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am77494
08-13-2014, 10:13 AM
I get the physics behind - if you lift and hold a mass stationary, then no work is being done on the object. (for the time that it is held)

However, is it true that the body is not doing any work internally i.e. does the heart need to do more work since the blood vessels are more constricted ? does the heart need to pump more blood to the muscles ? Why do weight lifters sweat a bunch and breathe faster when holding a weight stationary above their head ?

My understanding is that the body is actually doing a lot of work supplying energy to the muscles holding - that work is not conserved but converted to heat. Is this wrong ?


This is inspired by this thread - http://boards.straightdope.com/sdmb/showthread.php?t=730869

DrCube
08-13-2014, 11:16 AM
Well, you're not doing work (in the strict physics sense), but you're using energy and hence burning calories. It takes the force of your muscles to counteract the force of gravity on the weight, or else it would fall.

DrCube
08-13-2014, 11:18 AM
https://en.wikipedia.org/wiki/Isometric_exercise

lazybratsche
08-13-2014, 11:26 AM
On a microscopic scale, you're doing plenty of work. When you apply a force with your muscle at a constant position, the myosin in your muscle cells is continually grabbing onto actin filaments, pulling, letting go, and then slipping back. The myosin "power stroke" is applying force over distance, i.e. work. With sensitive enough equipment you can actually measure the work done by a single myosin protein (http://www.ncbi.nlm.nih.gov/pmc/articles/PMC3487478/?report=reader).

Chronos
08-13-2014, 01:55 PM
Quoth DrCube:

Well, you're not doing work (in the strict physics sense), but you're using energy and hence burning calories. It takes the force of your muscles to counteract the force of gravity on the weight, or else it would fall.
Both of these sentences are true, but they are completely unrelated to each other.

md2000
08-13-2014, 02:09 PM
A hovering helicopter or a rocket hovering on a flaming jet certainly is not using zero energy.

It's just being converted into heat and air currents rather than adding to the (gravitational) potential energy of the device.

A weight dragged horizontally against a frictional resistance should be using energy, even though the potential energy of the weight is not increased.

DrCube
08-13-2014, 02:10 PM
Both of these sentences are true, but they are completely unrelated to each other.

I'm not sure what you mean by that. I wasn't clear, I suppose, that if a table is holding up the weight, or some other rigid object, it neither does work nor uses energy. Simply counteracting an external force doesn't necessarily require an energy expenditure. But because of the way the human body is built -- to be flexible along certain joints and therefore require muscular contraction to stiffen the body against a load -- people do burn calories (i.e., use energy) while supporting a weight, even if they're performing no work on the load.

DSeid
08-13-2014, 08:21 PM
On a microscopic scale, you're doing plenty of work. When you apply a force with your muscle at a constant position, the myosin in your muscle cells is continually grabbing onto actin filaments, pulling, letting go, and then slipping back. The myosin "power stroke" is applying force over distance, i.e. work. With sensitive enough equipment you can actually measure the work done by a single myosin protein (http://www.ncbi.nlm.nih.gov/pmc/articles/PMC3487478/?report=reader).

Exactly which why the statement that no physics work is being done is false.

Various motor units are taking turns contracting, exerting a force over a unit distance, then relaxing and repeating. There is no external physics work being done upon the object but there is plenty of internal physics work being.

Calculating that amount of work is however a much complicated task than the issue of calculating external physics work. (How many times is each motor unit firing? How far is each contracting each time? With how much force? Varies with every motor unit involved and during the course of the task as some motor units fatigue and others are recruited.) But measuring the calories needed to keep it aloft compared to be at rest is likely a good ballpark.

Chronos
08-13-2014, 08:44 PM
Quoth DrCube:

I'm not sure what you mean by that. I wasn't clear, I suppose, that if a table is holding up the weight, or some other rigid object, it neither does work nor uses energy. Simply counteracting an external force doesn't necessarily require an energy expenditure. But because of the way the human body is built -- to be flexible along certain joints and therefore require muscular contraction to stiffen the body against a load -- people do burn calories (i.e., use energy) while supporting a weight, even if they're performing no work on the load.
But then again, human bodies also burn Calories when not supporting any load at all. It's misleading at best to say that Calories are being burned because of the supported load.

