# Spring energy **URL:** \ **Category:** Factual Questions\ **Created:** [May 20, 2015, 4:41pm UTC](https://boards.straightdope.com/t/spring-energy/720433 "2015-05-20T16:41:44Z")\ **Posts on this page:** 20\ **Page:** 2
**Author:** ![Machine\_Elf](https://avatars.discourse-cdn.com/v4/letter/m/82dd89/32.png) [@Machine\_Elf](https://boards.straightdope.com/u/Machine_Elf)\ **Post date:** [May 21, 2015, 7:22pm UTC](https://boards.straightdope.com/t/spring-energy/720433/21 "2015-05-21T19:22:45Z")
> [@naita](#): > > Well it all depends on the volume of water in the fall. Add more volume to the falls and a smaller percentage evaporates. But the falling water also gives up a lot of its energy to the air via aerodynamic drag. If this weren’t true, then you could conceivably have supersonic water impacting at the bottom, at which point you might see a significant temperature rise. ---
**Author:** ![am77494](https://avatars.discourse-cdn.com/v4/letter/a/d2c977/32.png) [@am77494](https://boards.straightdope.com/u/am77494)\ **Post date:** [May 21, 2015, 9:18pm UTC](https://boards.straightdope.com/t/spring-energy/720433/22 "2015-05-21T21:18:21Z")
One of my pet peeves about physics and chemistry education is that they are taught to students without any reference to real life (Physics teachers commonly make up spring constants or massless springs and Chemistry teachers deal exclusively in moles when it comes to reactions. Students end up having very little perspective of the magnitudes or scales involved. Back to OP : Take a real spring - for example PAC-5x2.0x900 from the catalog here ([Page 15](http://www.racingsprings.com/Multimedia/www.RacingSprings.com/Files/PAC_shock_Master_Catalog_2014.pdf)) If you depress this spring 20% then the energy stored in it is ~50 Joules If you dissolved this spring in Sulfuric Acid the energy released is ~1600 kilo Joules For comparison a gallon of gasoline contains ~ 130000 kilo Joules ---
**Author:** ![naita](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/naita/32/5862_2.png) [@naita](https://boards.straightdope.com/u/naita)\ **Post date:** [May 21, 2015, 9:24pm UTC](https://boards.straightdope.com/t/spring-energy/720433/23 "2015-05-21T21:24:45Z")
> [@Machine\_Elf](#): > > But the falling water also gives up a lot of its energy to the air via aerodynamic drag. If this weren’t true, then you could conceivably have supersonic water impacting at the bottom, at which point you might see a significant temperature rise. I think someone should just go out and do the experiment. Just don’t fall in when measuring the temperature at the top of the waterfall. And take a safe route down. ---
**Author:** ![Chronos](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/chronos/32/134_2.png) [@Chronos](https://boards.straightdope.com/u/Chronos)\ **Post date:** [May 21, 2015, 9:30pm UTC](https://boards.straightdope.com/t/spring-energy/720433/24 "2015-05-21T21:30:50Z")
But that just means that you’re adding the heat to the air, instead of adding it to the water at the bottom. The effect is the same. ---
**Author:** ![naita](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/naita/32/5862_2.png) [@naita](https://boards.straightdope.com/u/naita)\ **Post date:** [May 21, 2015, 9:32pm UTC](https://boards.straightdope.com/t/spring-energy/720433/25 "2015-05-21T21:32:47Z")
> [@Chronos](#): > > But that just means that you’re adding the heat to the air, instead of adding it to the water at the bottom. The effect is the same. Not if the effect we’re looking for is hotter water at the bottom. ---
**Author:** ![am77494](https://avatars.discourse-cdn.com/v4/letter/a/d2c977/32.png) [@am77494](https://boards.straightdope.com/u/am77494)\ **Post date:** [May 21, 2015, 9:35pm UTC](https://boards.straightdope.com/t/spring-energy/720433/26 "2015-05-21T21:35:11Z")
Here are the calcs to back up: 1\> Potential energy stored in spring : a\> Spring constant(K) of spring = 16.1(see Spring Rate) x 9.81 / 1e-3 = 160000 N/m So energy stored for 20% (1 inch or 25.4 mm compression) = 0.5x16000x25.4e-3^2 = 51.6 J ~ **50 J** 2\> Reacting spring with H2SO2 Moles in spring = Mass / Molecular weight = 870 / 56 ~ 16 mol Fe + H2SO4 —\> FeSO4 + H2 [Heat of formation](http://en.wikipedia.org/wiki/Standard_enthalpy_change_of_formation_(data_table)) of H2SO4 = -814 kJ/mol Heat of formation of Fe2SO4 = −929 kJ/mol Okay - I am approximating the heat of reaction neglecting heat of solution, temperature etc. etc. 929-814 = 115 kJ/mol is released in the reaction. I’ll approximate this to 100 kJ/mol So heat released by dissolving this spring = 16 mol x 100 kJ/mol = **1600 kJ** ---
