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QJEEM21Ph97f4: JEE Main 2021 Physics - Class 11 JEEM21Ph97f4 Solution: Two blocks (m = 0.5 kg and M = 4.5 kg) are arranged on a…

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]

JEE Main 2021 (Online) 17th March Morning Shift Physics - Laws of Motion Question 82 English

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JEE Main 2021 (Online) 17th March Morning Shift Physics - Laws of Motion Question 82 English Explanation
a=FM+m

f = ma = mFM+m

mFM+m ≤ μmg (for no slipping)

⇒ F ≤ μ(m + M)g

∴ Fmax = 37(0.5+4.5)×9.8 = 21 N

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