On a particle placed at origin a variable force `F=-ax`(where a is a positive constant) is applied. If U(0)=0, the graph between potential energy of particle U(x) and x is best represented by:-

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Answer

To solve the problem step by step, we will derive the potential energy function \( U(x) \) from the given force \( F = -ax \).

### Step 1: Understand the relationship between force and potential energy
The relationship between force \( F \) and potential energy \( U \) is given by:
\[ F = -\frac{dU}{dx} \]
This means that the force acting on the particle is equal to the negative gradient of the potential energy.

### Step 2: Substitute the given force into the equation
We are given that the force is:
\[ F = -ax \]
Substituting this into the equation gives:
\[ -ax = -\frac{dU}{dx} \]
This simplifies to:
\[ \frac{dU}{dx} = ax \]

### Step 3: Integrate to find potential energy
To find the potential energy \( U(x) \), we need to integrate the equation:
\[ dU = ax \, dx \]
Integrating both sides:
\[ U(x) = \int ax \, dx = \frac{a}{2} x^2 + C \]
where \( C \) is the constant of integration.

### Step 4: Use the initial condition to determine the constant
We are given that \( U(0) = 0 \). Substituting \( x = 0 \) into the potential energy equation:
\[ U(0) = \frac{a}{2} (0)^2 + C = 0 \]
This implies that:
\[ C = 0 \]
Thus, the potential energy function simplifies to:
\[ U(x) = \frac{a}{2} x^2 \]

### Step 5: Identify the shape of the graph
The equation \( U(x) = \frac{a}{2} x^2 \) represents a quadratic function, which is a parabola that opens upwards (since \( a \) is positive).

### Conclusion
The graph of potential energy \( U(x) \) versus \( x \) is an upward-opening parabola.