Question

Three blocks of masses $4kg,2kg,1kg$ respectively are in contact on a frictionless table as show in the figure. If a force of $14N$ is applied on the $4kg$ block, the contact force between the $4kg$ and the $2kg$ block will be:

• A. $2N$
• B. $6N$
• C. $8N$
• D. $14N$

Answer 1

The correct option is B $6N$

We know that, $F=ma$
$a=\frac{F}{m}=\frac{14}{7}=2m{s}^{-2}$
Hence, from the FBD of 4kg block
$14-N=4a$
$14-N=8$
$N=6N$