Question

A 100 kg roller coaster car is located at the top of a hill where the height is 20 m. What is the velocity when it reaches the bottom of the hill? 5.5 m/s 11.2 m/s 16.7 m/s 19.8 m/s

Answer 1

The correct answer is 19.8 m/s                                                               

Answer 2

Answer:

19.8 m/s

Explanation:

Assuming the car starts from rest, initially when it is at the top of the hill all the mechanical energy of the car is just potential energy, given by:

$U=mgh$

where

m = 100 kg is the mass

g = 9.8 m/s^2 is the gravitational acceleration

h = 20 m is the height

Substituting,

$U=(100 kg)(9.8 m/s^2)(20 m)=19,600 J$

At the bottom of the hill, all this energy has been converted into kinetic energy (because energy cannot be created or destroyed, but only transformed). Therefore, the kinetic energy of the car is:

$K=\frac{1}{2}mv^2=19,600 J$

where

m = 100 kg is the mass of the car

v is the speed of the car at the bottom of the hill

Re-arranging this equation, we can find the value of v:

$v=\sqrt{\frac{2K}{m}}=\sqrt{\frac{2(19,600 J)}{100 kg}}=19.8 m/s$