Problem 2 - Thomas Calculus in SI Units

Question

A spring has a natural length of \(10 \mathrm{cm} . \mathrm{An}\) 800-N force stretches the spring to \(14 \mathrm{cm}\). a. Find the force constant. b. How much work is done in stretching the spring from \(10 \mathrm{cm}\) to \(12 \mathrm{cm} ?\) c. How far beyond its natural length will a 1600 -N force stretch the spring?

Step-by-Step Solution

Answer

a. The force constant is 20000 N/m. b. 4 Joules of work is done. c. The spring is stretched 8 cm beyond its natural length.

Step 1: Understanding Hooke's Law

The force exerted by a spring is described by Hooke's Law, which states that the force needed to extend or compress a spring by some distance from its natural length is proportional to that distance. Mathematically, it is given by \( F = kx \), where \( F \) is the force applied, \( k \) is the spring constant, and \( x \) is the extension or compression from the natural length.

Step 2: Solving for the Force Constant

Given the information, the natural length of the spring is 10 cm and it is stretched to 14 cm by an 800-N force, so \( x = 14 \text{ cm} - 10 \text{ cm} = 4 \text{ cm} = 0.04 \text{ m} \). Using Hooke's Law \( F = kx \), we can solve for \( k \) as: \[ k = \frac{F}{x} = \frac{800 \text{ N}}{0.04 \text{ m}} = 20000 \text{ N/m} \]. The force constant \( k \) is 20000 N/m.

Step 3: Calculating Work Done

The work done to stretch the spring can be calculated using the formula \( W = \frac{1}{2} k x^2 \). The change in length \( x \) from 10 cm to 12 cm is 2 cm or 0.02 m. Plugging the values into the formula: \[ W = \frac{1}{2} \times 20000 \text{ N/m} \times (0.02 \text{ m})^2 = \frac{1}{2} \times 20000 \times 0.0004 = 4 \text{ J} \]. Thus, 4 Joules of work is done.

Step 4: Calculating Extension for 1600-N Force

To find how far a 1600-N force will stretch the spring, use the formula \( F = kx \). Rearranging this, we find \( x = \frac{F}{k} \). Substituting the known values: \[ x = \frac{1600 \text{ N}}{20000 \text{ N/m}} = 0.08 \text{ m} = 8 \text{ cm} \]. Therefore, a 1600-N force will stretch the spring 8 cm beyond its natural length.

Key Concepts

Spring Constant

Understanding the spring constant is crucial when dealing with Hooke’s Law. The spring constant, often denoted by the symbol \( k \), indicates the stiffness of a spring. The larger the spring constant, the stiffer the spring, which means it requires more force to stretch or compress it.
It is measured in Newtons per meter (\( \text{N/m} \)). Let’s break it down with an example.

• Say we have a spring that stretches 0.04 meters when a force of 800 Newtons is applied.
• Using Hooke's Law, we find the spring constant \( k \) as follows: \[ k = \frac{800 \text{ N}}{0.04 \text{ m}} = 20000 \text{ N/m} \]

This calculation tells us how resistant the spring is to being stretched. A higher \( k \) value, like 20000 \( \text{N/m} \), signifies a relatively stiff spring.

Work Done by Force

When you extend a spring, you're exerting force, and this force does work. The work done by the force while stretching the spring can be calculated using the formula: \( W = \frac{1}{2} k x^2 \). Understanding this formula is essential as it measures the energy needed to stretch the spring over a certain distance.

• The spring constant \( k \) indicates how much force is required for each unit of extension.
• \( x \) is the change in length of the spring from its natural length.

Let’s consider an example where we stretch a spring from its natural length of 10 cm to 12 cm, thus an extension of 0.02 meters.
By substituting these values into the equation, we calculate:
\[ W = \frac{1}{2} \times 20000 \text{ N/m} \times (0.02 \text{ m})^2 = 4 \text{ J} \]
This result shows us that 4 Joules of work is necessary to achieve this extension.

Extension of Spring

The extension of a spring is how much the spring stretches or compresses from its original length. Hooke's Law helps us to calculate this extension using the relationship \( F = kx \). Here, \( x \) is the extension, \( F \) is the force applied, and \( k \) is the spring constant.

• If a spring is naturally 10 cm long and stretches to 14 cm when an 800-N force is applied, its extension is 4 cm or 0.04 m.
• Using the same spring with a \( k \) of 20000 \( \text{N/m} \), and applying a different force of 1600 N, the equation becomes: \[ x = \frac{1600 \text{ N}}{20000 \text{ N/m}} = 0.08 \text{ m} \]

This means the spring will extend 8 cm beyond its natural length under a 1600-N force. The calculations show how forces relate to the extension, providing a practical understanding of Hooke’s Law.