Potential energy of a particle moving along x-axis is by
`U=(x^(3)/3-4x + 6)`.
here, U is in joule and x in metre. Find position of stable and unstable equilibrium.

Explanation

To solve the problem of finding the positions of stable and unstable equilibrium for the given potential energy function \( U(x) = \frac{x^3}{3} - 4x + 6 \), we will follow these steps:

### Step 1: Find the Force
The force \( F \) acting on the particle is related to the potential energy \( U \) by the equation:
\[ F = -\frac{dU}{dx} \]
We need to differentiate \( U \) with respect to \( x \).

### Step 2: Differentiate the Potential Energy
Differentiating \( U(x) \):
\[ \frac{dU}{dx} = \frac{d}{dx}\left(\frac{x^3}{3} - 4x + 6\right) = x^2 - 4 \]
Thus, the force can be expressed as:
\[ F = -\left(x^2 - 4\right) = -x^2 + 4 \]

### Step 3: Find the Equilibrium Positions
To find the equilibrium positions, we set the force \( F \) to zero:
\[ -x^2 + 4 = 0 \]
This simplifies to:
\[ x^2 = 4 \]
Taking the square root gives us:
\[ x = \pm 2 \text{ meters} \]

### Step 4: Determine Stability of Equilibrium
To determine whether these equilibrium positions are stable or unstable, we need to find the second derivative of the potential energy \( U \):
\[ \frac{d^2U}{dx^2} = \frac{d}{dx}(x^2 - 4) = 2x \]

### Step 5: Evaluate the Second Derivative at Equilibrium Positions
1. For \( x = 2 \):
\[ \frac{d^2U}{dx^2} = 2(2) = 4 \quad (\text{positive}) \]
This indicates that \( x = 2 \) is a position of **stable equilibrium**.

2. For \( x = -2 \):
\[ \frac{d^2U}{dx^2} = 2(-2) = -4 \quad (\text{negative}) \]
This indicates that \( x = -2 \) is a position of **unstable equilibrium**.

### Conclusion
The positions of equilibrium are:
- **Stable equilibrium** at \( x = 2 \) meters.
- **Unstable equilibrium** at \( x = -2 \) meters.

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