If I push my hand against a wall, the wall doesn't move, so there is no displacement and so no work done. But energy is being used to generate the force, so what is happening to the energy? Does that mean the total energy is being lost as heat? If so how would you calculate this energy? I though of a way to do it, but I'm not sure if its correct, and would like to know that. Work done i.e. W = F x s can be converted to a function of time and so it becomes W = F^2 * s^2 / 2m. And then if I knew the force I applied, the duration for which I applied it, and the mass of my hand, I can calculate the total energy lost as heat, based on the potential distance my hand would have gone if there was no wall. Will this give an accurate result for the energy being transferred to heat?
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1There is no work done, because as you pointed out $s=0$. It's analog to carrying a mass from the supermarket to your home instead of using a trolley. – Semoi Dec 06 '20 at 18:20
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Related: https://physics.stackexchange.com/q/1984/2451 and links therein. – Qmechanic Dec 06 '20 at 18:21
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but it is also true that I will get tired eventually, so energy is clearly being transformed – Neelim Dec 06 '20 at 18:26
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also I would like to know how to calculate this energy – Neelim Dec 06 '20 at 18:36
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This question was recently asked here https://physics.stackexchange.com/questions/594783/where-does-the-energy-get-transferred-when-we-push-the-wall/594789#594789 – Luke Pritchett Dec 06 '20 at 19:02
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See related questions which were suggested when you posted your question, such as When we push on a wall, can we say work is being done on the atomic particles in contact with our hand?. – sammy gerbil Dec 06 '20 at 21:16
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but those don't say how to calculate this energy – Neelim Dec 07 '20 at 12:35
2 Answers
remember, work done is defined as the (dot) product of force and distance moved by the point of application. Take the example of someone doing pushups. His hands clearly dont move during the process, so the work done by the hands on the floor is clearly 0. But its pretty obvious from observation that the person soon gets tired and therefore loses energy. Where did this energy come from and go?
Basically, whats happening is that although your hands are stationary, muscles all over the body ARE moving and exerting forces on each other. Thus, the work is being done by internal spring like muscular forces. Since muscles are doing the work, they are losing (chemical) energy, which is being converted the body's kinetic energy and heat.
Note: since no force exerted by the body on an external object is doing work, there is no external transfer of energy. Since internal muscular forces are doing work on objects inside the body, the transfer of energy is purely internal; no energy s atually leaving the body (at least initially)
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Why do you need to? I dont think its possible, given the shear amount of uncertainty – Vulgar Mechanick Dec 06 '20 at 18:40
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Bad analogy, because in your example your hands does a work $mgh$ raising your body against Earth gravity field and then again stopping your body from a free-fall with $g$ acceleration. – Agnius Vasiliauskas Dec 06 '20 at 19:28
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actually, the hands are not doing that work at all. The hands never move. – Vulgar Mechanick Dec 07 '20 at 00:49
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@OVERWOOTCH It's not the point about who exactly does this work. This doesn't change the fact that your analogy is invalid. In pushup case you must do work to conquer Earth gravity, and in OP case - you do NOT. Because in OP case force vector is perpendicular to Earth gravity field direction. While in push-ups case, these directions are colinear. – Agnius Vasiliauskas Dec 07 '20 at 08:23
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If it's not the point who exactly does the work, then don't say "your hands does a work". You can't just say something wrong and say it doesnt matter. Secondly, in each pushup cycle, gravity does the same amount of work on the way down as you do against it going up. If we only looked at this, then mechanical energy is consevrded, and you shouldn't get tired at all. This analogy is perfectly valid for what I am aiming. If you disagree, try answering the question yourself – Vulgar Mechanick Dec 07 '20 at 09:11
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I see no link between energy conservation and getting tired. Energy is being conserved in most cases. You can get tired simply by converting one form of energy to the other, this has nothing to do with conservation laws at all. – Agnius Vasiliauskas Dec 07 '20 at 09:40
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Your intuition is partially correct but also misleading you somewhat.
Consider a simpler example. Imagine a heavy weight sitting at rest on a hard concrete floor. The weight exerts a force on the floor but there is no displacement. In this example no work is done and clearly no energy is used. There's certainly no heat generated, otherwise we could get free heat just by putting heavy things on the ground!
On the other hand you know from experience that if you were to push against a wall for a long time you would get tired and you would need to eat food to replenish your energy.
This tells us that the rate of energy usage has something to do with the human body, not with force in general. See the answers linked in the comments above for an explanation of what's going on.
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