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A body slides over an inclined plane forming an angle of $45^{o}$ with the horizontal. The distance x travelled by the body in time t is described by the equation $x = k t^{2}$ , where $k = 1.732$ . The coefficient of friction between the body and the plane has a value $(\sqrt{6} = 2.45 a n d g = 9.8 m / s^{2})$

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$μ = 0.5$

$μ = 1$

$μ = 0.25$

$μ = 0.75$

answer is A.

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Detailed Solution

$S i n c e, x = k t^{2} = 1.732 t^{2} \Rightarrow x = \sqrt{3} t^{2} \Rightarrow x = 2 \sqrt{3} t \Rightarrow x = a = 2 \sqrt{3} m s^{- 1}$

$\Rightarrow a = g (\sin 45 - μ \cos 45) = 2 \sqrt{3} \Rightarrow g \frac{1}{\sqrt{2}} (1 - μ) = 2 \sqrt{3} \Rightarrow 1 - μ = \frac{2 \sqrt{6}}{g} \Rightarrow μ = 1 - \frac{2 \sqrt{6}}{g} = 1 - \frac{2 (2.45)}{9.8} \Rightarrow μ = 0.5$

Hence, the correct answer is (A)

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