Three blocks `A` , `B` and `C` of masses `4kg` , `2kg` and `1kg` respectively are in contact on a frictionless surface, as shown. If a force of `14N is applied on the `4kg` block, then the contact force between `A` and `B` is.

Options

A. 2N
B. 6N
C. 8N
D. 18N

Answer

B

Explanation

(b) Given,`m_(A)=4 kg,m_(B)=2 kg`
`implies m_(C)=1 kg`

So total mass (M) =4+2+1=7 kg
Now `F=Ma implies 14= 7a implies a=2ms^(-2)`
FBD of block A,

`F-F'=4a implies F'=14-4xx2 implies F'=6 N`