Question: In the Niagara Falls hydroelectric generating plant, the energy of falling water is converted into electricity. The height of the falls is about 50m. Assuming that the energy conversion is highly efficient, approximately how much energy is obtained from one kilogram of falling water? Therefore, approximately how many kilograms of water must go through the generators every second to produce a megawatt of power 1 X 10⁶W?

The phrase "energy" refers to a person's ability to perform tasks. The potential energy can be defined as kinetic energy, thermal energy, electrical energy, chemical energy, nuclear energy, and other types of energy. There's also the issue of heat and work (energy transmitted from one body to another).

• The mass of the water is m = 1 kg .
• The height of the water from falling is h = 50 m

The energy owned or obtained by an object as a result of a change in its position while in a gravitational field is known as gravitational potential energy.

The energy obtained from 1 kW of falling water by using the formula of the potential energy is where m denotes mass. The g is acceleration due to gravity that is given by , h stands for height.

The formula of work is, where Pis power and tis time.

Substitute and into the formula of work.

$$\begin{gathered} W={10}^6\times 1 \\ ={10}^6\text{J} \end{gathered}$$

Therefore, the work is ${10}^6\text{J}$.

The potential energy is given by U = mgh.

Substitute, m= 1 kg,$g=9.8{\text{m/s}}^2$and h= 50 m into the formula of potential energy.

$$\begin{gathered} U=1\times 9.8\times 50 \\ ={\text{490Js}}^{-\text{1}} \end{gathered}$$

Therefore, the potential energy of 1kg of water is 490 S^-1.

Divide the work by the potential energy to find the mass of the water.

Therefore, the mass of the water is and energy created from 1 kg is 490 S^-1 .