A 10 kg load is evenly lifted to a height of 20 m in 10 seconds using a winch. The efficiency of the process is equal to 50%; the power of the winch motor is equal to.
Given: m (mass of the load being lifted) = 10 kg; h (lifting height) = 20 m; t (duration of the winch) = 10 s; η (process efficiency) = 50% (0.5); the movement of the load is uniform.
Constants: g (acceleration due to gravity) ≈ 10 m / s2.
The engine power of the used winch can be expressed from the formula: η = Np / Nl = (A / t) / Nl = m * g * h / (Nl * t), whence Nl = m * g * h / (t * η).
Let’s calculate: Nl = 10 * 10 * 20 / (10 * 0.5) = 400 W.
Answer: The motor of the used winch has a power of 400 W.
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