Question

A 200 kg roller coaster is located at the top of a hill where the height is 30 m. What is the velocity when it reches the bottom of the hill?

Answer 1

Assuming it starts at rest, the roller coaster only has potential energy at the top of the hill, which is

`E_P=(200\,\mathrm{kg})(30\,\mathrm m)g`

When it reaches the bottom, its potential energy will have converted to kinetic energy,

`E_K=\frac12(200\,\mathrm{kg})v^2`

where
`v` is its velocity at that point. By the law of conservation of energy, assuming no loss of energy due to other sources (e.g. sound, heat), we have

`E_P=E_K\iff(6000\,\mathrm{kg}\cdot\mathrm m)g=(100\,\mathrm kg})v^2`

`\implies v=√(60g)\,(\rm m)/(\rm s)\approx24.2\,(\rm m)/(\rm s)`

Answer 2

A 200 kg roller coaster is located at the top of a hill where the height is 30 m. What is the velocity when it reches the bottom of the hill?

ANSWER 24.25