# 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

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