Problem 235 - Calculus Volume 1

Question

For the following exercises, find the work done. A spring has a natural length of 10 \(\mathrm{cm} .\) It takes 2 \(\mathrm{J}\) to stretch the spring to 15 \(\mathrm{cm} .\) How much work would it take to stretch the spring from 15 \(\mathrm{cm}\) to 20 \(\mathrm{cm} ?\)

Step-by-Step Solution

Answer

The work done to stretch the spring from 15 cm to 20 cm is approximately 3.11 J.

Step 1: Understand Hooke’s Law

Hooke's Law states that the force needed to stretch or compress a spring by some distance \(x\) from its natural length is proportional to \(x\). Mathematically, it is expressed as \(F(x) = kx\), where \(k\) is the spring constant.

Step 2: Determine the Spring Constant

Using the information that 2 J of work is required to stretch the spring from 10 cm to 15 cm, we can find \(k\). The work done in stretching from \(x_1\) to \(x_2\) is given by \(W = \int_{x_1}^{x_2} kx \, dx\). \[2 = \int_{0.1}^{0.15} kx \, dx = k \left[ \frac{x^2}{2} \right]_{0.1}^{0.15} = k \left( \frac{0.15^2}{2} - \frac{0.1^2}{2} \right)\]\[2 = k \left( \frac{0.0225}{2} - \frac{0.01}{2} \right) = k(0.005625)\]\(k = \frac{2}{0.005625} = 355.56 \, \text{N/m}\) approximately.

Step 3: Calculate Work from 15 cm to 20 cm

Now that we know \(k\), we calculate the work required to stretch the spring from 15 cm to 20 cm:\[ W = \int_{0.15}^{0.2} kx \, dx = 355.56 \left[ \frac{x^2}{2} \right]_{0.15}^{0.2} \]\[= 355.56 \left( \frac{0.2^2}{2} - \frac{0.15^2}{2} \right) = 355.56 \left( \frac{0.04}{2} - \frac{0.0225}{2} \right)\]\[= 355.56 \times 0.00875 = 3.11 \, \text{J}\] approximately.

Key Concepts

Hooke's Law

Hooke's Law is pivotal in understanding how springs behave when they are stretched or compressed. According to Hooke's Law, the force required to stretch or compress a spring is directly proportional to the amount it is stretched or compressed. This means if you pull a spring twice as far, it will require twice as much force, assuming you are within the spring's elastic limit bending the spring's behavior linearly.
The formula to express Hooke's Law is:

• \( F(x) = kx \)

Where:

• \( F(x) \): The force required to stretch or compress the spring.
• \( k \): The spring constant, a value that is unique to each spring.
• \( x \): The distance the spring is stretched or compressed from its natural length.

This law is incredibly useful in problems involving spring mechanics because it provides a straightforward way to calculate the force applied. It's important to remember, though, that Hooke's Law only applies within the elastic limit of the spring, and if deformed beyond a certain point, the spring will not return to its original shape.

Spring Constant

The spring constant \( k \) is an essential term in Hooke’s Law. It measures the stiffness of a spring; a stiffer spring will have a larger spring constant, whereas a more flexible one will have a smaller spring constant. Understanding this constant helps determine how much force you need to stretch or compress a particular spring.
The computation of the spring constant goes as follows:
Given that it requires 2 Joules to stretch the spring from 0.1 meters to 0.15 meters, this particular energy provides a clue to the spring's stiffness.
The spring constant can be derived when the work done is equated to the integral of the force, according to:

• \( W = \int_{x_1}^{x_2} kx \ \ ext{d}x \)

After performing the appropriate calculus, we approximate the spring constant using:

• \( k \approx 355.56 \, ext{N/m} \)

By calculating \( k \), once the spring constant is identified, it allows us to predict how much work or energy will be needed for other movements of the spring.

Integral Calculus

Integral calculus is a major branch of mathematical analysis that deals with the summation of parts to determine wholes. In physics, it is especially useful when calculating quantities that accumulate over a range, such as determining the area under a curve, which in practical terms can relate to total work done.
When considering problems involving work and energy, like stretching a spring, integral calculus can be employed to calculate the work needed to move from one point to another.
For springs, this is expressed as:

• \( W = \int_{x_1}^{x_2} kx \, dx \)

This integral sums up all the small amounts of work (force times distance) needed from an initial position \(x_1\) to a final position \(x_2\). This process is called integration, where we take tiny slices of movement over a distance and add them all up. In this context, it provides a powerful tool for precisely calculating the energy required to perform work on the spring, from one stretched position to another.