Two blocks of `2 kg` and `4 kg` are connected by a light string and kept on horizontal surface `A` force of `16 N` is acted on `4 kg` block horizontally as shown in figure. Besides ,it is given that coefficient of friction between `4 kg` and ground is `0.3` and between `2 kg` block and ground is `0.6`. Then frictional force between `2 kg` block ground is

Options

A. `12 N`
B. `6 N`
C. `4 N`
D. zero

Answer

C

Explanation

`(f_(2 kg))_(max) = mu_(2) m_(2) g = 0.6 xx 2 xx 10= 12 N`
`(f_(4 kg))_(max) = mu_(4) m_(4) g = 0.3 xx 4 xx 10= 12 N`
Net pulling force `F = 16 N`and
Net resistive force `F' = (f_(2 kg))_(max) + (f_(4 kg))_(max)`
`=24 N`
Since `F LT F'`, system will not move not free body diagrams of two blocks are as shown in below.

`4 kg T + 12 = 16 rArr T = 4 N`
`2 kg f_(2) = T = 4 N`