Calculate the power of an engine required to lift $10\ kg$ of coal per hour from a mine $360\ m$ deep. $( Take\ g=10\ ms^{-2})$.

Academic Physics NCERT Class 9

Here, mass $m=10\ kg.$, Time taken to lift the coal $t=1\ hr=60\times60=3600\ sec.$, Height $h=360\ m$

Energy used to lift the coal $W=mgh$

$=10\times10\times360$

$=360000\ Joule$

Thus, power used $P=\frac{W}{t}$

$=\frac{360000}{3600}$

$=100\ Watt$

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Updated on: 10-Oct-2022

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