A particle of mass 2/3 kgis subjected to a potential energy V ( x ) = (3x2 - 2x3)J.where x ≥ 0 and expressed in meters, thena)The position of the maxima is 1b)The position of the maxima is 0c)Maximum value of potential is 1.5 Jd)aximum value of potential is 1JCorrect answer is option 'A,D'. Can you explain this answer?

Question

A particle of mass 2/3 kgis subjected to a potential energy V ( x ) = (3x2 - 2x3)J.where x &ge; 0 and expressed in meters, thena)The position of the maxima is 1b)The position of the maxima is 0c)Maximum value of potential is 1.5 Jd)aximum value of potential is 1JCorrect answer is option 'A,D'. Can you explain this answer?

Answer

V { x ) = X²( 3 - 2x)
The position of maxima and minima are given by

x = 0.1

So. x=0 is a point of minima

x = 1 is a point of maxima

Answer

The potential energy of the particle is given by V(x) = 3x^2 - 2x^3 J.

To find the equilibrium positions of the particle, we need to find the points where the potential energy is at a minimum or maximum. This can be done by taking the derivative of the potential energy with respect to x and setting it equal to zero:

dV/dx = 6x - 6x^2 = 0

Factor out x:

x(6 - 6x) = 0

This equation has two solutions: x = 0 and x = 1.

To determine whether these points correspond to a minimum or maximum, we can take the second derivative of the potential energy:

d^2V/dx^2 = 6 - 12x

Plug in x = 0 and x = 1:

d^2V/dx^2 |x=0 = 6
d^2V/dx^2 |x=1 = -6

Since the second derivative is positive at x = 0, this point corresponds to a minimum. Since the second derivative is negative at x = 1, this point corresponds to a maximum.

Therefore, the equilibrium positions of the particle are x = 0 and x = 1.

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