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 1

Length of the pendulum, *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 × 95 × 1.5 × 9.8 / 100)^{1/2}

= 5.28 m/s

#### Solution 2

On releasing the bob of pendulum from horizontal position, it falls vertically downward by a distance equal to length of pendulum i.e., h = l = 1.5 m .

As 5% of loss in P.E. is dissipated against air resistance, the balance 95% energy is transformed into K.E. Hence,

`1/2mv^2 = 95/100 xx mgh`

`=>v = sqrt(2xx95/100 xx gh) = sqrt((2xx95xx9.8xx15)/100) = 5.3 ms^(-1)`