Calculate the work performed with uniformly accelerated lowering of a load weighing 200 kg to a height of 30 m in 16 s.

m = 200 kg.

g = 9.8 m / s2.

h = 30 m.

t = 16 s.

A -?

The work of force A is determined by the formula: A = F * S * сosα.

Since the load is lowered from a state of rest, we will find its acceleration a from the formula: h = a * t2 / 2.

a = 2 * h / t2.

a = 2 * 30 m / (16 s) 2 = 0.23 m / s2.

m * a = F – m * g.

The load is acted upon by a force F directed vertically upward, determined by the formula: F = m * (a + g).

∠α = 180 °, cos180 ° = -1.

A = – m * (a + g) * h.

A = – 200 kg * (0.23 m / s2 + 9.8 m / s2) * 30 m = – 60 180 J.

We express the work of the force of gravity Am by the formula: Am = m * g * h.

Am = 200 kg * 9.8 m / s2 * 30 m = 58800 J.

Answer: with uniformly accelerated lowering, gravity performs work Am = 58800 J, the force with which the rope is pulled does work A = – 60180 J.