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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