DSeid
08-13-2014, 08:53 PM
Just to help get a sense of how those motor units take turns (http://www.ncbi.nlm.nih.gov/pubmed/8299601).Single motor unit potentials were recorded from the brachial biceps muscle of eight male subjects, during isometric endurance experiments conducted at relative workloads corresponding to 10% and 40% of maximal voluntary contraction (MVC), respectively. The recordings from the 10% MVC experiment demonstrated a characteristic time-dependent recruitment. As the contraction progressed both the mean number of motor unit spikes counted and the mean amplitude of the spikes increased significantly (P < 0.01). This progressive increase in spike activity was the result of a discontinuous process with periods of increasing and decreasing activity. The phenomenon in which newly recruited motor units replace previously active units is termed "motor unit rotation" and appeared to be an important characteristic of motor control during a prolonged low level contraction. In contrast to the 10% MVC experiment, there was no indication of de novo recruitment in the 40% MVC experiment. Near the point of exhaustion a marked change in action potential shape and duration dominated the recordings.Each spike represents a contraction of a motor unit.

Also of note the ATP energy is actually burned during the relaxation segment of each motor unit contraction ... contraction is an allosteric change as actin binds to myosin and fibers slide across each other ... releasing it again takes energy from ATP.

Chronos, at rest calories are being burned because of the supported load. Muscles are never actually at rest. There are always some fraction of motor units contracting and others relaxing, each unit doing work over that time, even though, as in isometric contraction, no external work is performed. At rest fewer are contracting at a time is all.

rsa
08-13-2014, 08:54 PM
But then again, human bodies also burn Calories when not supporting any load at all. It's misleading at best to say that Calories are being burned because of the supported load.
Sure, balancing yourself and remaining upright requires a certain amount of energy. But balancing not only your own weight but also additional weight requires additional energy.

Snarky_Kong
08-14-2014, 06:32 AM
But then again, human bodies also burn Calories when not supporting any load at all. It's misleading at best to say that Calories are being burned because of the supported load.

Calories burned while supporting load - calories burning without supporting load = calories burned due to supporting load. No?

Quartz
08-14-2014, 07:13 AM
I get the physics behind - if you lift and hold a mass stationary, then no work is being done on the object. (for the time that it is held)

This is only true if you do it in zero gravity. On Earth you are working against the Earth's gravity.

DSeid
08-14-2014, 07:18 AM
So can we finally put this canard to rest? The High School physics question that some here are answering is about the external work done on the object - for the sake of High School physics homework there is no external work being done by the body on the 100 pound object being held stationary. Work equals force times distance. No distance, no work. Move on the question two about friction coefficients now.

However, the op asks the real question, not that High School homework one: is there any work being done internally? And the answer there is a resounded most definitely yes. Multiple motor units within various muscles are repetitively contracting and relaxing, producing forces across distances many multiple times during the period, using energy to elongate, and then contracting again. The muscles as a whole stay one length, producing force but moving no distance (or in most realities very little self-corrective distances); the object moves no distance; but that gross effect is the result of much physics work, force along distance, occurring within the muscles.

Machine Elf
08-14-2014, 07:21 AM
This is only true if you do it in zero gravity. On Earth you are working against the Earth's gravity.

Only while you are actually moving the weight. Work is force exerted across a distance (http://en.wikipedia.org/wiki/Work_(physics)); once you have raised the weight to its full height and begin holding it at a steady altitude, you have stopped delivering work to it.

Machine Elf
08-14-2014, 07:24 AM
So can we finally put this canard to rest?

Probably not - or at least not until we can school everyone on the physics concept of work (= F x D).

Ranger Jeff
08-14-2014, 08:18 AM
Perhaps the work done after lifting the 100 lb object and just being done while supporting it isn't measured in Foot Pounds or Newtons?:

DSeid
08-14-2014, 08:20 AM
So no actually answering the actual question asked because some of you can't go past question one of the High School Physics mid-term about external work performed on an object? (Not the question asked.) Can y'all get the idea that actual physics work (F x D), in the strict physics sense, is being done within the muscles? And that producing that work (and lost to heat) is where the extra energy needed to support the object goes.

Machine Elf
08-14-2014, 08:51 AM
Perhaps the work done after lifting the 100 lb object and just being done while supporting it isn't measured in Foot Pounds or Newtons?:

Work and energy use the same units, since work is just energy transferral from one body to another via mechanical force. So if not foot-pounds or newton-meters, you'll just be using some other unit that includes distance and force:

calories
electron-volts
ergs
horsepower-housr
joules
BTUs
ounce-inches

So no actually answering the actual question asked because some of you can't go past question one of the High School Physics mid-term about external work performed on an object? (Not the question asked.) Can y'all get the idea that actual physics work (F x D), in the strict physics sense, is being done within the muscles? And that producing that work (and lost to heat) is where the extra energy needed to support the object goes.

