Question

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

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

Hint:

For blocks in series on a frictionless surface, \(a = F/M_{total}\). The tension between any two blocks equals the product of the mass ahead (in direction of motion) and acceleration.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
For a system of blocks on a frictionless surface pulled by a force, the acceleration is common to all blocks. The tension in a string depends on the mass being pulled ahead of that string.
Step 2: Detailed Explanation:
Total mass \(= 10 + 6 + 4 = 20\) kg
Acceleration: \(a = \dfrac{F}{M_{total}} = \dfrac{40}{20} = 2\) m/s\(^2\)
\(T_2\) pulls \(m_1\): \[ T_2 = m_1 \times a = 10 \times 2 = 20 \text{ N} \]
Step 3: Final Answer:
The tension \(T_2 = 20 N\).