Question

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

Answer 1

P.E. = mgh = m x 9.8 x 1.5 J.

Since 5% of energy is lost, so

P.E. = m x 9.8 x 1.5 x 95/100

But, this energy is converted to K.E.

K.E. = P.E

1/2 mv² = m x 9.8 x 1.5 x 95/100

v = √{2 x 9.8 x 1.5 x 95/100} = 5.29 ms^-1