Question:
A block slides down an inclined surface of inclination $30^{\circ}$ with the horizontal. Starting from rest it covers $8 \mathrm{~m}$ in the first two seconds. Fins the coefficient of kinetic friction between the two.
Solution:
$N=m g \cos 30^{\circ}$
$F_{N}=m a$
$m g \sin 30-f f=m a$
$m g \sin 30-\mu_{k} N=m a$
$m g \sin 30-\mu_{k} m g \cos 30=m a$
$g \sin 30-\mu_{k} g \cos 30=a$
Now,
$u=0 ; s=8 m ; t=2 s$
$s=u t+\frac{1}{2} a t^{2}$
$8=0+\frac{1}{2}(a)(2)^{2}$
$a=4 \frac{m}{s^{2}}$
From (ii) and (iii)
$4=g \sin 30^{\circ}-\mu_{k} g \cos 30^{\circ}$
$\mu_{k}=0.11$
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