I'm trying to understand work and potential energy, specifically using the example of a human climbing against gravity to a high point, then falling under the force of gravity.

My understanding is that energy ($E$) is transferred to a system $S_1$ by another system $S_2$ doing positive work ($+\Delta W$) on system $S_1$. Potential energy $U_1$ of $S_1$ increases $+\Delta U_1$ as system $S_2$ does work $+\Delta W_2$ on $S_1$, while $S_1$ does negative work $-\Delta W_1$ on itself which results in $-\Delta U_2$. Is this understanding correct?

I then assign the variables: $S_1 =$ Body, $S_2 =$ Muscles/Metabolism. For the climb, $\Delta U_1 = zJ, \Delta U_2 = -zJ$. For the fall opposite, $\Delta U_1 = -zJ, \Delta U_2 = zJ$.

Does this appear correct?

  • $\begingroup$ Aren’t the muscles part of the body? $\endgroup$
    – Dale
    Commented Aug 24, 2021 at 23:58
  • $\begingroup$ @Dale can I treat the body being a container as one system, with the muscles as another system inside it? Unsure how to formulate this $\endgroup$
    – notaorb
    Commented Aug 25, 2021 at 0:10
  • $\begingroup$ How can a object do work on itself? Do you mean internal work? If so, the relationship between internal work and potential energy is $W_\mathrm{external} = -\Delta U_\mathrm{conservative}$ (this is the definition of potential energy). Here are some difficulties with your presentation: 1) you don't clearly make the distinction between internal and external work. 2) your object (body) is deformable. 3) the system(s) is(are) open (heat is generated). It seems that you do not have a clear understanding of Newton's Laws, and are applying that understanding to a very complex system. $\endgroup$
    – garyp
    Commented Aug 25, 2021 at 11:50
  • $\begingroup$ When learning, it's hard to know when one's understanding is adequate. But I think you are moving too fast. A hint of this is that you write $\Delta W$. Work is not a quantity that can change. It is an amount of energy. An object's energy can change by giving it an amount of energy either by work or heat. Slow down, sit back. make sure you understand the basics. As I said, it's hard to know when you've achieved understanding. A clue: potential energy is not defined for a single point-like object. If you understand this, and why I said "point-like", you are on your way. $\endgroup$
    – garyp
    Commented Aug 25, 2021 at 12:01

1 Answer 1


A person can do squats for hours and lower the potential energy stored in their metabolism. They are doing positive work against gravity, and then gravity is doing positive work on them, but they are not using that return work to increase their stored energy. It’s going as heat and as wear and tear which is just inefficient increase in entropy by breaking mechanical bounds and increasing the stored energy that was low when in the bound. Then they need to be repaired.

So the change in height potential energy is small compared to the work done. Trying to balance that way won’t work because along the way, work done on the person does not get stored (otherwise he would fall and suddenly have higher internal energy stores that he replenished by using the impact energy to re-up.)

Overall this is not a situation for applying the concept of mechanical work. The systems are too complex. Negative work is not returned. Inefficiencies just go off to the environment as heat and mechanical bound entropy. There are other forms of energy than mechanical. Work is good for situations where there aren’t, or where the conversions between types are clear.

Especially when processes are reversible https://en.m.wikipedia.org/wiki/Reversible_process_(thermodynamics)


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