Question

A uniform chain of length $L$ and mass $M$ is lying on a smooth table and one-third of its length is hanging vertically down over the edge of the table. If $g$ is the acceleration due to gravity, the work required to pull the hanging part onto the table is

Answer 1

Velocity, Acceleration due to gravity, and Work:

1. Velocity- It is the rate at which an object's position changes in a relation to the frame of reference and time.
2. Acceleration due to gravity- Acceleration of freely falling bodies caused by the gravitational force of the earth. Its value is 9.81 m/s².
3. Work- When a force is applied to a body across a distance, work is done.

Step 1: Given data

Length of chain, $L$

Length of chain hanging, $\frac{1}{3}L$

Mass of chain, $M$

Step 2: Formula used

$W=\int Fdl$

Where $W$ is work done, $F$ is the force applied, and $dl$ is the displacement length.

$F=mg$

Where $F$ is force, $m$ is mass, and $g$ acceleration due to gravity.

$m=\lambda l$

Where $m$is mass, $\lambda$ is linear mass density, and $l$ is length.

Step 3: Calculation

Work done is,

$W=\int Fdl$

$\Rightarrow W=\int Fdl$ …(i)

The force required to draw a body higher when it's under gravity is given as

$$\begin{gathered} F=mg \\ \Rightarrow F=Mg \end{gathered}$$

We calculate mass as

$$\begin{gathered} m=\lambda l \\ \Rightarrow M=\lambda l \end{gathered}$$

Replacing $M$ with $\lambda l$ in equation (i)

$F=\lambda lg$

Putting the value of $F$ in the equation of work done

$$\begin{gathered} W=\int \lambda gldl \\ W=\lambda g\int ldl \end{gathered}$$

As one-third of the chain is hanging

$l=\frac{1}{3}L$

So,

$W=\lambda g{\int }_0^{\frac{L}{3}}ldl$

$\Rightarrow W=\lambda g{{\left[\frac{l^2}{2}\right]}_0}^{\frac{L}{3}}$

$\Rightarrow W=\lambda g\frac{1}{2}\left[{\left(\frac{1}{3}L\right)}^2-0^2\right]$

$$\begin{gathered} \Rightarrow W=\lambda g\frac{1}{2}\left[{\left(\frac{1}{3}L\right)}^2-0^2\right] \\ \Rightarrow W=\lambda g\frac{1}{2}\left[\frac{1}{9}{L}^2\right] \\ \Rightarrow W=\frac{\lambda g{L}^2}{18} \end{gathered}$$

as, $M=\lambda L$

so,$W=\frac{MgL}{18}$

The work required to pull the hanging part onto the table, $W=\frac{MgL}{18}$.