A cubic block of mass m is sliding down on an inclined plane at 60° with an acceleration of g/2, the value of coefficient of kinetic friction is (a) √(3)/2 (b) √(2)/3 (c) 1 - √(3)/2 (d) √(3) - 1

Answer: (d). Sliding down a rough incline: a = g(sin θ - μcos θ). g/2 = g(√(3)/2 - μ/2) μ = √(3) - 1 So the answer is (d) √(3) - 1.

A cubic block of mass m is sliding down on an inclined plane at 60°… | JEE Main 2025 (7 Apr, Shift 1)

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Friction on an Incline

A cubic block of mass $m$ is sliding down on an inclined plane at $60^\circ$ with an acceleration of $\frac{g}{2}$ , the value of coefficient of kinetic friction is

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Answer: (d)

Sliding down a rough incline: $a = g(\sin\theta - \mu\cos\theta)$ .
$\dfrac{g}{2} = g\left(\dfrac{\sqrt{3}}{2} - \dfrac{\mu}{2}\right)$
$\mu = \sqrt{3} - 1$
So the answer is (d) $\sqrt{3} - 1$ .

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