Question

A combination of three blocks is shown in figure is pushed by a horizontal force of 30N on a frictionless surface. Force exerted by 2kg and 3kg is:

A. 20 N
B. 40 N
C. 80 N
D. 60 N

Answer 1

Hint: When two bodies are in contact with each other, they apply some force on one another which is equal and opposite, in accordance to Newton’s third law of motion. This force is known as contact force. This force is the indication that something is in contact with the other body.
Formula used:
 $F = ma$

Complete answer:
These types of problems could be easily solved by drawing the free body diagram (FBD) of the blocks and analyzing the forces which are being acted on the system.
But first of all, let’s find out the acceleration of each block. Since they are not separating from one another, the acceleration of all the blocks will be the same. Its value could be obtained by using:
$F = ma$
Since they are not being separated, hence we can take the three blocks as one system with mass (2 kg+ 1 kg + 3 kg) = 6 kg.
Also the net external force acting on the system is 30 N
Thus $a = \dfrac{30}{6} = 5\ ms^{-2}$

Now, drawing the FBD of 2kg and rest of the system:

Here, ‘R’ is the contact force.
Since on the system of (1 kg+ 3 kg) = 4 kg, no force other than ‘R’ is acting in horizontal direction, hence only contact force is responsible for its acceleration.
Thus using:$F = ma$, we get;
$R = (3+1) \times 5 = 4\times 5 = 20 N$

So, the correct answer is “Option A”.

Note:
It should be noted that if it were asked to find a force between 3kg and 1 kg block, we should have drawn FBD of 3 kg alone with 1 kg and 2 kg collectively as a system. This is the standard procedure for finding the contact force between any two blocks in a system.