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To calculate the calories burned doing 1 squat, wouldn't it simply involve the distance that your center of mass is moving and your weight? I read some other article that said your legs act as levers, so the calculation has to involve that. Then of course you have to come up with the efficiency of your body.

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No, there are many other factors. Think of someone holding a book by your reasoning no energy is consumed. –  jinawee Nov 20 '13 at 23:17
    
The human body is too complicated to reduce this to a simple physics question. What mechanics tells us isn't anywhere close to what your body will actually spend doing work. –  Brandon Enright Nov 20 '13 at 23:36
    
Related: physics.stackexchange.com/q/1984/2451 –  Qmechanic Nov 23 '13 at 7:50
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closed as off-topic by Brandon Enright, tpg2114, Dimensio1n0, Waffle's Crazy Peanut, Chris White Nov 25 '13 at 18:20

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1 Answer

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You are on the right track for the minimum energy required. It has to be at least that required to lift the weight over the height change.

However, there are so many other things envolved that this minimum energy per squat won't be a very useful figure, although the legs acting a levers has nothing to do with it. Either way, the body weight is raised, requiring energy. The efficiency of converting food calories into work to raise the body is going to be hard to nail down. Muscles also don't use up energy proportional to energy delivered, even if the efficiency is known. A lot, if not most, of the energy required to activate a muscle is more a function of the force it produces. For example, keeping yourself suspended at half-squat would be quite fatiguing and use up some calories, but isn't actually doing mechanical "work".

While your physics is right, I suspect the answer will be pretty meaningless in a real sense. The best way to determine calories/squat is to measure it. Metabolic activity can be measured reasonably well by measuring oxygen consumption. There are systems that do exactly that, often together with a treadmill to force the body to do a certain amount of biological work. Note that running on a flat surface is actually doing very little physics work.

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A minor correction: different parts of the body are raised different amounts. The feet and ankles don't really get any higher, the calves are lifted a bit, and so on. Of course, a minor correction to a "pretty meaningless" answer is well...pretty meaningless. –  User58220 Nov 21 '13 at 3:35
    
@User: Yes, your body weight below the hips has to be derated based on how much the center of mass of each piece actually moves upward. This changes the amount of work done a bit to raise yourself up from a squat, but that value is still a long way from telling you how many calories your body consumes in the process. –  Olin Lathrop Nov 21 '13 at 15:26
    
I don't like your sentence "isn't actually doing "work" in the physics sense." Consider changing this to "the usual classical mechanics sense," or to "isn't doing work on the box". Obviously if your muscles are heating up there's a change in energy. –  NeuroFuzzy Nov 22 '13 at 1:29
    
@Neuro: I think you're nit picking the and the intended meaning is clear enough (of course there is some energy conversion going on), but I edited to hopefully please more people. –  Olin Lathrop Nov 22 '13 at 14:32
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