What is the power of the electric motor if it is known that in 20s the electric motor lifts a load weighing 150kg to a height of 12m? The efficiency of the electric motor is 60%.
Given: t (duration of work) = 20 s; m (mass of the lifted load) = 150 kg; h (lifting height) = 12 m; η (efficiency of the motor used) = 60% (0.6).
Constants: g (acceleration due to gravity) ≈ 10 m / s2.
The power of the electric motor used can be expressed from the formula: η = Np / Nd = A / (t * Nd) = m * g * h / (t * Nd), whence Nd = m * g * h / (η * t).
Let’s calculate: Nd = 150 * 10 * 12 / (0.6 * 20) = 1500 W (1.5 kW).
Answer: The electric motor used has a power of 1.5 kW.
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