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A small block starts sliding down an inclined plane forming an angle $α$ with the horizontal. The friction coefficient depends on the distance $x$ covered by block as $μ = b x$ where b is a constant. Find its maximum speed during sliding

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$\sqrt{\frac{b \cos α}{g \sin^{2} α}}$

$\sqrt{\frac{g \sin^{2} α}{b \cos α}}$

$\sqrt{\frac{g \cos^{2} α}{b \sin α}}$

$\sqrt{\frac{b \sin α}{g \cos^{2} α}}$

answer is C.

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Detailed Solution

$a = gsinα - μgcosα$

$a = v \frac{dv}{dx}$

$\Rightarrow v = \sqrt{2 gx sinα - {bgx}^{2} cosα}$ .......(1)

Velocity is maximum when acceleration is zero.

$\begin{array}{l} Put a = 0 and μ=bx \\ \Rightarrow gsinα - μgcosα = 0 \\ \Rightarrow x = \frac{tanα}{b} \end{array}$

Using eqn (1),

$\Rightarrow v_{\max} = \sqrt{\frac{{gsin}^{2} α}{bcosα}}$

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