Haven't you heard how shitty Americans are at science? And now you (and the OP) want to complicate a fundamental question about mechanical work that most people already don't understand, by making them also think about how muscles work at a microscopic level? And you're surprised that people are struggling with it?

:D

It seems the OP's question has already been pretty thoroughly answered, thanks in part to your excellent explanations of how muscles work:

-No mechanical work is done on a mass held at constant altitude in earth's gravity.

-If the means of support for said mass is a living muscle that is actively contracting, the muscle uses chemical energy while it is exerting force; that energy is first converted to mechanical work which is dissipated, and ultimately manifests as heat, entirely within the muscle.

flight
08-14-2014, 08:52 AM
A hovering helicopter or a rocket hovering on a flaming jet certainly is not using zero energy.

Note that this is very different from the other examples as in this case simple obvious work is being done. Here it is just being done to the surrounding air and exhaust gasses.

GreatWyrmGold
08-14-2014, 12:39 PM
Exactly which why the statement that no physics work is being done is false.
The statement that no net work is done (what is usually meant) is true, however.

Note that this is very different from the other examples as in this case simple obvious work is being done.
The work being done (actually, the energy being expended) is merely more obvious. (And performed by a completely different mechanism, of course, but it was simply an analogy to show that energy can be spent without doing work.)

AnthonyElite
08-14-2014, 02:22 PM
Here's an addition to the question: how long would you have to hold it over your head to equate to the amount of work completed to get it up there to begin with?

AND:

How much additional force must be produced if someone is gently tickling your armpits with a feather?

HipGnosis
08-14-2014, 03:08 PM
It depends how (exactly) the weight is "held".
If you lay down and someone puts 100lbs of something on you, say taffy, as long as you lay still, you aren't doing work (though if enough of the taffy is on your chest, you're going to have to 'work' to breath).
This is akin to putting the 100 lbs of taffy in a box and putting the box on a table - no work done.

But if someone puts a 100 lb box of taffy in your arms, you have to work to hold onto the box. It's a different kind of 'work' than the mass*distance*time.
This kind of work was 'used' by Nuns that made students hold a small book in each hand at arms length for a bit of time... usually with a snarky remark like: 'Science says you don't work to hold the book in place, God says "the flesh is weak'''.

DrCube
08-14-2014, 04:37 PM
Your center of gravity is over the middle of your foot, basically the arches. If you hold anything away from that plane, you're creating a moment arm that your muscles must use strength to overcome in order to maintain balance. Same if it's a lopsided load, to your right or left, for example. If you are holding it directly over your center of gravity, you're basically only using grip strength and whatever strength it takes to keep your joints from coming out of their sockets. So the way you hold something has an enormous effect on the energy it takes to maintain its position.

This is why form is so important when lifting weights, and why some resistance exercises have larger effects than others. It's also why AnthonyElite's questions are going to be impossible to answer quantitatively.

Chronos
08-14-2014, 05:05 PM
As long as your center of gravity is over some part of your foot (strictly speaking, over the convex hull of your footprints), you don't need to expend energy to maintain that position. If your center of gravity isn't over your foot, then no amount of energy expenditure will prevent you from toppling over.

am77494
08-14-2014, 05:15 PM
As long as your center of gravity is over some part of your foot (strictly speaking, over the convex hull of your footprints), you don't need to expend energy to maintain that position. If your center of gravity isn't over your foot, then no amount of energy expenditure will prevent you from toppling over.

Not true. This is not static but dynamic equilibrium. Humans walking can be modeled as a series of falls where the step just stops it at the right moment. An inverted pendulum works on similar lines.

Chronos
08-14-2014, 06:24 PM
We're talking about standing, not walking.

EdwinAmi
08-14-2014, 06:44 PM
how long would you have to hold it over your head to equate to the amount of work completed to get it up there to begin with?

That depends on whether you're talking about the bare mechanical energy it took to lift it or the energy your body expended to lift it (which is more, due to inefficiencies in your muscles) and whether you're talking about the energy holding it up burned in the body, or burned in the body but adjusted to account for only the amount that comes out as mechanical ebergy

EdwinAmi
08-14-2014, 06:50 PM
no work is done ON the 100lb dumbbell. As you're standing, you're essentially acting as a table.

