# A hoop rolls down a 5.25 m high hill without slipping. What is the final speed of the hoop, in...

## Question:

A hoop rolls down a {eq}5.25\ m {/eq} high hill without slipping. What is the final speed of the hoop, in meters per second?

## Law of Conservation of Energy:

The law of conservation of energy states that the sum of the kinetic an potential energies of the system must be equal before and after a physical process has occurred. The kinetic energy is given as {eq}\displaystyle \frac{1}{2}mv^2 {/eq}, where {eq}\displaystyle m {/eq} is the mass of the object and {eq}\displaystyle v {/eq} is its velocity, while the potential energy is known to be {eq}\displaystyle mgh {/eq}, where {eq}\displaystyle g {/eq} is the acceleration due to gravity and {eq}\displaystyle h {/eq} is its height.

Assuming that the energy of the hoop is conserved, we can equate the kinetic energy, {eq}\displaystyle E =\frac{1}{2}mv^2 {/eq}, where {eq}\displaystyle m {/eq} is the mmass of the object and {eq}\displaystyle v {/eq} is its velocity, to the potential energy, {eq}\displaystyle E = mgh {/eq}, where {eq}\displaystyle g {/eq} is the gravitational constant and {eq}\displaystyle h {/eq} is the height of the object. We solve for {eq}\displaystyle v {/eq}.

{eq}\begin{align} \displaystyle E_{kinetic} &= E_{potential}\\ \frac{1}{2}mv^2 &= mgh\\ v^2 &= 2gh\\ v &= \sqrt{2gh}\\ &= \sqrt{2\cdot 9.81\rm{m/s^2}\cdot 5.25\ m}\\ &= 10.2\ \rm{m/s} \end{align} {/eq}