Problem 7 - Calculus of a Single Variable

Question

Hooke's Law In Exercises \(5-10\) , 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

To gives the final answer, the work done in stretching the spring 1 foot from its natural position using the calculated spring constant can be substituted into the formula and computed. This will give the solution to the problem.

Step 1: Calculate the spring constant (k)

Based on Hooke's Law, which has the form \(F = kx\), where \(F\) is the force applied, \(k\) is the spring constant and \(x\) is the displacement, we can calculate \( k = \frac{F}{x}\). Given that the force \(F\) is 20 pounds and the displacement \(x\) is 9 inches or \(\frac{3}{4}\) feet, we can input these values to find \(k\).

Step 2: Use the Work-Energy Theorem

The work done on stretching the spring from its natural position is given by the integral of the force over the distance. In this case, the formula becomes \(W = \frac{1}{2} k (x_2^2 - x_1^2)\), where \(x_1 = 0\) (the original position of the spring) and \(x_2 = 1\) foot. Subsitute the value of \(k\) from Step 1, \(x_1 = 0\) feet and \(x_2 = 1\) foot into this formula to find the work done.

Step 3: Compute the final value

Substitute the values into the formula and compute the final result for the work done in stretching the spring 1 foot from its natural position.

Key Concepts

Spring Constant

The spring constant, denoted by the symbol 'k', is a measure of how stiff a spring is. In Hooke's Law, given by the equation \( F = kx \), it relates the force applied to a spring (F) and the displacement caused in the spring (x). Finding the spring constant is crucial because it describes the spring's behavior under force. The higher the spring constant, the more force is needed to produce a certain amount of displacement.

In our exercise, the spring constant is found by rearranging Hooke's Law \( k = \frac{F}{x} \). When a force of 20 pounds stretches a spring by 9 inches (which is \( \frac{3}{4} \) feet), we substitute these values into the equation to find 'k'. This value of 'k' will then be used to determine the work required to stretch the spring further, leading us into the concept of the work-energy theorem.

Work-Energy Theorem

The work-energy theorem is a fundamental concept in physics that relates the work done on an object to its change in kinetic energy. However, in the context of springs and Hooke's Law, this theorem helps us calculate the work done in stretching or compressing a spring. The work done, denoted by 'W', is equal to the change in the spring's potential energy.

To calculate this work for a spring, we use the formula \( W = \frac{1}{2} k (x_2^2 - x_1^2) \), where 'k' is the spring constant, and \( x_1 \) and \( x_2 \) are the initial and final displacements of the spring from its equilibrium position, respectively. For our exercise, the spring is initially at its natural length (\( x_1 = 0 \)), and we want to calculate the work done to stretch it by 1 foot (\( x_2 = 1 \) foot). After finding 'k', we plug it into the formula along with the displacement values to determine the work needed.

Calculus Integration

Calculus integration is a mathematical tool that allows us to calculate the accumulated sum of quantities as they change over a continuous interval. In the study of physics, especially in the case of springs, integration comes into play when finding the work done to stretch or compress a spring from one position to another.

For a varying force, like the one exerted by a spring, the work done is not simply 'force times distance' because the force changes with displacement. Instead, we integrate the force over the range of displacement. In our exercise, we bypass the integral calculus since we use the Work-Energy theorem which has already integrated the spring force to give us the formula for the work done by the spring. The integral, which would be \( W = \int_{x_1}^{x_2} kx \, dx \), simplifies to the quadratic formula used in Step 2 involving the spring constant.