The bob of pendulum is released from a horizontal position . If the length of the pendulum is `1.5 m ` ,what is the speed with which the bob arrives at the speed with which the bob arrives at the lowermost point , given that it dissipated 5 % of its intial energy against air resistance ?

Explanation

Kinetic energy at point B = 95 % potential energy at point A ,
`1/2 mv^(2) = 0.95 xx mgh`
` :. 1/2 mv^(2) = 0.95 xx mgh`
` :. V^(2) = 2xx 0.95 xxgh`
` :. v = sqrt(2xx0.95 xx9.8 xx1.5) = 5.285 ms^(-1)= 5.3 ms^(-1)`