Assume you are pushing a heavy item. But it's not moving. So there is no work, and there should also be no dissipation of energy $P=F v=0$, since the forces don't work. But we see that it does demand energy to exert a force. For example, you need a motor with a batterie, or if I push myself, I'll get tired ! So where does the dissipation take place ?
If you push on a wall, it doesn't move. You have done no work in accordance with the physics definition of work. Similarly, if you hold a heavy weight and do not lift it, you do no physics work, even though you may tire due to the physical effort being made just holding the weight in place.
While you are not doing work as defined by physics, you are certainly burning calories as a result of your effort. You are doing, what Richard Feynman explains in his lectures, "physiological" work. He describes it as follows:
"What happens is that when a nerve impulse reaches a muscle fiber, the fiber gives a little twitch and then relaxes, so that when we hold something up , enormous volleys of nerve impulses are coming into the muscle, large numbers of twitches are maintaining the weight, while the others relax. We can see this of course: when we hold a heavy weight and get tired, we begin to shake. The reason is that the volleys are coming irregularly, and the muscle is tired and not reacting fast enough."
The implication of this is that the sum of all those muscle twitches (contractions and relaxations) requires energy, but that energy does not result in work being done external to the body.
Bottom line: Physiological effort does not always equate to mechanical work as defined by physics.
Hope this helps.