Welcome back to Professor Baker's Math Class! In this lesson, we are stepping out of the theoretical world and into the physical one. We are looking at Section 6-4: Applications of Integration - Work Problems. While we often use the word "work" to mean effort in daily life, in physics and calculus, it has a very specific definition relating force and displacement.

1. The Basics: Constant Force

If you push a book across a table with a constant force, calculating work is simple geometry. The formula is:

$$W = F \cdot d$$

Where $W$ is work, $F$ is force, and $d$ is distance. However, you must pay close attention to your units!

  • SI Metric System: Mass is in kilograms ($kg$), distance in meters ($m$), and force in Newtons ($N$).
    Note: If you are given mass ($kg$), you must convert it to force using Newton's Second Law ($F=ma$), where gravity $g \approx 9.8 m/s^2$. The resulting unit for Work is the Joule (J).
  • US Customary System: Force is usually given directly in pounds ($lb$) and distance in feet ($ft$). The unit for Work is the foot-pound (ft-lb).

2. The Calculus: Variable Force

What happens when the force changes as you move an object? Think about stretching a rubber band—it gets harder to pull the further you stretch it. A constant multiplication formula won't work here. Instead, we use a definite integral to sum up the work done over infinitesimally small distances.

If an object moves along the $x$-axis from $a$ to $b$, and the force applied at position $x$ is $f(x)$, the total work $W$ is:

$$W = \int_{a}^{b} f(x) \, dx$$

3. Application: Hooke's Law (Springs)

One of the most common applications of variable force in calculus is Hooke's Law. It states that the force required to maintain a spring stretched $x$ units beyond its natural length is proportional to $x$.

$$f(x) = kx$$

Here, $k$ is the spring constant, which represents how stiff the spring is. Let's look at a classic example from the class notes.

Example Scenario:
A force of $40 N$ is required to hold a spring that has been stretched from its natural length of $10 cm$ to a length of $15 cm$. How much work is done in stretching the spring from $15 cm$ to $18 cm$?

Solution:
First, we must ensure all units are in meters. The natural length is $0.10 m$. The stretched length is $0.15 m$, meaning the displacement $x = 0.05 m$.

  1. Find $k$: Using $f(x) = kx$, we know that $40 = k(0.05)$. Solving for $k$, we get $k = 800$. Therefore, our force function is $f(x) = 800x$.
  2. Set up the Integral: We want to find the work from $15 cm$ to $18 cm$. Relative to the natural length ($10 cm$), our lower limit is $0.05 m$ and our upper limit is $0.08 m$.
  3. Integrate:
$$W = \int_{0.05}^{0.08} 800x \, dx = \left[ \frac{800x^2}{2} \right]_{0.05}^{0.08} = 400x^2 \Big|_{0.05}^{0.08}$$ $$W = 400(0.08)^2 - 400(0.05)^2$$ $$W = 400(0.0064 - 0.0025) = 1.56 \text{ Joules}$$

4. Advanced Applications

The concept of slicing a problem into small pieces and integrating applies to other scenarios discussed in the notes, such as:

  • Lifting Cables: A heavy cable hanging from a building requires variable work because as you pull it up, there is less cable hanging, meaning the weight (force) decreases.
  • Pumping Liquids: Calculating the work required to empty a tank involves integrating the distance distinct "slices" of water must be lifted against gravity.

Mastering these work problems is a huge step in understanding how calculus models the physical universe. Keep practicing those integrals, and remember to check your units! Good luck!