A horizontal force 10 N is applied to a block A as shown in figure. The mass of blocks A and B are 2 kg and 3 kg respectively. The blocks slide over a frictionless surface. The force exerted by block A on block B is :

The correct answer is 6 N.

Step-by-Step Explanation:

1. Analyze the system and the given data:
From the provided image, we can identify the following components and values:
- An external horizontal force: $F=10\text{N}$ is applied towards the right on block A.
- Mass of block A: $m_A=2\text{kg}$.
- Mass of block B: $m_B=3\text{kg}$.
- The surface is frictionless, meaning there is no opposing force of friction from the ground.

2. Calculate the common acceleration of the system:
Since blocks A and B are in contact and pushed by the force $F$, they will move together with a common acceleration, $a$. We can treat both blocks together as a single system of total mass $M$:
$M=m_A+m_B=2\text{kg}+3\text{kg}=5\text{kg}$

Using Newton's second law of motion ($F=M\cdot a$), the common acceleration is:
$a=\frac{F}{m_A+m_B}$

$a=\frac{10\text{N}}{5\text{kg}}$

3. Calculate the force exerted by block A on block B:
Let us consider the free-body diagram of block B. The only horizontal force acting on block B is the normal contact force exerted on it by block A, which we can call $F_{\text{AB}}$.
Applying Newton's second law specifically to block B:
$F_{\text{AB}}=m_B\cdot a$

Substitute the known values into the equation:
$F_{\text{AB}}=3\text{kg}\cdot 2{\text{m/s}}^2=6\text{N}$

Thus, the force exerted by block A on block B is 6 N.