A particle of mass 10^-2 kg is moving along the positive X-axis under the influence of a force F(x)

Q: Wait, why is the force negative if $k$ and $x^2$ are positive?

The negative sign in $F(x) = -\frac{k}{2x^2}$ explicitly indicates the *direction* of the force. Since $x$ is positive (the particle is on the positive X-axis), the force points towards the negative X-direction (towards the origin).

Q: Why can't we use $v^2 = u^2 + 2as$ here?

That equation is strictly for *constant* acceleration. Here, the force (and thus acceleration) depends on $x$ ($F \propto \frac{1}{x^2}$). As the particle moves, the force changes, so you must use integration (Work-Energy Theorem).

Q: When taking the square root for velocity, why did we choose the negative root?

Mathematically, $v^2 = 1$ gives $v = 1$ or $v = -1$. Physically, the particle started from rest and was pulled towards the origin by a negative force. Therefore, it must be moving in the negative X-direction, making its velocity negative.

Q: How did you know to substitute $x = \sin^2\theta$?

The expression $\sqrt{\frac{x}{1-x}}$ is a classic algebraic form. The denominator $1-x$ strongly hints at the Pythagorean identity $1 - \sin^2\theta = \cos^2\theta$. Substituting $x = \sin^2\theta$ perfectly eliminates the square root.

Q: What happens to the particle as it gets infinitely close to $x=0$?

As $x \to 0$, the force $F \propto \frac{1}{x^2}$ approaches negative infinity. The work done (and thus kinetic energy) would also approach infinity. In a purely mathematical sense, it would reach infinite speed at the origin!

Question

A particle of mass $10^{-2} 	ext{ kg}$ is moving along the positive X-axis under the influence of a force $F(x) = -\frac{k}{2x^2}$ where $k = 10^{-2} 	ext{ Nm}^2$. At time $t = 0$, it is at $x = 1.0 	ext{ m}$ and its velocity $v = 0$.
(a) Find its velocity when it reaches $x = 0.5 	ext{ m}$.
(b) Find the time at which it reaches $x = 0.25 	ext{ m}$.

Sigma Percentile · Accepted Answer

### The Journey to the Origin: A Tale of Work, Energy, and Calculus

Imagine a particle resting peacefully on the positive X-axis at $x = 1.0 	ext{ m}$. Suddenly, it experiences a force pulling it towards the origin. But this isn't just any constant force; it's a force that grows stronger as the particle gets closer to the origin, given by the inverse-square law $F(x) = -\frac{k}{2x^2}$.

Because the force is variable, we cannot rely on our standard kinematic equations like $v^2 = u^2 + 2as$. Those are strictly reserved for constant acceleration. Instead, we must summon the elegant power of the Work-Energy Theorem and Calculus.

### Part A: The Work-Energy Theorem

To find the velocity of the particle when it reaches $x = 0.5 	ext{ m}$, we need to calculate the work done by this variable force. The Work-Energy Theorem states that the net work done on an object equals its change in kinetic energy:

$$W = \Delta K = \frac{1}{2}mv_f^2 - \frac{1}{2}mv_i^2$$

Since the particle starts from rest, $v_i = 0$, meaning the work done will directly equal the final kinetic energy. Let's set up the integral for the work done as the particle moves from $x = 1.0 	ext{ m}$ to $x = 0.5 	ext{ m}$:

$$W = \int_{1.0}^{0.5} F(x) dx = \int_{1.0}^{0.5} \left( -\frac{k}{2x^2} ight) dx$$

Pulling the constant $-\frac{k}{2}$ out of the integral, we are left with integrating $x^{-2}$, which is simply $-\frac{1}{x}$.

$$W = -\frac{k}{2} \left[ -\frac{1}{x} ight]_{1.0}^{0.5} = \frac{k}{2} \left( \frac{1}{0.5} - \frac{1}{1.0} ight)$$

Evaluating the limits, we get $W = \frac{k}{2} (2 - 1) = \frac{k}{2}$.

Now, we equate this work to the kinetic energy:

$$\frac{1}{2}mv^2 = \frac{k}{2} \implies v^2 = \frac{k}{m}$$

Given that $k = 10^{-2} 	ext{ Nm}^2$ and $m = 10^{-2} 	ext{ kg}$, the ratio $\frac{k}{m}$ is exactly $1$. Therefore, $v^2 = 1$, which gives $v = \pm 1 	ext{ m/s}$. Since the force is pulling the particle in the negative X-direction, we must choose the negative root.

