I guess it depends somewhat on the setup, but I do see now that the individual molecules are carrying away some of the energy.
Your example supposes that the spring is at a fixed length and that it’s able to maintain that length with the force/tension varying as it is eaten away. Right at the end, you’d have essentially 0 force because the spring had become so weak.
My example is thinking of a fixed force (a 1lb weight). In my scenario, I do see now that you’d lose some potential energy as the spring is dissolved - illustrated by the fact that the weight would move as the spring becomes weaker - but because my scenario maintains a constant force, you’d never eat the spring away to nothing, and there would still be 1 lb of force from beginning to end. The end would happen when the spring’s strength was reduced so far that it could no longer support 1 lb of force and it would snap/crush/bend under the weight.
Even so, as you dissolve the spring, it will stretch (or compress) if you keep the same load on it. So, you are still giving up energy to the solution as it dissolves.
Beowulf yes frequently, drill strings and casing strings are just giant tubular springs and wall loss to corrosion is a concern, you loose wall and don’t change the load the stress in the material rises and then breaks.
In your example the whole spring is supporting X tension
That tension induces say 100 Ksi of stress in the whole material.
Cut the spring in half , the stress in the material is staying the same, but as each spring is now thinner it will only support half of the tension as you say, which does not contradict any thing I said.
Now if you start to slowly take away material from each of those two springs - but they are constrained to hold the same tension, the stress in the remaining material rises as there is less material remaining. Eventually the stress will exceed the mechanical properties of the material and it will break.
If the spring is free to move ( say a spring holding a weight) with a fixed load, as the material gets thinner the stress rises and the materiel will stretch ( Youngs mod plot (stress / strain) )
This implies that the spring is wound, and then the ends are held in that position (say, by some plastic that won’t dissolve). Nowhere does it say that the tension on the spring must remain constant.
So he winds a spring and puts it into the bath if you don’t hold it it unwinds immediately returning all the energy to kinetic energy.
A spring must be restrained if it is going to hold tension
That’s true, it must be restrained.
Say the restraints are in a fixed position.
What can be said about the force on them as the spring dissolves?
(Hint: it’s not constant).
We have a spring we put 100lbs weight under a spring, it extends from 2 inches to 3 inch.
We are going to have to hold that extension at three inches as per the “keep the restraints in a fixed position”
The OP did specify “so the tension is not released” so he was advocating keeping the 100lbs constant, but we can manage.
Hookes law F= k X
F is the tension or force (100lbs) ,
k is the spring constant or stiffness
X is the displacement (one inch).
K is the spring constant , which is a function of the materials’ properties such as Elastic modulus, poisson Ratio, and the physical dimensions of the spring.
When we stretch this spring we also induce a stress into the material. The stress ( force per unit area aka psi) in the material depends how much the material is straining ( strain = % increase in length not the same at the spring stretch unless you are into straight tube springs ) and the youngs modulus of the material. Let’s say we have caused 4000 psi of stress in a material with a yield strength of 6000 psi.
Now we dissolve the material so k is going to reduce as we are reducing the physical dimensions of the spring ( diameter).
If k reduces and we want to keep X constant – then we as the experimenters have to reduce the force down, so we have to remove some of the weight, to keep the extension constant. The is no dissolving atoms carrying tension stress energy or anything away, you must remove some of the load to keep the X the same. If we remove that mass to keep X the same we keep the stress in the material constant
Now let’s keep the force or tension constant at 100lbs
Back to Hooke
F= k X
We dissolve material from the spring, k drops, we are keeping the force constant ( ie not removing any weight) the spring will extend and the stress in the spring may rise as the spring elongates . s the material gets thinner X keeps going up, the stress in the material keeps going up and eventually the stress exceeds the yield strength of the material and the spring breaks. Again no need for the dissolving atoms to carry away any energy , the energy is just simply taken up in the extension of the spring.
Now let’s keep both load and extension the same ( which is what I was envisioning)
We will take a spring and stretch it one inch between two posts that won’t move so X is fixed as well. We have applied 100lbs of force to do this, the post won’t move so the 100 lbs force / tension will remain constant. IF we look at a cross section of the spring say it is 0.025 square inches and has a stress of 4000 psi the cross section is taking 100lbs force.
Now we dissolve material.
The k drops but we are holding F and X constant, things seam to get squirrelly but all that happens is the stress in the material rises
We now reduce the cross section to 0.02 the stress rises to 5,000 psi, we reduce cross section a bit more and the stress rises more ( tension on the spring is staying the same ) eventually the stress in the material exceeds yield strength of the material and the spring breaks.
