Question about potential energy: gravity and a vertical spring

[itchybrain]

Let's say a mass is gently laid on top of a massless spring. The spring compresses.

There is a change in the height of the mass. Therefore, there is a change in the gravitational potential energy: a decrease.

The compressed spring now has potential energy (it has gained energy).

The change in the height of the mass is equal to the spring compression (delta-h in both cases).

Is it fair to say that, because of conservation of energy, the change in gravitational potential energy becomes elastic potential energy?

I.e.: in this particular case, does m*g*(delta-h)=(1/2)*[k*(delta-h)sq] ?

 

[Delta2]

Summary:: Let's say a mass is gently laid on top of a massless spring.

Is it fair to say that, because of conservation of energy, the change in gravitational potential energy becomes elastic potential energy?

I.e.: in this particular case, does m*g*(delta-h)=(1/2)*[k*(delta-h)sq] ?

I.e.: in this particular case, does m*g*(delta-h)=(1/2)*[k*(delta-h)sq] ?

This holds only if delta-h equals the maximum compression of the spring. for delta-h not maximum (that is we are in an inbetween position) there is a kinetic energy term and the correct equation is $$mgx=\frac{1}{2}kx^2+\frac{1}{2}mv(x)^2$$ where x the coordinate of the inbetween position, and $v(x)$ the velocity of the mass at that position.

 

[itchybrain]

Thank you. That does make sense. The kinetic energy must be accounted for.

Maximum compression must be a theoretical case, right? I.e. the spring will always be in the in-between position for any given mass?

If, in the case of maximum compression, m*g*(delta-h)=(1/2)*[k*(delta-h)sq] hold true: then k = 2mg/h .

Which we know is not true, because (from force diagrams) mg = kh ; and k = mg/h.

 

[Delta2]

No maximum compression is not a theoretical case, we can prove that the spring will have a maximum compression . The maximum compression is not the position h for which mg=kh, we can prove this too. It is instead the position h for which mgh=1/2kh^2.

 

[itchybrain]

In the case of maximum compression, is the spring constant (for any given spring) different than in the in-between cases?

I just don't see how k=mg/h (from force diagram) and k=2mg/h (from energy conservation at maximum compression) can both simultaneously be true.

 

[kuruman]

At maximum compression $h$ and the mass at rest, the force diagram correctly gives $k=\dfrac{mg}{h}$. Mechanical energy is not conserved when you gently lower the mass onto the spring because your hand also does non-conservative work on the mass. If you want to conserve energy, it will have to be total energy and you have to account for the work done by your hand.

Your hand force does positive work because it is down and the displacement is also down. The combined spring + Earth force is up until the mass reaches maximum compression where this combined force becomes zero. Lowering the mass gently means that the mass has practically zero constant velocity which means that the down force exerted by the hand always matches the up Earth + spring force. Thus, the work done of the hand on the mass and the work done by the Earth + spring force are equal in magnitude and opposite in direction. The net work done by all the forces is zero, and the mass reaches the maximum compression with zero velocity (the hand sees to that) where it stays at rest.

 

[itchybrain]

Yes.

My assumption that the gravitational potential energy would (conservatively) transfer to elastic potential energy was correct. I just failed to account for the additional energy going into the system ("work-of-the-hand energy"). That accounts for the difference between the 2mgh and the mgh.

Thank you.

 

[kuruman]

My assumption that the gravitational potential energy would (conservatively) transfer to elastic potential energy was correct.

It's not an assumption it's the truth: the force of gravity is a conservative force.

 

[jbriggs444]

It's not an assumption it's the truth: the force of gravity is a conservative force.

I am not sure that I get what you are trying to say here.

Yes, gravity is a conservative force. That means that the work done by the force along a path can always be determined by subtracting the potential at the two ends of the path. That has nothing to do with whether some other external force on the same object (e.g. from a spring) is or is not also conservative.

Like money, energy is fungible. We cannot always track where a particular bit of input energy winds up. Nor do we need to. It is enough that the books balance at the end.

 

[kuruman]

I am not sure that I get what you are trying to say here.

It was a minor point about verb use. One does not assume that the force of gravity is conservative, one knows that it is because there are no instances when it isn't.

 

[jbriggs444]

It was a minor point about verb use. One does not assume that the force of gravity is conservative, one knows that it is because there are no instances when it isn't.

I do not read the assumption being made as being about whether gravity is a conservative force. The assumption we are invited to visit is whether energy is "transferred" from gravitational potential to elastic potential. We need three things for that assumption to hold.

1. That gravity is a conservative force -- so that there is a pool of "gravitational potential energy" to tap.
2. That elastic force is conservative -- so that there is a pool of "elastic potential energy" we can deposit into.
3. There is a transfer from one to the other -- that gravity does work on the object, and that the elastic force extracts work from the object.

Ideally we do not want any other fingers on the scale while we are looking at the energy-in versus energy-out balance. But such is not to be in this case.

With [counter-factually] no other fingers on the scale and no change in kinetic energy, we could have invoked energy conservation and the work-energy theorem and confidently declared that work in from gravity = work out to spring. That equation does not require either gravity or the spring force to be conservative.

 

[kuruman]

I do not read the assumption being made as being about whether gravity is a conservative force. The assumption we are invited to visit is whether energy is "transferred" from gravitational potential to elastic potential. We need three things for that assumption to hold.

1. That gravity is a conservative force -- so that there is a pool of "gravitational potential energy" to tap.
2. That elastic force is conservative -- so that there is a pool of "elastic potential energy" we can deposit into.
3. There is a transfer from one to the other -- that gravity does work on the object, and that the elastic force extracts work from the object.

Ideally we do not want any other fingers on the scale while we are looking at the energy-in versus energy-out balance. But such is not to be in this case.

With [counter-factually] no other fingers on the scale and no change in kinetic energy, we could have invoked energy conservation and the work-energy theorem and confidently declared that work in from gravity = work out to spring. That equation does not require either gravity or the spring force to be conservative.

Now it's my turn not to get why you are saying all this because I agree with all of the above. I think we are probably talking past each other and that is not productive. I liked the FBD with the fingers on the scale though.

 

[Delta2]

In the case of maximum compression, is the spring constant (for any given spring) different than in the in-between cases?

I just don't see how k=mg/h (from force diagram) and k=2mg/h (from energy conservation at maximum compression) can both simultaneously be true.

The spring constant is the same for all the positions. It is the variable $h$ that changes. In one case it is the compression at the position of force balance (mg=kh) and at the other case it is the maximum compression h' for which it hold that 2mg=kh'. From these two equations you can conclude that h'=2h that is the maximum compression is double the compression at the position of force balance.

P.S The forces don't balance at the position of maximum compression (equation $mg=kh'$ is not correct), unless you add an external force, like the force from hand for example ($mg+F_{hand}=kh'$).

P.S2 I think now I understand your logic, you took it to be that for every position h, it is a position of force balance. This might be true but we need a variable (that varies as the position varies) external force like the force from hand, which we shouldn't forget to add in the equation.