(D) The work done $(W)$ to lift the coal is given by $W = mgh$, where $m = 8000 \ kg$, $g = 10 \ m \ s^{-2}$, and $h = 108 \ m$.$W = 8000 \times 10 \t

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(D) The work done $(W)$ to lift the coal is given by $W = mgh$, where $m = 8000 \ kg$, $g = 10 \ m \ s^{-2}$, and $h = 108 \ m$.$W = 8000 \times 10 \times 108 = 8.64 \times 10^6 \ J$.The time taken $(t)$ is $1 \ hour = 3600 \ s$.The useful power output $(P_{out})$ is $P_{out} = W / t = (8.64 \times 10^6) / 3600 = 2400 \ W = 2.4 \ kW$.Given the efficiency $(\eta)$ is $80 \% = 0.8$, the input power $(P_{in})$ is $P_{in} = P_{out} / \eta$.$P_{in} = 2.4 / 0.8 = 3 \ kW$.Thus, the power of the crane is $3 \ kW$.

Class 11 Physics Work, Energy, Power and Collision Power

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$A$ crane of efficiency $80 \%$ is used to lift $8000 \ kg$ of coal from a mine of depth $108 \ m$. If the time taken by the crane to lift the coal is one hour, then the power of the crane (in $kW$) is (Acceleration due to gravity $= 10 \ m \ s^{-2}$)

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$A$ crane of efficiency $80 \%$ is used to lift $8000 \ kg$ of coal from a mine of depth $108 \ m$. If the time taken by the crane to lift the coal is one hour, then the power of the crane (in $kW$) is (Acceleration due to gravity $= 10 \ m \ s^{-2}$)

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$A$ pump is required to lift $1000 \, kg$ of water per minute from a well of depth $10 \, m$ and eject it with a speed of $10 \, m/s$. The horsepower of the engine needed is: (Assume $g = 10 \, m/s^2$)

$A$ car of mass $m$ is driven with acceleration $a$ along a straight level road against a constant external resistive force $R$. When the velocity of the car is $V$,the rate at which the engine of the car is doing work will be

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