An object, previously at rest, slides 7 m down a ramp, with an incline of (pi)/6 , and then slides horizontally on the floor for another 24 m. If the ramp and floor are made of the same material, what is the material's kinetic friction coefficient?

Apr 21, 2016

Here length of the ramp (l)= 7m

Angle of inclination of the ramp $\theta = \frac{\pi}{6}$
Height of the object from horizontal floor,$h = l \sin \theta$
If mass of the body is $m k g$
The initial gravitational potential energy of the body $P E = m g h = m g l \sin \theta = 7 m g \sin \theta J$

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 kinetic friction
work done against frictional force when sliding down the ramp ${W}_{1} = {F}_{r} \times l = \mu m g \cos \theta \times 7 J$

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 24m along floor
${W}_{2} = {F}_{f} \times 24 = 24 \mu m g J$

Now applying conservation of mechanical energy we can write

The initial KE being zero
Initial PE = total work done against frictional force $= {W}_{1} + {W}_{2}$
$\therefore 7 m g \sin \theta = {W}_{1} + {W}_{2}$
$\implies 7 m g \sin \theta = 7 \mu m g \cos \theta + 24 \mu m g$
$\implies \mu = \frac{7 \sin \theta}{24 + 7 \cos \theta} = \frac{7 \sin \left(\frac{\pi}{6}\right)}{24 + 7 \cos \left(\frac{\pi}{6}\right)}$
$= \frac{7 \times \frac{1}{2}}{24 + 7 \times \frac{\sqrt{3}}{2}} = \frac{3.5}{33} \approx 0.105$