When unsuccessfully trying to lift a weight, where does my energy go to?

[khan]

Let's say I am strong enough to lift a 100 kg dumbbell, but not strong enough for a 200 kg one.
If I lift the 100 kg dumbbell, I suppose the chemical energy of my body tranforms into the potential energy of the lifted dumbbell. Suppose I'm trying to lift the 200 kg one. I'm pulling as hard as I would be if I were lifting the 100 kg one, so the loss of my energy should be the same. But the dumbbell stays on the ground so there is no gain in its potential energy. My question is, where does my energy go? It seems to me that on my part, nothing changes, I'm pulling as hard as I can in both cases, sweating, pressing the floor, whatever. So what accounts for the difference in the change of potential energy in the case of the lighter dumbbell and no change of potential energy in the case of the heavier one?

 

[Lnewqban]

Welcome, khan!
Your body uses energy to produce only contraction of the muscles.
There is no work produced if the length of the muscle does not change.

Please, see:
https://en.wikipedia.org/wiki/Muscle_contraction

 

[jbriggs444]

The human body is not a particularly efficient machine. And its efficiency is not even constant. The energy goes into heating your muscles. If you are not budging the weight, the efficiency is 0%. If you are climbing down stairs, your efficiency even goes negative. You burn energy absorbing energy.

 

[PeroK]

Let's say I am strong enough to lift a 100 kg dumbbell, but not strong enough for a 200 kg one.
If I lift the 100 kg dumbbell, I suppose the chemical energy of my body tranforms into the potential energy of the lifted dumbbell. Suppose I'm trying to lift the 200 kg one. I'm pulling as hard as I would be if I were lifting the 100 kg one, so the loss of my energy should be the same. But the dumbbell stays on the ground so there is no gain in its potential energy. My question is, where does my energy go? It seems to me that on my part, nothing changes, I'm pulling as hard as I can in both cases, sweating, pressing the floor, whatever. So what accounts for the difference in the change of potential energy in the case of the lighter dumbbell and no change of potential energy in the case of the heavier one?

I wonder how successful you would be at body-building if you worked only with weights that you couldn't move?

 

[jbriggs444]

I wonder how successful you would be at body-building if you worked only with weights that you couldn't move?

https://en.wikipedia.org/wiki/Isometric_exercise

 

[gleem]

Isometrics was a muscle building technique advocated by Charles Atlas in the 50's

 

[Merlin3189]

I wonder how successful you would be at body-building if you worked only with weights that you couldn't move?

I immediately thought of Charles Atlas and Dynamic Tension, but when I checked, that is not quite isometric exercise. Dynamic Tension was 100% inefficient use of muscles, since they worked against each other, but there was movement.
Isometric exercises are supposed to build muscle strength, so one might think they'd increase muscle mass as well.

 

[Merlin3189]

Back to OP. The phenomenon is not unusual. If you used a simple DC motor powered from a fixed voltage (and other types) to lift a load, when you exceed stall torque, it actually consumes more power than when lifting a load, but all the power goes into resistive heating of the windings and none into useful work.

Another simple eg. of inefficiency: you can block the hose of a vacuum cleaner so that it can no longer do useful work moving air. Again all the electrical energy then goes int heat. What fascinated me about this was that the motor, far from stalling, speeds up and draws less electrical power, but still ends up with 0 efficiency as an air pump.

 

[khan]

The human body is not a particularly efficient machine. And its efficiency is not even constant. The energy goes into heating your muscles. If you are not budging the weight, the efficiency is 0%. If you are climbing down stairs, your efficiency even goes negative. You burn energy absorbing energy.

I'm not sure if I understand you. Are you trying to say that the energy that would go into lifting the weight goes instead into heating my muscles? So my muscles are heating more if I'm not unable to lift the weight? That would explain where the energy goes to, but seems quite unintuitive...
And what exactly do you mean by "you burn energy absorbing energy"?
Thanks!

 

[PeroK]

And what exactly do you mean by "you burn energy absorbing energy"?
Thanks!

It's tiring to walk down stairs! Or to walk downhill. Whereas, on a bike you can go downhill without expending any energy.

