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Question:

Two blocks A and B of masses m_A = 1 kg and m_B = 3 kg are kept on the table as shown in figure. The coefficient 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)

16 N

12 N

40 N

8 N

Solution:

Correct option is A. 16 N
a_Amax = μg = 2 m/s^2
F - 8 = 4 × 2
F = 16 N