# Spring Energy Thought Experiment

1. Jul 3, 2014

### blintaro

Hi guys,

I work at the tutoring center of my university, and recently a student came in with a question that has been troubled me ever since.

Say you have a certain mass M a distance Δχ above a massless spring. When asked how much energy this system has, you would say it has only gravitational potential energy related to the mass's distance above your P = 0 line, as the mass is not moving, and the spring itself is at rest in equilibrium.

After you drop the mass, considering the precise moment where it latches inelastically onto the spring, wouldn't the mass now not only have that original potential energy (now part converted to kinetic) but also spring energy related to the distance of the latched mass from the new equilibrium point? So have we somehow "added" energy? (Further: would't the mass accelerate faster than g downward toward the new equilibrium point?)

2. Jul 3, 2014

### Nathanael

Wouldn't the kinetic energy gained not completely account for the change in potential energy? (I don't understand why it is assumed that it would)

I would imagine that the amount of spring energy gained would be the difference between ΔGPE and ΔKE

(The spring is applying a force on the object the whole time that the object compresses the spring, so why would all of the ΔGPE become KE?)

3. Jul 3, 2014

### blintaro

Hey Nathanael, thanks for the reply.

Well I was trying to think of exactly the moment when the mass became attached to the spring but has not yet "pushed down", so the moment the mass and the spring become a single system.

It seems like after the mass and spring become a single system, the spring has a new equilibrium point lower to the ground than it originally had, so wouldn't the spring would apply a force in the direction of the new equilibrium, causing the mass to actually accelerate downward? It seems like once this mass reaches the new equilibrium point (considering this new point to have zero potential energy and zero spring energy) it would have a velocity greater than it would have if it had just fell the same distance, which according to conservation of energy cannot happen... Sorry if this reply is a bit rushed and doesn't answer your question, gotta go catch a train but I will be back! I appreciate your time and effort!

4. Jul 3, 2014

### Nathanael

Ah ok, sorry, I misunderstood you originally.

This is where it seems wrong to me. The equillibrium point would be lower, but why does that mean that the spring would apply a force downward?

It seems to me that you're treating the new equillibrium point as if it were the new "unstretched/uncompressed length" of the spring, but the "unstretched/uncompressed length" of the spring is still what it oringially was (and so, in my eyes, the spring wouldn't accelerate the object downwards).

(I'm not very good at physics, so sorry if I'm not helping, but I still enjoy thinking/talking about it.)

5. Jul 4, 2014

### blintaro

On the contrary sir! I was just wondering what others would think about it.

The point about the mass accelerating toward the equilibrium seems wrong to me too because it's just ridiculous to think you can drop something on a spring and the impact won't slow the mass down, but actually speed it up... But then perhaps that intuition only comes from the idea of momentum, where a moving mass collides inelastically with a static mass and together they must subsequently have a slower velocity. But the spring in this case is "massless".

It's been awhile since I was in the class, but I thought that a spring system always tends to its equilibrium point with force F=-kx. So the new "unstretched" length on the spring/mass system would be at the new equilibrium point...

I definitely think something is wrong in my thinking because it just doesn't feel right.

6. Jul 4, 2014

### ModusPwnd

If I understand your scenario right...

The mass in motion does indeed have three forms of energy to be considered. Its kinetic, its spring potential and its gravitational potential. If the mass is dropped from a point where p=0 to a point where, for example, p=-1 then the mass kinetic and spring potential must equal 1. If the mass is dropped in the same fashion without the spring then the entire energy amount of 1 is in kinetic. If the mass becomes attached to the spring sometime after falling then at that moment the spring will slow down its acceleration in such a way that the kinetic and spring potential equal the original gravitational potential. The energy that gets tied up in spring potential is "robbed" from the kinetic. This can also be seen from F=mg-kx=ma where an additional "F" (-kx) in the opposite direction of "g" lowers the net "a".

Does that provide any insight?

7. Jul 4, 2014

### Nathanael

Ah, ok. I don't quite understand this concept of massless. How can the spring even have potential energy at all if it's massless? Because, if it's massless, (to me it seems that) it shouldn't be able to apply a force on the object, because then an equal and opposite force is on the spring. But how can there be a force on the spring if it's massless? (F=ma=0)

"Massless" is indeed a peculiar concept to me.

