A particle, which is constrained to move along x-axis, is subjected to a force in the same direction

Q: Why do we assume $U(0) = 0$?

Potential energy is relative. We can choose our reference point anywhere. Unless specified otherwise, it is standard convention to set the potential energy at the origin to zero to simplify the math. The shape of the curve remains exactly the same; it would just shift vertically if we chose a different reference.

Q: Why does the curve go up initially if the force has a negative sign?

For very small $x$, the force $F(x) \approx -kx$. A negative force means it points towards the origin (it is a restoring force). If you move a particle against a restoring force, you have to do work against it, which gets stored as potential energy. Hence, the potential energy increases as you move away from the origin.

Q: How do we know it is a maximum and not a minimum at $x = \sqrt{\frac{k}{a}}$?

You can use the second derivative test! $\frac{d^2U}{dx^2} = -\frac{dF}{dx} = -(-k + 3ax^2) = k - 3ax^2$. At $x = \sqrt{\frac{k}{a}}$, the second derivative is $k - 3a(\frac{k}{a}) = -2k$. Since $k$ is positive, the second derivative is negative, confirming it is a local maximum.

Q: What is the physical meaning of the curve crossing the x-axis?

The point where $U(x) = 0$ (other than the origin) is where the particle has the exact same potential energy as it did at the start. If the particle was released from rest at the origin, it could never reach this point because it would need extra kinetic energy to climb over the potential barrier (the maximum peak).

Q: Why does the $x^4$ term dominate for large $x$?

Think about the numbers! If $x=10$, $x^2 = 100$, but $x^4 = 10,000$. As $x$ grows, higher powers grow exponentially faster than lower powers. So, no matter what the constants $k$ and $a$ are, eventually, the $x^4$ term will become vastly larger than the $x^2$ term, dictating the overall downward direction of the graph.

Question

A particle, which is constrained to move along $x$-axis, is subjected to a force in the same direction which varies with the distance $x$ of the particle from the origin as $F(x) = -kx + ax^3$. Here, $k$ and $a$ are positive constants. For $x \ge 0$, the functional form of the potential energy $U(x)$ of the particle is

Options:

(A) Graph (a)

(B) Graph (b)

(C) Graph (c)

(D) Graph (d)

Sigma Percentile · Accepted Answer

### The Dance of Force and Energy

Imagine a particle constrained to move along a straight line, the $x$-axis. It isn't moving freely; it is being pushed and pulled by a force that changes depending on its exact position. The force is given by the equation $F(x) = -kx + ax^3$, where $k$ and $a$ are positive constants. Our mission is to visualize the potential energy landscape, $U(x)$, that this particle lives in.

To do this, we must first understand the profound connection between force and potential energy. In physics, a conservative force is simply the negative spatial gradient of potential energy. In mathematical terms, this means $F = -\frac{dU}{dx}$.

Think of potential energy as a hilly landscape. The particle always wants to roll down the hill to reach a state of lower energy. The steepness of the hill (the gradient) determines the strength of the force pushing it down.

### The Mathematical Translation

To find the total potential energy at any point $x$, we need to sum up all the infinitesimal changes in energy as the particle moves. We achieve this by integrating the force function. Rearranging our fundamental equation, we get $dU = -F \, dx$.

Assuming the standard convention that the potential energy at the origin is zero ($U(0) = 0$), we can set up our definite integral from $0$ to $x$:

$$ U(x) = -\int_0^x F(x) \, dx $$

Substituting our specific force function into the integral, we get:

$$ U(x) = -\int_0^x (-kx + ax^3) \, dx $$

Now, we perform the integration step-by-step. The integral of $-kx$ is $-\frac{kx^2}{2}$, and the integral of $ax^3$ is $\frac{ax^4}{4}$. Applying the negative sign from outside the integral, we arrive at our master equation for the potential energy:

$$ U(x) = \frac{kx^2}{2} - \frac{ax^4}{4} $$

### Decoding the Blueprint

This equation is the blueprint for our graph. Let's analyze its behavior to sketch the curve. At the origin ($x = 0$), the potential energy is clearly zero. But what happens as we take a tiny step away from the origin?

For very small values of $x$, the $x^2$ term is significantly larger than the $x^4$ term. Therefore, the $\frac{kx^2}{2}$ term dominates the behavior of the function. Since $k$ is a positive constant, the energy starts off positive and curves upwards, resembling a standard parabola.

Does it keep rising forever? To find out, we look for turning points where the slope of the curve is zero ($\frac{dU}{dx} = 0$). This occurs exactly where the force is zero. Setting $-kx + ax^3 = 0$, we find a turning point at $x = \sqrt{\frac{k}{a}}$. Because the curve was initially rising, this point represents a local maximum—the peak of our potential energy hill.

Finally, we must consider what happens at very large distances. For large values of $x$, the $x^4$ term grows exponentially faster than the $x^2$ term. The $-\frac{ax^4}{4}$ term becomes a giant, overpowering the positive $x^2$ term. This drags the potential energy down, forcing the curve to cross the $x$-axis and plunge towards $-\infty$.

### The Final Verdict

Putting all these clues together, we have a complete picture of the potential energy landscape. The graph must start at the origin $(0,0)$, rise to a positive peak, and then fall back down, eventually crossing the $x$-axis into negative territory.

When we compare this journey to the given options, only one graph perfectly matches this exact sequence of events. Graph (d) starts at the origin, goes up to a maximum, and then goes down. Thus, Graph (d) is the correct representation of the particle's potential energy.

Select Answer:

Visualized Solution

The Force Function

Force and Potential Energy

Setting up the Integral

Integrating the Force

Behavior Near the Origin

Finding the Peak

Behavior at Large Distances

Conclusion

The Sigma Insight: Kinetic Energy, Potential Energy and Power

The Dance of Force and Energy

The Mathematical Translation

Decoding the Blueprint

The Final Verdict

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