In 1 minute, the transport lifts a load weighing 300 kg to a height of 8 m. Transport efficiency is 60%.

| August 8, 2021 | education

In 1 minute, the transport lifts a load weighing 300 kg to a height of 8 m. Transport efficiency is 60%. Determine the strength of the current passing through the electric motor of the vehicle if the voltage in the network is 380 V.

t = 1 min = 60 s.

m = 300 kg.

g = 10 m / s2.

h = 8 m.

U = 380 V.

Efficiency = 60%.

I -?

Let’s write down the definitions for the efficiency of the hoist: efficiency = Ap * 100% / Az.

We will express the useful work Ap by the formula: Ap = m * g * h.

The expended work of the current in the motor of the hoist Az is expressed according to the Joule-Lenz law by the formula: Az = I * U * t.

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

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

I = 300 kg * 10 m / s2 * 8 m * 100% / 60% * 380 V * 60 s = 1.8 A.

Answer: in the hoist motor, the current strength is I = 1.8 A.



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