A uniform rope of mass `m` per unit length, hangs vertically from a support so that the lower end just touches the table top shown in figure. If it is released, show that at the time a length `y` of the rope has fallen, the force on the table is equivalent to the weight of the length `3y` of the rope.

Explanation

`v=sqrt(2gy)`
Thrust force downwards
`F=lambdav^2` (`lambda=mass//l eng th`)
or `F=m(2gy)` or `F=2mgy`
`y` length is lying on table. So its weight
`W=(ym)g`
Total force on table `=F+W=(3mgy)`
`="weight of length 3y of the rope"`