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 37. 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]

When the friction between blocks is limiting , they attain maximum acceleration, amax=μg=37×9.8=4.2 ms-2 Thus, for combine system, Fmax=(m+M)amax=5×4.2=21 N

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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 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 $9.8 m s^{- 2}$ ]

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answer is 21.

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Detailed Solution

$W h e n t h e f r i c t i o n b e t w e e n b l o c k s i s l i m i t i n g, t h e y a t t a i n m a x i m u m a c c e l e r a t i o n, a_{m a x} = μ g = (\frac{3}{7}) \times 9.8 = 4.2 m s^{- 2} T h u s, f o r c o m b i n e s y s t e m, F_{m a x} = (m + M) a_{m a x} = 5 \times 4.2 = 21 N$