Absolutely none of the tension in the spring is carried away by the dissolving atoms.
Converts to heat. Pretty much all it can do, if it isn’t released in little bits of metal flying off at some speed.
Take a compressed spring, and encase it in concrete. The spring slowly rusts. It can never “release.” But when it has all rusted away…where did the potential energy go? Released as heat.
Take a compressed spring…and compress it even farther, well beyond the point of nonlinearity and Hooke’s Law. Compress it so tightly it welds into a solid piece of iron. It obviously cannot release. You’ll get heat.
Take a compressed spring, locked in place by pins, and melt it in a furnace (the pins holding it in place are of an alloy that won’t melt at the furnace’s temperature.) The spring never releases. Energy? >> heat.
Mix vinegar and baking soda inside a steel cannister, so it can’t physically expand. No expansion? Chemical energy >> heat.
Drive a car real fast at a giant concrete wall… Watch ocean waves smash against a concrete sea-wall… Hammer an iron horseshoe on an anvil… Heat.
Kinda off topic…but I must ask. If a compressed (wound) spring was anchored at both ends against a pair of piezo -electric crystals, would it generate current?
Or, another way to ask, why does it take a continual input of energy to compress a small spring, and keep it compressed, with your fingers, say? (I know I’m missing something elemental here)(how to get that energy back, is what I think Im asking)
oh I checked a bit of google before posting, I seem to be behind the knowledge curve…:smack: dang…another idea bites it.
Yes, but the OP is bad at physics and your interpretation turns the question into something akin to the “plane on a threadmill”-question. Lots of interpretations with varying degrees of realism and a totally uninteresting discussion where people think they agree on the premisses, but actually don’t.
The OP quite obviously doesn’t mean to have a situation where we remove energy by external manipulation, but wants an explanation for a situation that can be naively interpreted as as “energy vanishes”-scenario.
So we apologise for not being rigorous and start by stating “well as the spring dissolves, and the end points of the spring don’t move, k will change and the tension will change, as that’s where the energy is stored”. But your approach does not work in the “near real world” where the OP wants his answer. If the spring is under tension every atom in it will carry some extra energy, and that energy will be carried away when that atom leaves the spring. Manipulating the load on the spring can’t remove all the tension carried by particular atoms and distribute it among the others.
Could you give an insight as to what the OP meant when he said "so that the tension in the spring is not released "
It’s a spring, it is storing energy, it is under tension (total load), the material in the spring is under a certain stress and the spring has been extended. Those are what you need to know to look at the problem.
We take a spring, hang it from a post and hang a weight off it then put the whole thing into a bath of acid. Can we have an insoluble support post and insoluble weight, that’s not unreasonable. The spring us under a certain tension, the spring material is in a certain level of stress, there is energy being stored in the spring, I think we can all agree on that. Don’t confuse stress in the material with tension.
If you draw a stress / strain graph as the spring stretches as we hang the weight, basically we plot the stress in the material as we strain (stretch)it, the area under the graph is strain energy per unit volume. In this case let us say it is a nice straight line from the origin to what ever stress and strain numbers we hit, so the area of that triangle is the strain energy per unit vol. So a straight lne from the origin of the graph to say 5 on the Y (stress) axis and 0.3 on the strain. Area under that is the strain energy.
Now we dissolve part of the spring, we now have less spring material.
The weight is the same, but less spring material to support it, so the stress rises, and the spring stretches. Of if you wil you are hanging the same wight off a thnner spring so it stretches.
If we continue to plot that on our stress strain graph our stress will rise quite dramatically ( with cross sectional loss) and the strain goes up as it stretches. We will not imagine spherical cow materials that don’t increase in length with an increase in stress. We have continued to plot our line upwards to the right, that additional area under the stress strain graph is the additional strain energy per unit volume. As we have less volume ( we lost material) the net effect is the total strain energy (hence elastic energy) is going to stay the same.
As you dissolve more material, the stress rises, the strain goes up, we have even more strain energy per unit volume of material, but les material so the net energy stays the same.
Eventually the stress level in the spring material exceeds the ultimate tensile strength of the material and the spring breaks and you get all you energy as the weight drops and the spring pings back.
Any spring system , torsional , clock springs, compression, steel bars ,leaf springs can all be thought of in exactly the same way as in this scenario, it’s a spring under a load , which stretches the spring ( strain in the material) which stores energy and causes a stress in the material . If you think through them and at any time think ‘ well that tension must be decreasing’ , that’s fine, there can be all sorts of ways of the load being removed, but pull it straight back to the weight hanging from the spring model. If you reduce the tension on the spring you must have removed some of the mass hanging which had some gravitational potential energy and the reduction in spring energy is accounted for by taking that weight away .
