Question

The potential energy of a particle varies with position x according to the relation U(x) = 2x^4-27x. The point x=3/2 is of...
A) unstable equilibrium
B) Stable equilibrium
C) Neutral equilibrium
D) None of these
If u r brainliest solve dis sum​

Answer 1

Answer:

D is correct because the potential energy varies with height.No states of equlibrium

Answer 2

Explanation:

first, we have to differentiate u i. e potential energy into force

U(x) =2x^4-27x

we know that,

F= -du/dt = 8x^3 - 27

putting the value of x into it , we get

F= 8(3/2)^3 - 27 = 0

As we know in equilibrium, if we get Fnet = 0 then we have double differentiate the equation,

F = d2U/dx^2

F = 24x^2 - 0

F = 24x^2

Again put the value of x in the equation,

F = 24(3/2) ^2

F = 36>0

lastly we get the force is greater than 0 , so it will be in stable equilibrium

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