A chain of length `l = 80 cm` and mass m = 2 kg is hanging from the end of plane so that the length `l_(o)` of the vertical segment is 50 cm as shown in the figure. The other end of the chain is fixed by a nail. At a certain instant, the nail is pushed out, what is the velocity of the chain at the moment it completely slides off the plane ? Neglect the friction.

Explanation

We assume the zero potential energy level at the horizontal plane. The initial and final configuration of the chain are shown in the figure. Initially, `KE_("in")=0`
`U_("in")=0+((m)/(l)l_(0))g(-(l_(0))/(2))`
or `U_("in")=-(ml_(0)^(2))/(2l)g`
Note that the part of chain lying over the table has zero potential energy.

Finally, `KE_(f)=(1)/(2)mv^(2)`
Where v is the final velocity of the chain. and `U_(f)=-mg.(1)/(2)`
Using the law of energy conservation
`KE_(f)+U_(f)=KE_("in")+U_("in")`
`(1)/(2)mv^(2)-mg.(l)/(2)=0-(ml_(0)^(2)g)/(2l)`
or `v=sqrt((g)/(l)(l^(2)-l_(0)^(2)))`
Putting `l = 0.8m, l_(0) = 0.5m, g = 10 m//s^(2)`, we get
`v=sqrt(5.1)m//s`
or `v=2.23 m//s`