Problem 7 - Essential Calculus

Question

Hooke's Law In Exercises 3-10, use Hooke's Law to determine the variable force in the spring problem. A force of 20 pounds stretches a spring 9 inches in an exercise machine. Find the work done in stretching the spring 1 foot from its natural position.

Step-by-Step Solution

Answer

The work done in stretching the spring 1 foot from its natural position is 13.335 pound-feet

Step 1: Calculate the constant k from Hooke's Law

First, use Hooke's Law to find the constant of proportionality \(k\). Rearrange Hooke's Law to get \(k=\frac{F}{x}\). We fill in the provided values from the problem which are F=20 pounds and x=9 inches. But before we substitute the values, it is important to note that the units of measurement must be consistent. Here x is given in inches, and we are asked to find work done when the spring is stretched 1 foot. So, convert x from 9 inches into feet. There are 12 inches in a foot, hence \(x=\frac{9}{12}=0.75 (ft)\). Now we can substitute: \(k=\frac{20}{0.75} = 26.67 (pounds/ft)\)

Step 2: Use the constant k to find the force required to stretch the spring 1 foot

Now, use the constant k to find the force required to stretch the spring 1 foot. Substitute \((F=kx)\) with the computed value of k (26.67 pounds/ft) and x (1 foot): \(F = 26.67 * 1 = 26.67 (pounds)\)

Step 3: Calculate the work done

Finally, calculate the work done in stretching the spring 1 foot from its natural position. The work done is given by \(\frac{1}{2}Fx\), substitute F=26.67 pounds and x=1 foot: Work =\(\frac{1}{2} * 26.67 * 1 = 13.335 (pound-feet)\)

Key Concepts

Constant of Proportionality

In the context of Hooke's Law, the constant of proportionality, often represented by the letter \(k\), plays a crucial role. Hooke's Law states that the force \(F\) needed to stretch or compress a spring by some distance \(x\) is directly proportional to that distance. This relationship is mathematically expressed as \(F = kx\). This means the force required changes linearly with the distance the spring is stretched or compressed.

To determine \(k\), you need known values of force and distance from the problem. The formula to calculate \(k\) is \(k = \frac{F}{x}\). It is important to ensure that the units used are consistent, especially when converting distances. For example, if the problem measures distance in inches but requires calculations in feet, a conversion is necessary (as one foot equals 12 inches).

By determining \(k\), you can predict how much force will be needed for various displacements. This ability to calculate force for different springs and distances makes the concept of the constant of proportionality particularly valuable in physics and engineering.

Spring Force Calculation

Once the constant of proportionality \(k\) is known, calculating the force required for a specific extension of the spring becomes straightforward. As per Hooke's Law, \(F = kx\), where \(F\) is the force, \(k\) is the spring constant from earlier calculations, and \(x\) is the distance the spring is stretched or compressed from its natural length.

Using this relationship, you can compute the force for any given stretch or compression, provided you have the value of \(k\). For instance, if \(k = 26.67\) pounds/foot as calculated, and you need to stretch the spring by one foot, the force required is simply \(26.67 \times 1 = 26.67\) pounds.

This formula allows you to make quick calculations for various scenarios, such as changes in the distance or different spring constants, making it a versatile tool in physics problems related to springs and elastic materials.

Work in Physics

In physics, work is a measure of energy transfer that occurs when a force moves an object over a distance. When working with springs, work is calculated using the formula \( \text{Work} = \frac{1}{2}Fx \). This formula is derived from the integration of Hooke's Law, capturing the variable force applied across the distance the spring is stretched or compressed.

In the given exercise, to calculate the work done when stretching a spring 1 foot, we substitute \(F = 26.67\) pounds and \(x = 1\) foot into the equation. This yields \( \frac{1}{2} \times 26.67 \times 1 = 13.335 \) pound-feet.

Understanding work in the context of springs can apply to various real-world situations, such as designing exercise equipment or calculating energy needs in mechanical systems. By quantifying the energy required to manipulate these systems, physics provides valuable insights into efficiency and functionality.