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
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Answers
Answered by
0
Answer:
D is correct because the potential energy varies with height.No states of equlibrium
Answered by
1
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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