Question

Two blocks of masses 4 kg and 2 kg are attached by an inextensible light string as shown in figure. Both the blocks are pulled vertically upwards by a force F = 120 N.

Find (a) the acceleration of the blocks, (b) tension in the string. (Take g = 10m/s²)

Answer 1

(a) The acceleration of the blocks is $10\, \text{m/s}^2$. (b) The tension in the string is $40\, \text{N}$.

Answer 2

The problem involves two blocks connected by a string, being pulled vertically upwards. We need to find the acceleration of the blocks and the tension in the string.

Given:

• Mass of the top block, $m_1 = 4\, \text{kg}$
• Mass of the bottom block, $m_2 = 2\, \text{kg}$
• Applied upward force, $F = 120\, \text{N}$
• Acceleration due to gravity, $g = 10\, \text{m/s}^2$

(a) Acceleration of the blocks

We can consider the two blocks as a single system since they move together with the same acceleration.

1. Calculate the total mass of the system: $M = m_1 + m_2 = 4\, \text{kg} + 2\, \text{kg} = 6\, \text{kg}$

2. Identify the forces acting on the system:

   • Upward applied force, $F = 120\, \text{N}$
   • Downward gravitational force on the total mass, $Mg = (6\, \text{kg}) \times (10\, \text{m/s}^2) = 60\, \text{N}$

3. Apply Newton's Second Law for the entire system: Let $a$ be the upward acceleration of the system. Net Force = Total Mass × Acceleration $F - Mg = Ma$ $120\, \text{N} - 60\, \text{N} = (6\, \text{kg}) \times a$ $60\, \text{N} = 6a$ $a = \frac{60}{6} = 10\, \text{m/s}^2$

(b) Tension in the string

To find the tension in the string, we need to analyze the forces acting on one of the blocks individually. It's usually simpler to analyze the block that has fewer unknown forces acting on it. Let's consider the bottom block (2 kg).

1. Draw the Free Body Diagram (FBD) for the 2 kg block ($m_2$):

   • Upward force: Tension in the string, $T$
   • Downward force: Gravitational force on $m_2$, $m_2g = (2\, \text{kg}) \times (10\, \text{m/s}^2) = 20\, \text{N}$
   • The block is accelerating upwards with $a = 10\, \text{m/s}^2$.

2. Apply Newton's Second Law for the 2 kg block: Net Force = Mass × Acceleration $T - m_2g = m_2a$ $T - 20\, \text{N} = (2\, \text{kg}) \times (10\, \text{m/s}^2)$ $T - 20\, \text{N} = 20\, \text{N}$ $T = 20\, \text{N} + 20\, \text{N}$ $T = 40\, \text{N}$

Visual Representation:

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Explanation of the solution:

1. Acceleration (a): Treat both blocks as a single system with total mass $M = m_1 + m_2 = 6\, \text{kg}$. The net upward force is $F - Mg = 120\, \text{N} - (6\, \text{kg} \times 10\, \text{m/s}^2) = 120\, \text{N} - 60\, \text{N} = 60\, \text{N}$. Applying Newton's second law, $F_{net} = Ma$, we get $60\, \text{N} = 6\, \text{kg} \times a$, which yields $a = 10\, \text{m/s}^2$.

2. Tension (T): Consider the forces on the 2 kg block ($m_2$). The upward force is the tension $T$, and the downward force is its weight $m_2g = 2\, \text{kg} \times 10\, \text{m/s}^2 = 20\, \text{N}$. Applying Newton's second law for this block, $T - m_2g = m_2a$. Substituting the value of $a$, we get $T - 20\, \text{N} = 2\, \text{kg} \times 10\, \text{m/s}^2$, which simplifies to $T - 20\, \text{N} = 20\, \text{N}$. Therefore, $T = 40\, \text{N}$.