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

Given , h = 1.5 m

Kinetic energy of bob at point ` B = 1/2 mv^(2)`
Potential energy of bob at point `A = mgh `
As 5 % of potential energy is disipated against air resistance .
Then , `1/2 mv^(2) = 95/100 mgh`
`v^(2) = (2xx95)/100 gh`
`v = sqrt((2xx95xxgh)/100)`
`v = sqrt((2xx95xx9.8 xx1.5)/100)`
` = 5.28 " m "s^(-1)`