# The crane lifts uniformly accelerated load weighing 1140kg from rest to a height of 10m. The electric motor of the crane

**The crane lifts uniformly accelerated load weighing 1140kg from rest to a height of 10m. The electric motor of the crane is powered from the 380V mains and at the end has an efficiency of 60%. The current in the motor winding is 102A. Determine the time of lifting the load.**

Given:

m = 1140 kilograms is the mass of the cargo;

h = 10 meters – the height to which the load is lifted;

U = 380 Volts – motor voltage;

I = 102 Amperes – electric motor current;

n = 60% = 0.6 – efficiency of the electric motor;

g = 10 m / s ^ 2 – acceleration of gravity.

It is required to determine t (seconds) – the time of lifting the load.

Let’s find the total power of the electric motor:

W = U * I = 380 * 102 = 38760 watts.

Then the useful power will be equal to:

W useful = W * n = 38760 * 0.6 = 23256.

To lift a load, you need to perform work equal to:

A = F * h = m * g * h = 1140 * 10 * 10 = 114000 Joules.

Then the time of lifting the load will be equal to:

t = A / W usable = 114000/23256 = 4.9 seconds.

Answer: the load will be lifted in 4.9 seconds.