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  4. Two blocks ( m =0.5 kg and M =4.5 kg ) are arranged on a horizontal frictionless table as shown in 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 ms -2 ] <img class=img-fluid question-image alt=image src=https://cdn.tardigrade.in/img/question/physics/d93d476cb19628c5db97f6709d6dc15e-.png />

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Q. Two blocks $( m =0.5\,kg$ and $M =4.5\,kg )$ are arranged on a horizontal frictionless table as shown in 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 $9.8 \,ms ^{-2}$ ]
image

JEE MainJEE Main 2021Laws of Motion

Solution:

$a_{\max }=\mu g=\frac{3}{7} \times 9.8$
$F=(M+m) a_{\max }=5\, a_{\max }$
$=21$ Newton