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

The Correct Option is D

To determine the contact force between the 4\, kg block and the 2\, kg block, let's first understand the physics of the problem.

The total system consists of three blocks with masses 4\, kg, 2\, kg, and 1\, kg, placed in contact on a frictionless table. A force of 14\, N is applied to the 4\, kg block.

Since the table is frictionless and the force is applied horizontally, let's compute the total acceleration of the entire system first:

1. Calculate the total mass of the system:
   M_{\text{total}} = 4\, kg + 2\, kg + 1\, kg = 7\, kg
2. Using Newton's second law, compute the acceleration (a) of the system:
   F = M_{\text{total}} \cdot a
   14\, N = 7\, kg \cdot a
   a = \frac{14}{7} = 2\, m/s^2
3. The acceleration of the entire system is 2\, m/s^2.

Now, let's calculate the contact force between the 4\, kg block and the 2\, kg block:

4. Consider only the 2\, kg and 1\, kg blocks. Since these blocks experience the same acceleration, and their total mass is 3\, kg, the force required to accelerate them is:
   F_{\text{contact}} = M_{\text{23}} \cdot a
   = (2\, kg + 1\, kg) \cdot 2\, m/s^2
   = 3\, kg \cdot 2\, m/s^2
   = 6\, N
5. Thus, the contact force between the 4\, kg and the 2\, kg block is 6\, N.

Therefore, the correct answer is 6\, N, making the correct option: 6 N.