Two blocks m=0.5 kg and M=4.5 kg
Question:

Two blocks $(\mathrm{m}=0.5 \mathrm{~kg}$ and $\mathrm{M}=4.5 \mathrm{~kg})$ 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__________$\mathrm{N}$.

(Round off to the NearestInteger)

[Take g as $9.8 \mathrm{~ms}^{-2}$ ]

Solution:

(21)

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

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

$=21$ Newton

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