The bob of a 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 lowermost point, given that it dissipated 5% of its initial energy against air resistance?
Given,
Length of the pendulum, l = 1.5 m
Energy dissipated against air resistance, E = 5%
Let m be the mass of the bob and v be its velocity at the lowermost point.
At the extreme end:
Potential energy of the bob, P = mgl
Kinetic energy of the bob, K = 0 (∵ bob is at rest)
Total energy of bob, E = P + E
⇒ E = mgl ……………(1)
At the lowermost point:
Potential energy of the bob, P = 0
Kinetic energy of the bob, K = (1/2)mv2
Total energy of the bob, E = P + K
⇒ E = (1/2)mv2
During the motion from an extreme end to the lowermost point, the bob loses 5% of its energy, i.e., it has 95% of the energy.
So, (1/2)mv2 = (95/100)mgl
⇒ v2 = 95gl/50
⇒ v = 
⇒ v = 
⇒ v = 5.28 m s-1