Problem 12 - Thomas Calculus

Question

Force of attraction When a particle of mass \(m\) is at \((x, 0),\) it is attracted toward the origin with a force whose magnitude is \(k / x^{2}\) . If the particle starts from rest at \(x=b\) and is acted on by no other forces, find the work done on it by the time it reaches \(x=a\) \(0 < a < b .\)

Step-by-Step Solution

Answer

The work done is \( k \left( \frac{1}{b} - \frac{1}{a} \right) \).

Step 1: Identify The Formula for Work Done

The work done by a force when it moves an object can be calculated using the integral:\[ W = \int F(x) \, dx \]where \( F(x) \) is the force as a function of \( x \).

Step 2: Substitute the Force Function into the Work Formula

The given force is \( F(x) = \frac{k}{x^2} \). We substitute this into the work formula:\[ W = \int_{b}^{a} \frac{k}{x^2} \, dx \]

Step 3: Integrate the Force Function

Evaluate the integral:\[ \int \frac{k}{x^2} \, dx = -\frac{k}{x} + C \]where \( C \) is the constant of integration. Since we have definite limits, we ignore the constant \( C \) for this calculation.

Step 4: Calculate The Definite Integral

Evaluate the definite integral from \( b \) to \( a \):\[ W = \left[ -\frac{k}{x} \right]_{b}^{a} = -\frac{k}{a} - \left( -\frac{k}{b} \right) \]

Step 5: Simplify the Result

Simplifying the expression from the definite integral, we have:\[ W = -\frac{k}{a} + \frac{k}{b} \]\[ W = \frac{k}{b} - \frac{k}{a} \]

Step 6: Final Result

The work done by the force as the particle moves from \( x = b \) to \( x = a \) is given by:\[ W = k \left( \frac{1}{b} - \frac{1}{a} \right) \]

Key Concepts

Integral Calculus

Integral calculus is an essential tool in mathematics that helps us compute quantities when they accumulate. It is particularly useful in physics to calculate quantities like area, volume, and other physical properties.
A fundamental idea in integral calculus is the integration process, as shown in:

• The indefinite integral, which includes a constant of integration, represents a family of functions.
• The definite integral gives a specific numerical value, representing the net accumulation of quantities over a specific interval.

In this problem, the integral calculus helps us calculate the work done by a force acting over a distance. This is done by evaluating the definite integral of the force function, which allows us to find the exact work done from one point to another.

Force of Attraction

Force of attraction refers to the pull experienced by objects due to their masses, such as gravitational force. In the given exercise, the force is specified as inversely proportional to the square of the distance, represented as \( F(x) = \frac{k}{x^2} \).
This type of force is common in fields like gravity and electrostatics, where the force decreases as the distance increases. Here, the constant \( k \) represents the strength of the attraction,
a critical factor in calculating the work done as the particle moves closer to the origin.
Understanding this attraction force's nature helps in identifying how energy changes as distances reduce, which is important in determining the work done.

Particle Mechanics

Particle mechanics is the branch of physics dealing with the movement and interaction of particles under forces. It explores dynamics such as motion, force, and energy.
Within this context, we consider how particles move under specific forces, like the attractive force described in our problem.
We are particularly interested in initial and final states of motion:

• The particle starts at rest, which implies zero initial kinetic energy.
• The work done on the particle causes it to move from an initial position \( x=b \) to a final position \( x=a \).

This movement under force application connects with the work-energy principle stating that work done by forces equals the change in kinetic energy, reflecting particle behavior's real-world implications.

Definite Integral

A definite integral computes the accumulation of a quantity over a specific interval. It's essential in finding the work done in our exercise, as it calculates the total force exerted on the particle across a given distance.

• The boundaries \( a \) and \( b \) represent the initial and final positions of the particle.
• The force equation \( \frac{k}{x^2} \) is integrated from \( x=b \) to \( x=a \).
• We perform this integration to find the total work done as the particle moves.

This process involves calculating the integral of the force function to determine the precise work done over the interval, resulting in \( W = k \left( \frac{1}{b} - \frac{1}{a} \right) \). Each step in evaluating this integral helps clarify the underlying physical concepts.

Physics Applications

The principles of physics apply integrally to many natural phenomena and engineering problems. Calculating work done using force and integral calculus is just one application under a broader physics framework.

• This understanding helps solve real-world problems, such as energy transfer in systems where forces act over distances.
• Engineering projects, like designing mechanical systems, require calculating work to ensure efficiency and functionality.
• Kinetic and potential energies, tied to work concepts, play a role in determining system behaviors.

By applying these principles, learners connect mathematical practices to tangible physics concepts, reinforcing their understanding and ability to solve practical challenges.