A body is made to move up along an inclined plane of inclination 30^o and the coefficient of friction is 0.5, then its retardation is
(g - acceleration due to gravity)

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$(\frac{2 + \sqrt{3}}{4}) g$

$(\frac{2 + \sqrt{3}}{2}) g$

$(\frac{2 - \sqrt{3}}{4}) g$

$(\frac{2 - \sqrt{3}}{2}) g$

answer is A.

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Detailed Solution

The forces acting are friction and component of weight.

$F = m g \sin θ + μ m g \cos θ = \frac{m g}{2} + \frac{m g \sqrt{3}}{4}$

$F = (\frac{2 + \sqrt{3}}{4}) m g = m a$

So, $a = (\frac{2 + \sqrt{3}}{4}) g$

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