A cubic block of mass m is sliding down on an inclined plane at 6 0 ∘ with an acceleration of 2 g , the value of coefficient of kinetic friction is

mg sin 6 0 ∘ − μ mg cos 6 0 ∘ = ma g sin 60 − μg cos 60 = 2 g 2 3 − 2 μ = 2 1 μ = 3 − 1

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PhysicsNLMNLM/Impulse/FrictionMedium2 minPYQ_2025

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A cubic block of mass $m$ is sliding down on an inclined plane at $6 0^{\circ}$ with an acceleration of $\frac{g}{2}$ , the value of coefficient of kinetic friction is

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Answer:

Solution:

$mg sin 6 0^{\circ} - μ mg cos 6 0^{\circ} = ma$
$g sin 60 - μg cos 60 = \frac{g}{2}$
$\frac{3}{2} - \frac{μ}{2} = \frac{1}{2}$
$μ = 3 - 1$

Stream:JEESubject:PhysicsTopic:NLMSubtopic:NLM/Impulse/Friction

⏱ 2mℹ️ Source: PYQ_2025

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