Question

Find the power of an engine required to lift ${10}^5$$kg$ of coal per hour from a mine of $360m$ deep.

Answer 1

Step 1: Given data:

Mass, $m={10}^{5}kg$

Height, $h=360m$

Time, $t=1h\Rightarrow 60\times 60=3600s$

Assumption: $($acceleration due to gravity, $g=10m/s^2$$)$

Step 2: Finding the work done with mass, height and acceleration due to gravity:

Work done is given by,

$W=mgh$

Where $m$ is mass, $g$ is acceleration due to gravity and $h$ is height.

$W={10}^5\times 10\times 360$

$=3600\times {10}^{5}J$

Step 3: Finding the power with work done and time:

Power is given by,

$P=\frac{W}{t}$,where $P$ is power, $W$ is work done and $t$ is time.

$P=\frac{3600\times {10}^5}{3600}$

$P={10}^{5}watt$

Hence power of an engine required to lift ${10}^5$$kg$ of coal per hour from a mine of $360m$ deep is${10}^{5}watt$.