Question

Two blocks of masses $4\text{ kg}$ and $2\text{ kg}$ are connected by a heavy string and placed on rough horizontal plane. The $2\text{ kg}$ block is pulled with a constant force $F$. The coefficient of friction between the blocks and the ground is $0.5$. The value of $F$ so that tension in the string is constant throughout during the motion of the blocks is

• A. $40\text{ N}$
• B. $30\text{ N}$
• C. $50\text{ N}$
• D. $60\text{ N}$

Hint:

In these problems, always calculate the total friction first. If the applied force $F$ is equal to the total friction, the acceleration is zero, and the tension $T$ will simply be equal to the friction on the rear block.

Answer 1

The Correct Option is B

Step 1: Calculate Frictional Forces: The frictional force is given by $f = \mu mg$. Let $g = 10\text{ m/s}^2$.

• Friction on $4\text{ kg}$ block ($f_1$): $0.5 \times 4 \times 10 = 20\text{ N}$

• Friction on $2\text{ kg} $ block ($f_2$): $0.5 \times 2 \times 10 = 10\text{ N}$

• Total Friction ($f_{total}$): $20 + 10 = 30\text{ N}$

Step 2: Equation of Motion for the System: Applying Newton's second law ($F_{net} = ma$) to the combined system ($4\text{ kg} + 2\text{ kg} = 6\text{ kg}$): $$F - f_{total} = (m_1 + m_2)a$$ $$F - 30 = 6a \quad \dots \text{(Eq. 1)}$$

Step 3: Analyze the $4\text{ kg}$ block (Rear block): The only forward force on the $4\text{ kg}$ block is the tension $T$, and the backward force is friction $f_1$: $$T - f_1 = m_1 a$$ $$T - 20 = 4a \quad \dots \text{(Eq. 2)}$$