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(B) The work done $W$ in lifting a mass $m$ to a height $h$ against the gravitational force is given by $W = mgh$.Power $P$ is defined as the rate of

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$10^{5} 	ext{ W}$

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(B) The work done $W$ in lifting a mass $m$ to a height $h$ against the gravitational force is given by $W = mgh$.Power $P$ is defined as the rate of doing work,$P = \frac{W}{t}$.Given values are: $m = 10^{5} 	ext{ kg}$,$g = 10 	ext{ m s}^{-2}$,$h = 360 	ext{ m}$,and $t = 1 	ext{ hour} = 3600 	ext{ s}$.Substituting these values into the formula:$P = \frac{10^{5} 	imes 10 	imes 360}{3600}$$P = \frac{10^{6} 	imes 360}{3600}$$P = 10^{5} 	ext{ W}$.Thus,the power required is $10^{5} 	ext{ W}$.

Calculate the power of an engine required to lift $10^{5} \text{ kg}$ of coal per hour from a mine $360 \text{ m}$ deep. [Take $g = 10 \text{ m s}^{-2}$]

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