Problem 3 - Calculus: Early Transcendentals

Question

A variable force of \( 5x^{-2} \) pounds moves an object along a straight line when it is \( x \) feet from the origin. Calculate the work done in moving the object from \( x = 1 ft \) to \( = 10 ft \).

Step-by-Step Solution

Answer

The work done is \( \frac{9}{2} \) foot-pounds.

Step 1: Identify the Problem

We need to calculate the work done by a variable force when moving an object along a straight line from one point to another. The force given is a function of the position of the object, measured from the origin.

Step 2: Determine the Work Formula

The work done by a variable force along a straight line is given by the integral of the force over the distance. The formula for work is \( W = \int_{a}^{b} F(x) \, dx \), where \( a \) and \( b \) are the initial and final positions, respectively, and \( F(x) \) is the force as a function of \( x \).

Step 3: Setup the Integral

Substitute the given force function \( F(x) = 5x^{-2} \) and the limits of integration \( a = 1 \) and \( b = 10 \) into the work formula: \[ W = \int_{1}^{10} 5x^{-2} \, dx \].

Step 4: Integrate the Function

Integrate the function \( 5x^{-2} \):\( \int 5x^{-2} \, dx = 5 \cdot \left(-x^{-1}\right) = -\frac{5}{x} \).Now, apply the limits of integration: \[ W = \left[-\frac{5}{x}\right]_{1}^{10} \].

Step 5: Apply the Limits of Integration

Evaluate the antiderivative at the bounds:\[ W = -\frac{5}{10} - \left(-\frac{5}{1}\right) \], which simplifies to\[ W = -\frac{1}{2} + 5 = \frac{9}{2} \].

Step 6: Conclude the Calculation

The work done is \( \frac{9}{2} \) foot-pounds. The calculations confirm that this is the amount of work required to move the object from 1 ft to 10 ft while subject to the variable force.

Key Concepts

Variable Force

A variable force is one that changes depending on the position of an object. In many physics problems, force isn't constant; it varies as the object moves through space. For instance, if a force depends on the distance from a point, like in the equation given: \( F(x) = 5x^{-2} \), it means the force decreases as the object moves away from the origin.
This kind of force is often seen in gravitational or electromagnetic fields where the intensity changes with distance.

• Variable forces require special consideration because the force applied isn't the same as the object moves.
• Understanding how the force varies is crucial for calculating the work done.

By integrating such forces over the distance, we can find out the total work done.

Integration

Integration is a mathematical operation used to find the accumulation of quantities, such as area under a curve. In the context of work and force, integration helps compute how much total force was applied over a certain distance.
When dealing with a variable force, integration is essential as it allows us to sum seemingly continuous and infinitesimally small amounts of work over a distance. This is done by using the integral of the force function over the path of motion.

• The integral \( \int F(x) \, dx \) gives the area under the force curve, representing total work done.
• It's like adding up all tiny elements of work along the path to get a complete picture.

By mastering integration, we can tackle a variety of problems involving variable forces.

Work Done

Work done in physics is the process of moving an object by applying a force along a path. When the force is variable, we can't just multiply force by distance as we would in simpler cases. Instead, we use integration to calculate the work done from one position to another.
In the problem, work done is computed as the integral of the force function from the starting point \( a \) to the endpoint \( b \). This tells us how much energy is required to move the object under the influence of a varying force.

• Work depends on both the force applied and the distance over which it is applied.
• The units for work are typically foot-pounds or joules, indicating energy transferred.

Understanding the concepts behind calculating work helps apply physics to real-world situations.

Definite Integral

A definite integral calculates the net value of a function over a specific interval, often representing physical quantities such as area, volume, or work. For a function \( F(x) \), the definite integral from \( a \) to \( b \) - written as \( \int_a^b F(x) \, dx \) - represents the accumulated value of the function between two points.
In the work problem, we use a definite integral to find the total work done by integrating the force function \( 5x^{-2} \) from \( x = 1 \) to \( x = 10 \). The limits \( a \) and \( b \) provide clear start and end points for calculations.

• The definite integral provides an exact numerical value representing the total accumulation of force over the distance.
• It's crucial for solving problems where sums of small parts need precise calculation.

This technique is a cornerstone in applied calculus, allowing us to solve practical problems involving continuously changing quantities.