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 mechainical energy of the particle is `2 J` . Then , the maximum speed (in m//s) isA. `1//sqrt(2)`B. `2`C. `3//sqrt(2)`D. `sqrt(2)`

The potential energy of a `1 kg` particle free to move along the x- ax

1.	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 mechainical energy of the particle is `2 J` . Then , the maximum speed (in m//s) isA. `1//sqrt(2)`B. `2`C. `3//sqrt(2)`D. `sqrt(2)`
Answer» Correct Answer - C `U =(x^(4))/(4)-(x^(2))/(2),(dU)/(dx) =x^(3) -x = 0 rArrx =0, x = pm 1` `(d^(2)U)/(dx^(2)) = 3x^(2) -1(d^(2)U)/(dx^(2)) = +ve` for `x =pm1` `U(pm 1)=-(1)/(4)` `K_(max) =Ulmin=T.E = 2J` `K_(max) =(9)/(4)` `K_(max) =(1)/(2) mv^(2) , v = (3)/(sqrt(2))`.

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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 mechainical energy of the particle is `2 J` . Then , the maximum speed (in m//s) isA. `1//sqrt(2)`B. `2`C. `3//sqrt(2)`D. `sqrt(2)`

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