However, you are burning energy. Your body has to burn energy to maintain muscles in contraction mode which provides the forces necessary to kep you from falling over and dropping the weight. On the microscopic level, even when your limbs aren't moving, if they're contracing, the little fibers are "twitching", with this lever mechanism repeatedly stroking a fiber to get it to stay short, so there actually is motion on the microscopic level that could help one understand that work is in fact done, though not on the weight.

Here's a good analogy. You could hold something up with a jet of water also, keeping it at the same height. Your pump would have to keep working to keep the pressure up, even though no work is done ON the stationary weight. Of course, most hydraulic type mechanisms have a valve to prevent backflow so that whetever motion is done cant be undone, and the fluid can hold up the weight just fine. But it would be possible to create a hydraulic system without such a valve, where you'd have to continue pumping

In that analogy, the pump expends energy and does work, but it doesn't do anywork ON the weight it's holding up

DSeid
08-14-2014, 08:00 PM
The statement that no net work is done (what is usually meant) is true, however. ...

Not so sure. Would you accept no net work was done after I performed 100 150 pound bench presses leaving the bar in the same position it began in? No net work in exactly that same way, within motor units.

DSeid
08-14-2014, 08:06 PM
As long as your center of gravity is over some part of your foot (strictly speaking, over the convex hull of your footprints), you don't need to expend energy to maintain that position. If your center of gravity isn't over your foot, then no amount of energy expenditure will prevent you from toppling over.Absolutely untrue. Really a very silly statement.

Muscles need to be constantly firing, contracting and relaxing, motor units producing force across distances, expending energy, to maintain a vertical static position. Relax the muscles of a person standing directly over the center of gravity and they would fall to the ground.

EdwinAmi
08-14-2014, 08:29 PM
the problem with the term "net" energy is that it's a loose term.

The whole issue with no work being done when something isn't moving always has to be described as no work being done on the object. That's what I was told to say - it's work done on a thing. Which of course says nothing about energy burned/used up in the whole system/situation

DrCube
08-14-2014, 08:54 PM
As long as your center of gravity is over some part of your foot (strictly speaking, over the convex hull of your footprints), you don't need to expend energy to maintain that position. If your center of gravity isn't over your foot, then no amount of energy expenditure will prevent you from toppling over.
This is not quite true. You have to use muscles to maintain your center of gravity AND to counteract the moment of a weight held out at arm's length. If an uncentered weight was somehow attached to your body, you'd still have to use your muscles to maintain your center of gravity in a position over your midfoot, but not to counteract the moment on your shoulder.

The issue is that your body is not a rigid object. You have to use strength (aka "force") to maintain its position under load. So if you have a heavy backpack, relative to your bodyweight, you have to lean forward to keep your COG over your midfoot. If your muscles weren't using energy holding your body in that position, you'd topple over, like you said. Human balance is basically just strength. The stronger you are, i.e., the more force your muscles can generate, the more capable you are of maintaining your COG.

Try standing forever and tell me it doesn't use energy. :)

DrCube
08-14-2014, 08:58 PM
Not so sure. Would you accept no net work was done after I performed 100 150 pound bench presses leaving the bar in the same position it began in? No net work in exactly that same way, within motor units.
No net work on the barbell against gravity, because gravity is a conservative force, meaning the bar loses going down whatever energy you gave it going up. Now, your muscles on the other hand....

DSeid
08-14-2014, 10:59 PM
No net work on the barbell against gravity, because gravity is a conservative force, meaning the bar loses going down whatever energy you gave it going up. Now, your muscles on the other hand....

So help me further understand what you mean by net work -

I carry 120 pounds 100 yards forward and return the 100 yards to the same position. Work done? Same position, net distance travelled is zero, work = zero?

Not quite what is going on inside motor units though ...

I pull a 200 pound weight horizontally compressing a spring and release it, repeat 100 times, having the weight end at the same position it began in. Net work done? Zero? Do springs count like gravity as a conservative force?

A bit more like what is going on inside multiple individual motor units continuously during an isometric contraction of the complete muscle, except that the energy is actually used during the lengthening phase while the force is produced on the contraction and it is less a spring than fibers that slide across each other and back again.