 

[jbriggs444]

I'm not sure if I understand you. Are you trying to say that the energy that would go into lifting the weight goes instead into heating my muscles? So my muscles are heating more if I'm not unable to lift the weight? That would explain where the energy goes to, but seems quite unintuitive...
And what exactly do you mean by "you burn energy absorbing energy"?
Thanks!

Simply tensing your muscles involves individual fibers repetitively clenching and releasing to produce the desired tension. Regardless of whether any motion results, this consumes chemical energy.

If you descend a flight of stairs, this still happens. So your fibers are being intermittently tensed, consuming chemical energy even while the descent of the stairs is providing mechanical energy.

Energy is still conserved, of course. The excess shows up as heat.

 

[khan]

Back to OP. The phenomenon is not unusual. If you used a simple DC motor powered from a fixed voltage (and other types) to lift a load, when you exceed stall torque, it actually consumes more power than when lifting a load, but all the power goes into resistive heating of the windings and none into useful work.

Another simple eg. of inefficiency: you can block the hose of a vacuum cleaner so that it can no longer do useful work moving air. Again all the electrical energy then goes int heat. What fascinated me about this was that the motor, far from stalling, speeds up and draws less electrical power, but still ends up with 0 efficiency as an air pump.

Thanks, these are great examples!
So you think that my muscles are heated more if I'm trying to lift the weight but am not able to? That would explain where the energy goes. But it would also mean that the more work you do, the less you sweat :) It makes sense, but still seems weird...

 

[jbriggs444]

Were you planning to lower the weight after you lift it? If so, does it really matter that you moved it? The work done on the weight was zero.

 

[khan]

It's tiring to walk down stairs! Or to walk downhill. Whereas, on a bike you can go downhill without expending any energy.

It is tiring because I'm slowing my descent. If I just fell from the stairs the potential energy I would loose would got to my kinetic energy and then to the vibrations of the floor. If I'm walking down I move more slowly and don't stomp so much, so some of the potential energy I loose somehow stays inside me. Is that right? If so, what is the form of the energy? Is it heat again? And only heat?

 

[khan]

Were you planning to lower the weight after you lift it? If so, does it really matter that you moved it? The work done on the weight was zero.

Well, I didn't go that far in my planning :)
But thanks so much, I think I get it now. The answer seems to be just heat, heat, and heat again.

 

[jbriggs444]

It is tiring because I'm slowing my descent.

All things being equal, that should be the opposite of tiring. That should "charge your batteries".

But muscles do not do regenerative braking. So your batteries do not get recharged.

If I just fell from the stairs the potential energy I would loose would got to my kinetic energy and then to the vibrations of the floor. If I'm walking down I move more slowly and don't stomp so much, so some of the potential energy I loose somehow stays inside me. Is that right? If so, what is the form of the energy? Is it heat again? And only heat?

Yes, it all goes to heat. The chemical energy you consume keeping your muscles tense and the potential energy that is used up both go into heating your body.

 

[russ_watters]

So you think that my muscles are heated more if I'm trying to lift the weight but am not able to? That would explain where the energy goes. But it would also mean that the more work you do, the less you sweat :) It makes sense, but still seems weird...

We're not biologists here, but I don't think that's correct. I think the energy/heat dissipated is related to the force generated and largely independent of the mechanical power generated.

It's also worth noting that for most exercises the net mechanical output work is zero anyway. You lift a weight and then you put it back down where it started.

 

[jbriggs444]

I think the energy/heat dissipated is related to the force generated and largely independent of the mechanical power generated.

I am in agreement here. As a practical matter, any real effect is likely to disappear into the experimental noise.

A quick trip to Google suggests a figure of as much as 30% for aerobic efficiency of a cyclist (counting just the delta from unloaded versus loaded cycling). Reference here. The number is higher than I'd expected.

If one were able to measure for heat gain in the body against O₂ consumption, the result would presumably be that the work produced moving the bike is indeed subtracted from the heat gained in the body. Anything else would violate energy conservation.

[The experiment here is not directly on point. It is comparing loaded versus unloaded muscle use rather than loaded immobile versus loaded moving. The figure of merit I was after was efficiency. For that, it is a good reference]