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The potential energy of a 4kg particle f...

The potential energy of a `4kg` particle free to move along the x-axis varies with x according to following relationship : `U(x) = ((x^(3))/(3)-(5x^(2))/(2)+6x+3)` Joules, where `x` is in meters. If the total mechanical energy of the particle is `25.5` Joules, then the maximum speed of the particle is `x m//s`, find `x`

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The correct Answer is:
3
The potential …………..
`U(x) = (x^(3))/(3)-(5x^(2))/(2)+6x+3`
`(dU)/(dx) = x^(2) - 5x+6=0`
`x = 2, 3`
`(d^(2)U)/(dx^(2)) = 2x -5`
At `x = 2m, (d^(2)U)/(dx^(2)) = -1`, which is a negative value this `U` is maximum
At `x = 3 m, (d^(2)U)/(dx^(2)) = +1`, which is a positive value thus `U` is minimum
`U_(min) = U(x=3)=7.5J`
`T.M.E = 25.5J`
`K_(max)+U_(min)= 25.5J`
`K_(max) = 25.5-7.5=18J`
`(1)/(2)(4)V^(2)max=18`
`implies V_(max)^(2) = 9 implies V_(max) = 3m//s`
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