Assuming it starts at rest, the roller coaster only has potential energy at the top of the hill, which is
[tex]E_P=(200\,\mathrm{kg})(30\,\mathrm m)g[/tex]
When it reaches the bottom, its potential energy will have converted to kinetic energy,
[tex]E_K=\dfrac12(200\,\mathrm{kg})v^2[/tex]
where [tex]v[/tex] 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
[tex]E_P=E_K\iff(6000\,\mathrm{kg}\cdot\mathrm m)g=(100\,\mathrm kg})v^2[/tex]
[tex]\implies v=\sqrt{60g}\,\dfrac{\rm m}{\rm s}\approx24.2\,\dfrac{\rm m}{\rm s}[/tex]
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
