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(A) The acceleration $a$ of a block sliding down a rough inclined plane is given by:$a = g \sin 	heta - \mu g \cos 	heta = g(\sin 	heta - \mu \cos

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$\frac{1}{5\sqrt{3}}$

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(A) The acceleration $a$ of a block sliding down a rough inclined plane is given by:$a = g \sin 	heta - \mu g \cos 	heta = g(\sin 	heta - \mu \cos 	heta)$Given: $u = 0$,$s = 8 \ m$,$t = 2 \ s$,$	heta = 30^{\circ}$,$g = 10 \ m/s^2$.Using the equation of motion $s = ut + \frac{1}{2}at^2$:$8 = 0 	imes 2 + \frac{1}{2} 	imes a 	imes (2)^2$$8 = 2a \implies a = 4 \ m/s^2$Now,substitute the expression for $a$:$4 = 10(\sin 30^{\circ} - \mu \cos 30^{\circ})$$4 = 10(\frac{1}{2} - \mu \frac{\sqrt{3}}{2})$$4 = 5 - 5\sqrt{3}\mu$$5\sqrt{3}\mu = 5 - 4 = 1$$\mu = \frac{1}{5\sqrt{3}}$

$A$ block slides down on 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 friction. (Take $g = 10 \ m/s^2$)

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