NTA Abhyas JEE Main2020PhysicsWork, Power and EnergyPractice
A bead of mass m can slide without friction on a fixed circular horizontal ring of radius 3 R having a centre at the point C . The bead is attached to one of the ends of spring of spring constant k . Natural length of spring is R and the other end of the spring is fixed at point O as shown in the figure. If the bead is released from position A , then the kinetic energy of the bead when it reaches point B is
Options
- A25 2 kR 2
- B9 2 kR 2
- C8 kR 2
- D12 kR 2
Correct answer
C. 8 kR 2
Step-by-step solution
Given, Mass of bead = m Radius of circular horizontal ring = 3 R Natural length of spring = R According to conservation of energy, KE i + PE i = KE f + PE f ⇒ 0 + 1 2 k OA - R 2 = KE f + 1 2 k OB - R 2 ⇒ 0 + 1 2 k 5 R - R 2 = KE f + 1 2 k R - R 2 ⇒ KE f = 8 kR 2