# An object, previously at rest, slides 15 m down a ramp, with an incline of pi/3 , and then slides horizontally on the floor for another 6 m. If the ramp and floor are made of the same material, what is the material's kinetic friction coefficient?

Apr 9, 2016

$\approx 0.96$

#### Explanation:

Here length of the ramp (l)= 15m

Angle of inclination of the ramp $\theta = \frac{\pi}{3}$
Height of the object from horizontal floor,$h = l \sin \theta$
If mass ofthe body is $m k g$
The initial gravitational potential energy of the body $P E = m g h = m g l \sin \theta = 15 m g \sin \theta J$
The normal reaction acting on the body when it is sliding down the ramp is ${N}_{r} = m g \cos \theta$
and the corresponding frictional force ${F}_{r} = \mu {N}_{r} = \mu m g \cos \theta$
where $\mu =$coefficient of kinetc friction
work done against frictional force when sliding down the ramp ${W}_{1} = {F}_{r} \times l = \mu m g \cos \theta \times 15 J$

when the body slides on horizontal force,then normal reaction ${N}_{f} = m g$ and corresponding frictional force ${F}_{f} = \mu {N}_{f} = \mu m g$
work done against frictional force when sliding 6m along floor
${W}_{2} = {F}_{f} \times 6 = 6 \mu m g J$
Now applying conservation of mechanical energy we can write

The initial KE being zero
Initial PE = total work done against frictional force $= {W}_{1} + {W}_{2}$
$\therefore 15 m g \sin \theta = {W}_{1} + {W}_{2}$
$\implies 15 m g \sin \theta = 15 \mu m g \cos \theta + 6 \mu m g$
$\implies \mu = \frac{15 \sin \theta}{6 + 15 \cos \theta} = \frac{5 \sin \left(\frac{\pi}{3}\right)}{2 + 5 \cos \left(\frac{\pi}{3}\right)}$
$= \frac{5 \times \frac{\sqrt{3}}{2}}{2 + 5 \times \frac{1}{2}} = \frac{2.5 \times \sqrt{3}}{4.5} = \frac{5}{9} \times \sqrt{3} = \approx 0.96$