Solution:
Retardation \(a = g(\sin\theta + \mu\cos\theta) = 10(\sin 30^\circ + \frac{1}{\sqrt{3}}\cos 30^\circ) = 10(0.5 + 0.5) = 10\text{ m/s}^2\). Using \(v^2 = u^2 - 2as ⇒ 0 = 1^2 - 2(10)s\), we get \(s = 0.05\text{ m} = 5\text{ cm}\).
A body of mass 1 kg has velocity \(1\text{ ms}^{-1}\), up an inclined plane of angle of \(30^\circ\) to the horizontal. The friction coefficient is \(\frac{1}{\sqrt{3}}\). The distance the body travels before stopping is (\(g = 10\text{ m s}^{-2}\)):
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