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The figure alongside shows a simple pendulum of mass 200 g. It is displaced from the mean position A to the extreme position B. The potential energy at position A is zero. At position B the bob is raised by 5 m.
(a) What is the potential energy of the pendulum at position B ?
(b) What is the total mechanical energy at point C?
(c) What is the speed of the bob at position A when released from position B ? (Take g = 10 ms-2)

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Answer

height (h) = 5 m,

Mass (m) = 200 g = 0.2 kg,

gravity (g) = 10 ms-2

(i) Potential energy UB at B is given by

UB = m × g × h

Substituting the values we get,

UB = 0.2 × 10 × 5 = 10 J

Hence, the potential energy of the pendulum at the position B = 10 J

(ii) Total mechanical energy at point C = 10 J

The total mechanical energy is the same at all points of the path due to the conservation of mechanical energy.

(iii) At A, Bob has only kinetic energy which is equal to potential energy at B,

Therefore,

1\over 2m(vA)2 = UB

1\over 2 × 0.2 × (vA)2 = 10

⇒ (vA)2 = 10\over 0.1 = 100

⇒ vA = √100 = 10 m s-1

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