Problem 16 - Master Resource Book in JEE Main Physics

Question

A uniform chain of length \(L\) and mass \(M\) overhangs a horizontal table with its two-third part on the table. The friction coefficient between the table and the chain is \(\mu\). The work done by the friction during the period the chain slips off the table is (a) \(-\frac{1}{4} \mu \mathrm{Mg} \mathrm{L}\) (b) \(-\frac{2}{9} \mu \mathrm{Mgl}\) (c) \(-\frac{4}{9} \mu \mathrm{Mg} L\) (d) \(-\frac{6}{7} \mu \mathrm{Mg} L\)

Step-by-Step Solution

Answer

The work done by friction is \(-\frac{2}{9} \mu M g L\). Option (b) is correct.

Step 1: Identify the Part Hanging Off the Table

The chain is of total length \( L \), and two-thirds of it is on the table. It means that one-third of the chain is hanging off the table.

Step 2: Calculate the Mass of the Overhanging Part

The total mass of the chain is \( M \), and since one-third of the chain is hanging, the mass of the hanging part is \( \frac{M}{3} \).

Step 3: Calculate the Mass on the Table

The mass of the chain on the table is two-thirds of the total mass. Thus, it is \( \frac{2M}{3} \).

Step 4: Compute the Force Exerted by Section on the Table

The normal force due to the chain's mass on the table is equal to the weight of the section of the chain lying on the table, given by \( \frac{2M}{3}g \).

Step 5: Calculate the Frictional Force

The frictional force \( f \) opposing the motion is \( \mu \times \) Normal Force, which is \( \mu \times \frac{2M}{3}g \). Thus, \( f = \frac{2}{3} \mu M g \).

Step 6: Determine Initial and Final Positions

Initially, the chain is static with one-third hanging off, and finally, it completely slips off the table with nothing on it.

Step 7: Calculate Displacement of the Chain's Center of Mass

The center of mass of the chain moves from a position \( \frac{L}{2} \) of the hanging part to \( L \) length off in the negative direction, thus displacing by \( \frac{L}{3} \).

Step 8: Compute the Work Done by Friction

The work done by the friction force when the chain completely slips is negative due to the opposing direction: \[ W = f \cdot \text{displacement} = \left(\frac{2}{3}\mu M g \right) \left( \frac{L}{3} \right) = -\frac{2}{9} \mu M g L. \]

Step 9: Choose the Correct Answer

From earlier calculation, the correct answer, based on the work done, is option (b) \(-\frac{2}{9} \mu M g L\).

Key Concepts

Work Done by Friction

The concept of work done by friction is crucial in understanding various mechanical problems, including those involving a chain slipping off a table. Friction is a force that resists the motion of two surfaces in contact. When dealing with physics problems like the one presented, it's important to comprehend how friction impacts the movement of objects.

In this scenario, as the chain slides off the table, friction works against the motion. The work done by friction is calculated using the formula:

• \( W = f \cdot d \)

Where:

• \( W \) is the work done by friction,
• \( f \) is the frictional force,
• \( d \) is the displacement of the chain's center of mass.

Given that friction is opposing the motion, the work done is negative. This negative sign indicates that energy is being removed from the chain, slowing it down as it slides. In the solved problem, the frictional force is multiplied by the displacement of the chain’s center of mass during the slipping process to compute the total work done by friction. Thus, understanding the components of the frictional force and the displacement are key to solving such problems.

Center of Mass

The center of mass of an object is a point representing the mean position of the matter in a body. It's a fundamental concept in mechanics, particularly when analyzing the movement of objects such as chains or ropes.

In this exercise regarding the chain, knowing the position of the center of mass allows us to determine how it moves as the chain slips off the table. Initially, the chain's center of mass is stationed, but as the chain begins to slide, this point also shifts. The initial position of the center of mass of the hanging part is calculated at \( \frac{L}{2} \).

After the chain completely leaves the table, the center of mass moves to a position opposite to its initial one, showing a total displacement of \( \frac{L}{3} \).

Understanding how the center of mass travels helps in calculating displacement, which is essential for determining the work done by forces such as friction. This movement translates the physical shift of the chain into measurable values that feed into calculating energy changes and work.

Chain Problems in Mechanics

Chain problems, as introduced here, are a common topic in mechanics that often involve tangling with concepts like friction, gravity, and motion. These problems typically require a comprehensive understanding of how forces interact with a continuous body, such as a chain.

In mechanics, when dealing with chain problems, one often has to consider different sections separately - for instance, the section on the table and the section hanging off. This distinction can significantly alter the forces applied on different parts and likewise the resulting movement.

Key factors to consider in chain problems include:

• Calculating separate masses for different segments of the chain.
• Understanding the forces acting on each section, like gravitational force and friction.
• Analyzing the motion resulting from these forces as portions of the chain move off supportive surfaces.

In solving the problem, you learned how the interplay of these elements causes work to be done by friction, influencing how energy gets distributed and expended as the chain moves.