**The velocity at $x = 0.5 	ext{ m}$ is $-1 	ext{ m/s}$.**

### Part B: The Kinematic Differential Equation

Now for the real challenge: finding the exact time it takes to reach $x = 0.25 	ext{ m}$. To find time, we first need velocity as a general function of position, $v(x)$. We repeat our Work-Energy integration, but this time, we integrate from $x = 1.0$ to an arbitrary position $x$:

$$W = \frac{k}{2} \left[ \frac{1}{x} ight]_{1.0}^{x} = \frac{k}{2} \left( \frac{1}{x} - 1 ight)$$

Equating this to kinetic energy and substituting $\frac{k}{m} = 1$, we get:

$$v^2 = \frac{1-x}{x}$$

Taking the negative square root (because it's moving left), we find our velocity function:

$$v = -\sqrt{\frac{1-x}{x}}$$

Velocity is fundamentally the rate of change of position, $v = \frac{dx}{dt}$. This gives us a beautiful, yet intimidating, differential equation:

$$\frac{dx}{dt} = -\sqrt{\frac{1-x}{x}}$$

### The Calculus Climax

To solve for time, we separate the variables, moving all $x$ terms to one side and $dt$ to the other:

$$dt = -\sqrt{\frac{x}{1-x}} dx$$

Integrating both sides from the initial state ($t=0, x=1.0$) to the final state ($t=?, x=0.25$):

$$\int_{0}^{t} dt = -\int_{1.0}^{0.25} \sqrt{\frac{x}{1-x}} dx$$

This integral is a classic candidate for **trigonometric substitution**. The term $1-x$ in the denominator screams for the Pythagorean identity. Let's substitute $x = \sin^2	heta$. This means the differential $dx$ becomes $2\sin	heta\cos	heta d	heta$.

The square root simplifies beautifully:

$$\sqrt{\frac{\sin^2	heta}{1-\sin^2	heta}} = \frac{\sin	heta}{\cos	heta}$$

Multiplying this by our new $dx$, the integrand transforms into:

$$\left( \frac{\sin	heta}{\cos	heta} ight) (2\sin	heta\cos	heta d	heta) = 2\sin^2	heta d	heta$$

Using the double-angle identity, $2\sin^2	heta = 1 - \cos(2	heta)$.

We must also update our limits. When $x = 1.0$, $\sin^2	heta = 1$, so $	heta = \frac{\pi}{2}$. When $x = 0.25$, $\sin^2	heta = \frac{1}{4}$, so $\sin	heta = \frac{1}{2}$, which means $	heta = \frac{\pi}{6}$.

Our time integral is now a straightforward trigonometric evaluation:

$$t = -\int_{\pi/2}^{\pi/6} (1 - \cos 2	heta) d	heta$$

Integrating yields $	heta - \frac{\sin 2	heta}{2}$. Applying the limits:

$$t = -\left[ \left( \frac{\pi}{6} - \frac{\sin(\pi/3)}{2} ight) - \left( \frac{\pi}{2} - \frac{\sin(\pi)}{2} ight) ight]$$

$$t = -\left( \frac{\pi}{6} - \frac{\sqrt{3}}{4} - \frac{\pi}{2} ight) = \frac{\pi}{3} + \frac{\sqrt{3}}{4}$$

Substituting the approximate values $\pi \approx 3.1416$ and $\sqrt{3} \approx 1.732$, we get:

$$t \approx 1.047 + 0.433 = 1.48 	ext{ s}$$

**The particle reaches $x = 0.25 	ext{ m}$ at exactly $1.48 	ext{ s}$.** A perfect harmony of physics and calculus!

Visualized Solution

Analyzing the Setup

Work-Energy Theorem

Setting up the Integral

Evaluating the Work Done

Velocity at $x = 0.5 m$

General Velocity Function $v (x)$

Expression for $v (x)$

Kinematic Relation

Separating Variables

Trigonometric Substitution

Transforming the Integral

Changing the Limits

Evaluating the Time Integral

Final Calculation

The Sigma Insight: Work Done by Forces

The Journey to the Origin

Part A

Part B

The Calculus Climax

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