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NTA Abhyas JEE Main2020PhysicsWork, Power and EnergyPractice

The potential energy of 1 kg particle free to move along the X-axis is given by U = x 4 4 - x 2 2 J . The total mechanical energy of the particle is 2 J . Maximum speed of the particle is

Options

  1. A4 2
  2. B1 2
  3. C3 2
  4. D2

Correct answer

C. 3 2

Step-by-step solution

U = x 4 4 - x 2 2 (given) For maxima or minima of PE = d U d x = ( x 3 - x ) ∴ x x 2 - 1 = 0 ⇒ x = 0 or ± 1 d 2 U d x 2 = 3 x 2 - 1 at x = ± 1 d 2 U d x = +   v e    i . e . ,    a t    x = ± 1 , P.E. is minimum U = - 1 4 E = K m a x + V m i n ⇒ 2 = K m a x - 1 4 ∴ 1 2 m v m a x 2 = 9 4 ∵ m = 1   kg (given) ⇒ v m a x = 3 2 m   s - 1  

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