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A man in a gym lifts a weight and then puts it down where it was before.

1) What can be say about the work done by the man to the weight?

2) Can it be equal in absolute value to the work made by the gravitational force (that is $0$)?

3) Is it actually equal to the amount of physical energy spent by the man for this physical performance?

4) Can we quantify this energy given the mass and the height?

5) Is the energy spent by the man really independent on the time used for this action?

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  • $\begingroup$ Remember, work is the transfer of energy. $W_{man}$ is zero because there is no net change in mechanical energy of the system aka the weight. $\endgroup$ – user111965 Apr 6 '16 at 18:58
  • $\begingroup$ So why the man is spending/losing energy? $\endgroup$ – Marco Disce Apr 6 '16 at 18:59
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    $\begingroup$ Because the man does work on the weight to lift it; thus changing the weight's potential energy. Then, gravity does negative work on the system when it gets put down and the net work done on the system becomes zero. By the way $W_{net}$ should be used instead of $W_{man}$ in my previous comment. $\endgroup$ – user111965 Apr 6 '16 at 19:07
  • $\begingroup$ If the weight falls down we need to consider also the work of the forces that stop the motion on the ground in the sum, but usually the man would do this work instead of letting the weight fall. $\endgroup$ – Marco Disce Apr 6 '16 at 19:36
  • $\begingroup$ That is correct. $\endgroup$ – user111965 Apr 6 '16 at 19:41
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1) What can be say about the work done by the man to the weight?

Unlike gravity, the force exerted by the man is in general not constant, does not depend only on position of the body (you may apply different force on the weight at the same height $h$ on the way up vs. on the way down), and is not conservative (does not arise from a potential). When the force is conservative, like gravity, you can conclude that its work is $0$ when the body returns to the initial place. This is because the work of such a force measures the change in potential energy, like $mgh$ for gravity. But the force exerted by the man doesn't arise from any useful notion of potential energy and you cannot conclude that its work equals $0$ just because the weight returns to the same place.

2) Can it be equal in absolute value to the work made by the gravitational force (that is $0$)?

Despite what I wrote above, we can conclude that the work done by the man is $0$, but only if the weight, on return to its initial location, has the same speed as in the beginning (that is, $0$). This is true for the following reason. The work of the resultant force acting on the body (the sum of all forces) always equals the change in kinetic energy. If the kinetic energy didn't change, then the total work is $0$. The total work is the sum of the work by gravity and by the man. The work by gravity is $0$ for a different reason, as explained above (gravity is conservative). So the work by the man must be $0$ too.

So if the man carefully brings the weight to the ground, letting it touch the ground gently and settle, the work will be zero. If he throws the weight down or lets it fall, or moves it violently, the work done by the man will not be $0$. It will be equal to the weight's kinetic energy, $mv^2/2$, at the moment of impact with the ground.

3) Is it actually equal to the amount of physical energy spent by the man for this physical performance?

No. Our bodies are built from soft tissues, and even the strong parts, the bones, are connected by tissues that cannot withstand large forces passively and still maintain the amount of control we desire. When we lift the weight, we generate the force needed to counteract gravity by contracting and relaxing many fibers in our muscles constantly. If this contracting/relaxing was purely mechanical, it might do just a little work (not $0$ because some kinetic energy will be lost to friction, but almost). But it's not, it's chemical and requires spending chemical energy on contracting of which only a part is given back when relaxing.

4) Can we quantify this energy given the mass and the height?

No, that depends a lot on muscle effectiveness, particular movements, etc. Biologists attempt to measure those things - for example, there are articles on the effectiveness of running. But it;'s hard.

5) Is the energy spent by the man really independent on the time used for this action?

Not in general, no.

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  • $\begingroup$ "chemical energy on contracting of which only a part is given back when relaxing": do you mean some chemical energy is produced when the muscle relaxes? $\endgroup$ – L. Levrel Apr 7 '16 at 17:52
  • $\begingroup$ I phrased it badly; what I had in mind was the storage of elastic energy by tendons. The muscles (I think - I'm not an expert here and only repeat what I read somewhere once) only burn energy, either contracting or relaxing; but tendons store and return energy very efficiently, and depending on the kind of movement can be responsible for a large part of even the majority of energy transfer. $\endgroup$ – AnatolyVorobey Apr 7 '16 at 21:04
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1&2) Let's suppose there are only two forces in play: the one exerted by the man, and weight.

The sum of works done on the body equals the variation of kinetic energy (this is a theorem of mechanics; I fail to find its English name). Since the body has no velocity at the start and at the end, the sum of works is zero.

The net work done by the man on the body equals the variation of mechanical energy (this is another theorem of mechanics). Since the body is at the same height, and has no velocity, at the start and at the end, the work done by the man is zero.

3&4&5) Biological energy consumed is an entirely different problem. You can see this clearly since the "mechanical" work done by the man is zero, although with a heavy weight this is a performance!

