Question

Calculate the change in PE of 8 million kg of water dropping 50 m over Niagara falls

Answer 1

Answer:
`392(10)^(7) J`

Step-by-step explanation:

The gravitational potential energy
`PE` is the energy that a body or object possesses, due to its position in a gravitational field. That is why this energy depends on the relative height of an object with respect to some point of reference and associated with the gravitational force.

In the case of the Earth, in which the gravitational field is considered constant, the value of the gravitational potential energy is given by:

`PE=-mgh` (1)

Where:

`m=8(10)^(6) kg` ia the mass of water

`g=9.8 m/s^(2)` is the acceleration due gravity

`h` is the height of the object

Now, if we want to calculate the change in gravitational potential energy
`\Delta PE` we have to use the following equation:

`\Delta PE=PE_(f)-PE_(o)` (2)

Being
`PE_(o)=-mgh_(o)` the initial potential energy where
`h_(o)=50 m` is the initial height and
`PE_(f)=-mgh_(f)` the final potential energy where
`h_(f)=0 m` is the final height in the waterfall.

Solving:

`\Delta PE=-mgh_(f)-(-mgh_(o))` (3)

`\Delta PE=mg(-h_(f)+h_(o))` (4)

`\Delta PE=(8(10)^(6) kg)(9.8 m/s^(2))(-0 m+500 m)` (5)

Finally:

`\Delta PE=392(10)^(7) J` (6)