Two blocks A and B of masses `m_(A) = 1kg and m_(B) = 3kg` 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 ________ (in N): [Take `g = 10 m//s^(2)`]

Options

A. 12 N
B. 8 N
C. 16 N
D. 40 N

Answer

C

Explanation

For 1 kg block,
`a _(max) = ( mu (1) g )/( (1)) = 2 ms ^(-2)`
So, `F _(max) - f _("ground") = m _("total") xx a _(max)`
`F _(max) - mu ( 3 +1) g = (3 +1) 2`
` F _(max) = 16 N`