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

Acceleration of system of blocks is
`a=(F)/(m_(A)m_(B)+m_(C))=(14)/(4+2+1)=2m//s^(2)`
Taking the block `B` and `C` as system
`N=(m_(B)+m_(C))=(2+1)xx2=6N` .