**Author:** ![hondodoc](https://avatars.discourse-cdn.com/v4/letter/h/bcef8e/32.png) [@hondodoc](https://boards.straightdope.com/u/hondodoc)\ **Post date:** [May 26, 2015, 2:23pm UTC](https://boards.straightdope.com/t/spring-energy/720433/27 "2015-05-26T14:23:07Z")
But I don’t see how the potential energy is included in the math and how it would be different if not compressed. ---
**Author:** ![naita](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/naita/32/5862_2.png) [@naita](https://boards.straightdope.com/u/naita)\ **Post date:** [May 26, 2015, 3:15pm UTC](https://boards.straightdope.com/t/spring-energy/720433/28 "2015-05-26T15:15:43Z")
> [@hondodoc](#): > > But I don’t see how the potential energy is included in the math and how it would be different if not compressed. It would be negligible. For that particular spring and 20 % compression **am77494** calculates 50 J of energy from potential energy in the spring vs. 1.6 million joules released simply by the dissolution if the spring in sulphuric acid. The difference between compressed and uncompressed is 50 J. Ignoring that the 1.6 million is a rough approximation and definitely not done to this precision: compressed …1,600,050 J Uncompressed…1,600,000 J Or was it something else you didn’t grasp? ---
**Author:** ![hondodoc](https://avatars.discourse-cdn.com/v4/letter/h/bcef8e/32.png) [@hondodoc](https://boards.straightdope.com/u/hondodoc)\ **Post date:** [May 26, 2015, 3:38pm UTC](https://boards.straightdope.com/t/spring-energy/720433/29 "2015-05-26T15:38:28Z")
I still don’t understand how the potential energy is converted into heat. ---
**Author:** ![Mangetout](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/mangetout/32/19_2.png) [@Mangetout](https://boards.straightdope.com/u/Mangetout)\ **Post date:** [May 26, 2015, 6:33pm UTC](https://boards.straightdope.com/t/spring-energy/720433/30 "2015-05-26T18:33:43Z")
> [@hondodoc](#): > > I still don’t understand how the potential energy is converted into heat. It’s been explained several times. The mechanical tension is released in the form of particles of metal departing the spring more energetically than they would if the spring was uncompressed. As each particle ‘pushes’ off into the solution, it stirs the particles it encounters and this stirring heats it. ---
**Author:** ![Trinopus](https://avatars.discourse-cdn.com/v4/letter/t/2bfe46/32.png) [@Trinopus](https://boards.straightdope.com/u/Trinopus)\ **Post date:** [May 27, 2015, 12:31am UTC](https://boards.straightdope.com/t/spring-energy/720433/31 "2015-05-27T00:31:49Z")
> [@sbunny8](#): > > . . . Hypothetically, in a game of Jenga, if you could remove all the blocks of wood simultaneously then the tower would never fall, but that’s simply not possible. Extremely elegant analogy, and best answer in the thread. ---
**Author:** ![naita](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/naita/32/5862_2.png) [@naita](https://boards.straightdope.com/u/naita)\ **Post date:** [May 27, 2015, 9:18am UTC](https://boards.straightdope.com/t/spring-energy/720433/32 "2015-05-27T09:18:49Z")
> [@sbunny8](#): > > Hypothetically, in a game of Jenga, if you could remove all the blocks of wood simultaneously then the tower would never fall, but that’s simply not possible. > [@Trinopus](#): > > Extremely elegant analogy, and best answer in the thread. I should think that it would be a trivial task to make a “remove all the Jenga blocks simultaneously”-device. ---
**Author:** ![Mangetout](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/mangetout/32/19_2.png) [@Mangetout](https://boards.straightdope.com/u/Mangetout)\ **Post date:** [May 27, 2015, 8:11pm UTC](https://boards.straightdope.com/t/spring-energy/720433/33 "2015-05-27T20:11:52Z")
> [@naita](#): > > I should think that it would be a trivial task to make a “remove all the Jenga blocks simultaneously”-device. I have one. It’s called a tube of glue. ---
