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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

A

`(3)/(sqrt2)m//s`

B

`sqrt2m//s`

C

`(1)/(sqrt2)m//s`

D

`2 m//s`

Text Solution

Verified by Experts

The correct Answer is:
A
At max. speed , `KE` is max. and `PE` is min. `PE`
is min at `x=+-1` so `PE_(min)=-(1)/(4)`
`KE_(max)=TE_(min)-PE_(min)rArr(1)/(2)mV_(max)^(2)=(9)/(4)rArrV_(max)=(3)/(sqrt2)m//s`
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