The potential energy of a 2 kg particle, free to move along the x -axis is given by V ⁡ ( x ) = ( x 4 4 - x 2 2 ) J . The total mechanical energy of the particle is 2 J then, the maximum speed (in m s - 1 ) is

Step-by-step solution

Total energy E T = 2 J It is fixed. For maximum speed, kinetic energy is maximum The potential energy should, therefore, be minimum ∵ V ⁡ ( x ) = x 4 4 - x 2 2 or d ⁡ V ⁡ d ⁡ x = 4 x 3 4 - 2 x 2 = x ( x 2 - 1 ) For V to be minimum, d ⁡ V ⁡ d ⁡ x = 0 ∴ x ( x 2 - 1 ) = 0 , ⁡ or ⁡ x = 0 , ⁡ ± 1 at x = 0 , V ( x ) = 0 At x = ± 1 , ⁡ V ⁡ ( x ) = - 1 4 J ∴ ( Kinetic energy ) max = E ⁡ T ⁡ - V ⁡ min or ∴ ( Kinetic energy ) max = 2 + 1 4 = 9 4 J or 1 2 m ⁡ ν m ⁡ 2 = 9 4 ν m 2 = 9 × 2 m × 4 or ν m ⁡ 2 = 9 × 2 m ⁡ × 4 = 9 ×