Two blocks of masses 2 kg and 4 kg are in close contact on a frictionless horizontal table. A horizontal force of 18 N is applied to the larger mass. What is the force at the surface of contact between the blocks?

Options

A. 4N
B. 5N
C. 6N
D. 8N

Explanation

To solve the problem step by step, we will follow these instructions:

### Step 1: Identify the total mass of the system
The two blocks have masses of 2 kg and 4 kg. Therefore, the total mass \( M \) of the system is:
\[ M = 2 \, \text{kg} + 4 \, \text{kg} = 6 \, \text{kg} \]

### Step 2: Calculate the acceleration of the system
A horizontal force of 18 N is applied to the larger mass (4 kg). According to Newton's second law, the acceleration \( a \) of the system can be calculated using the formula:
\[ F = M \cdot a \]
Rearranging gives:
\[ a = \frac{F}{M} = \frac{18 \, \text{N}}{6 \, \text{kg}} = 3 \, \text{m/s}^2 \]

### Step 3: Analyze the smaller block (2 kg)
Next, we need to find the force at the surface of contact between the two blocks. We will analyze the forces acting on the smaller block (2 kg). The only horizontal force acting on this block is the contact force \( F_c \) exerted by the larger block.

Using Newton's second law for the 2 kg block:
\[ F_c = m \cdot a \]
Where:
- \( m = 2 \, \text{kg} \)
- \( a = 3 \, \text{m/s}^2 \)

Substituting the values:
\[ F_c = 2 \, \text{kg} \cdot 3 \, \text{m/s}^2 = 6 \, \text{N} \]

### Step 4: Conclusion
The force at the surface of contact between the blocks is:
\[ \boxed{6 \, \text{N}} \]