# What is the final velocity of a solid disk that rolls without slipping down a 6.8 m high hill?...

## Question:

What is the final velocity of a solid disk that rolls without slipping down a 6.8 m high hill? Assume that it started from rest.

## Conservation of Energy:

The law of conservation of energy states that the sum of the potential, {eq}\displaystyle E_{pot} {/eq}, and kinetic {eq}\displaystyle E_{kin} {/eq}, must be equal before and after an occurrence of a physical phenomenon, wherein the system has no external forces acting on it.

Determine the velocity of the solid disk by equating the potential energy while it is at rest, {eq}\displaystyle E = mgh {/eq}, where {eq}\displaystyle m {/eq} is the mass of the object, {eq}\displaystyle g {/eq} is the acceleration due to gravity, and {eq}\displaystyle h {/eq} is the height, to the kinetic energy at the surface, {eq}\displaystyle E = \frac{1}{2}mv^2 {/eq}, where {eq}\displaystyle v {/eq} is the velocity of the disk. We solve for {eq}\displaystyle v {/eq} and plug in the given values to find the answer.

{eq}\begin{align} \displaystyle mgh &= \frac{1}{2}mv^2\\ 2gh &= v^2\\ \sqrt{2gh} &= v\\ \sqrt{2\cdot 9.8\ \rm{m/s^2}\cdot 6.8\ m} &= v\\ 11.5\ \rm{m/s} &= v \end{align} {/eq}