JEE Main201910 Apr 2019Evening ShiftPhysicsLaws of MotionActual
Two blocks A and B of masses m A = 1 k g and m B = 3 k g are kept on the table as shown in figure. The coefficients of friction between A and B is 0.2 and between B and the surface of the table is also 0.2 . The maximum force F that can be applied on B horizontally, so that the block A does not slide over the block B is : [Take g = 10 m / s 2 ]
Options
- A16 N
- B12 N
- C40 N
- D8 N
Correct answer
A. 16 N
Step-by-step solution
Maximum possible acceleration of 1   k g block a = µ m g m = µ g = 2   m / s 2 For 1   g block not to slide the maximum acceleration of both blocks should be F m a x - µ m g = m a m a x F m a x - 0.2 × 4 × 10 = 4 × 2 F m a x -   8 = 8 F m a x = 16   N