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Physics - NLM Question with Solution | TestHub

PhysicsNLMSimple Application of NLMEasy2 minQB

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A block of mass $m$ is kept on an inclined plane of a lift moving down with an acceleration of $2 m s^{- 2}$ . What should be the coefficient of friction for the block to move down with constant velocity relative to lift? The angle of inclination of the wedge is $30 °$ . $g = 10 m s^{- 2}$

Options:

Answer:

Solution:

$g_{eff} = g - 2 = 8 {m s}^{- 2}$

$m (8) \sin 30^{o} = μ . m (8) \cos 30^{o} \Rightarrow μ = \tan 30^{o}$

$\Rightarrow m (g - a) sin θ = μ m (g - a) cos θ$

$\Rightarrow μ = tan θ = tan 3 0^{o} = \frac{1}{\sqrt{3}}$

Stream:JEESubject:PhysicsTopic:NLMSubtopic:Simple Application of NLM

⏱ 2mℹ️ Source: QB

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