Isn't heat production a kind of work?

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The discussion highlights the distinction between work done on a weight versus work done on surrounding air molecules when holding a heavy object. While textbooks claim no work is accomplished on the weight due to gravity, microscopic work occurs as the body exerts force on air molecules, causing movement. This input work stabilizes muscles but results in zero output work on the weight itself, leading to a 0% efficiency scenario. The conversation emphasizes the need for clearer explanations in educational materials regarding these nuances of work and energy. Overall, the complexities of work in physics extend beyond simple definitions, warranting further clarification in course content.
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One often read in coursebooks that no work is done when (a living human is) holding a (heavy) weight. But is there not microscopic work done when you are standing there shaking like jelly. You have become a little power plant and power plants do work, don't they? There is a certain power, energy per time unit, and creating energy is work, isn't it?
 
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The textbooks are specific to work being done on the weight, not the air molecules around you. You are correct, work is being done on the air molecules because you are applying a force to them and they are moving.

In the case of the weight you are applying a force however it is not moving since gravity is working destructively against you, no work is accomplished.
 
It's a difference between input and output. It takes a lot of input work to hold your muscles steady, but you are doing no output work on the weight you are holding. You might say that in that situation, your body is 0% efficient.
 
Ok, so there is not only work done, but even useful work. Because air molecules of higher temperature will rise up and overcome gravitational potential. And therefore create a force times distance, which is work by definition. But there is no work done on the weight.

I wish the books could be more clear on this point. Thanks for replies.
 
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