**Author:** ![beowulff](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/beowulff/32/542_2.png) [@beowulff](https://boards.straightdope.com/u/beowulff)\ **Post date:** [May 27, 2015, 8:34pm UTC](https://boards.straightdope.com/t/spring-energy/720433/34 "2015-05-27T20:34:08Z")
> [@Mangetout](#): > > I have one. It’s called a tube of glue. Not the _[Kragle](http://2.bp.blogspot.com/-MNaaTw_8nU4/UyOmk6WkAGI/AAAAAAAABPA/D8PUZAcWM-I/s1600/kragle1.jpg)_! ---
**Author:** ![dracoi](https://avatars.discourse-cdn.com/v4/letter/d/90db22/32.png) [@dracoi](https://boards.straightdope.com/u/dracoi)\ **Post date:** [May 27, 2015, 8:41pm UTC](https://boards.straightdope.com/t/spring-energy/720433/35 "2015-05-27T20:41:24Z")
Wait, are we accepting the “micro-spoing of each atom” theory? That just makes no sense to me. The way I see it, the spring is pulled to a certain tension. Let’s call that 1 lb of force. At the beginning of the experiment, the wire in the spring has a certain strength. Let’s call that 5 lbs of strength. The tension on the spring is not released until it’s, well, released. Each individual molecule is not carrying away tension. So at some point when about 80% of the spring has dissolved, we have 1 lb of force on a wire with 1 lb of strength. At that point, it will break somewhere and you’ll have a sproing that releases all the potential energy required to create the 1 lb of force. ---
**Author:** ![beowulff](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/beowulff/32/542_2.png) [@beowulff](https://boards.straightdope.com/u/beowulff)\ **Post date:** [May 27, 2015, 9:04pm UTC](https://boards.straightdope.com/t/spring-energy/720433/36 "2015-05-27T21:04:06Z")
> [@dracoi](#): > > Wait, are we accepting the “micro-spoing of each atom” theory? > > That just makes no sense to me. > > The way I see it, the spring is pulled to a certain tension. Let’s call that 1 lb of force. At the beginning of the experiment, the wire in the spring has a certain strength. Let’s call that 5 lbs of strength. The tension on the spring is not released until it’s, well, released. Each individual molecule is not carrying away tension. So at some point when about 80% of the spring has dissolved, we have 1 lb of force on a wire with 1 lb of strength. At that point, it will break somewhere and you’ll have a sproing that releases all the potential energy required to create the 1 lb of force. Let’s say you have a spring with an un-compressed length of 2", and you put it in a plastic holder, compressing it to 1". Say that puts the spring under 100 lbs. of compression. Now, if you put the assembly in acid, and it dissolves uniformly, are you saying that the spring will be under 100 lbs of compression to the very end? I don’t think so. ---
**Author:** ![Mangetout](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/mangetout/32/19_2.png) [@Mangetout](https://boards.straightdope.com/u/Mangetout)\ **Post date:** [May 27, 2015, 11:00pm UTC](https://boards.straightdope.com/t/spring-energy/720433/37 "2015-05-27T23:00:42Z")
Agreed - the tension of the spring is distributed throughout the metal (maybe not uniformly due to variability of material, and because of the shape of things, but if 80% of the spring is dissolved, then some of the stored energy is also gone. Imagine a spring made from 10mm thick wire and another spring the same shape made from 2mm thick wire, All other things being equal, they won’t require the same force to compress them. Now let’s imagine that the acid dissolving the spring does so very uniformly - dissolving the 10mm wire spring until it’s the same as the spring made of 2mm wire - some of the stored energy must have gone. ---
