Problem 9 - Basic Technical Mathematics with Calculus

Question

The gravitational force (in \(1 \mathrm{b}\) ) of attraction between two objects is given by \(F=k / x^{2},\) where \(x\) is the distance between the objects. If the objects are \(10 \mathrm{ft}\) apart, find the work required to separate them until they are \(100 \mathrm{ft}\) apart. Express the result in terms of \(k\).

Step-by-Step Solution

Answer

The work required is \( \frac{9k}{100} \).

Step 1: Define Work Formula in Terms of Force

The work done to move an object under the influence of a force is given by the integral of the force over the distance. Here, the work \(W\) to separate the objects from \(x_1 = 10\) ft to \(x_2 = 100\) ft is given by the formula:\[ W = \int_{x_1}^{x_2} F(x) \, dx = \int_{10}^{100} \frac{k}{x^2} \, dx \]

Step 2: Set Up the Integral for Work

Substitute the force equation \( F(x) = \frac{k}{x^2} \) into the work integral:\[ W = \int_{10}^{100} \frac{k}{x^2} \, dx \]

Step 3: Evaluate the Integral

The integral \( \int \frac{1}{x^2} \, dx = -\frac{1}{x} \) leads to:\[ W = \left[-\frac{k}{x}\right]_{10}^{100} \]Now apply the limits of integration.

Step 4: Apply Limits of Integration

Substitute the limits into the evaluated integral:\[ W = \left(-\frac{k}{100}\right) - \left(-\frac{k}{10}\right) \]Simplify the resulting expression.

Step 5: Simplify the Result

Simplify the expression obtained:\[ W = -\frac{k}{100} + \frac{k}{10} = \frac{k}{10} - \frac{k}{100} \]\[ W = \frac{10k}{100} - \frac{k}{100} = \frac{9k}{100} \]

Step 6: Present the Final Answer

The work required to separate the objects from 10 ft to 100 ft is: \[ W = \frac{9k}{100} \]

Key Concepts

Work Done

When you think about work done, imagine yourself pushing a box across the floor. Work is done whenever a force moves something over a distance. In physics, work done is a way to describe the energy transferred by a force acting over a distance. This is especially important in situations like moving objects under the influence of gravitational forces. If you want to calculate this work mathematically, you use the integral of force over the distance the object is moved.

For example:

• When two objects separated by a distance are attracted by gravitational force, work is done if they move apart.
• In mathematical terms, you write the work (\( W \)) as an integral: \[ W = \int F(x) \, dx \]
• This integral sums up the tiny bits of work over each little slice of distance between the start and end points.

So, for our exercise, where the gravitational force varies inversely with the square of the distance, we evaluate this integral from when the objects are 10 ft apart to when they are 100 ft apart.

Integral Calculus

Integral calculus is a fundamental tool in mathematics that helps you find the total amount of something when you know the rate at which it changes. It's like finding the total distance traveled by a car if you know its speed every moment in time. For physics problems, integral calculus becomes vital when dealing with continuous quantities, like forces that change over a range of distances.

In the problem of gravitational force, we use integral calculus to calculate the total work done by considering the continuous force across a span of distances (from 10 ft to 100 ft).

• The force function,\( F(x) = \frac{k}{x^2} \), changes with distance \( x \).
• To find the work done, we evaluate the integral of this force function over variable \( x \) from 10 to 100.
• This involves applying integration, where \( \int \frac{1}{x^2} \, dx \) results in a simplified form, \( -\frac{1}{x} \).

Remember, this work result tells us the energy required to pull the objects apart against gravity.

Distance in Physics

Distance in physics is more than just a measure of space or length; it's crucial because it often affects how forces work. When dealing with forces like gravity, the effect changes not just with how hard you pull but also with how far apart things are. The gravitational force, an essential part of many physics problems, is inversely proportional to the square of the distance between objects. This means:

• As distance increases, the gravitational force decreases rapidly.
• This concept is captured in the formula \( F = \frac{k}{x^2} \), demonstrating that double the distance results in one-fourth of the force.
• When calculating work done, the change in force with respect to distance is crucial since the integral accounts for this varying force across the distance.

In our exercise, moving from 10 ft to 100 ft meant computing the work done against a force lessening over a larger distance, which gave us the result \( W = \frac{9k}{100} \). Understanding this relationship in physics allows you to predict how force and distance can impact energy and motion.