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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 mechanical energy of the particle is 2 J. Then, the maximum speed (in m/s) is 

A

2

B

`3/(sqrt2)`

C

`sqrt(2)`

D

`1/(sqrt2)`

Text Solution

Verified by Experts

The correct Answer is:
B
Total energy `E_t = 2J` . It is fixed .
For maximum speed. Kinetic energy is maximum.
The potential energy should therefore be minimum.
As `V (x) = (x^4)/(4) - (x^2)/(2)`
`:. (dV)/(dx) = (4x^3)/(4) - (2x)/(2) = x^3 - x = x(x^2 -1)`
For V to be minimum, `(dV)/(dx) = 0`
`:. x(x^2 - 1) = 0 , "or " x = 0, pm 1`
At `x = 0, V(x) = 0` and at `x = pm 1, V(x) = -1/4 J`
`:. ("Kinetic energy")_("max") = E_T - V_("min")`
or `("Kinetic energy")_("max") = 2 - (-1/4) = 9/4 J`.
or `1/2 mv_("max")^(2) = 9/4 " or " v_("max")^(2) = (9 xx 2)/(m xx 4) = (9 xx 2)/(1 xx 4) = 9/2`
`:. v_("max") = 3/(sqrt2) m//s` .
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Knowledge Check

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