Problem 3 - Thomas Calculus in SI Units

Question

A force of 2 \(\mathrm{N}\) will stretch a rubber band \(2 \mathrm{cm}(0.02 \mathrm{m}) .\) Assuming that Hooke's Law applies, how far will a 4-N force stretch the rubber band? How much work does it take to stretch the rubber band this far?

Step-by-Step Solution

Answer

The rubber band stretches 0.04 m and the work done is 0.08 J.

Step 1: Understanding Hooke's Law

Hooke's Law states that the force required to stretch or compress a spring is directly proportional to the distance it is stretched or compressed. It can be expressed as \( F = kx \), where \( F \) is the force applied, \( x \) is the extension, and \( k \) is the spring constant. We are given a force of 2 N stretching the rubber band to 0.02 m.

Step 2: Calculate the Spring Constant

From Hooke's Law, we can determine the spring constant \( k \) using the given initial conditions. Plug \( F = 2 \, \text{N} \) and \( x = 0.02 \, \text{m} \) into the equation \( F = kx \). Thus, \( k = \frac{F}{x} = \frac{2}{0.02} = 100 \, \text{N/m} \).

Step 3: Calculate the Extension for a 4-N Force

With the spring constant \( k = 100 \, \text{N/m} \) known, use Hooke's Law to find the extension caused by a 4 N force. Substitute \( F = 4 \, \text{N} \) into \( F = kx \), leaving \( 4 = 100x \). Solving for \( x \), we get \( x = \frac{4}{100} = 0.04 \, \text{m} \). Thus, the rubber band will stretch 0.04 m.

Step 4: Calculate Work Done to Stretch the Rubber Band

The work done to stretch a spring (or a rubber band, in this case) can be calculated using the formula \( W = \frac{1}{2}kx^2 \). Here, \( k = 100 \, \text{N/m} \) and \( x = 0.04 \, \text{m} \). Plugging in these values, we get \( W = \frac{1}{2}(100)(0.04)^2 = \frac{1}{2}(100)(0.0016) = 0.08 \, \text{J} \).

Key Concepts

Spring Constant

The spring constant, often symbolized by the letter 'k', is a fundamental property of springs and elastic materials that tells us about their stiffness. It measures how much force is needed to stretch or compress a material by a certain amount. In general, a higher spring constant indicates a stiffer material, while a lower constant is indicative of a more flexible material.

Hooke's Law defines the relationship between the force applied to a spring and the amount it stretches or compresses, described by the equation:

• \( F = kx \)

where \( F \) is the force applied, \( k \) is the spring constant, and \( x \) is the displacement. By rearranging this formula, we can solve for the spring constant:

• \( k = \frac{F}{x} \)

In the original exercise, a 2 N force stretches the rubber band 0.02 m. This information allows us to calculate the spring constant as 100 N/m, which tells us that for every meter the rubber band is stretched, it requires 100 N of force.

Work Done

The concept of work done in stretching a spring involves transferring energy to the spring, causing it to deform. In mechanics, the 'work' done by a force is defined as the product of the force applied and the displacement over which it acts, but for springs, the situation is a bit different because the force can vary along the displacement.

For springs following Hooke's Law, the work done \( W \) to stretch or compress a spring from the initial position to a final position is given by the formula:

• \( W = \frac{1}{2}kx^2 \)

where \( k \) is the spring constant and \( x \) is the displacement from the equilibrium position. This formula arises because the force is not constant; it increases linearly from zero up to \( kx \).

Applying this to the exercise, the work done to stretch the rubber band by 0.04 meters using a force of 4 N is calculated as 0.08 joules. This work is the energy stored in the rubber band as potential energy.

Elastic Potential Energy

Elastic potential energy is the energy stored in elastic materials as they are deformed, such as stretching or compressing. This energy is the work done on the spring and is capable of being released, returning the material to its original shape.

For a spring, the elastic potential energy \( E \) is given by the same expression used for the work done:

• \( E = \frac{1}{2}kx^2 \)

This highlights that the elastic potential energy is dependent on both the spring constant and the square of the displacement. A key insight here is that potential energy increases with either a stiffer spring (higher \( k \)) or a larger displacement (greater \( x \)).

In the context of our rubber band example, when the band is stretched 0.04 meters, the stored energy is 0.08 joules. This energy will potentially be released as kinetic energy if the rubber band returns to its unstretched state.