An engine can pump 40,000 liters of water to the vertical height of 35 meters in 5 minutes. Calculate the gravitational potential energy of water at given height.

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Solution

Density of water $= 10^{3} k g m^{- 3} : 1000 l i t e r s = 1 m^{3}$
Volume of water $=$ 40,000 liters
$= \frac{40, 000}{1000} = 40 m^{3}$
Mass of water, $m = d e n s i t y \times v o l u m e$
$= 1000 \times 40$
$= 40, 000 k g$
$P E = m g h$
$= 40, 000 \times 9.8 \times 35 = 1.37 \times 10^{7} J$

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An engine can pump 40,000 liters of water to the vertical height of 35 meters in 5 minutes. Calculate the gravitational potential energy of water at given height.

Density of water =103kgm−3:1000liters=1m3Volume of water = 40,000 liters=40,0001000=40m3Mass of water, m=density×volume=1000×40=40,000kgPE=mgh=40,000×9.8×35=1.37×107J

Density of water $= 10^{3} k g m^{- 3} : 1000 l i t e r s = 1 m^{3}$
Volume of water $=$ 40,000 liters
$= \frac{40, 000}{1000} = 40 m^{3}$
Mass of water, $m = d e n s i t y \times v o l u m e$
$= 1000 \times 40$
$= 40, 000 k g$
$P E = m g h$
$= 40, 000 \times 9.8 \times 35 = 1.37 \times 10^{7} J$