Question

Three blocks A, B and C are pulled on a horizontal smooth surface by a force of 80 N as shown in figure

The tensions T1 and T2 in the string are respectively:

• A. 40N, 64N
• B. 60N, 80N
• C. 88N, 96N
• D. 80N, 100N

Answer 1

Answer: (A)

40N, 64N

Answer 2

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 $T_2$:

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