Question

You are riding on a roller coaster that starts from rest at a height of 25.0 m and moves down a frictionless track to a height of 3.00 m.

How fast are you moving when you arrive at the 3.00-m height?- 22.1 m/s
- 20.8 m/s
- 23.4 m/s
- 14.7 m/s

Answer 1

To solve this problem, we need to use both energy conservation and potential kinetic equations.

When the energy accumulated from a certain height is released, it becomes 'motion' energy or kinetic energy. Mathematically this can be expressed as

PE = KE

`mg\Delta h = (1)/(2)mv^2`

Where

m = mass

g = Gravitational acceleration

v = Velocity

`\Delta h =` Change of the height

We know that the body, based on a reference system where the floor is the zero coordinate, starts from being 25 meters high to fall to 3 meters high, so the total difference in height would be

`\Delta h = 25-3`

We also have to

`g = 9.8m/s^2`

Using the previous equation we have to:

`mg\Delta h = (1)/(2)mv^2`

`g\Delta h = (1)/(2)v^2`

`v = √(2g\Delta h)`

Replacing

`v = √(2(9.8)(25-3))`

`v = 20.76m/s\approx 20.8m/s`

The correct answer is 20.8m/s