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Calculate the potential energy associated with 1 m^3 of water at 607 feet tall taking the mass of 1 m^3 of water to be 1000 kg

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Answer:

The potential energy associated with the given mass equals 1814.98 kilo Joules.

Explanation:

We know that for a object of mass 'm' standing at a height of 'h' meters above the surface of earth the potential energy associated with the object is given by


P.E=mass* g* h

where

'g' is acceleration due to gravity.

Since it is given that mass of 1 cubic meter of water is 1000 kilograms that the mass associated with given quantity of water is also 1000 kilograms since the volume is 1 cubic meter.

The height is given as 607 feet =
{607}* 0.3048=185.0136meters

Applying the values in the above equation we get


P.E=1000* 9.81* 185.0136=1814.98kJ

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