Problem 10 - Calculus of a Single Variable

Question

Hooke's Law In Exercises \(5-10\) , use Hooke's Law to determine the variable force in the spring problem. Seven and one-half foot-pounds of work is required to compress a spring 2 inches from its natural length. Find the work required to compress the spring an additional one-half inch.

Step-by-Step Solution

Answer

The work required to compress the spring an additional one-half inch is approximately 1.875 foot-pounds.

Step 1: Find the spring constant

Given that 7.5 foot-pounds of work is required to compress the spring 2 inches, you can use the work-energy theorem to calculate the spring constant. You have the equation \( W = \frac{1}{2}kx^2 \), which can rearranged to \( k = \frac{2W}{x^2} \). Substituting the given values gives \( k = \frac{2*7.5}{2^2} = 7.5 \) pounds per foot.

Step 2: Find the Work required for extra compression

Next, you can use the work-energy theorem again to find the work required to compress the spring an additional one-half inch. Let's denote the original length compressed as \( x1 = 2 \) inches and the extra length as \( x2 = 2.5 \) inches. We then need to calculate the work done \( W2 \) from \( x1 \) to \( x2 \) which can be calculated using the integral of Hooke’s Law, \( W = \frac{1}{2}k(x2^2 - x1^2) \). Substituting the known values gives \( W = \frac{1}{2}*7.5*(2.5^2 - 2^2) \approx 1.875 \) foot-pounds.

Key Concepts

Spring Constant

The spring constant, often represented by the letter \( k \), is a measure of a spring's stiffness. It tells us how much force is needed to compress or extend the spring by a certain distance. The greater the spring constant, the stiffer the spring.

To find the spring constant, we use the formula related to Hooke's Law:

• \( W = \frac{1}{2}kx^2 \)

Where \( W \) is the work done, \( k \) is the spring constant, and \( x \) is the compression length.

If we rearrange it to solve for \( k \), we find:

• \( k = \frac{2W}{x^2} \)

This formula allows us to compute the spring constant if we know the work done on the spring and the distance it was compressed or extended.

In this exercise, calculating with a work of 7.5 foot-pounds and a compression of 2 inches gives \( k = 7.5 \) pounds per foot.

Work-Energy Theorem

The work-energy theorem states that the work done on an object is equal to the change in its kinetic energy. For springs, this theorem helps us calculate the work required to change the compression.

In this context, we use a specific form of this theorem:

• \( W = \frac{1}{2}k(x2^2 - x1^2) \)

Here, \( W \) is the work done to compress or extend the spring, \( k \) is the spring constant, \( x1 \) is the initial compression, and \( x2 \) is the final compression.

This equation allows us to calculate how much more work is needed when the spring is compressed beyond its original position. It shows how energy changes as we apply a force to the spring.

Compression

Compression refers to the act of pressing a spring together. When a spring is compressed, it stores potential energy. This potential energy is then released as the spring returns to its natural state.

In physics exercises involving springs, understanding how to calculate the amount of work required for further compression is crucial. In the given problem, the spring is already compressed by 2 inches, and we need to calculate the work needed to compress it an additional half-inch.

Using the formula \( W = \frac{1}{2}k(x2^2 - x1^2) \) allows us to understand the changes in potential energy during the compression process, vital for evaluating different states of the spring.

Variable Force

In real-life spring problems, the force exerted by the spring is not always constant. It varies with the amount of compression or extension. This variable force highlights the non-linear nature of spring behavior described by Hooke's Law.

Hooke's Law states that the force exerted by a spring is proportional to its displacement. Mathematically represented as:

• \( F = kx \)

Where \( F \) is the force, \( k \) is the spring constant, and \( x \) is the displacement from the natural length.

This equation shows that the more you compress a spring, the larger the force it exerts. Thus, when solving problems, recognizing this variable nature of force aids in applying the correct equations and understanding spring behavior deeply.