The bob A of a simple pendulum is released from a horizontal position A as shownin in figure. If the length of the pendulum is 1.5m , what is the speed with which the bob arrives at the lowermost point B, given that it dissipates `5%` of its initial energy against air resistance ?

Explanation

Here, `h=1.5m , upsilon=?`
Energy dissipated `=5%`
Taking B as the lowest position of the bob, its potential energy at B is zero. At the horizontal position A, total potential energy of the bob is mgh
In going from A to B, P.E. of the bob is converted into K.E.
Energy converted `=95% (mgh)`
It `upsilon` is velocity acquired at B, then K.E. `=(1)/(2) m upsilon^(2)=(95)/(100)mgh`
`upsilon=sqrt((95)/(100)xx2gh)=sqrt((19)/(20)xx2xx9.8xx1.5)=5.28ms^(-1)`