I recently answered a similar question. It was about the work done by a man while carriying a 50Kg suitcase over his head, traveling an horizontal distance of 15m.
The answer I gave about the work done was:
If your height is, lets say, 1.80m, then the total mechanical work would be the work done while lifting the suitcase, plus some “extra” work due to the inefficiencies of the human body machinery internals.
So, that should amount to not less than mgh = 50*10*1.8 = 900 Joules.
Imagine the following case:
John lifts 10 50Kg potato sacks from the ground to the platform of a truck, which is at a height of 1.5m.
He drives to the destination, and afterwards John downloads the 10 potato sacks, very carefully and gently putting them on the ground … he doesn’t want to spoil the potatos.
John’s boss says: Well John, for every potato sack your body, as a system, did a negative mechanical work when lifting the sack, but afterwards you body did a positive mechanical work while setting down the sack, and the net balance is that you did a zero amount of mechanical work at the end. I only pay you for driving the truck!
But John, who likes mechanic artifacts, quickly uses a piece of paper and pencil and draws this invention:
When you turn on the machine, a “chemical” motor inside the A block of the machine pushes the piston B, so the hydraulic fluid pushes the piston C, and the machine lifts the mass M to a certain height h.
When you turn off the machine, the motor in the A block of the machine stops doing any force to the B piston, but some brake-claws in the C piston at F, driven by a microcontroller device, begin to do enough normal force over the cylinder surface, as to stop (by friction) the movement of the mass M exactly at the starting point 1 (so, the mass M lands over the starting 1 position at zero speed, with no kinetic energy).
John’s reasonig is the following:
The machine is the mechanical equivalent of my body workings.
When I lifted the mass, the machine did a negative mechanical work, which was more or less mgh plus some “extra” work also to displace upwards the gravity center of the hydraulic fluid mass inside the cylinder.
But when I set down the mass (turned off the motor), the machine did no net mechanical work. The internal friction forces did a negative mechanical work (always opposed to the displacement), which exactly compensated the positive mechanical work from the push of the mass over the machine platform.
What do you think?
P.D. To avoid any further arguments with his boss, John decided to just drop the potato sacks to the ground in the future, because in that scenario was clearer that he did not any mechanical work during the setting down phase.