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

The potential energy of a 1kg particle free to move along x-axis is gi

1.	The potential energy of a 1kg particle free to move along x-axis is given by V(x)=((x^(4))/(4)-(x^(2))/(2))J. The total mechanical energy of the particle is 2J. Then, the maximum speed (in m/s) is :
Answer» `3//SQRT(2)` `sqrt(2)` `1//sqrt(2)` 2 Solution :At mean POSITION X = 0, PE. U = 0 `THEREFORE` At `x=0`, K.E. = Total energy `(1)/(2)mv^(2)=2impliesv=sqrt((2xx2)/(1))=2MS^(-1)`

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

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