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