We are given the following details:

• A force of 18 lb stretches a spring 4 inches beyond its natural length.
• We need to calculate the work $W$ done in stretching the spring from its natural length to 9 inches beyond its natural length.

Step 1: Use Hooke's Law

The force required to stretch or compress a spring is given by Hooke's Law: $F(x) = kx$ where:

• $F(x)$ is the force,
• $k$ is the spring constant,
• $x$ is the displacement from the natural length.

We know $F(4) = 18$ lb, meaning: $18 = k \times 4$ Solving for $k$: $k = \frac{18}{4} = 4.5 \text{ lb/in}.$

Step 2: Work Calculation

The work done to stretch the spring is the integral of the force over the displacement: $W = \int_0^9 F(x) \, dx = \int_0^9 kx \, dx.$ Substitute $k = 4.5$: $W = \int_0^9 4.5x \, dx.$ Now, integrate:

= 4.5 \times \frac{81}{2} = 4.5 \times 40.5 = 182.25 \text{ in-lb}.$$ ### Step 3: Convert to ft-lb Since there are 12 inches in a foot, we convert the work from in-lb to ft-lb: $$W = \frac{182.25}{12} \approx 15.19 \text{ ft-lb}.$$ Thus, the work done in stretching the spring from its natural length to 9 inches beyond its natural length is approximately **15.19 ft-lb**. Do you need more details or have any questions? --- Here are some related questions to deepen understanding: 1. What is Hooke's Law and how is it applied in other spring problems? 2. How would the problem change if the spring constant $$k$$ were different? 3. How do you calculate the spring constant if given a different displacement and force? 4. What are the units of work in the metric system and how would this problem be solved in those units? 5. How would you calculate the work if the spring were stretched from 4 in. to 9 in. instead? **Tip**: Always check if units need conversion, especially when combining quantities measured in different systems like inches and feet.