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The potential energy of a 2 kg particle ...

The potential energy of a `2 kg` particle free to move along the `x-` axis is given by
`V(x)[(x^(4))/(4)-(x^(2))/(2)]jou l e`
The total mechanical energy of the particle is `0.75 J`. The maximum speed of the particle `(`in `m//s)` is `:`

Text Solution

Verified by Experts

When speed is maximum, KE is maximum.
PE is minimum.
As `V(x)=((x^(4))/( 4)-(x^(2))/( 2))J=` minimum
`:. (dV(x))/( dx)=0`
`4 (x^(3))/( 4)-(2x)/(2)=0 :. x=0 ` or `x=+-1`
`V_(min)("at" x=+- 1 ) =(1)/(4)-(1)/(2)=-(1)/(4)=-0.25J`
As `K_(max)+V_(min)=` Mechanical energy
`:. K_(max)=ME-V_(min)`
`=(1)/(2) m upsilon_(max)^(2)=0.75-(-0.25)=1J`
`upsilon_(max)=sqrt((2xx1)/( m))=sqrt((2)/(2))=1m//s`
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