The potential energy of a 1 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 &#828