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)
- 2.0 m/s
- 0.2 m/s
- 14.1 m/s
- 1.41 m/s
Answer
(4) 1.41 m/s
Solution
At the equilibrium (mean) position P.E. = 0 (reference), so the whole energy is kinetic.
12mv2 = 0.02 J, with m = 20 g = 0.020 kg
⇒ v2 = 2 0.020.020 = 2
⇒ v = 2 1.41 m/s
Answer: (4)