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The potential energy of a 1 kg particle ...

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) is

Text Solution

Verified by Experts

For maximum vlaue of `U, (dU)/(dx) = 0`
`:. (4 x^(3))/(4) - (2x)/(2) = 0` or `x = 0, x = pm 1`.
At `x = 0, (d^(2) U)/(dx^(2)) = -1` and At `x = pm 1, (d^(2) U)/(dx^(2)) = 2`
Hence `U` is minimum at `x = pm 1` with value
`U_(min) = (1)/(4) -(1)/(2) = -(1)/(4) J`
`K_(max)+U_(min)=E` or`K_(max)-(1)/(4) = 2` or `K_(max) = (9)/(4)`
`rArr (1)/(2) mv^(2) = (9)/(4) rArr v_(max) = (3)/(sqrt(2))ms^(-1)`.
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