Calculate the power of an engine required to lift $10^{5} k g$ of coal per hour from a mine $360 m$ deep. Consider $g = 10 m s^{- 2}$ .

1 k W

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10 k W

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100 k W

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1000 k W

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Solution

The correct option is A $100 k W$
$m = 10^{5} k g, g = 10 m s^{- 2}, h = 360 m, t = 1 h = 3600 s$
The work needed in lifting a mass $m$ to a height $h$ against the force due to gravity is $W = m g \times h$
And, power, $P = \frac{W}{t} = \frac{m g h}{t}$
$∴ P = \frac{10^{5} \times 10 \times 360}{3600} W = 10^{5} W = 100 k W$

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