In the diagram shown coefficient of friction between 2 kg block and 4 kg block is `mu =0.1 ` and between 4 kg and ground is `mu =0.2 ` then answer the following quwstions .(tis time)

Force of friction on 2 kg block at `t=(1)/(2)` s will be

Options

A. 1 N towards left
B. 1 N towards right
C. 0N
D. 2 N towards left

Answer

B

Explanation

Maximum available value of friction between 2 kg and 4 kg block is
`(f_(1))_("max") = 0.1 xx 2 xx 10 = 2N`
Similarly, maximum available value of friction between 4 kg and ground is
`(f_(2))_("max") = 0.2 xx 6 xx 10 = 12 N`
At `= (1)/(2) s F = 1N`, the whole system is in equilibrium and free body diagrams of different blocks are as shown in figure below