Animated Solution for Physics - Work, Energy, and Power: A uniform chain of length $2\text{ m}$ is kept on a table such that a length of $60\text{ cm}$ hangs freely from the edge of the table. The total mass of the chain is $4\text{ kg}$. What is the work done in pulling the entire chain on the table?

Imagine you are standing by a well, pulling up a heavy rope. As you pull more and more rope up, it gets easier, right? That's because the amount of rope hanging down—and thus the weight you are pulling against gravity—is constantly decreasing.

This classic physics problem about a hanging chain operates on the exact same principle. We have a uniform chain resting on a smooth table, with a portion of it dangling off the edge. Our mission is to calculate the total work required to pull that hanging portion completely back onto the table.

Let's break down the given parameters. We have a chain with a total length $L=2\text{ m}$ and a total mass $M=4\text{ kg}$. A length of $l=0.6\text{ m}$ is hanging freely from the edge of the table.

To find the work done, we need to understand what we are working against. Since the table is smooth (frictionless), the only force we are fighting is gravity acting on the hanging part of the chain. The part of the chain already on the table doesn't require any work to move horizontally because there is no friction.

We could solve this using calculus by integrating the work done on each tiny segment of the chain. However, there is a much more elegant and faster way: The Center of Mass (COM) method.

The work done in lifting an object is simply the change in its potential energy. For a uniform object like our chain, we can pretend that its entire mass is concentrated at a single point—its center of mass.

Therefore, the work done $W$ is given by: where $m$ is the mass of the hanging part, $g$ is the acceleration due to gravity, and $h$ is the vertical distance the center of mass needs to be raised to reach the table level.

Step 1: Find the mass of the hanging part ($m$) Since the chain is uniform, its mass is distributed evenly. We first find the linear mass density $\lambda$ (mass per unit length):

Now, we multiply this density by the length of the hanging part to find its mass:

Step 2: Find the depth of the center of mass ($h$) For a uniform rod or chain, the center of mass is located exactly at its geometric center. Since the hanging part has a length of $0.6\text{ m}$, its center of mass is exactly halfway down:

Step 3: Calculate the Work Done Now we just plug these values into our potential energy formula. Let's take $g=10{\text{ m/s}}^2$ for simplicity (as is standard in such problems unless specified otherwise):

What if the table wasn't smooth? If there was a coefficient of kinetic friction ${\mu }_k$ between the chain and the table, the problem becomes significantly more complex. As you pull the chain, the mass on the table increases, which increases the normal force, and consequently, the frictional force increases linearly. You would then have to integrate the variable frictional force over the distance pulled, in addition to the work done against gravity. This is a fantastic extension to test your calculus skills in physics!