An object, previously at rest, slides 3 m down a ramp, with an incline of pi/8 , and then slides horizontally on the floor for another 4 m. If the ramp and floor are made of the same material, what is the material's kinetic friction coefficient?

Feb 22, 2018

The coefficient of friction of the material is $= 0.17$

Explanation:

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

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

The length of the ramp is $s = 3 m$

The angle of the incline is $\theta = \frac{1}{8} \pi$

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 $\left(u\right) m {s}^{-} 1$

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 $d = 4 m$

The deceleration is calculated with Newton's Second Law of Motion

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

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

We apply the equation of motion

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

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

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

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

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

The acceleration due to gravity is $g = 9.8 m {s}^{-} 2$

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

$= \frac{9.8 \cdot 3 \cdot \sin \left(\frac{1}{8} \pi\right)}{9.8 \cdot 4 + 9.8 \cdot 3 \cdot \cos \left(\frac{1}{8} \pi\right)}$

$= \frac{11.25}{66.36}$

$= 0.17$