Problem 10 - 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 \( W = k \left( \frac{1}{b} - \frac{1}{a} \right) \).

Step 1: Identify the Force Acting on the Particle

The force acting on the particle due to the attraction towards the origin is given by the formula: \( F(x) = \frac{k}{x^2} \), where \( k \) is a constant. This force is directed towards the origin along the x-axis.

Step 2: Set Up the Integral for Work

Work done by a force is calculated by integrating the force over the displacement. Hence, the work done, \( W \), on the particle as it moves from \( x = b \) to \( x = a \) is given by: \( W = \int_{b}^{a} F(x) \, dx = \int_{b}^{a} \frac{k}{x^2} \, dx \).

Step 3: Evaluate the Integral

To find the work done, evaluate the integral: \( \int_{b}^{a} \frac{k}{x^2} \, dx = k \int_{b}^{a} x^{-2} \, dx \). Integrating, we have: \( \int x^{-2} \, dx = -x^{-1} \), so the result is: \( k[-x^{-1}] \bigg|_{b}^{a} \).

Step 4: Compute the Definite Integral

Substitute the limits \( x = b \) and \( x = a \) into the antiderivative: \( W = k \left( -\frac{1}{a} + \frac{1}{b} \right) \). Thus, \( W = k \left( \frac{1}{b} - \frac{1}{a} \right) \).

Step 5: Finalize the Work Done Expression

The work done on the particle by the time it reaches \( x = a \) is given by the expression: \( W = k \left( \frac{1}{b} - \frac{1}{a} \right) \). Since \( b > a \), this result is positive, confirming that work was indeed done on the particle as it moved towards the origin.

Key Concepts

Force of Attraction

The concept of force of attraction is quite fascinating. It describes how objects are pulled towards each other due to various factors like mass and distance. In this exercise, the force of attraction is directed towards the origin. Specifically, for a particle of mass, the force magnitude is given by the formula:

• \( F(x) = \frac{k}{x^2} \)

Here, \( k \) is a constant, and the force diminishes as the particle moves away from the origin, which can be seen in the inverse square law form (i.e., \( 1/x^2 \)).
The essence of forces like gravity follow this inverse square law, meaning they become weaker with the square of the distance.

Integral Calculus

Integral calculus is used to find quantities such as areas, volumes, and in this context, work done by a force. When a force acts along a path, integral calculus helps in calculating the total work done over that path.

• The process involves setting up an integral, which is a mathematical description of summing up an infinite number of infinitesimally small quantities.

The integral for work done in this case is: \[ W = \int_{b}^{a} F(x) \, dx = \int_{b}^{a} \frac{k}{x^2} \, dx \]This integral tells us we are summing up the force contributions as the particle moves from point \( b \) to point \( a \).
It’s like adding up all the tiny pieces of work done along its journey.

Mechanics

Mechanics is a branch of physics that deals with motion and the forces that cause it. This exercise falls under the branch of classical mechanics, where the motion of particles is analyzed under the influence of forces.

• The particle in this problem starts at rest, meaning its initial velocity is zero.
• The only force acting on it is the attractive force towards the origin, given by \( F(x) = \frac{k}{x^2} \).
• This force gets translated into work done as the particle moves along the x-axis.

Understanding the principles of mechanics allows us to set up and solve these problems, providing insights into how objects move under various forces.

Definite Integral

The definite integral plays a crucial role in calculating the work done in this problem. Unlike an indefinite integral, which gives a general antiderivative, a definite integral computes the net value over an interval.

• The definite integral here is expressed as:

\[ W = k\left( \frac{1}{b} - \frac{1}{a} \right) \]When evaluating, you use the limits of integration \( b \) and \( a \) to find the actual work done. This is the net work considering the entire path from \( b \) to \( a \).
The result being positive indicates that work was done by the force of attraction to move the particle closer to the origin.