__Question 6.18__:
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?

__Solution__:*l*= 1.5 m

Mass of the bob =

*m*

Energy dissipated = 5%

According to the law of conservation of energy, the total energy of the system remains constant.

__At the horizontal position:__

Potential energy of the bob,

*E*

_{P}=

*m*g

*l*

Kinetic energy of the bob,

*E*

_{K}= 0

Total energy =

*m*g

*l*… (

*i*)

__At the lowermost point (mean position):__

Potential energy of the bob, E

_{P}= 0

Kinetic energy of the bob, E

_{K}= (1/2)mv

^{2}

Total energy E

_{x}= (1/2)mv

^{2}....(ii)

As the bob moves from the horizontal position to the lowermost point, 5% of its energy gets dissipated.

The total energy at the lowermost point is equal to 95% of the total energy at the horizontal point, i.e.,

(1/2)mv

^{2}= (95/100) mgl

∴ v = (2 X 95 X 1.5 X 9.8 / 100)

^{1/2}

= 5.28 m/s

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