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 on the table is ?

Answer 1

cut an element dx at x distance from the vertical hanging part of chain.

mass of element , dm = (M/L)dx
now element is located at x distance above from the lowest point of hanging chain.

so, work done required to pull the hanging part on the table ,
$W=dm.g.x\\\\W=\frac{M}{L}dx.g(x)\\\\W=\frac{Mg}{L}\int\limits^{L/3}_0xdx\\\\=\frac{Mg}{L}\{\frac{x^2}{2}\}^{L/3}_0\\\\=\frac{Mg}{L}\frac{L^2}{9\times2}$

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

Answer 2

Mark me brainliest if it helps

Explanation: