A block of mass m is placed on a smooth wedge of inclination θ . The whole syste | TestHub

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A block of mass $m$ is placed on a smooth wedge of inclination $θ$ . The whole system is accelerated horizontally so that the block does not slip on the wedge. The force exerted by the wedge on the block ( $g$ is acceleration due to gravity) will be

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Answer:

A

Solution:

$N = m a \sin θ + m g \cos θ$ …(i)

Also, $m g \sin θ = m a \cos θ$ …(ii)

From (1) and (2), $a = g \tan θ$

$\begin{array}{l} ∴ N = m g \frac{\sin^{2} θ}{\cos θ} + m g \cos θ [substituting t h e v a l u e o f a] \\ ∴ N = \frac{m g}{\cos θ} (\sin^{2} θ + \cos^{2} θ) = \frac{m g}{\cos θ} \\ or, N = \frac{m g}{\cos θ} \end{array}$

Stream:NTA_ABHYAS_NEETSubject:PhysicsTopic:NLMSubtopic:FBD & Equilibrium

⏱ 2mℹ️ Source: PYQ_2020

Physics - NLM Question with Solution | TestHub

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