Question

how much force is exerted to lift 750 kg of water in the well against the force of gravity? How much work is done to lift this water 0.05m? How does this compare to the work done by the man in the treadmill?

Answer 1

1) Force required to lift the water (its weight)
F = m g = 750 kg × 9.8 m/s² = 7 350 N (≈ 7.35 × 10^3 N).

2) Work to lift that water 0.05 m
W = F d = (7 350 N) × (0.05 m) = 367.5 J ≈ 3.7 × 10^2 J.

3) Comparison with a man on a treadmill
- If the treadmill is flat and the man runs in place, his center of mass does no net vertical displacement relative to the ground, so the net mechanical work against gravity is essentially zero (neglecting small vertical oscillations). Muscles still expend metabolic energy, but that energy is mostly dissipated as heat rather than doing external gravitational work.
- If the treadmill (or the man) raises his body by 0.05 m, the work against gravity is W = m_man g 0.05. For example, a 75 kg man: W = 75×9.8×0.05 ≈ 36.8 J, which is about one-tenth of the work to lift the 750 kg of water 0.05 m.

If you have a specific man’s mass, treadmill incline, or time/power values, I can give a direct numerical comparison.