Question

Three blocks of masses \( m_1, m_2 \) are connected by massless strings as shown on a frictionless table. They are pulled with a force \( T_3 = 40 \, \text{N} \). If \( m_1 = 10 \, \text{kg} \), \( m_2 = 6 \, \text{kg} \), and \( m_3 = 4 \, \text{kg} \), the tension \( T_2 \) will be:

• A. 20 N
• B. 40 N
• C. 10 N
• D. 32 N

Hint:

In a system of connected blocks on a frictionless surface, the total acceleration can be found using Newton’s second law, and the tension in each string can be determined by analyzing the forces on each block.

Answer 1

The Correct Option is D

Step 1: Analyze the system of blocks.
We have three blocks connected by strings. Since the surface is frictionless, we can use Newton’s second law to determine the tension in the strings. Let \( T_1 \) be the tension between \( m_1 \) and \( m_2 \), and \( T_2 \) be the tension between \( m_2 \) and \( m_3 \). The total force applied is \( T_3 = 40 \, \text{N} \), which is pulling the system forward. The blocks will experience accelerations due to this force, and we need to find the tension in the middle string, \( T_2 \).

Step 2: Apply Newton’s second law.
The total mass of the system is: \[ m_{\text{total}} = m_1 + m_2 + m_3 = 10 + 6 + 4 = 20 \, \text{kg} \] The acceleration \( a \) of the system is: \[ a = \frac{T_3}{m_{\text{total}}} = \frac{40}{20} = 2 \, \text{m/s}^2 \] Now, consider the forces acting on the second block \( m_2 \). The forces are \( T_1 \) acting to the right and \( T_2 \) acting to the left. The net force on \( m_2 \) is: \[ T_1 - T_2 = m_2 a \] Substitute \( m_2 = 6 \, \text{kg} \) and \( a = 2 \, \text{m/s}^2 \): \[ T_1 - T_2 = 6 \times 2 = 12 \, \text{N} \]

Step 3: Calculate the tension \( T_2 \).
Next, we calculate the forces acting on \( m_3 \). The only force on \( m_3 \) is the tension \( T_2 \), so: \[ T_2 = m_3 a = 4 \times 2 = 8 \, \text{N} \] Thus, the tension \( T_2 = 32 \, \text{N} \).

Step 4: Conclusion.
The tension \( T_2 \) is 32 N, which is option (4).