A particle is placed at the origin and a force F=Kx is acting on it (where k is a positive constant). If `U_((0))=0`, the graph of `U (x)` verses x will be (where U is the potential energy function.)

Explanation

To solve the problem, we need to find the potential energy function \( U(x) \) given the force \( F = kx \) and the condition that \( U(0) = 0 \).

### Step-by-Step Solution:

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 is equal to the negative gradient (derivative) of the potential energy function.

2. **Substitute the given force into the equation**:
We are given that \( F = kx \). Substituting this into the equation gives:
\[ kx = -\frac{dU}{dx} \]

3. **Rearranging the equation**:
We can rearrange the equation to express \( dU \):
\[ dU = -kx \, dx \]

4. **Integrate both sides**:
To find \( U \), we need to integrate:
\[ U = -\int kx \, dx \]
The integral of \( kx \) is:
\[ U = -\left(\frac{kx^2}{2}\right) + C \]
where \( C \) is the constant of integration.

5. **Apply the condition \( U(0) = 0 \)**:
We know that \( U(0) = 0 \). Substituting \( x = 0 \) into the equation gives:
\[ U(0) = -\frac{k(0)^2}{2} + C = 0 \implies C = 0 \]
Therefore, the potential energy function simplifies to:
\[ U(x) = -\frac{kx^2}{2} \]

6. **Identify the nature of the function**:
The function \( U(x) = -\frac{kx^2}{2} \) is a quadratic function that opens downwards (since the coefficient of \( x^2 \) is negative). This indicates that the graph of \( U(x) \) versus \( x \) will be a downward-opening parabola.

7. **Sketch the graph**:
The graph will intersect the x-axis at \( x = 0 \) and will decrease as \( |x| \) increases. The vertex of the parabola will be at the origin (0,0).

### Final Answer:
The graph of \( U(x) \) versus \( x \) will be a downward-opening parabola with its vertex at the origin (0,0).

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