Question

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:

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

Answer 1

The correct option is D $6N$

$\text{Given :}{M}_A=4kg,M_B=2kg,M_C=1kg$

$\text{Step 1 : Acceleration of all the blocks}$

From figure, we see that the blocks A, B and C will move with a common acceleration a.

By Newton's Second Law, Acceleration $=\frac{F_{netExternal}}{M_{total}}$

$\Rightarrow a=\frac{F}{M_A+M_B+M_C}$

$a=\frac{14}{4+2+1}=2m/s^2$

$\text{Step 2 :Draw free body diagram for block A [Refer figure]}$

$\text{Step 3 : Apply newton's law on block A :}$
$\Sigma F=ma$

$\Rightarrow 14-N_{AB}=4a_A$

Substitute the value of $a_A=a=2m/s^2$ :

$\Rightarrow N_{AB}=14-8=6N$

Hence, the correct answer is option D.