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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 - Physics

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?

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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, EP = mgl

Kinetic energy of the bob, EK = 0

Total energy = mg… (i)

At the lowermost point (mean position): Potential energy of the bob, EP = 0

Kinetic energy of the bob, EK = (1/2)mv2

Total energy Ex = (1/2)mv2    ....(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)mv2 = (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)`

  Is there an error in this question or solution?
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NCERT Class 11 Physics Textbook
Chapter 6 Work, Energy and Power
Q 18 | Page 137
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