# Water drops over 49 m high Niagara Falls at the rate of 6.0 * 10^6 kg/s. If all the energy of the falling water could be harnessed by a hydroelectric power plant, what would be the plant's power output?

##### 1 Answer
Nov 13, 2015

$2881.2 \text{MW}$

#### Explanation:

The potential energy is given by:

$P E = m g h$

In 1 second the potential energy of $6 \times {10}^{6} \text{kg}$ of water is:

$6 \times {10}^{6} \times 9.8 \times 49 = 2881.2 \times {10}^{6} \text{J}$

This means that this amount of energy is being produced every second.

This is what power output refers to. 1 Watt = 1 Joule per second.

So the plant's power output is :

$2881.2 \times {10}^{6} \text{W}$

or :

$2881.2 \text{MW}$