potential enrgy of a particle along x-axis varies as, `U=-20 + (x-2)^(2)`, where U is in joule and x in meter. Find the equilibrium position and state whether it is stable or unstable equilibrium.

Explanation

To solve the problem, we need to find the equilibrium position of the particle and determine whether it is in stable or unstable equilibrium based on the given potential energy function.

### Step-by-Step Solution:

1. **Identify the Potential Energy Function**:
The potential energy \( U \) is given by:
\[ U = -20 + (x - 2)^2 \]

2. **Find the Force**:
The force \( F \) acting on the particle can be found using the relationship:
\[ F = -\frac{dU}{dx} \]
We first need to differentiate \( U \) with respect to \( x \).

3. **Differentiate the Potential Energy**:
Differentiate \( U \):
\[ \frac{dU}{dx} = \frac{d}{dx}[-20 + (x - 2)^2] \]
Using the power rule, we get:
\[ \frac{dU}{dx} = 0 + 2(x - 2) \cdot \frac{d}{dx}(x - 2) = 2(x - 2) \]

4. **Set the Force to Zero for Equilibrium**:
For equilibrium, we set the force \( F \) to zero:
\[ F = -\frac{dU}{dx} = -2(x - 2) = 0 \]
Solving this equation gives:
\[ 2(x - 2) = 0 \implies x - 2 = 0 \implies x = 2 \text{ meters} \]

5. **Determine the Nature of Equilibrium**:
To determine whether the equilibrium at \( x = 2 \) is stable or unstable, we need to check the second derivative of the potential energy:
\[ \frac{d^2U}{dx^2} = \frac{d}{dx}[2(x - 2)] = 2 \]
Since \( \frac{d^2U}{dx^2} > 0 \), this indicates that the potential energy is at a minimum at \( x = 2 \).

6. **Conclusion**:
Therefore, the equilibrium position is at \( x = 2 \) meters, and it is a stable equilibrium position because the potential energy is at a minimum.

### Final Answer:
- **Equilibrium Position**: \( x = 2 \) meters
- **Nature of Equilibrium**: Stable equilibrium