Question

A chain of mass $m=0.80kg$ and length $l=1.5m$ rests on a rough-surfaced table so that one of its ends hangs over the edge. The chain starts sliding off the table all by itself provided the overhanging part equals $\eta =\frac{1}{3}$ of the chain length. The total work performed by the friction forces acting on the chain by the moment it slides completely off the table is $-X$. Write the value of $10X$.

Answer 1

From the initial condition of the problem the limiting friction between the chain lying on the horizontal table equals the weight of the overhanging part of the chain, i.e.

$\lambda \eta lg=k\lambda (1-\eta )lg$ (where $\lambda$ is the linear mass density of the chain)
So, $k=\frac{\eta }{1-\eta }$........ (1)
Let (at an arbitrary moment of time) the length of the chain on the table is $x$. So the net friction force between the chain and the table, at this moment

$f_r=kN=k\lambda xg$......... (2)
The differential work done by the friction forces
$dA=\overrightarrow{f_r}\cdot d\overrightarrow{r}=-f_{r}ds=-k\lambda xg(-dx)=\lambda g\left(\frac{\eta }{1-\eta }\right)xdx$........ (3)
(Note that here we have written $ds=-dx$., because $ds$ is essentially a positive term and as the length of the chain decreases with time, $dx$ is negative)
Hence, the sought work done
$A={\int }_{(1-\eta )l}^0\lambda g\frac{\eta }{1-\eta }xdx=-(1-\eta )\eta \frac{mgl}{2}=-1.3J$