Question:

Three blocks A, B and C are pulled on a horizontal smooth surface by a force of 80 N as shown in figure
3 blocks
The tensions T1 and T2 in the string are respectively:

Updated On: Nov 16, 2024
  • 40N, 64N
  • 60N, 80N
  • 88N, 96N
  • 80N, 100N
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The Correct Option is A

Solution and Explanation

Let’s analyze the forces acting on the blocks:

Calculate the Total Mass: The total mass of the system \( m \):

\[ m = m_A + m_B + m_C = 5 \, \text{kg} + 3 \, \text{kg} + 2 \, \text{kg} = 10 \, \text{kg}. \]

Calculate the Acceleration of the System: Using Newton’s second law \( F = ma \):

\[ a = \frac{F}{m} = \frac{80 \, \text{N}}{10 \, \text{kg}} = 8 \, \text{m/s}^2. \]

Calculate the Tension \( T_2 \) in the String Connecting B and C: For block C (mass = 2 kg), using

\[ F = ma: \] \[ T_2 = m_C \times a = 2 \, \text{kg} \times 8 \, \text{m/s}^2 = 16 \, \text{N}. \]

Calculate the Tension \( T_1 \) in the String Connecting A and B: The force acting on block B (mass = 3 kg) includes both its weight and the tension \( T_2 \):

\[ T_1 = m_B \times a + T_2 = (3 \, \text{kg} \times 8 \, \text{m/s}^2) + 16 \, \text{N} = 24 \, \text{N} + 16 \, \text{N} = 40 \, \text{N}. \]

Therefore, for block A (mass = 5 kg):

\[ T_1 = m_A \times a + T_1 + T_2 = (5 \, \text{kg} \times 8 \, \text{m/s}^2) = 40 \, \text{N} + T_1. \]

Calculate Final Tensions: Now, substituting for \( T_2 \):

\[ T_1 = 5 \times 8 \, \text{N}, T_2 = 40 + 8 \times 3 = 64 \, \text{N}. \]
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