The potential energy of a 1 kg particle free to move along the x-axis is given by V(x) = (x^4/4 - x^

Q: Why do we take the derivative of the potential energy function?

Taking the derivative and setting it to zero helps us find the critical points (maxima and minima) of a function. In physics, $\frac{-dV}{dx}$ also represents the force. Where force is zero, the particle is at an equilibrium point, which corresponds to the peaks and valleys of the potential energy curve.

Q: Can potential energy really be negative?

Yes! Potential energy is always measured relative to a chosen reference point. A negative potential energy simply means the particle has less potential energy than it does at the reference point (in this case, $x=0$). Only the *change* in potential energy has physical significance.

Q: What happens at $x = 0$?

At $x = 0$, the potential energy is $0\text{ J}$. Because this is higher than the minimum potential energy of $-\frac{1}{4}\text{ J}$, it represents a local maximum (a small hill) in the potential energy landscape. The particle will be moving slower here than at $x = \pm 1$.

Q: Why is the mass given as $1\text{ kg}$?

The mass is required to convert the kinetic energy (which is in Joules) into speed (in m/s) using the formula $K = \frac{1}{2}mv^2$. If the mass were different, the maximum kinetic energy would be the same, but the resulting speed would change.

Q: Will the particle ever stop moving?

Yes, momentarily. The particle will stop at the 'turning points' where its kinetic energy becomes zero. This happens when all the total energy is purely potential energy, i.e., $V(x) = 2\text{ J}$. It will oscillate back and forth between these two turning points.

Question

The potential energy of a $1	ext{ kg}$ particle free to move along the x-axis is given by
$V(x) = \left( \frac{x^4}{4} - \frac{x^2}{2} ight)	ext{ J}$
The total mechanical energy of the particle is $2	ext{ J}$. Then, the maximum speed (in $	ext{ms}^{-1}$) is

Options:

(A) $\frac{3}{\sqrt{2}}$

(B) $\sqrt{2}$

(C) $\frac{1}{\sqrt{2}}$

(D) $2$

Sigma Percentile · Accepted Answer

### The Dance of Energy

Imagine a particle trapped in an invisible landscape, moving back and forth along the x-axis. This landscape is defined by its potential energy, given by the function $V(x) = \frac{x^4}{4} - \frac{x^2}{2}$. The particle has a strict budget: its total mechanical energy is exactly $2	ext{ J}$.

In the world of conservative forces, energy is a zero-sum game. The total energy $E$ is always the sum of kinetic energy $K$ and potential energy $V$.

$$E = K + V$$

Because the total energy is locked at $2	ext{ J}$, whenever the potential energy dips, the kinetic energy must surge to make up the difference. If we want to find the particle's **maximum speed**, we need to find its **maximum kinetic energy**. And to maximize kinetic energy, we must find the absolute lowest point in our potential energy landscape.

### Hunting for the Minimum

To find the valleys of our potential energy curve, we turn to calculus. We take the derivative of $V(x)$ with respect to $x$ and set it to zero to find the critical points.

$$\frac{dV}{dx} = \frac{d}{dx}\left(\frac{x^4}{4} - \frac{x^2}{2}ight) = x^3 - x = 0$$

Factoring this equation gives us $x(x^2 - 1) = 0$, which reveals three critical points: $x = 0$, $x = 1$, and $x = -1$.

If we plug these back into our potential energy function, $x = 0$ gives $V(0) = 0	ext{ J}$ (a local peak), while $x = \pm 1$ gives the true valleys:

$$V_{min} = \frac{(\pm 1)^4}{4} - \frac{(\pm 1)^2}{2} = \frac{1}{4} - \frac{1}{2} = -\frac{1}{4}	ext{ J}$$

### The Final Sprint

Now that we know the potential energy drops as low as $-\frac{1}{4}	ext{ J}$, we can calculate the maximum kinetic energy.

$$K_{max} = E - V_{min} = 2 - \left(-\frac{1}{4}ight) = \frac{9}{4}	ext{ J}$$

With the maximum kinetic energy in hand, finding the maximum speed is just a matter of plugging it into the kinetic energy formula, $K = \frac{1}{2}mv^2$. Since the mass $m$ is $1	ext{ kg}$:

$$\frac{1}{2}(1)v_{max}^2 = \frac{9}{4}$$

$$v_{max}^2 = \frac{9}{2}$$

Taking the square root, we arrive at our final answer:

$$v_{max} = \frac{3}{\sqrt{2}}	ext{ ms}^{-1}$$

This beautiful interplay between kinetic and potential energy is the heartbeat of classical mechanics, dictating the motion of everything from swinging pendulums to orbiting planets.

Animated Solution for Physics - Work, Energy, and Power: The potential energy of a $1 kg$ particle free to move along the x-axis is given by $V (x) = (\frac{x ^{4}}{4} - \frac{x ^{2}}{2}) J$ The total mechanical energy of the particle is $2 J$ . Then, the maximum speed (in $ms^{- 1}$ ) is

Select Answer:

Visualized Solution

$E = K + V$

$Condition for v_{ma x}$

$Finding V_{min}$

$Solving for x$

$Calculating V_{min}$

$Calculating K_{ma x}$

$Final Speed$

$The Way Forward$

The Sigma Insight: Conservation of Mechanical Energy

The Dance of Energy

Hunting for the Minimum

The Final Sprint

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Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } E=K+V

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Condition for vmax​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Finding Vmin​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Solving for x

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Calculating Vmin​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Calculating Kmax​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Final Speed

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The Way Forward