Question

A uniform chain of length $16m$ and mass $540g$ whose $\frac{2}{3^{rd}}$ part of its length is hanging vertically down over the edge of the table and remaining part on the smooth table. Consider $g$ is $10m{s}^{-2}$ , then the work required to pull the hanging part on to the table is $\frac{x}{10}J$ . Find $x$ .

Answer 1

Answer: 192

The length of hanging part is $\frac{2L}{3}$ , so its mass would be $\frac{2M}{3}$ and will act at the centre of gravity G of hanging part, i.e., at a distance $\frac{L}{3}$ below the surface of table.
$\therefore$ Work done in pulling the hanging part on the
table $=$ increase in P.E.
$=mgh$
$=\frac{2M}{3}\times g\times \frac{L}{3}$
$=\frac{2MgL}{9}=\frac{2\times 0.54\times 10\times 16}{9}$
Alternate Method
$F=\frac{M}{L}(x)g$
$\therefore W=Fdx=\frac{M}{L}(x)gdx$
$\therefore$ Total work done to pull the $\frac{2^{\text{rd }}}{3}$ part on the table,
$W=\underset{0}{\overset{\frac{2L}{3}}{\int }}\frac{M}{L}gxdx=\frac{M}{L}g\underset{0}{\overset{\frac{2L}{3}}{\int }}xdx$
$=\frac{M}{L}g{\left[\frac{x^2}{2}\right]}_0^{\frac{L}{3}}=\frac{M}{L}g\left[\frac{{\left(\frac{2L}{3}\right)}^2}{2}\right]$
$=\frac{Mg}{L}\times \frac{4L^2}{18}$
$\therefore W=\frac{2MgL}{9}$
$\therefore W=\frac{2\times 0.54\times 10\times 16}{9}$
$\therefore W=\frac{192}{10}J$
So, $x=192$ .