Problem 10 - Single Variable Calculus

Question

A force of \(8 \mathrm{lb}\) is required to stretch a spring 2 in. beyond its natural length. Find the work required to stretch the spring 3 in. beyond its natural length.

Step-by-Step Solution

Answer

The work required to stretch the spring 3 inches beyond its natural length is \(10 \mathrm{lb-in}\).

Step 1: Find the spring constant (k)

We are given F = 8 lbs and x = 2 in. We can use Hooke's Law formula to find the spring constant (k): \[F = kx\] \[k = \frac{F}{x}\] \[k = \frac{8 \mathrm{lb}}{2 \mathrm{in}} = 4 \frac{\mathrm{lb}}{\mathrm{in}}\] The spring constant k is \(4 \frac{\mathrm{lb}}{\mathrm{in}}\).

Step 2: Calculate the work (W)

Now we need to calculate the work required to stretch the spring further. We need to find the work done from x = 2 in. to x = 3 in: \[W = \int_{2}^{3} F(x)dx\] We know that F(x) = kx, where k = \(4 \frac{\mathrm{lb}}{\mathrm{in}}\). So, we can plug this into the formula for work: \[W = \int_{2}^{3} 4x dx\]

Step 3: Evaluate the integral

Now we can evaluate the integral to find the work required to stretch the spring: \[W = \int_{2}^{3} 4x dx\] \[W = 4\int_{2}^{3} x dx\] \[W = 4 \left[\frac{1}{2}x^2\right]_2^3\] Evaluate the expression at the limits: \[W = 4\left(\frac{1}{2}(3^2) - \frac{1}{2}(2^2)\right)\] \[W = 4\left(\frac{1}{2}(9 - 4)\right)\] \[W = 4\left(\frac{1}{2}(5)\right)\] \[W = 4(2.5)\] \[W = 10\] The work required to stretch the spring 3 inches beyond its natural length is 10 lb-in.

Key Concepts

Hooke's Law

Hooke’s Law is a fundamental principle in physics and is used to describe how springs behave under load. It states that the force required to stretch or compress a spring is directly proportional to the distance it is stretched or compressed. The formula for Hooke's Law is given by:
\[F = kx\]where:

• \(F\) is the force applied to the spring (measured in pounds in this case).
• \(k\) is the spring constant (measured in pounds per inch).
• \(x\) is the displacement of the spring from its natural length (measured in inches).

The spring constant \(k\) is a measure of the stiffness of the spring. A larger \(k\) means that more force is needed to stretch or compress the spring by the same amount. By rearranging Hooke's Law, \(k\) can be found by\(\[k = \frac{F}{x}\] \)which enables us to calculate the spring constant when the force and displacement are known. In the exercise, we found \(k = 4 \frac{\mathrm{lb}}{\mathrm{in}}\), indicating a relatively firm spring.

Integral Calculus

Integral Calculus is a branch of mathematics focused on finding the total amount of something when the rate of change is known, often used to compute quantities like work, area, and volume. In the context of physics, it helps in determining the work done by variable forces, such as stretching a spring.

To calculate the work done in stretching the spring from 2 inches to 3 inches beyond its natural length, we use the formula for work \(W\):
\[W = \int_{a}^{b} F(x) \, dx\]Here, \(a\) and \(b\) represent the initial and final positions of the spring, and \(F(x)\) is the force function. Given that \(F(x) = kx\), we integrate:\[W = \int_{2}^{3} 4x \, dx\]The process requires evaluating this integral, which involves finding the antiderivative of \(4x\) and then applying the limits of integration. This yields the total work done in moving the spring over the specified distance.

By using definite integrals, we effectively sum up all the tiny bits of work done over small increments, giving us the total work to be 10 lb-inches.

Spring Constant Calculation

Calculating the spring constant is essential because it quantifies the spring's rigidity, determining how much force is needed for a specific stretch or compression. The spring constant (\(k\)) is derived from Hooke's Law. In our exercise:

• We have \(F = 8\) pounds, which is the force needed to stretch the spring 2 inches beyond its natural length.
• The displacement \(x\) is 2 inches.

Plugging these values into the formula \(k = \frac{F}{x}\) gives us:\[k = \frac{8 \, \text{lb}}{2 \, \text{in}} = 4 \, \frac{\text{lb}}{\text{in}}\]This result tells us how much force is required per unit of length to stretch this spring. A higher \(k\) value implies a steeper force requirement for the same amount of stretch, indicating a stiffer spring. Calculating \(k\) is crucial for determining the spring's influence in mechanical systems and ensuring accurate force application in practical situations.