The statement that no net work is done (what is usually meant) is true, however.
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.)
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?
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 massdistancetime.
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’’’.
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.
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.
We’re talking about standing, not walking.
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
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
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.
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.
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
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. 
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.
“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.
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 force, friction is not.
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?
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.
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?
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.