# The crane lifts a load weighing 6 tons in 2 minutes to a height of 12 m.

**The crane lifts a load weighing 6 tons in 2 minutes to a height of 12 m. At the same time, the current of the electric motor is 51 A, voltage is 380 V. What is the efficiency of the crane?**

m = 6 t = 6000 kg.

g = 10 m / s2.

h = 12 m.

t = 2 min = 120 s.

I = 51 A.

U = 380 V.

Efficiency -?

Coefficient of efficiency The efficiency of the crane shows what part of the expended work Az goes into useful Ap.

Efficiency = Ap * 100% / Az.

The useful work Ap will be expressed by the formula: A = m * g * h, where m is the mass of the load lifted by the crane, g is the acceleration of gravity, h is the height of the load.

The expended work Az is expressed by the formula: Az = I * U * t, where I is the current in the electric motor of the crane, U is the voltage of the current, t is the operating time of the crane.

Efficiency = m * g * h * 100% / I * U * t.

Efficiency = 6000 kg * 10 m / s2 * 12 m * 100% / 51 A * 380 V * 120 s = 31%.

Answer: the crane has an efficiency of 31%.