Potentials which depend on positions of two particles

AI Thread Summary
The discussion centers on the nature of conservative forces, particularly in the context of springs and two linked masses. It is established that the force exerted by a spring is conservative because it depends solely on the configuration, allowing for the work done to be recoverable by reversing the deformation path. The potential energy associated with a spring belongs to the spring itself, as it is stored in the interparticle forces within it, regardless of whether masses are attached. When considering two masses linked by a spring, energy is conserved for the entire system, as the total kinetic energy and potential energy of the spring remain constant. Understanding these principles clarifies the distinction between conservative forces and non-conservative forces like friction.
BrunoIdeas
Messages
14
Reaction score
0
Hello. I have been asked if it is correct to say that the force of a spring is conservative.

So, for a mass attached to a spring on the roof it's OK.
Now consider two masses linked by a spring. Now the potential will depend on both coordinates. It is now not so clear to me any of the properties of conservative forces.
Closed loop integral = 0 and so forth. Neither the fact rot F = 0 => consrevative

Any comments will be appreciated.
 
Physics news on Phys.org
It's conservative because any force that depends only on configuration is always conservative. That's because if the force depends only on configuration, then the work done by that force will always be expressible in terms of a potential energy function, via the line integral of that force over deformations of the configuration. The simplest way to see this, especially for 1D deformations like that of a spring, is to simply reverse the path-- if the force depends only on the 1D deformation, then you can get all the work back just by reversing the path of the deformation. If you can always get the work back, it's conservative. Note this doesn't require that either end of the spring be locked down, the work done on the endpoints depends only on the force and the change in the stretch of the spring, because that change is the difference in work on the two ends (given that the force on the two ends is equal and opposite, as per Newton's third law).
 
Ken G said:
It's conservative because any force that depends only on configuration is always conservative. That's because if the force depends only on configuration, then the work done by that force will always be expressible in terms of a potential energy function, via the line integral of that force over deformations of the configuration. ... then you can get all the work back just by reversing the path of the deformation. If you can always get the work back, it's conservative. Note this doesn't require that either end of the spring be locked down, the work done on the endpoints depends only on the force and the change in the stretch of the spring, because that change is the difference in work on the two ends (given that the force on the two ends is equal and opposite, as per Newton's third law).

I am so grateful for your answer. I particularly apreciate you've given me the intuiton about understanding conservative forces as position dependent and the notion of getting work back. Using this one evidently sees why friction is not, and probably why it conflicts with time invariance or reversal of Newton's eqs.

I however do not fully understand what is after the bold text.

Once again thanks.
 
I think you understand completely, the stuff after the bolded text was in there just because it sounded like you could understand this better if one end of the spring was locked down, but if both ends were free to move around it wasn't so clear. I was trying to show that it doesn't matter if either end is locked down, what matters is that if the configuration of the spring is all you need to know the force it is applying at both ends, then the net work done only depends on the configuration (i.e., the stretch) of the spring, and re-extracting that work is also only dependent on returning to the original configuration, even if the spring has waffled around in the mean time.
 
Fine. Thanks, I get that. Now I would like to ask what about energy conservation in such cases.
1) Consider a mass attached to a spring hooked from the ceiling with no gravity (or with it, it does not matter). We say that energy is conserved because forces are conservative.
Question: Does the potential energy belong to the spring, to the mass, or to the system.

I understand this question may be misleading but answer me as if I knew nothing.


2) Consider the two masses linked by a spring. Is energy conserved? I understand/guess that for the system it it, not for the masses, but I cannot give good arguments. Could you help me?


Thanks Ken G, and everyone.
 
BrunoIdeas said:
Question: Does the potential energy belong to the spring, to the mass, or to the system.
The potential energy belongs to the spring. It is stored in interparticle forces within the spring. If you remove the mass completely, all the potential energy is still in that spring, as anyone has discovered to their chagrin if they did remove the mass suddenly!
2) Consider the two masses linked by a spring. Is energy conserved? I understand/guess that for the system it it, not for the masses, but I cannot give good arguments. Could you help me?
Energy is conserved if there are two free masses linked by a spring. The sum of the kinetic energies of the two masses, plus kx2/2 for the spring, will always be the same. I'm not sure what aspect of this is where your question is focused.
 
The rope is tied into the person (the load of 200 pounds) and the rope goes up from the person to a fixed pulley and back down to his hands. He hauls the rope to suspend himself in the air. What is the mechanical advantage of the system? The person will indeed only have to lift half of his body weight (roughly 100 pounds) because he now lessened the load by that same amount. This APPEARS to be a 2:1 because he can hold himself with half the force, but my question is: is that mechanical...
Some physics textbook writer told me that Newton's first law applies only on bodies that feel no interactions at all. He said that if a body is on rest or moves in constant velocity, there is no external force acting on it. But I have heard another form of the law that says the net force acting on a body must be zero. This means there is interactions involved after all. So which one is correct?
Thread 'Beam on an inclined plane'
Hello! I have a question regarding a beam on an inclined plane. I was considering a beam resting on two supports attached to an inclined plane. I was almost sure that the lower support must be more loaded. My imagination about this problem is shown in the picture below. Here is how I wrote the condition of equilibrium forces: $$ \begin{cases} F_{g\parallel}=F_{t1}+F_{t2}, \\ F_{g\perp}=F_{r1}+F_{r2} \end{cases}. $$ On the other hand...
Back
Top