trying to turn the system around, is this correct?

PeroK said:
The potential energy in the spring is the potential energy of the particle.
Just to clarify, are you saying that if a particle affected by the force from a spring has 100 joules in potential energy, the spring also has 100 joules in potential energy?
 
Elias Waranoi said:
So the equation is describing the change in potential energy for a mass that is affected by a conservative force? And since me pressing my thumb on a spring is not a conservative force I just get trash when trying to turn the system around, is this correct?Just to clarify, are you saying that if a particle affected by the force from a spring has 100 joules in potential energy, the spring also has 100 joules in potential energy?

Yes, all the PE is stored in the spring's mechanism, independent of the mass. If you hold the spring and remove the mass, then the mass loses all its PE!

You might then ask where all the PE goes if there is no mass? That's where the assumption about a massless spring is exposed. It must go to KE of the spring.
 
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For simple comparison, I think the same thought process can be followed as a block slides down a hill, - for block down hill, simple starting PE of mgh to final max KE 0.5mv^2 - comparing PE1 to max KE2 would result in finding the work friction did through the process. efficiency is just 100*KE2/PE1. If a mousetrap car travels along a flat surface, a starting PE of 0.5 k th^2 can be measured and maximum velocity of the car can also be measured. If energy efficiency is defined by...

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