Question

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

• A. 8 N
• B. 18 N
• C. 2 N
• D. 6 N

Answer 1

Answer: (D)

6 N

Answer 2

We know that, $F=m a$
$a=\frac{F}{m}=\frac{14}{7}=2\, ms ^{-2}$

Hence, from the figure
$14-N=4 a$
$14-N =8$
$N =6\, N$