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  • If the maximum load carried by an elevator is 1400 kg (600 kg – Passengers + 800 kg - elevator), which is moving up with a uniform speed of <img alt="3\; m\; s^{-1}" src="https://entranc

If the maximum load carried by an elevator is 1400 kg (600 kg – Passengers + 800 kg - elevator), which is moving up with a uniform speed of 3\; m\; s^{-1} and the frictional force acting on it is 2000 N, then the maximum power used by the motor is __________ kW(g=10ms^{-1})

Option: 1

48


Option: 2

-


Option: 3

-


Option: 4

-


Answers (1)

best_answer

Tension \: in \: the\: string \Rightarrow 16000 \mathrm{~N}

\text { Maximum power }=(\mathrm{F})(\mathrm{V})

$$ \begin{aligned} & =16000 \times 3 \\ & =48000 \\ & =48 \mathrm{kw} \end{aligned}

 

Posted by

Rishabh

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