Question

Three blocks of masses \(2\ \mathrm{kg}\), \(3\ \mathrm{kg}\) and \(5\ \mathrm{kg}\) are connected to each other with light string and are then placed on a frictionless surface as shown in the figure. The system is pulled by a force \(\mathrm{F} = 10\ \mathrm{N}\), then tension \(\mathrm{T_1}\) is

• A. 1 N
• B. 5 N
• C. 8 N
• D. 10 N

Hint:

In connected systems, find total acceleration first, then analyze forces on sections.

Answer 1

The Correct Option is C

To find the tension \( T_1 \) in the string connecting the 2 kg and 3 kg blocks, we need to analyze the forces acting on the blocks. The system is pulled by a force \( F = 10\ \mathrm{N} \) on a frictionless surface. Let's consider the entire system as a whole first to determine the acceleration, then focus on the individual blocks to find the tension.

Step 1: Calculate the Total Acceleration

The total mass of the system is the sum of all three blocks:

\(m_{\text{total}} = 2\ \mathrm{kg} + 3\ \mathrm{kg} + 5\ \mathrm{kg} = 10\ \mathrm{kg}\)

Using Newton’s second law, the net force is equal to the total mass times the acceleration:

\(F = m_{\text{total}} \cdot a\)

Substituting the given values, we have:

\(10\ \mathrm{N} = 10\ \mathrm{kg} \cdot a\)

Solving for acceleration \( a \):

\(a = \frac{10\ \mathrm{N}}{10\ \mathrm{kg}} = 1\ \mathrm{m/s^2}\)

Step 2: Calculate the Tension \( T_1 \)

Now consider the 2 kg block alone. The only forces acting in the horizontal direction are the tension \( T_1 \) and the force due to acceleration:

The force due to acceleration of block 2 kg is:

\(F_{2\ \mathrm{kg}} = 2\ \mathrm{kg} \cdot 1\ \mathrm{m/s^2} = 2\ \mathrm{N}\)

Since the 2 kg block is being pulled by the 3 kg and 5 kg system, the remaining force on the combined 3 kg and 5 kg blocks should be:

\(T_1 = F - F_{2\ \mathrm{kg}} = 10\ \mathrm{N} - 2\ \mathrm{N} = 8\ \mathrm{N}\)

Thus, the tension \( T_1 \) in the string connecting the 2 kg and 3 kg blocks is 8 N.