You say "a spring system always tends to its equilibrium point." But that doesn't necessarily mean that the spring is what pulls it down towards the equilibrium point (right?).

That makes sense to me that it's the "unstretched" point of the spring-mass-system... but the spring itself shouldn't be pulling down (because the spring itself is still compressed relative to it's own "unstretched" length)

I apologize if I'm bothersome. Perhaps I should be reading up on the subject. (But... it's far more fun to learn through discussion, haha)

8. Jul 4, 2014

### Nathanael

I don't understand why you need to wait for the mass to touch the spring for it to be considered a single system. It seems perfectly valid to me to consider it a single system before they touch, and on a similar note, it seems perfectly valid to me to consider them seperate systems after they've touched.

So even while the mass is falling (and hasn't touched the spring) the equillibrium point of the system is where ever it is. Yet, the spring is not contracting towards that equilibrium, because the spring (as a seperate system) has a different equillibrium position.

(I'm assuming by "spring energy" you're speaking of potential spring energy? because I know of no other kind)
(And I assume when you say "potential energy" you're speaking of gravitational?)
This part really throws me off, why would the spring have zero potential energy? I understand that this positiion is where the system finds equillibrium, but there would still be potential (spring) energy, right?

As I said in my previous post, I've yet to really learn about equillibrium, but if I had to guess I would say that zero potential energy is not a requirement of equillibrium.

Let's take another mini thought experiment:
If you "turned off" gravity what would happen to the system? Wouldn't the spring expand (towards it's own equillibrium length) pushing the mass up in the sky, giving it kinetic energy? Isn't this kinetic energy a manifestation of the potential spring energy? If the spring energy were actually zero, how would the mass gain kinetic energy?

(A pedantic sidenote: I think EVERYTHING (that's not "attatched") would float up if gravity were turned off, because the normal force would likely accelerate objects up for a brief amount of time (I'm guessing). But I don't think this has much relevance because in the above paragraph the mass would gain additional kinetic energy from the spring, moreso than an identical mass that is not resting on a spring)

Then again......... Everything I just said could probably be undermined by the fact that the spring is "massless" (I don't know if it is or not because I can't quite conceptually grasp "massless")

9. Jul 4, 2014

### olivermsun

The spring is attached to something. If you're compressing a 'massless' spring between a heavy mass and the floor (say, in a scale), then the mass is pushing down on the spring which is pushing down on the floor, and the floor is pushing back up on the spring and the spring is pushing back up on the mass.

All the potential energy in the spring is stored in the compression of the spring according to Hooke's Law, F = -kx.

10. Jul 4, 2014

### nasu

Yes, it's not right.:)
The spring force and elastic potential energy are due to the material of the spring being distorted.
So the force and energy depend on how much is this distortion, the compression or stretching of the material itself. It has nothing to do with the equilibrium position due to other external fields, like gravity.
So when the falling body latches to the spring, the spring is still un-distorted, so there is no elastic force. The force starts to appear as the spring is compressed.

The displacement from the equilibrium position does not give the elastic force (as F=-k*x) alone but the restoring force, which is the balance between gravity and elastic force.

Assuming that the equilibrium is somewhere at distance d below the un-compressed position,
we will have
kd=mg.

If the body is at x above this position, the spring is compressed by d-x so the elastic force will be
Fe= -k(d-x) -upward
and the weight will be mg - downward.
So the net force will be
mg-k(d-x)=kx - downward, towards the equilibrium position. This is the restoring force.

11. Jul 4, 2014

### blintaro

Hello everyone who posted, thank you very much I've read all your replies and I think I understand what's wrong.

So yeah, I was equating the state of zero spring potential energy to be the state of force equilibrium in the spring. So basically a false conflation of force and energy. I was thinking that at equilibrium of the spring, the entire energy of the system would be kinetic, but this is not so, it will definitely have spring potential at this point. In fact, would it be correct to say at no point will the spring/mass system have entirely kinetic energy?

Anyways, thanks everyone! I drew a messy graphic while figuring this out that I'm including here for your possible viewing pleasure.

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