Problem 23 - Calculus Early Transcendentals

Question

Calculate the work required to stretch the following springs 0.5 m from their equilibrium positions. Assume Hooke's law is obeyed. a. A spring that requires a force of \(50 \mathrm{N}\) to be stretched \(0.2 \mathrm{m}\) from its equilibrium position. b. A spring that requires \(50 \mathrm{J}\) of work to be stretched \(0.2 \mathrm{m}\) from its equilibrium position.

Step-by-Step Solution

Answer

Based on the given information and calculations, the work required to stretch the first spring 0.5 m from its equilibrium position is 31.25 J. While for the second spring, the work required is 312.5 J.

Step 1: Part a: Calculate the spring constant for the first spring

To find the spring constant (k) for the first spring, we will use Hooke's law, which states that the force acting on a spring is proportional to the displacement (x) from the equilibrium position: \[F = kx\] We know that in this case, a force of 50N is required to stretch the spring 0.2 m from its equilibrium position. Using these values, we can write the equation as: \[50 \mathrm{N} = k \times 0.2 \mathrm{m}\] Now, we can solve for the spring constant k: \[k = \frac{50 \mathrm{N}}{0.2 \mathrm{m}} = 250 \frac{\mathrm{N}}{\mathrm{m}}\]

Step 2: Part a: Calculate the work required for the first spring

To calculate the work required to stretch the first spring 0.5 m, we can use the work-energy theorem, which states that work is equal to the change in potential energy stored in the spring: \[W = \frac{1}{2}kx^2\] Now that we know the spring constant k from the previous step, and we know the desired displacement (x = 0.5 m), we can plug these values into the equation and calculate the work: \[W = \frac{1}{2} \times 250 \frac{\mathrm{N}}{\mathrm{m}} \times (0.5 \mathrm{m})^2 = \frac{1}{2} \times 250 \frac{\mathrm{N}}{\mathrm{m}} \times 0.25 \mathrm{m^2} = 31.25 \mathrm{J}\] Therefore, the work required to stretch the first spring 0.5 m from its equilibrium position is 31.25 J.

Step 3: Part b: Calculate the spring constant for the second spring

For the second spring, we are given the work required to stretch it 0.2m from its equilibrium position, which is 50J. To find the spring constant, we can again use the work-energy theorem equation in the form: \[W = \frac{1}{2}kx^2\] Now we can plug in the known values for work and displacement to solve for k: \[50 \mathrm{J} = \frac{1}{2}k \times (0.2 \mathrm{m})^2\] \[k = \frac{50 \mathrm{J} \times 2}{(0.2 \mathrm{m})^2} = \frac{100 \mathrm{J}}{0.04 \mathrm{m^2}} = 2500 \frac{\mathrm{N}}{\mathrm{m}}\] The spring constant for the second spring is 2500 N/m.

Step 4: Part b: Calculate the work required for the second spring

Now that we know the spring constant for the second spring, we can calculate the work required to stretch it 0.5 m from its equilibrium position, using the work-energy theorem: \[W = \frac{1}{2}kx^2\] Inserting the values for k and x in the equation: \[W = \frac{1}{2} \times 2500 \frac{\mathrm{N}}{\mathrm{m}} \times (0.5 \mathrm{m})^2\] \[W = \frac{1}{2} \times 2500 \frac{\mathrm{N}}{\mathrm{m}} \times 0.25 \mathrm{m^2} = 312.5 \mathrm{J}\] Therefore, the work required to stretch the second spring 0.5 m from its equilibrium position is 312.5 J.

Key Concepts

Hooke's Law

In simple terms, Hooke's Law helps us understand how springs behave when they're stretched or compressed. It states that the force required to change the length of a spring is directly proportional to the amount of stretch or compression. This can be described with the equation:
\[ F = kx \]
where:

• \( F \) is the force applied to the spring,
• \( k \) is the spring constant (a measure of the spring's stiffness),
• \( x \) is the displacement from the spring's equilibrium position.

Understanding this relationship is crucial because it lets us quantify how much force is needed to pull or push a spring by a certain distance. If you know the spring constant and the displacement, you can easily determine the force utilizing this wonderful law.

Spring Constant

The spring constant, represented as \( k \) in our equations, is a crucial factor in describing a spring's behavior. It tells us how stiff or flexible a spring is. The unit for spring constant is \( \frac{\text{Newton}}{\text{meter}} \), signifying how many Newtons it takes to extend or compress the spring by one meter.

Imagine a very stiff spring; it has a high spring constant because it requires more force to stretch. Conversely, a loose, more flexible spring would have a lower spring constant.

A helpful way to determine this constant is by rearranging Hooke's Law:

• Measure the force needed to stretch the spring and the distance stretched,
• Calculate \( k \) using \( k = \frac{F}{x} \).

This value remains constant for a given spring and helps predict how the spring will react to various forces.

Potential Energy

Potential energy in the context of springs is stored energy due to position changes, like stretching or compressing. Imagine pulling a spring—you're storing energy in it as potential energy. This energy has the capability to do work due to the spring's restored position when let go.

The potential energy (PE) in a spring is calculated with:\[ PE = \frac{1}{2}kx^2 \]
Here:

• \( PE \) is the potential energy,
• \( k \) is the spring constant,
• \( x \) is how far the spring is stretched or compressed from its normal position.

This equation shows us that potential energy increases with the square of the displacement, meaning the further you stretch it, the more energy is stored.

Work-Energy Theorem

The work-energy theorem connects the concepts of work and energy, allowing us to explore how the force applied to a spring influences its energy transformation. It states that the work done on an object is equal to the change in its kinetic energy. When applied to springs, we often talk about potential energy changes instead.

For springs, the work done when stretching or compressing a spring is calculated by:\[ W = \frac{1}{2}kx^2 \]
where \( W \) is the work done:

• To stretch or compress it, resulting in a change in potential energy,
• Reflected as the energy stored in the spring.

This theorem is powerful as it links force, displacement, and energy, providing a complete picture of how energy systems work in springs. It enables us to determine exactly how much work is required to store a specific amount of energy in the spring by knowing the spring constant and how far we want to stretch or compress the spring.