Question

8.
In a certain field, the potential energy is U = ax? –
bx, where a and b constants. The particle is in stable
equilibrium at x equal to
(1) Zero
13b​

Answer 1

Answer:

u=at-ut8yfyfydfxrdzdztchvubuvu

Answer 2

Answer:

$x=\dfrac{a}{3b}$

Explanation:

Lets take

Potential energy given as

U=a x²-b x³

For stable equilibrium

$\dfrac{\partial^2 U}{\partial x^2}=0$

$\dfrac{\partial U}{\partial x}=2ax-3bx^2$

$\dfrac{\partial^2 U}{\partial x^2}=2a-6bx$

So

2 a - 6 b x= 0

a - 3 b x = 0

$x=\dfrac{a}{3b}$

At $x=\dfrac{a}{3b}$ system will be in equilibrium position.