An object, previously at rest, slides 5 m down a ramp, with an incline of (3pi)/8 , and then slides horizontally on the floor for another 15 m. If the ramp and floor are made of the same material, what is the material's kinetic friction coefficient?

Jul 20, 2017

The coefficient of kinetic friction is $= 0.27$

Explanation:

At the top of the ramp, the object possesses potential energy.

At the bottom of the ramp, part of the potential energy is converted into kinetic energy and the work done by the frictional force.

The ramp is $s = 5 m$

The angle is $\theta = \frac{3}{8} \pi$

Therefore,

$P E = m g h = m g \cdot s \sin \theta$

${\mu}_{k} = {F}_{r} / N = {F}_{r} / \left(m g \cos \theta\right)$

${F}_{r} = {\mu}_{k} \cdot m g \cos \theta$

Let the speed at the bottom of the ramp be $u m {s}^{-} 1$

so,

$m g \cdot s \sin \theta = \frac{1}{2} m {u}^{2} + s \cdot {\mu}_{k} \cdot m g \cos \theta$

$g s \sin \theta = \frac{1}{2} {u}^{2} + s {\mu}_{k} g \cos \theta$

${u}^{2} = 2 g s \left(\sin \theta - {\mu}_{k} \cos \theta\right)$

On the horizontal part,

The initial velocity is $= u$

The final velocity is $v = 0$

The distance is $d = 15 m$

The deceleration is calculated with Newton's Second Law

$F = F {'}_{r} = {\mu}_{k} m g = m a$

$a = - {\mu}_{k} g$

We apply the equation of motion

${v}^{2} = {u}^{2} + 2 a d$

$0 = 2 g s \left(\sin \theta - {\mu}_{k} \cos \theta\right) - 2 {\mu}_{k} g d$

$2 {\mu}_{k} g d = 2 g s \sin \theta - {\mu}_{k} 2 g s \cos \theta$

$2 {\mu}_{k} g d + {\mu}_{k} 2 g s \cos \theta = 2 g s \sin \theta$

${\mu}_{k} \left(2 g d + 2 g s \cos \theta\right) = 2 g s \sin \theta$

${\mu}_{k} = \frac{g s \sin \theta}{g d + g s \cos \theta}$

$= \frac{9.8 \cdot 5 \cdot \sin \left(\frac{3}{8} \pi\right)}{9.8 \cdot 15 + 9.8 \cdot 5 \cdot \cos \left(\frac{3}{8} \pi\right)}$

$= \frac{45.27}{165.75}$

$= 0.27$