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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

2

B

`3//sqrt(2)`

C

`sqrt(2)`

D

`1//sqrt(2)`

Text Solution

Verified by Experts

The correct Answer is:
B
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