EdwinAmi, the op started out expressing a clear understanding of the physics concept of work performed on an object and that holding an object stationary overhead performs no work, and asked if any work was being done internally. Why posters here can't get past making clear that the op gets that concept that the op stated clearly was understood is beyond me.

The op asked about the work being performed internally within the body, is there any and if so what sort? Answering that there is no work being performed on the object is not answering the question. The GQ answer is slightly more complicated but really not too hard to comprehend.

DrCube
08-15-2014, 08:55 AM
"Work", in the physical sense, is a type of energy transfer that requires (is defined by, actually) a force acting on a mass over a certain distance. "Work", in the colloquial sense, just means a transfer of energy. We might say that a toaster "works" on your bread by transferring heat to it. But that's not work in the F . D sense (http://en.wikipedia.org/wiki/Work_%28physics%29).

In your second example, though, moving an object horizontally and back, net work is performed. If you work against gravity (up and back down), you get back going down what you put in pushing it up. If you work against the friction of the ground, that's not the case. Gravity is a conservative (http://en.wikipedia.org/wiki/Conservative_force) force, friction is not.

DSeid
08-15-2014, 11:21 AM
For the purposes of this thread work is f x d (or the integral of to be more precise).

Let us imagine a frictionless surface. How does the spring one count then? Net work or not?

Chronos
08-15-2014, 12:10 PM
I'm going to need a more detailed description of what you're doing in the spring example, because it's probably not going to behave how you're expecting if it's on a frictionless table.

DSeid
08-15-2014, 01:13 PM
Okay.

A person uses x amount of force to compress a weight against a spring of some given stiffness coefficient k a given distance d on a frictionless surface. Releases. Waits until the weight settles into rest again (energy dissipating as heat, thanks second law). Repeats 100 times. Leaves with the weight in the same position as (s)he found it.

Work done in a strict physics is?

scr4
08-15-2014, 01:36 PM
If you push the spring to compress it, then yes, you've done work on the spring. In this case, force is not constant, so the amount of work done will be the integral of f(x)*dx over the distance. The energy is stored in the spring.

If you then release the spring and let it bounce back and forth until it stops, the spring has used the stored energy and converted it to heat and movement of air (which eventually turns into heat).

If, instead, you keep your hands on the spring and let it slowly extend to the original length, the spring has done work on you. The energy is dissipated in your muscles as heat.

DSeid
08-15-2014, 01:45 PM
Hands off other than during the multiple times compressing it. Spring ends in same position. Net work done?

scr4
08-15-2014, 01:49 PM
Hands off other than during the multiple times compressing it. Spring ends in same position. Net work done?

I'm not sure why "net work" would be a meaningful number here. But if you repeatedly do work on a spring, and the spring doesn't do any work on you, then obviously "net work" is positive (non-zero).

Chronos
08-15-2014, 02:04 PM
Either the system isn't frictionless, or it's not settling to rest. If it settles to rest, then ultimately, work is being done against whatever the force is that's making it settle. If it's frictionless, then work is done on the mass to get it oscillating, and the energy remains in the oscillating mass-spring system.

scr4
08-15-2014, 02:15 PM
I'm not sure why "net work" would be a meaningful number here. But if you repeatedly do work on a spring, and the spring doesn't do any work on you, then obviously "net work" is positive (non-zero).

By this I was referring to net work done by you, on the spring.

If you mean net of all work done on/by the spring, then it'll be zero for an ideal spring (i.e. no internal friction) if it ends up stationary at the same length as in the beginning. As Chronos notes, this will not happen if the spring itself and its environment are frictionless.

DSeid
08-15-2014, 02:46 PM
I'm not sure why "net work" would be a meaningful number here. But if you repeatedly do work on a spring, and the spring doesn't do any work on you, then obviously "net work" is positive (non-zero).
The comment was made that motor units are doing no "net work" since they end up in the same position and I am trying to clarify my understanding of what is meant by that.

DSeid
08-15-2014, 03:01 PM
Either the system isn't frictionless, or it's not settling to rest. If it settles to rest, then ultimately, work is being done against whatever the force is that's making it settle. If it's frictionless, then work is done on the mass to get it oscillating, and the energy remains in the oscillating mass-spring system.
I don't think so.

Spring is compressed/stretched/deformed and that deformation of the material cause some energy loss as heat - again, second law. Same as a vertical spring with a weight pulled down and released even in a vacuum - it won't be a perpetual motion machine.

