Two blocks ( m = 0 . 5 kg and M = 4 . 5 kg ) are arranged on a horizontal frictionless table as shown in the figure. The coefficient of static friction between the two blocks is 3 7 . Then the maximum horizontal force that can be applied on the larger block so that the blocks move together is N . (Round off to the Nearest Integer) [Take g as 9 . 8 m s - 2 ]

a max = μ g = 3 7 × 9 . 8 F = ( M + m ) a max = 5 a max = 21 N

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PhysicsNLMFrictionEasy2 minPYQ_2021

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Two blocks ( $m = 0.5 kg$ and $M = 4.5 kg$ ) are arranged on a horizontal frictionless table as shown in the figure. The coefficient of static friction between the two blocks is $\frac{3}{7} .$ Then the maximum horizontal force that can be applied on the larger block so that the blocks move together is $N .$ (Round off to the Nearest Integer) [Take $g$ as $9.8 m s^{- 2}$ ]

Answer:

Solution:

$a_{\max} = μ g = \frac{3}{7} \times 9.8$

$F = (M + m) a_{\max} = 5 a_{\max}$

$= 21 N$

Stream:JEESubject:PhysicsTopic:NLMSubtopic:Friction

⏱ 2mℹ️ Source: PYQ_2021

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