Question

A block slides down an inclined surface of inclination ${30}^{\circ }$ with the horizontal. Starting from rest it covers 8 m in the first two seconds. Find the coefficient of kinetic friction between the two.

Answer 1

From the free body diagram,
$R-mgcos\theta =0\Rightarrow R=mgcos\theta$
For the block u = 9, S = 8, t = 2 sec,
From $S=ut+\frac{1}{2}a{t}^{2}8=0+\frac{1}{2}a{2}^2\Rightarrow a=4m/s^2$
Again $\mu R+ma-mg\sin \theta =0$

From equation $\left(i\right)\mu mg\cos \theta +ma-mg\sin \theta$
$\Rightarrow m(\mu g\cos \theta +a-g\sin \theta )=0\Rightarrow \mu \times 10\times \cos {30}^{|circ}=gsin{30}^{\circ }=0\Rightarrow \mu \times 10\cos {30}^{\circ }=g\sin {30}^{\circ }-a\Rightarrow \mu \times 10\frac{\sqrt{3}}{2}=10\times \left(\frac{1}{2}\right)-4\Rightarrow \left(5\sqrt{3}\right)\mu =1\Rightarrow \mu =\left(\frac{1}{5\sqrt{3}}\right)=0.11$
Hence co-efficient of kinetic friction between the two is 0.11