On edit I see the comment about "internal friction" - not so idealized. There is some heat loss. I am trying to make a rough analogy to muscle function here.

scr4
08-15-2014, 03:37 PM
The comment was made that motor units are doing no "net work" since they end up in the same position and I am trying to clarify my understanding of what is meant by that.

When you compress the spring, you are doing work on the spring.

When you let the spring uncompress while still applying a force, the spring is doing work on you.

If the spring ends up where it started, then the net work is zero - you've got back all the energy you put into the spring.

DSeid
08-15-2014, 04:51 PM
And again, letting go of the spring without still applying force, "hands off" when not compressing it. I've received no energy back. The energy the spring lost was to heat loss due to material deformation in this case.

Again, I am only focusing on "net work" because it was claimed that such is "what is usually meant" when discussing work.

Hence we have established that do 100 reps of bench presses of 150 pounds does no physics work, in that usual sense, if the barbell is returned to its original position. (Gravity is conservative.) Individual motor units inside complete muscles function a bit differently, a slight bit closer to what I am attempting to set up as my hypothetical with the spring (albeit still a very imperfect analogy) and I am trying to get closer to understanding if by these terms there actually is no work being done by muscle fibers within the frame of reference of a body really ever, even though the body as a whole can perform work on objects.

Quartz
08-15-2014, 05:58 PM
Only while you are actually moving the weight. Work is force exerted across a distance (http://en.wikipedia.org/wiki/Work_(physics)); once you have raised the weight to its full height and begin holding it at a steady altitude, you have stopped delivering work to it.

Yes, but the OP asks about expending calories. And under Earth's gravity, he most certainly is.

scr4
08-15-2014, 06:26 PM
And again, letting go of the spring without still applying force, "hands off" when not compressing it. I've received no energy back. The energy the spring lost was to heat loss due to material deformation in this case.
I feel I'm repeating myself. If you push a spring to compress it, then let go, yes, you have done a non-zero amount of net work.

Individual motor units inside complete muscles function a bit differently, a slight bit closer to what I am attempting to set up as my hypothetical with the spring (albeit still a very imperfect analogy) and I am trying to get closer to understanding if by these terms there actually is no work being done by muscle fibers within the frame of reference of a body really ever, even though the body as a whole can perform work on objects.
I really don't see the analogy here.

If you are looking at a real-world situation (e.g. not ignoring friction), and consider holding a weight and moving it up and down, the individual muscle fibers do non-zero net work, because they are working against various frictional losses.

But I don't think that's why your muscles get tired when holding a weight in the air. They get tired because muscles burn energy just to keep contracted, like a helicopter that uses fuel just to stay in one spot. Or a car on an uphill slope using the clutch to stay still.

DSeid
08-15-2014, 07:25 PM
And talk about repeating one's self. No that is not how muscles work when engaged in an isometric contraction as in holding a weight overhead. They do not simply statically "keep contracted." Each muscle consists of multiple motor units each made up of variable numbers of muscle fibers Different motor units also have different sorts of muscle fibers with different contraction patterns. A muscle, even at apparent rest, is constantly firing some number of these motor units repetitively. A muscle engaged in significant isometric activity is firing these individual motor units more often and more of them at a time and also rotating through them as some fatigue. All of the fibers within any given motor unit fire in response to one motor neuron's command, contracting (by sliding fibers across each other) and producing force across a distance, and then with the addition of energy in the form of ATP elongating again to the same original length. That's where the increased caloric need comes in. And in energy lost as heat. We are far from perfectly efficient machines.

They do return to the same length after each contraction. The process of holding a weight steady overhead for ten minutes requires many thousands upon thousands of muscle fibers taking turns contracting, producing force across distance, and then returning to the same length, many many many thousands of times. But they do end at the same lengths on average. By your take on the spring analogy this seems to nevertheless count as net work.

Anyway a stationary aloft helicopter is also doing work, real physics work. It is moving a mass of air some distance which is what produces the force that keeps its weight aloft. I cannot comment too intelligently about the clutch example but I suspect highly that it is not like either of the other two.

scr4
08-15-2014, 07:37 PM
I still don't see what analogy you were trying to set up by talking about friction-less springs. Muscles are far from being friction-less.

DSeid
08-15-2014, 08:55 PM
Chronos was in response to my question focusing on the work as the work needed to overcome the friction, which was immaterial to the pertinent points of the crude and imperfect analogy. The major issue in muscle fiber function is not overcoming friction. Loss as heat, yes.