The key thing to understand is the OP has no idea what the word “tension” means in physics. or engineering. He’s used tension and energy as synonyms. You and I and several others know they’re not even remotely synonyms.
So we first have to surmise / guess which half-formed idea is in the OP’s head. And then explain, using only simple words carefully defined, what he really intended to ask and then using those same simple words give the answer to that question.
IMO … he meant: Compress a spring & lock the ends in place. Dissolve the spring but not the holding mechanism in acid. Something (force, energy, power, oomph?? I [OP] don’t know which) was injected into and then stored in the spring when I compressed it. Where did that something, whatever it is, go?
That is the question. IMO our task is to educate about what *was *injected into the spring, how the spring reacted to store that something and why, and last of all what happens once the acid gets involved.
OK
Can we agree if we take an extension / tension graph, Delta L ( extension) on X axis and tension(force) on Y axis, and as we apply the force stretching the spring we move from the origin to your point of 10N and 0.01 extension. Let’s call that point A (0.001,10), the gradient of that line is k =1000nM
Can we also agree the area under that line we have drawn from O ( 0,0) to A ( 0.001 , 10) is the energy stored in the spring. If we can’t look here.
That’s for an ideal spring, in real life, that line is probably not a nice straight line and what we are doing is integrating under the line as the tension and extension rise to get energy. As we have an ideal spring, the area is a triangle and the energy works out to be a nice 1/2F Delta L^2 ( which is ½ k X^2, all the same). But the thing to remember is the area under that line is the energy in the spring. In our starting case 0.05J
Now we dissolve the spring as per your numbers we move to point B which is coordinates (0.0125,10) , basically a horizontal move to the right.
As we move to the right we are adding area under our graph , so energy is going into the spring as it stretches. Where does that come from, well in the case of the suspended weight so Gravitational potential Energy is being given up and going into spring energy.
The total area under the graph is the energy stored. So the energy in the spring is the original area under the line to point A ( ½ kX^2) PLUS the area under the rectangle created from going from A to B , which is another 10 x 0.0025 ( 0.0125m -0.01m) or the 0.025J of work you calculated
So total energy in our spring is 0.05+ the extra 0.025 work so 0.075J. The area under the graph going from O(0,0) to A to B.
But what about the fact the spring constant is 800 and according to our calculation the energy in a spring is ½ K X^2. You have to remember that that was derived from integrating under the line on the extension/tension graph. So the formula (1/2 kX^2) assumes we started with a spring of k 800 and moved from (0,0) along the line k=800 to point B (0.0125,10) doing work and accumulating energy in the spring as we move along that line from O to B. The final area under that triangle is indeed 0.0625J
Compare the area under the graph going from O to A to B , it is indeed more than going straight from O to B, so the dissolved spring has more energy than just stretching an ideal spring of k= 800 from O to point B. they do not have the same energy.
Now we are at point B. Let’s relax the spring and go all the way back to (0,0) . Indeed we will follow the straight line from B back to O, after all we all agree the new spring constant is 800, because it is a thinner spring and all. Now the energy we give back is the area under the O to B triangle ( 0.0625J), but the spring has 0.075J in it. That difference in energy is given out, as many have said, as heating of the spring.
Important to note, when we are at B, following the dissolve spring route (O A B) the energy in the spring is 0.075 (O to A to B area) , we only give up that energy as we relax the tension and extension and move back along B to O line, with the delta in the areas being given up as heat. The Energy is not carried away when the spring is dissolving Ie moving from A to B, in fact as the string stretches we are getting more energy into the spring, 0.025J of work in this case.
It should be noted that the OP doesnt mention that the anchoring points of the spring are fixed at either end. Doesnt that create two completely different calculations? One for a fixed spring, and one for an unfixed (anchored) spring.
I would guess that a spring fixed only at one end, would indeed break as it gets thinner in a acid bath, as the load is constant, whereas a fixed spring, blocked in place, would just give off heat and not snap.
None of this makes any difference.
You can add complications all you want, but the simple fact is that the energy in a spring is stored in the intermolecular bonds. If you remove even one molecule, you have reduced the amount of energy stored in the spring, and that energy ends up as heat.
Do you agree the area under the tension extension graph as you stretch the spring is the energy stored in the spring?
This is not a complication but a very simple way of describing the energy in a spring - see link high school physics.
(how to get that energy back, is what I think Im asking)