Two blocks ( m = 0 . 5   kg and M = 4 . 5   kg ) are arranged on a horizontal frictionless table…

Two blocks (m=0.5 kg and M=4.5 kg) are arranged on a horizontal frictionless table as shown in the 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 m s-2]

Solution

amax=μg=37×9.8

F=(M+m)amax=5amax

=21 N

Asked in: JEE Main 2021 (17 Mar Shift 1)

Practice more Laws of Motion questions on Aicharya