Q50 / 92 · NTA Abhyas (JEE Main) 2020 · Physics · Work Power Energy

The potential energy of a 1 kg particle, free to move along the x-axis, is given by V⁡(x)=(x44-x22) J. The total mechanical energy of the particle is 2 J then, the maximum speed (in m s-1) is

  • A
    2
  • Correct answer
    B
    3 / 2
  • C
    2
  • D
    1 / 2

Solution

Total energy ET =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=(14)=94 J 

or 12m⁡vm⁡2=94 

vm2=9×2m×4

or vm⁡2=9×2m⁡×4=9×21×4

∴vm⁡=32 m s-1

Hence, the correct answer is option: 'B'.

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