What is the final speed of the hoop rolling down a 4.25 m high hill?

What is the final speed of the hoop, in meters per second?

Determining the Final Speed of the Hoop

When the hoop rolls down the 4.25 m high hill without slipping, its potential energy at the top of the hill is converted into kinetic energy at the bottom of the hill. The potential energy (PE) of the hoop at the top of the hill can be calculated using the formula:

PE = mgh

Where:
m = mass of the hoop
g = acceleration due to gravity (approximately 9.8 m/s^2)
h = height of the hill (4.25 m)

The kinetic energy (KE) of the hoop at the bottom of the hill is given by the formula:

KE = (1/2)mv^2

Where:
v = final speed of the hoop

According to the principle of conservation of energy, the potential energy at the top of the hill is equal to the kinetic energy at the bottom of the hill. Therefore, we can set the equations for potential energy and kinetic energy equal to each other:

mgh = (1/2)mv^2

Simplifying the equation and canceling out the mass (m) from both sides, we get:

gh = (1/2)v^2

Solving for the final speed (v), we rearrange the equation:

v^2 = 2gh

Taking the square root of both sides, we get:

v = √(2gh)

Substituting the given values, we have:

v = √(2 * 9.8 * 4.25)

Calculating this expression, we find that the final speed of the hoop is approximately 9.8 m/s.

← Understanding flux through a cylinder Driving a motor vehicle understanding reaction time →