An elevator weighing 1.2 * 10 to the 3rd degree kg evenly rises to a height of 15m in 30s. The voltage at the terminals

An elevator weighing 1.2 * 10 to the 3rd degree kg evenly rises to a height of 15m in 30s. The voltage at the terminals of the electric motor is 380V, its efficiency is 90%. Determine the amperage in the electric motor.

It all comes down to the work that needs to be spent on lifting the elevator with the “handles” and the work that the elevator motor does.
Accordingly
A1 = Fh – the work that the elevator does during the ascent. Also, useful work.
A2 = Pt – the work done by the elevator motor. Also, full work.
Moreover, F = mg, P = UI
efficiency is the ratio A1 / A2
knowing all this, we make the equation
mgh = UIt * (efficiency)
from where
I = mgh / (Ut * (efficiency))
substituting given
I = 1.2 * 1000 * 9.8 * 15 / (380 * 30 * 0.9) = 17.2 A