JEE Mains · Physics · STD 11 - 4.2 friction
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
- A \(\sqrt{3}-1\)
- B \(\frac{\sqrt{3}}{2}\)
- C \(\frac{\sqrt{2}}{3}\)
- D \(1-\frac{\sqrt{3}}{2}\)
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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