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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

`sqrt2`

B

`1//sqrt2`

C

2

D

`3//sqrt2`

Text Solution

Verified by Experts

The correct Answer is:
D
`V= ((x^(4))/(4)- (x^(2))/(2)) J`
For max. KE, PE should be minimum
`(dV)/(dx)= 0 rArr x^(3)- x= 0 rArr x = pm 1, x= 0`
`(d^(2)V)/(dx^(2))= 3x^(2)-1`
`(d^(2)V)/(dx^(2)) gt 1` (minimum) `V_("min") = [(1^(4))/(4)- (1^(2))/(2)] = - (1)/(4)J`
`(KE)_("max")`= Total energy `-V_("min")`
`rArr (1)/(2) mv_(m)^(2)= 2- (- (1)/(4))= (9)/(4)`
`rArr v_(m)^(2)= (9)/(2) rArr v_(m)= (3)/(sqrt2)`
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