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This question already has an answer here:

Please read till the end, before you answer. I'll start off with the examples to get to the point.

Drop a block from 1 meter and it will have speed v when it reaches the ground, with total work done on it w. Drop it from 2 meters and it won't have twice the speed, since it has less time to gain speed. Surprisingly, 2w work is done to it.

Another example. I know it's weird but it's important

A 1 Ton block of lead in space is at rest. A human is standing beside the lead block. The human is going to use chemical potential energy 2000 J to do work on the lead block and move it and displace it 2000 meters, using a force of 1 N on the block, Neglecting the force he uses to move himself to be able to push the block. How much time has he spent doing force?

Since force 1 N is done to on a 1 ton(1000 kg) block, and acceleration is force over mass, the acceleration is 1/1000. using the distance traveled by acceleration formula, the time needed is 2000 seconds. He has increased the momentum of the object from 0 to 2 thousand.

In another scenario, A human used 2000 J chemical potential energy using force 1 N on a 1 ton block of lead, but this time it was moving with a speed 50 m/s. The amount of time he would need to move it would of be less than 20 seconds, because of the block's speed. The increase in momentum of the object must be less than 20. Surprisingly, In both scenarios the human used chemical potential energy 2000 J, which is absurd, because the one in the first scenario must have used more energy.

Tl;Dr 2 humans consume the same amount of joules but the first uses so much "effort" and the second guy uses no effort but in the end, same joules consumed.

This is what I'm trying to say. The amount of work you have done is affected significantly by the initial speed of the object. What consumes more food(energy) , adding 2, 000 momentum to an object or 20? Physics tries to tell me they use the same energy?I don't understand how in each of the scenarios the human consumed the same amount of Joules even though he clearly would be so much more tired after pushing the block at rest than the moving block . This absurdity is slowly driving me insane. Where did I mess up?

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marked as duplicate by knzhou, John Rennie newtonian-mechanics Sep 25 '18 at 7:33

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ I edited your title to better reflect what you're asking; feel free to change it back if you disagree. $\endgroup$ – knzhou Sep 28 '17 at 12:26
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    $\begingroup$ Work is a way to combine "effort" and "result" into one parameter. You can do a huge effort on pushing a wall but it doesn't go anywhere. Effort wasted. Noone comes to you and says "nice work you did there". On the other hand, you can easily push a balloon very far. But that requires as good as no force. No effort. Noone comes to you impressed and says "nice work you did there". When effort is done and it gives a proper result, as when you lift your grand piano to the second floor, then we can all agree that you did some "nice work there". $\endgroup$ – Steeven Dec 29 '17 at 14:35
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Kinetic energy scales with the square of the velocity; momentum scales linearly with velocity.

This is because the POWER you need supply to an object when you exert a certain force scales with the force and velocity (P = v x F), while the momentum scales with the force and time : M = F x t

The integral of power with time shows you more power is applied at the later stages of the acceleration (when the velocity is greatest) - you are adding more "units of energy per unit time". But momentum increases at a constant rate.

Another way to see this is to express the energy as a function of momentum. To do this we first write

$$\begin{align}E &= \frac12 m v^2\\ &= \frac{(mv)^2}{2m}\\\\ \rm{now}\\ p &= m v\\\\ \rm{so}\\ E &= \frac{p^2}{2m}\end{align}$$

So energy scales with momentum squared.

Perhaps the relationships make more sense now.

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  • $\begingroup$ I'm not concerned about the relationsgip between energy and momentum. I'm saying that in two examples a human consumed the same amount of Joules but a very different "effort". $\endgroup$ – user170541 Sep 28 '17 at 15:34
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    $\begingroup$ A change of momentum requires force, and time. From a human biology perspective, "applying force" is something that "takes work" even when you are not doing work in the physics sense (example: standing still with a 200 pound back pack on your back is tiring). This is because there are loss mechanisms in the muscle; and that's why your intuition about these things can easily go astray. But you know that it's harder work to push against a certain force while sprinting, than it is to apply the same force while standing still. Does that help? $\endgroup$ – Floris Sep 28 '17 at 17:27
  • $\begingroup$ That's precisely what I don't get. Why does it take more energy to apply force on a moving object than it is to apply the same force on an object at rest? $\endgroup$ – user170541 Apr 1 '18 at 16:24
  • $\begingroup$ @user170541 do you do work when you leave an object sitting still on the ground? Answer is no. When you lift the object, the act of keeping the muscles under tension requires certain biological processes that you experience as “tiring”. But if you lift the same object with a rope and pulley, then tie a know in the rope, you no longer need to “do work” - it will stay in the same position without your effort. That’s how physics looks at it. $\endgroup$ – Floris Apr 1 '18 at 22:01
  • $\begingroup$ That's not the issue anymore. I know you don't don't do work just by cancelling the gravitation force. The thing is, I think that if it takes x Joules to speed some object at rest to speed s, it would also take x Joules to speed it from speed s to 2s. What I don't get is why it takes more work to speed it up from s to 2s $\endgroup$ – user170541 Apr 2 '18 at 11:15
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Work is a way to combine "effort" and "result" into one parameter.

  • You can do a huge effort on pushing a wall but it doesn't go anywhere. Effort wasted. Noone comes to you and says "nice work you did there". -On the other hand, you can easily push a balloon very far. But that requires as good as no force. No effort. Noone comes to you impressed and says "nice work you did there".
  • I When effort is done and it gives a proper result, as when you lift your grand piano to the second floor, then we can all agree that you did some "nice work there".

In your examples you seem to equate work with speed. But don't equate it to speed, equate it to kinetic energy. Then it all fits.

Let me go through your examples one by one:

Drop a block from 1 meter and it will have speed v when it reaches the ground, with total work done on it w. Drop it from 2 meters and it won't have twice the speed, since it has less time to gain speed. Surprisingly, 2w work is done to it.

You are equating speed with work. Don't. The speed is not doubled, no, but the gain in kinetic energy is.

There is an energy conservation law but it does not include speed.

A 1 Ton block of lead in space is at rest. A human is standing beside the lead block. The human is going to use chemical potential energy 2000 J to do work on the lead block and move it and displace it 2000 meters, using a force of 1 N on the block, [...]

In another scenario, A human used 2000 J chemical potential energy using force 1 N on a 1 ton block of lead, but this time it was moving with a speed 50 m/s. [...]

The increase in momentum of the object must be less than 20. Surprisingly, In both scenarios the human used chemical potential energy 2000 J.

Again, you equate speed (or momentum) with work. The gains in speed are different in either case. But the gains in kinetic energy are equal. Again, don't equate speed to work - no law ties them together. But equate kinetic energy to work - that is the energy conservation law.

You even mentioned the fact about energy yourself in the next sentence:

[...] which is absurd, because the one in the first scenario must have used more energy.

No, the one in the first scenario has not spent more energy. You are looking at the speed gains, which correctly is larger in the first scenario. The energy (kinetic energy gain) is the same!

For this to make sense, remember how the formula for kinetic energy looks:

$$K=\frac12 mv^2$$

Doubling speed does not double kinetic energy. They do not follow each other linearly. So doubling the work done will double the kinetic energy gain due to $W=\Delta K$, but this does not double the speed.

And the difference in time taken that you also emphasize is likewise not important. It doesn't matter for work how long time it took to do a certain job - all that matters is effort and result (force and displacement).

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  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$ – David Z Apr 6 '18 at 8:00

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