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

$mg\Delta h = \frac{1}{2}mv^2$

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:

$mg\Delta h = \frac{1}{2}mv^2$

$g\Delta h = \frac{1}{2}v^2$

$v = \sqrt{2g\Delta h}$

Replacing

$v = \sqrt{2(9.8)(25-3)}$

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

The correct answer is 20.8m/s