Energy transfer involved in lowering a weight

Question

Suppose I lift a weight of 1kg through 1metre, say from the floor to my waist height. The Potential Energy or work done against gravity is:

PE=mgh= 1kg x 9.81m/s^2 x 1m = 9.81J or approximately 10 Joules.

My question is: what happens to that Potential Energy when I return the weight to the floor? Clearly, if I simply release the weight from its 1m height, it will fall back to the floor and all of the Potential Energy will be converted into Kinetic Energy (until it hits the floor at speed). But what if, instead of dropping the weight, I slowly lower it and bring it gently to rest on the ground?

Is it correct to say that those 10Joules of Potential Energy are transferred to my body and converted into other forms, e.g. heat? (If not, then what happens to the Potential Energy?)

Lelouch · Answer

If the process is quasistatic, no energy is lost as heat, which is obviously impossible in the real case.

Wolphram jonny · Answer

Suppose you move the body down at constant speed, as small as you like, then the net force on the weight will be zero, that is, the force upwards you make will be exactly the same as the weight, $mg$. This upwards force makes negative work, as the displacement is opposite to its direction, and this work is exactly equal to the loss of potential energy, $mgh$. The potential energy was transferred to you in the form of work, if it becomes stored or dissipated is a matter of biology at this point, perhaps the muscles have some mechanism to store part of the energy, but most likely a large portion will be dissipated as heat.

Answered by Wolphram jonny on December 21, 2020

Energy transfer involved in lowering a weight

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