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 ms^-1 and the frictional force acting on it is 2000N, then the maximum power used by the motor is ______kW. ( $g = 10 m / s^{2}$ )

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answer is 48.

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Detailed Solution

Net force exerted by string on elevator.
$\begin{array}{l} T = (600 + 800) \times 10 + 2000 \\ T = 16000 N \end{array}$
Power delivered by motor
$\begin{array}{l} P = \vec{T} \cdot \vec{v} = 16000 \times 3 = 48000 watt \\ P = 48 k W \end{array}$

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