Problem 9 - 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. Eighteen foot-pounds of work is required to stretch a spring 4 inches from its natural length. Find the work required to stretch the spring an additional 3 inches.

Step-by-Step Solution

Answer

The solution involves calculating the integral for the work done, which will be the final answer.

Step 1: Calculate k

First, use the given work (18 foot-pounds) and stretch distance (4 inches) to find the spring constant. The work, W, is given by the integral of F dx from 0 to x. Thus, we have \[W = \int_0^x F dx = \int_0^x kx dx\]; replacing the given values, we get\[18 = \int_0^{1/3} kxdx\]; which simplifies to \[18 = \frac{1}{2}k * (1/3)^2\]; from this equation, we solve for k.

Step 2: Solve for k

Rearranging the equation we got above to solve for k, we get \[k = 2 * \frac{18}{(1/3)^2} \]Solve this equation to find the value of k.

Step 3: Find the work done to stretch the spring an additional 3 inches

Now with the calculated k, we can find the work done to stretch the spring a further 3 inches. Use the formula for work \[W = \int_{a}^{b} F dx\]Our initial stretching position (a) is 1/3 ft (given 4 inches = 1/3 ft). The final stretching position (b) after an additional stretch of 3 inches (or 1/4 ft) is 1/3+1/4=7/12 ft. Thus, we have \[W = \int_{1/3}^{7/12} kx dx\] Evaluate this integral to get the work done for stretching additional 3 inches.

Step 4: Solve for Work

Calculate the integral derived above to find the work done to stretch the spring an additional 3 inches. The result will be the required work done.

Key Concepts

Spring Constant

When dealing with springs, understanding the spring constant, denoted by \( k \), is crucial. The spring constant measures the stiffness of a spring. The larger the \( k \), the stiffer and less flexible the spring is. Hooke's Law, which is represented as \( F = kx \), relates the force exerted by the spring \( F \) to the spring constant \( k \) and the displacement \( x \) from the natural length of the spring.
To find \( k \), use the work done on the spring, which is known to be 18 foot-pounds in this instance, along with the initial stretch distance of 4 inches. By setting up the integral of \( kx \), and solving for \( k \) gives us insight into how resistant the spring is to stretching.

Work Calculation

The work done on a spring is calculated via the formula \( W = \int_0^x F \, dx \). Here, \( F \) is the variable force applied, which can change as the spring stretches.
To compute the initial work of 18 foot-pounds used to stretch 4 inches (or \( \frac{1}{3} \) feet), set up the integral \( W = \int_0^{1/3} kx \, dx = 18 \). This equation shows that the work required to stretch a spring can be looked at as an integral of an increasing force over a stretch distance.
The computation involves simplifying the integral to \( \frac{1}{2}k \left( \frac{1}{3} \right)^2 \) and solving for \( k \). Knowing \( k \) allows us to assess any further stretching efforts.

Integral Calculus

Integral calculus plays a vital role in determining the work needed to stretch a spring by considering variable forces. The main idea is to add up all the small contributions of force over a distance to calculate total work.
The integral \( \int F \, dx \) is solved from the starting position to the final position (i.e., \( \int_{a}^{b} kx \, dx \)). By calculating these integrations, we obtain the work done over specific distances. This method helps in isolating the specific energy input needed for additional stretches.

Variable Force

The concept of variable force is essential in spring problems. Unlike constant forces, the force required to stretch or compress a spring varies with distance. As the spring stretches more, the force increases. Thus, the relationship \( F = kx \) indicates how force grows with increased displacement \( x \).
To find the work done from the initial to the new stretched length, calculate how this force varies within those points. By determining the integral of \( kx \) over the changes in distance, we achieve a comprehensive view of the variable nature of force in Hooke's Law applications.

Stretch Distance

The stretch distance \( x \) in spring problems signifies how far a spring is pulled from its original position. It impacts not only the force applied due to the spring constant \( k \) but also the resultant work done.
In the exercise, the spring is initially stretched by 4 inches and then by an additional 3 inches. Calculating work done over these stretches requires assessing the stretch incrementally, mapping \( x \) within the limits of integration.
Understanding the total change in stretch, such as moving from \( \frac{1}{3} \) ft to \( \frac{7}{12} \) ft, helps in setting up the necessary integrals and computations for evaluating precise work for each stretch increment.