A 60 HP electric motor lifts an elevator having a maximum total load capacity of 2000 kg. If frictional force on the elevator is 4000 N, the speed of the elevator at full load is close to _____ $m / s (Take 1 HP = 746 W, g = 10 {ms}^{- 2})$ (Round off to the nearest integer)

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answer is 2.

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Detailed Solution

$\begin{array}{l} F = mg + f \\ F = 24000 N \end{array}$
Power P= F.V
$\begin{array}{l} V = \frac{P}{F} = \frac{60 \times 746}{24000} \\ = 2 m / s \end{array}$

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