A uniform cable of mass M and length L is placed on a horizontal surface such that its (1/n)th part

Q: Why don't we consider the potential energy of the part of the cable that is already on the table?

Because that part moves horizontally. Its vertical height doesn't change, so its gravitational potential energy remains constant. Work is only done against gravity to lift the hanging part.

Q: Why is the initial potential energy taken as negative?

Potential energy is relative. It's standard practice to set the final destination (the table surface) as the reference level ($U=0$). Since the hanging part is initially below this level, its potential energy is negative.

Q: Can I solve this using integration instead of the center of mass?

Absolutely! You can take a small element $dx$ at a distance $x$ from the top, find its mass $dm = (M/L)dx$, and integrate the work $dW = (dm)gx$ from $x=0$ to $x=L/n$. You will get the exact same result.

Q: What if the table had friction? Would the work done be the same?

No. If there was friction, you would have to do extra work to overcome the frictional force acting on the part of the cable sliding on the table. The total work would be the change in potential energy plus the work done against friction.

Q: Why is the center of mass at $L/2n$ and not $L/n$?

$L/n$ is the total length of the hanging part. For a uniform object, the center of mass is exactly in the middle. Half of $L/n$ is $L/2n$.

Question

A uniform cable of mass $M$ and length $L$ is placed on a horizontal surface such that its $\left(\frac{1}{n}\right)$th part is hanging below the edge of the surface. To lift the hanging part of the cable upto the surface, the work done should be

Options:

(A) $\frac{2MgL}{n^2}$

(B) $nMgL$

(C) $\frac{MgL}{n^2}$

(D) $\frac{MgL}{2n^2}$

Sigma Percentile · Accepted Answer

### The Setup: A Dangling Dilemma

Imagine a heavy, uniform cable resting on a perfectly smooth table. A fraction of it, exactly $\frac{1}{n}$th of its total length $L$, is dangling over the edge. Our mission is to calculate the work required to pull this hanging section completely back onto the table.

At first glance, you might think about calculating the varying force required to pull the chain as it gets shorter and shorter. While that's a valid approach using calculus, there is a much more elegant and faster way to solve this using the principles of energy.

### The Elegant Shortcut: Center of Mass

To find the work done, we don't need to complicate things with forces. We can use the work-energy theorem. The work we do against gravity will simply be equal to the change in the gravitational potential energy of the cable.

$$W_{	ext{ext}} = \Delta U = U_f - U_i$$

Let's set the table surface as our reference level, meaning any mass on the table has a potential energy of zero ($U=0$). We only need to focus on the hanging part, because the part already on the table doesn't change its vertical position, so its potential energy remains constant.

### Crunching the Numbers

Since the cable is uniform, its mass is distributed evenly. Therefore, the mass of this dangling section, let's call it $m'$, is simply the total mass $M$ divided by $n$.

$$m' = \frac{M}{n}$$

Now, where does the entire mass of this hanging section effectively act? At its center of mass! For a uniform straight cable, the center of mass lies exactly at its geometric center. So, it's located halfway down the hanging length, which is at a depth of $h$ below the table.

$$h = \frac{1}{2} 	imes \left(\frac{L}{n}ight) = \frac{L}{2n}$$

Let's calculate the initial potential energy. Since the center of mass is below our reference level, the potential energy will be negative.

$$U_i = -m'gh = -\left(\frac{M}{n}ight)g\left(\frac{L}{2n}ight) = -\frac{MgL}{2n^2}$$

When the cable is pulled completely onto the table, its center of mass reaches the reference level, making the final potential energy zero ($U_f = 0$). The work done is the final energy minus the initial energy.

$$W = U_f - U_i = 0 - \left(-\frac{MgL}{2n^2}ight) = \frac{MgL}{2n^2}$$

**And there we have it! The work done is $\frac{MgL}{2n^2}$.**

### The Calculus Alternative

While the center of mass approach is incredibly fast, it's always good to know the calculus method. You could take a tiny element of length $dx$ at a distance $x$ from the top. Its mass would be $dm = \left(\frac{M}{L}ight)dx$. The work done to lift just this tiny piece to the table is $dW = (dm)gx$. Integrating this from $x=0$ to $x=\frac{L}{n}$ will yield the exact same beautiful result!

Animated Solution for Physics - Work, Energy, and Power: A uniform cable of mass $M$ and length $L$ is placed on a horizontal surface such that its $(\frac{1}{n})$ th part is hanging below the edge of the surface. To lift the hanging part of the cable upto the surface, the work done should be

Select Answer:

Visualized Solution

\text{Analyzing the Hanging Cable}

\text{Work-Energy Theorem}

\text{Mass of Hanging Section}

\text{Locating the Center of Mass}

\text{Initial Potential Energy } (U_i)

\text{Calculating Work Done}

\text{Alternative Method: Integration}

The Sigma Insight: Kinetic Energy, Potential Energy and Power

The Setup

The Elegant Shortcut

Crunching the Numbers

The Calculus Alternative

Similar Questions

A constant power delivering machine has towed a box, which was initially at rest, along a horizontal straight line. The distance moved by the box in time $t$ is proportional to

A body at rest is moved along a horizontal straight line by a machine delivering a constant power. The distance moved by the body in time $t$ is proportional to

A body is moved along a straight line by a machine delivering constant power. The distance moved by the body in time $t$ is proportional to

A body is moved along a straight line by a machine delivering a constant power. The distance moved by the body in time $t$ is proportional to

An automobile of mass $m$ accelerates starting from origin and initially at rest, while the engine supplies constant power $P$ . The position is given as a function of time by

A particle of mass $m$ is moving in a circular path of constant radius $r$ such that its centripetal acceleration $a_{c}$ is varying with time $t$ as $a_{c} = k^{2} r t^{2}$ , where $k$ is a constant. The power delivered to the particle by the force acting on it is

A body of mass $m$ is accelerated uniformly from rest to a speed $v$ in a time $T$ . The instantaneous power delivered to the body as a function of time, is given by

A body of mass $2 kg$ is driven by an engine delivering a constant power of $1 J/s$ . The body starts from rest and moves in a straight line. After $9 s$ , the body has moved a distance (in m) ......... .