The sum of kinetic energy and potential energy of a simple pendulum bob is $0.02$ joule. The speed of the simple pendulum bob at equilibrium position is approximately :
(Consider mass of the bob $= 20$ g)

Question

The sum of kinetic energy and potential energy of a simple pendulum bob is 0.02 joule. The speed of the simple pendulum bob at equilibrium position is approximately : (Consider mass of the bob = 20 g)

Quantrex Academy · Accepted Answer

Correct answer: B. 1.41 m/s

Total energy of the simple pendulum is given as

E = 0.02

J.

Mass of the bob,

m = 20

g

= 0.02

kg.

At the equilibrium position, the potential energy is zero and the total energy is purely kinetic.

E = \dfrac{1}{2} m v^2

0.02 = \dfrac{1}{2} \times 0.02 \times v^2

v^2 = 2

v = \sqrt{2} \approx 1.41

m/s.

Answer:

1.41

m/s

The sum of kinetic energy and potential energy of a simple pendulum...

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