Problem 3 - Calculus Early Transcendentals: Pearson New International Edition

Question

A force of \(0.6\) newton is required to keep a spring with a natural length of \(0.08\) meter compressed to a length of \(0.07\) meter. Find the work done in compressing the spring from its natural length to a length of \(0.06\) meter. (Hooke's Law applies to compressing as well as stretching.)

Step-by-Step Solution

Answer

The work done in compressing the spring is \(0.012\) joules.

Step 1: Understand Hooke's Law

Hooke's Law states that the force needed to compress or extend a spring by a distance \( x \) from its natural length is proportional to that distance: \( F = kx \), where \( k \) is the spring constant. Here, the spring is compressed from its natural length, so \( x \) will be the difference between the compressed length and the natural length.

Step 2: Calculate the Spring Constant

Given that a force of \( 0.6 \) newtons is required to compress the spring from its natural length of \( 0.08 \) meters to \( 0.07 \) meters, \( x = 0.08 - 0.07 = 0.01 \) meters. According to Hooke’s Law, \( F = kx \). We can solve for \( k \) by rearranging the equation: \( k = \frac{F}{x} = \frac{0.6}{0.01} = 60 \) N/m.

Step 3: Set Up the Work Done Integral

The work done on a spring is given by \( W = \int kx \, dx \). We're going to find the work done in compressing the spring from its natural length \( L = 0.08 \) meters to \( L = 0.06 \) meters, that is, \( x \) varies from \( 0 \) to \( 0.02 \) meters.

Step 4: Evaluate the Integral

Set up the integral \( W = \int_0^{0.02} 60x \, dx \). First, compute the antiderivative: \( \int 60x \, dx = 30x^2 \). Now, evaluate the definite integral: \( W = \left[ 30x^2 \right]_0^{0.02} = 30(0.02)^2 - 30(0)^2 = 30(0.0004) = 0.012 \) joules.

Key Concepts

Spring Constant

When dealing with springs, one critical concept is the spring constant, denoted by \( k \). This constant tells us how stiff a spring is.
The larger the spring constant, the stiffer the spring and the more force it takes to compress or extend it. Hooke's Law helps us find this constant, which says that the force \( F \) required to change the length of a spring is directly proportional to the displacement \( x \):

• Formula: \( F = kx \).
• Force and displacement are linearly related.
• \( k = \frac{F}{x} \).

Knowing \( k \) is crucial for figuring out how a spring behaves under different forces, and it's essential for calculating other aspects like work done.

Work Done

Work done is essentially the energy transferred when an object is moved over a distance by an external force. In the context of springs, it's useful to understand that work is done when compressing or stretching a spring.
The amount of work can be calculated using the formula:

• \( W = \int F \, dx \).

In our case, because the force changes as the spring compresses or stretches, we often use an integral to find the exact work done over a specific displacement.
This helps us understand the energy required to compress or extend the spring.

Integral Calculation

The integral calculation is fundamental in problems involving variable force, such as springs. When force is not constant across the displacement, we employ calculus to compute the work done.
For springs, we express this as:

• \( W = \int kx \, dx \).

When setting up this integral, one needs to consider:

• Limits of integration: represent the initial and final position of the spring.
• Constant \( k \): previously found from Hooke's Law.

The solution provides the total work done, calculated by evaluating the integral, which in this exercise involved finding the antiderivative and then computing the values at the specified limits.

Force and Displacement

Force and displacement are two pivotal concepts in understanding spring mechanics. Displacement refers to the change in position of the spring from its rest position. Force, on the other hand, is the push or pull required to cause that displacement.

• Hooke's Law links these two concepts: \( F = kx \).

This relationship means that the force required and the displacement are directly proportional: the more you want to compress or extend, the more force you need.
Understanding this interaction helps when dealing with the energy involved (work done) because they dictate how much effort or energy is necessary to manipulate the spring.