Consider a uniform chain of mass `m` and length `L`, the chain is released when hanging length is `L_0`. Find the speed with which the chain leaves the table.

Explanation

`K_1=0`,`U_1=-(1)/(2)((m)/(L)L_0)gL_0`
`K_2=(1)/(2)mv^2`,`U_2=-(mgL)/(2)`
By mechanical energy conservation
`K_1+U_1=K_2+U_2`
`0-(mgL_0^2)/(2L)=(1)/(2)mv^2-(mgL)/(2)`
`v^2=gL-(gL_0^2)/(L)=(g(L^2-L_0^2))/(L)`
`v=sqrt((g(L^2-L_0^2))/(L))`