**Author:** ![Precambrianmollusc](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/precambrianmollusc/32/63_2.png) [@Precambrianmollusc](https://boards.straightdope.com/u/Precambrianmollusc)\ **Post date:** [May 27, 2015, 11:20pm UTC](https://boards.straightdope.com/t/spring-energy/720433/38 "2015-05-27T23:20:57Z")
> [@hondodoc](#): > > I have a question about an experiment that seems to violate the law of conservation and energy. I’m hopeful that you will be able to dispel my misunderstanding. > > Suppose you take a metal spring wound to its capacity, thereby storing potential energy in the spring. Suppose then that you immerse the spring in a bath of acid so that it dissolves into the acid in a uniform way so that the tension in the spring is not released because one part dissolves faster than the rest of the spring. > > What then has become of the potential energy in the spring? > > Brian in Texas When you wind the spring up to a certain tension , you induce a certain stress in the spring material. As the spring is slowly dissolved you are loosing material , the tension has to be maintained and so the stress in the remaining material goes up. Eventually the stress will exceed the tensile strength of the material and it will yield and move and release the energy as kinetic and some energy loss to deforming the material. Just the same as if you put a load on a bar and so the bar under compression. A certain stress is generated in the steel, as material is dissolved off the bar gets thinner, the stress level goes up in the remaining material to support the load until the stress level exceeds the mechanical properties of the bar and it buckles. The dissolving atoms of the spring are not carrying away any mechanical stress. If you are down to one single string of unobtanium atoms wound up, they still have the same tension, just a really high stress state, when these are dissolved and the bonds break, they will fly off in whatever direction they were being pulled in due to the tension so the energy from the tension goes into kinetic energy. ---
**Author:** ![beowulff](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/beowulff/32/542_2.png) [@beowulff](https://boards.straightdope.com/u/beowulff)\ **Post date:** [May 27, 2015, 11:28pm UTC](https://boards.straightdope.com/t/spring-energy/720433/39 "2015-05-27T23:28:54Z")
> [@Precambrianmollusc](#): > > When you wind the spring up to a certain tension , you induce a certain stress in the spring material. > As the spring is slowly dissolved you are loosing material , the tension has to be maintained and so the stress in the remaining material goes up. > Eventually the stress will exceed the tensile strength of the material and it will yield and move and release the energy as kinetic and some energy loss to deforming the material. Just the same as if you put a load on a bar and so the bar under compression. A certain stress is generated in the steel, as material is dissolved off the bar gets thinner, the stress level goes up in the remaining material to support the load until the stress level exceeds the mechanical properties of the bar and it buckles. > > The dissolving atoms of the spring are not carrying away any mechanical stress. > If you are down to one single string of unobtanium atoms wound up, they still have the same tension, just a really high stress state, when these are dissolved and the bonds break, they will fly off in whatever direction they were being pulled in due to the tension so the energy from the tension goes into kinetic energy. OK, this is completely wrong. Have you never worked with springs? If I wind up a spring (say a mainspring in a watch), the spring is under a certain tension, and is being prevented from unwinding by the ends of the spring being kept in a fixed position. If I then cut the spring in half (lengthwise, so it doesn’t unwind) I don’t end up with two springs at the same tension - I end up with two springs each with HALF the tension of the original. ---
**Author:** ![Precambrianmollusc](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/precambrianmollusc/32/63_2.png) [@Precambrianmollusc](https://boards.straightdope.com/u/Precambrianmollusc)\ **Post date:** [May 27, 2015, 11:33pm UTC](https://boards.straightdope.com/t/spring-energy/720433/40 "2015-05-27T23:33:40Z")
the tension is the mechanical load held The stress of the material is what you need to consider If you hold the same tension on a spring ( the mechanical load) but remove material the stress in the material must rise, until you exceed the mechanical properties of the material [Previous page](https://boards.straightdope.com/t/spring-energy/720433.md?page=1) [Next page](https://boards.straightdope.com/t/spring-energy/720433.md?page=3)