Problem 9 - Essential Calculus

Question

Hooke's Law In Exercises 3-10, 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

Therefore, the work required to stretch the spring an additional 3 inches is approximately 7.1595 foot-pounds in addition to the initial work done, calculated using Hooke's Law.

Step 1: Convert All Distances to the Same Unit

Convert 4 inches to feet, as the work is given in foot-pounds. Note that 1 foot equals 12 inches, so \(4 \text { inches }=\frac{4}{12}=0.3333 \text { feet }\). The new displacement becomes 0.3333 feet.

Step 2: Find the Spring Constant (k)

Since the formula of work is \(W=\frac{1}{2}kx^2\), you can solve for \(k\) by rearranging this formula in terms of \(k\), to give \(k=\frac{2W}{x^2}\). Substituting the given values (where \(W = 18 \text { foot-pounds}\) and \(x=0.3333 \text { feet}\)) into the formula to get the spring constant, \(k=\frac{2*18}{(0.3333)^2}\) which simplifies to give approximately \(k=324.3243 \text { foot-pounds/feet}\).

Step 3: Compute Work for Additional Stretch

Now we need to compute the work required to stretch the spring an additional 3 inches (which should also be converted to feet, becoming \(0.25 \text { feet}\)). The new total stretch will be \(0.3333+0.25=0.5833 \text { feet}\). Substituting values into the formula \(W=\frac{1}{2}kx^2\), we find the total work done to stretch the spring by \(0.5833 \text { feet}\). Subtracting the initial work of 18 foot-pounds gives the work done to stretch the spring the additional 3 inches.

Key Concepts

Spring Constant

In the realm of physics, when dealing with springs, the spring constant ( \(k\) ) is critical. It is a measure of a spring's stiffness. According to Hooke's Law, the force required to stretch or compress a spring is proportional to the displacement of the spring from its natural length. This relationship is expressed by the equation \(F = kx\) , where \(F\) is the force applied, \(k\) is the spring constant, and \(x\) is the displacement.
To find this spring constant, we can use the work-energy principle. In this case, work is given in foot-pounds, and the displacement is converted to feet. Once you know how much work is required to stretch a spring a certain distance, you can calculate the spring constant, \(k\).
Following the formula \(W = \frac{1}{2} k x^2\) , rearrange to find \(k = \frac{2W}{x^2}\) . By substituting the known values into this equation, you will determine the stiffness of the spring in specific units (such as foot-pounds/feet). This constant remains the same as long as the physical conditions of the spring are unchanged.

Work-Energy Principle

The work-energy principle is a cornerstone of physics, especially helpful in analyzing spring mechanics. It states that work done on an object is equal to the change in energy of that object. For a spring, this principle relates the work ( \(W\) ) performed on a spring to alter its length, to the energy stored within the spring.
When a spring is stretched or compressed from its natural position, it stores potential energy.Using the formula \(W = \frac{1}{2} k x^2\) calculated from Hooke's Law, one can determine how much work is required to achieve a particular stretch or compression.
In a practical sense, knowing the work-energy relationship allows you to solve for unknowns, such as additional work required when extending a spring further. Simply plug in the new displacement, and subtract any initial work.

Unit Conversion

Unit conversion is an essential skill in solving physics problems, especially when working with measurements like inches and feet, or different units of force and energy. In our problem, distances given in inches need to be converted to feet, because the work is expressed in foot-pounds.
Knowing the relationship between units is key; here, 1 foot equals 12 inches. Thus, to convert inches to feet, divide the number of inches by 12. For instance, 4 inches translates to \(\frac{4}{12} = 0.3333\) feet.
Being consistent with units ensures accuracy in your results. It's especially important in complex problems where multiple measurements interact under various physical laws.

Calculus Problem Solving

Calculus is indispensable when dealing with physics problems like those involving springs. It allows you to handle problems involving changes, such as varying forces over a displacement.
The work done by a variable force, as in a spring, can be calculated using calculus. For a spring, this often involves integrating the force function over a displacement range.
In our problem, while the step-by-step solution involved basic algebra, understanding calculus concepts adds depth. Integrals can compute exact work done with variable forces. They encompass the summation of infinite, minute increments of work over a distance, which is ideal for complex spring scenarios.

• Understand derivative functions to relate force to displacement.
• Apply integration to compute work across varying distances.

Mastering these calculus concepts enhances problem-solving precision and capability, allowing students to tackle even the most advanced spring-related challenges.