Elephant talks about bringing the mass down ... by raising another of the lower mass.

I have the impression that you forget that the latter takes the potential energy of gravity during its ascent.

The kinetic energy acquired by the heaviest mass is therefore much lower than that which would be acquired by free fall ...

Let 1 be the index for mass 1,

Let 2 be the index for mass 2

We assume the altitude h for the descending mass, and 0 for the rising. At the end of the movement, the altitudes are swapped.

The system evolves to

**constant mechanical energy**. The mechanical energy balance gives:

m1 gh = 1/2 (m1 v1² + m2 v2²) + m2 gh

As v1 = v2 by inextensibility of the cable ...

**(m1 + m2) v² = 2gh x (m1-m2)**

From where v² = 2gh (m1-m2) / (m1 + m2)

When m2 = 0, we find the famous expression "v² = 2gh" of free fall.