Animated Solution for Physics - Work, Energy, and Power: 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 $U(x)$ versus $x$ will be (where, $U$ is the potential energy function)

Imagine you are observing a tiny particle resting peacefully at the origin of a coordinate system. Suddenly, a mysterious force begins to act on it. This force isn't constant; it changes depending on where the particle is. The problem tells us that the force is given by $F=kx$, where $k$ is a positive constant.

This means that the further the particle moves away from the origin, the stronger the force becomes. If it moves to the right (positive $x$), the force pushes it further to the right. If it moves to the left (negative $x$), the force pushes it further to the left. It's like an anti-spring!

Our mission is to uncover the hidden landscape of potential energy, $U(x)$, that governs this particle's motion. We are given a crucial clue: at the origin, the potential energy is zero, meaning $U(0)=0$.

To find the potential energy, we need to pull out a fundamental tool from our physics arsenal. The relationship between a conservative force and potential energy is one of the most beautiful connections in mechanics. It is given by the equation:

This equation tells us that force is the negative gradient of potential energy. In simpler terms, a particle always "wants" to move towards lower potential energy, just like a ball rolling down a hill. The steeper the hill, the stronger the force.

To find $U(x)$, we need to work backwards. We need to integrate the force. Let's rearrange our master equation to isolate the tiny change in potential energy, $dU$:

This tells us that the work done by the force over a tiny distance $dx$ results in a decrease in potential energy.

Now, we are ready to integrate. We will add up all these tiny changes from our starting point (the origin, where $x=0$) to some arbitrary position $x$.

This is the mathematical equivalent of walking along the path and keeping a running tally of the energy changes.

Let's substitute the specific force given in our problem, $F=kx$, into our integral:

Now, we execute the integration. The left side is straightforward. The integral of $dU$ is simply $U$, evaluated from $0$ to $U(x)$:

The right side requires us to integrate $x$ with respect to $x$. The power rule for integration tells us that the integral of $x^1$ is $\frac{x^2}{2}$. Since $k$ is a constant, it just comes along for the ride:

Equating both sides, we get:

We remember our initial clue: $U(0)=0$. Substituting this in, we arrive at our final, elegant expression for the potential energy:

Look at this final equation: $U(x)=-\frac{k{x}^2}{2}$. What does it look like geometrically?

Because $x$ is squared, we know it's a parabola. Because of the negative sign in front, we know it's an inverted, or downward-opening, parabola. And because there are no other terms, its vertex is perfectly centered at the origin $(0,0)$.

This perfectly matches the graph shown in option (a). The particle is sitting at the top of a potential energy "hill". Any slight nudge will cause it to roll down the hill, accelerating as the force $F=kx$ pushes it further away from the origin. This is a classic example of unstable equilibrium!