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

The Correct Option is A

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 \( T_2 \):

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

Answer 2

This problem asks us to find the tensions, \(T_1\) and \(T_2\), in the strings connecting three blocks of different masses. The blocks are pulled by a horizontal force on a smooth surface, causing the entire system to accelerate together.

Concept Used:

The solution is based on Newton's Second Law of Motion, which states that the net force acting on an object is equal to the product of its mass and acceleration (\(F_{\text{net}} = ma\)).

For a system of connected bodies moving together, we can apply this law in two ways:

1. To the entire system: The net external force on the system equals the total mass of the system times the common acceleration.
2. To individual components: The net force on any part of the system equals the mass of that part times its acceleration. Internal forces, such as tension, are treated as external forces when considering an individual component.

Step-by-Step Solution:

Step 1: List the given values and calculate the total mass of the system.

Mass of block A, \( m_A = 5 \, \text{kg} \)
Mass of block B, \( m_B = 3 \, \text{kg} \)
Mass of block C, \( m_C = 2 \, \text{kg} \)
Applied external force, \( F = 80 \, \text{N} \)

The total mass of the system is:

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

Step 2: Calculate the common acceleration (\(a\)) of the system.

Since the blocks are connected and pulled by a single external force \(F\) on a smooth surface, they all move with the same acceleration. Applying Newton's Second Law to the entire system:

\[ F = M_{\text{total}} \times a \] \[ 80 \, \text{N} = (10 \, \text{kg}) \times a \]

Solving for the acceleration \(a\):

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

Step 3: Calculate the tension \(T_1\).

To find the tension \(T_1\), we consider the forces acting on block A. The only horizontal force acting on block A is the tension \(T_1\), which pulls it to the right. Applying Newton's Second Law to block A:

\[ F_{\text{net on A}} = T_1 = m_A \times a \]

Substituting the values of \(m_A\) and \(a\):

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

Step 4: Calculate the tension \(T_2\).

To find the tension \(T_2\), we can consider the forces acting on the subsystem consisting of blocks A and B together. This subsystem, with a combined mass of \(m_A + m_B\), is pulled to the right by the tension \(T_2\). Applying Newton's Second Law to this subsystem:

\[ F_{\text{net on (A+B)}} = T_2 = (m_A + m_B) \times a \]

Substituting the values of the masses and acceleration:

\[ T_2 = (5 \, \text{kg} + 3 \, \text{kg}) \times (8 \, \text{m/s}^2) = (8 \, \text{kg}) \times (8 \, \text{m/s}^2) = 64 \, \text{N} \]

Final Computation & Result:

The calculated values for the tensions are:

Tension \(T_1 = 40 \, \text{N}\)

Tension \(T_2 = 64 \, \text{N}\)

Therefore, the tensions \(T_1\) and \(T_2\) in the strings are 40 N and 64 N, respectively.