# The electric motor of the crane with a power of P = 1.47 kW lifts the load at a speed of modulus = 50 mm / s.

**The electric motor of the crane with a power of P = 1.47 kW lifts the load at a speed of modulus = 50 mm / s. Find the mass of the load to be lifted if the efficiency of the crane motor is 60%.**

To find out the required mass of the lifted load, we use the formula (we assume that the load is lifted at a constant speed): ηel = Pp / Rzatr = Ft * V / Pcr = mx * g * V / Pcr, whence we express: mx = Pcr * ηel / ( g * V).

Const: g – acceleration due to gravity (g ≈ 9.81 m / s2).

Data: Pcr – crane power (Pcr = 1.47 kW = 1.47 * 10 ^ 3 W); ηel – the efficiency of the electric motor (ηel = 60% = 0.6); V is the speed of lifting the load (V = 50 mm / s; in the SI system V = 0.05 m / s).

Let’s perform the calculation: mx = Pcr * ηel / (g * V) = 1.47 * 10 ^ 3 * 0.6 / (9.81 * 0.05) ≈ 1.8 * 10 ^ 3 kg.

Answer: The mass of the load to be lifted should be 1.8 tons.