Problem 374 - JEE Physics

Question

Two blocks of mass \(8 \mathrm{~kg}\) and \(4 \mathrm{~kg}\) are connected a heavy string Placed on rough horizontal Plane, The \(4 \mathrm{~kg}\) block is Pulled with a constant force \(\mathrm{F}\). The co-efficient of friction between the blocks and the ground is \(0.5\), what is the value of \(F\), So that the tension in the spring is constant throughout during the motion of the blocks? \(\left(\mathrm{g}=10 \mathrm{~ms}^{-2}\right)\) (A) \(40 \mathrm{~N}\) (B) \(60 \mathrm{~N}\) (C) \(50 \mathrm{~N}\) (D) \(30 \mathrm{~N}\)

Step-by-Step Solution

Answer

The force required to keep the tension constant throughout the motion is \(40 \mathrm{~N}\). The correct answer is (A) \(40 \mathrm{~N}\).

Step 1: Calculate the friction force on each block

We know that the friction force is given by: \(f = \mu N\), where \(f\) is the friction force, \(\mu\) is the coefficient of friction, and \(N\) is the normal force on the object. For both blocks, the normal force is equal to the weight of the block i.e., \(mg\). The friction force on the 8 kg block is: \(f_1 = (0.5) (8) (10) = 40 N\), and the friction force on the 4 kg block is: \(f_2 = (0.5) (4) (10) = 20 N\).

Step 2: Analyze forces acting on the 8 kg block

According to Newton's second law, the net force acting on an object is equal to the product of its mass and acceleration, i.e., \(F_{net} = ma\), where \(F_{net}\) is the net force, \(m\) is the mass of the object, and \(a\) is the acceleration. For the 8 kg block, the net force is the difference between the tension in the string (T) and the friction force acting on it. We have: \(F_{net} = T - f_1\).

Step 3: Analyze forces acting on the 4 kg block

For the 4 kg block, the net force is the difference between the force applied (F) and the friction force acting on it, and also the tension acting in the opposite direction. The tension T is in the negative direction with respect to the net force F acting on the 4 kg block, so we have: \(F_{net} = -T + F - f_2\).

Step 4: Apply the condition of constant tension

For the tension in the string to remain constant throughout the motion, the acceleration of both blocks must be equal. Therefore, by equating their net forces we have: \[(T - f_1) = -(-T + F - f_2)\]

Step 5: Solve for F

Now, we will solve this equation to find F: \begin{align*} T - f_1 &= -(-T + F - f_2)\\ 2T - (f_1 + f_2) &= F\\ \end{align*} We know that T is also the force required to overcome the friction acting on the 8 kg block, therefore: \[T = f_1\] Substituting this value of T into the equation, we get: \begin{align*} 2f_1 - (f_1 + f_2) &= F\\ F &= 40 N\\ \end{align*} So, the force required is 40 N. The correct answer is (A) \(40 \mathrm{~N}\).

Key Concepts

Friction

Friction is a force that opposes the motion of objects. When two surfaces interact, friction acts on them. In this exercise, the friction force is calculated using the formula:

• \( f = \mu N \)

where \( f \) is the friction force, \( \mu \) is the coefficient of friction, and \( N \) is the normal force. On a horizontal plane, the normal force equals the weight (mass \( \times \) gravity) of each block.
The given coefficient of friction \( \mu = 0.5 \) means that half of the normal force opposes the motion as friction.
For the 8 kg block, the friction force \( f_1 \) is \( 40 \mathrm{~N} \), and for the 4 kg block, \( f_2 \) is \( 20 \mathrm{~N} \). These forces must be overcome by the applied force \( F \) to move the blocks.

Tension

Tension is the pulling force transmitted along a string, rope, or cable when forces act at an end. In block systems, the tension in the connecting string ensures that both blocks undergo the same acceleration. For this exercise, the tension \( T \) must remain constant.

• In the 8 kg block, the tension \( T \) offsets its friction \( f_1 \) leading to the equation \( T - f_1 = ma \).
• In the 4 kg block, the external force \( F \) minus tension and friction results in \( F - T - f_2 = ma \).

To achieve constant tension, the accelerations of both blocks need to match. Through equalizing net forces, the problem becomes solvable.
Overall, tension acts centrally in managing how forces are balanced, ensuring a unified motion of blocks.

Block Systems

Block systems involve multiple connected objects that move together. Forces applied to one block affect the others through tension in a connecting string. In this specific system:

• The blocks are placed on a horizontal plane connected by a heavy string.
• When the 4 kg block is pulled, both blocks are set in motion.
• It involves analyzing how forces distribute across the system and understanding how tension links the motion of the blocks.

Understanding the concepts of block systems helps unravel how distributed forces influence motion. Importantly, ensuring both blocks have the same acceleration simplifies solving for applied forces in the system.

Acceleration

Acceleration refers to the rate of change of velocity with time. In this exercise, both blocks accelerate due to the applied force \( F \).
Newton's Second Law forms the basis:

• Net force \( F_{net} = ma \)

For each block, acceleration arises from the net forces acting upon them. Constant acceleration is achievable by equating the forces from both blocks:

• The 8 kg block's force equation: \( T - f_1 = ma \).
• The 4 kg block's force equation includes the applied force: \( F - T - f_2 = ma \).

Ultimately by equating their accelerations,
you solve for the force \( F \) needed. This guarantees a steady motion, ensuring the dynamics between friction, tension, and acceleration come together harmoniously.