NEETPhysicsWork, Power and Energy
What happens to a particle of total energy 1 J moving under the potential V(x) = (0.5 N/m^(-1))x^2/2 when it reaches x = ± 2 m?
Options
- AIt will continue to move indefinitely in the same direction.
- BIt will come to a stop at x = ± 2 m and remain there.
- CIt will turn back when it reaches x = ± 2 m.
- DIt will accelerate when it reaches x = ± 2 m.
Correct answer
C. It will turn back when it reaches x = ± 2 m.
Step-by-step solution
Correct Option is : (C) It will turn back when it reaches x = ± 2 m.