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Solved 2024 Specimen Paper ICSE Class 10 Physics — Question 9

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Question

Question 6(iii)

The figure below 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 the position A is zero. At the position B the pendulum bob is raised by 5 m.

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 the position A is zero. At the position B the pendulum bob is raised by 5 m. What is the potential energy of the pendulum at the position B? What is the total mechanical energy at point C? What is the speed of the bob at the position A when released from B? ICSE 2024 Specimen Physics Solved Question Paper.

(a) What is the potential energy of the pendulum at the position B?

(b) What is the total mechanical energy at point C?

(c) What is the speed of the bob at the position A when released from B?

(Take g = 10 ms-2 and given that there is no loss of energy.)

Answer

Given,

h = 5 m

m = 200 g = 0.2 kg

g = 10 ms-2

(i) Potential energy UB at B is given by

UB = m x g x h

Substituting the values we get,

UB = 0.2 x 10 x 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 same at all points of the path due to conservation of mechanical energy.

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

Therefore,

12×m×(vA)2=UB0.5×0.2×(vA)2=10(vA)2=100.1⇒vA=100⇒vA=10 m s−1\dfrac{1}{2} \times m \times (v_A)^{2} = U_B \\[0.5em] 0.5 \times 0.2 \times (v_A)^{2} = 10 \\[0.5em] (v_A)^{2} = \dfrac{10}{0.1} \\[0.5em] \Rightarrow v_A = \sqrt{100} \\[0.5em] \Rightarrow v_A = 10 \text{ m s}^{-1}