A ball tied to a rope making an angle `30^(@)` with the horizontal is held so that the rope is just If the ball is released, calculate the heat generated in joule when it crosses `30^(@)` down the horizontal (dotted)Given `m=Kg,L=2m,g=10m//s^(2)`.

Explanation

The speed just after jerking
`v=sqrt(2gL) cos 30^(@)=sqrt((3gL)/(2))`
`therefore` heat=10 sec in kinetic energy
`=(1)/(2)m(2gL-(3gL)/(2))`
`=(mgL)/(4)=5J`