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

 

$\text{ Here, }{M}_A=4 \mathrm{kg},M_B=2 \mathrm{kg},M_C=1 \mathrm{kg},F=14 \mathrm{N}$

$\text{ Net mass, }M=M_A+M_B+M_C=4+2+1=7 \mathrm{kg}$

$\text{ Let }a\text{ be the acceleration of the system. }$

$\text{ Using Newton's second law of motion, }$

$\begin{array}{l}F=Ma\\ 14=7a \therefore a=2 \mathrm{m}{ \mathrm{s}}^{-2}\end{array}$

Let F' be the force applied on block A by block B i.e. the contact force between A and B. Free body diagram for block A

$\text{ Again using Newton's second law of motion, }$

$\begin{array}{l}F-F^{\text{'}}=4a\\ 14-F^{\text{'}}=4\times 2\\ 14-8=F^{\text{'}}\\ \therefore F^{\text{'}}=6 \mathrm{N}\end{array}$