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 ?
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.
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.
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.
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.
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.
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.
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.
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.
Only while you are actually moving the weight. Work is force exerted across a distance; 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.
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.
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:
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?
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.
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.
