An object, previously at rest, slides 8 m down a ramp, with an incline of π/6 , and then slides horizontally on the floor for another 5 m. If the ramp and floor are made of the same material, what is the material's kinetic friction coefficient?

Nov 27, 2016

$\frac{4}{5 + 4 \sqrt{3}}$

Explanation:

Calling the incline angle $\theta = \frac{\pi}{6}$,

${d}_{1} = 8 , {d}_{2} = 5$ and ${\mu}_{k}$ the friction coefficient, we have

${d}_{1} m g \sin \theta = {\mu}_{k} m g {d}_{1} \cos \theta + \frac{1}{2} m {v}_{0}^{2}$

Here the left term represents the potential energy regarding the low level floor. In the right side we have the work done by the friction force and the kinetic energy associated to the remaining speed ${v}_{0}$

After the ramp we have

$\frac{1}{2} m {v}_{0}^{2} = {\mu}_{k} m g {d}_{2}$ which represents the work consumed by the friction forces.

Substituting into the first equation we have

${d}_{1} m g \sin \theta = {\mu}_{k} m g {d}_{1} \cos \theta = {\mu}_{k} m g {d}_{2}$

solving for ${\mu}_{k}$ we get

${\mu}_{k} = \frac{{d}_{1} \sin \theta}{{d}_{2} + {d}_{1} \cos \theta} = \frac{4}{5 + 4 \sqrt{3}}$