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JEE Mains · Physics · STD 11 - Friction
  • JEE Mains 2025
  • Easy
  • MCQ

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

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Answer & Solution

Correct Answer

(A) \(\sqrt{3}-1\)

Step-by-step Solution

Detailed explanation


\(\mathrm{mg} \sin 60^{\circ}-\mu \mathrm{mg} \cos 60^{\circ}=\mathrm{ma}\)
\(g \sin 60-\mu g \cos 60=\frac{g}{2}\)
\(\frac{\sqrt{3}}{2}-\frac{\mu}{2}=\frac{1}{2}\)
\(\mu=\sqrt{3}-1\)
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