Our muscles use so-called "molecular motors". I'm not a specialist of this field, but I know that these have the property of constantly "slipping", they cannot be in a "locked" state. As a consequence, they need power to sustain a force even if they don't move (at the macroscopic scale): this is common knowledge that you get exhausted by simply holding a big weight, whereas in physics words there is no mechanical work done (but the consumed energy ends up heating the muscles; the 1st law of thermodynamics is not violated!).

There's no law giving "biological energy" consumed as a function of force applied that I know of. Maybe are there some empirical models.

The energy spent is of course dependent on the time used: if the movement is really slow, the man holds the weight for a very long time, and this is very exhausting!

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The correct statement is - the total work done on the weight is zero i.e. the total energy of the weight before and after the experiment is same. However, when the man is lifting the weight he is obviously working against gravity. More importantly, when he is lowering the weight, he is still working against gravity, as gravity would rather lower the weight in a 'free fall'. To stop the free fall, the man is indeed applying a force in the upward direction while lowering the weight, only the magnitude of the force (on an average) is lower than the force of gravity, which is why the weight is lowered.

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  • $\begingroup$ So we have $W_{\text{man}}=-W_{\text{grativty}}$ and both are nonzero? Can we quantify them given the mass and the height? $\endgroup$ – Marco Disce Apr 6 '16 at 19:23
  • $\begingroup$ When moving down the weight, if initial and final velocities are $0$, then the average force applied by the man(w.r.t time or position) is exactly equal to the weight of the object. The same applies when lifting up. $\endgroup$ – Tofi Apr 6 '16 at 19:34
  • $\begingroup$ If we assume that $F_{\text{man}}=-mg$ then it seems to me that we will end up with a zero total work, isn't it? $\endgroup$ – Marco Disce Apr 6 '16 at 19:40
  • $\begingroup$ One needs to decouple the work done by the man from the work done on the weight and the work done by gravity. Work done on the weight is the same as the work done by gravity, in fact that is how one defines the work done on a body. The work done by the man is not, in any way, getting transferred to the body. The work done by the man is the force applied by him times the displacement - that's it. $\endgroup$ – Subhayan Apr 6 '16 at 19:46
  • $\begingroup$ Isn't work always done on something that moves? Otherwise which displacement would we have to consider? Displacement of what? $\endgroup$ – Marco Disce Apr 6 '16 at 20:01
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I recently answered a similar question. It was about the work done by a man while carriying a 50Kg suitcase over his head, traveling an horizontal distance of 15m.

The answer I gave about the work done was:

If your height is, lets say, 1.80m, then the total mechanical work would be the work done while lifting the suitcase, plus some “extra” work due to the inefficiencies of the human body machinery internals.

So, that should amount to not less than mgh = 50*10*1.8 = 900 Joules.

Imagine the following case:

John lifts 10 50Kg potato sacks from the ground to the platform of a truck, which is at a height of 1.5m.

He drives to the destination, and afterwards John downloads the 10 potato sacks, very carefully and gently putting them on the ground … he doesn’t want to spoil the potatos.

John’s boss says: Well John, for every potato sack your body, as a system, did a negative mechanical work when lifting the sack, but afterwards you body did a positive mechanical work while setting down the sack, and the net balance is that you did a zero amount of mechanical work at the end. I only pay you for driving the truck!

But John, who likes mechanic artifacts, quickly uses a piece of paper and pencil and draws this invention:

patend pending

  • When you turn on the machine, a “chemical” motor inside the A block of the machine pushes the piston B, so the hydraulic fluid pushes the piston C, and the machine lifts the mass M to a certain height h.

  • When you turn off the machine, the motor in the A block of the machine stops doing any force to the B piston, but some brake-claws in the C piston at F, driven by a microcontroller device, begin to do enough normal force over the cylinder surface, as to stop (by friction) the movement of the mass M exactly at the starting point 1 (so, the mass M lands over the starting 1 position at zero speed, with no kinetic energy).

John’s reasonig is the following:

The machine is the mechanical equivalent of my body workings.

When I lifted the mass, the machine did a negative mechanical work, which was more or less mgh plus some “extra” work also to displace upwards the gravity center of the hydraulic fluid mass inside the cylinder.

But when I set down the mass (turned off the motor), the machine did no net mechanical work. The internal friction forces did a negative mechanical work (always opposed to the displacement), which exactly compensated the positive mechanical work from the push of the mass over the machine platform.

What do you think?

P.D. To avoid any further arguments with his boss, John decided to just drop the potato sacks to the ground in the future, because in that scenario was clearer that he did not any mechanical work during the setting down phase.

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Lets not consider the extra energy consumed by biological reactions in a human body. Imagin a robot is doing this. The total work done on the weight is zero(if problem is simplified to constant velosity and not existing friction) because W robot=0 and W mg=0 in one complete receprocation. BUT the energy spent by the robot is ∫|f.dx| . i.e in both raising and lowering ways the positive amount of work is integrated and equals to total energy consumed by the robot.

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  • $\begingroup$ This is incorrect. A robot can recover energy when lowering the weight. $\endgroup$ – Chris Jan 2 '19 at 23:11

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