To solve this problem, we can apply Hooke's Law and the work formula for a spring. Here's the process:

Step 1: Hooke's Law

Hooke's Law is given as: $F = kx$ where:

• $F$ is the force applied (in pounds),
• $k$ is the spring constant (in pounds per foot),
• $x$ is the displacement of the spring from its natural length (in feet).

We are given: $F = 4 \, \text{pounds}, \quad x = 0.1 \, \text{feet}.$

Substituting these values, we solve for $k$: $4 = k(0.1) \implies k = \frac{4}{0.1} = 40 \, \text{pounds/foot}.$

Step 2: Work Done to Stretch the Spring

The formula for the work done on a spring is: $W = \int_{x_1}^{x_2} kx \, dx$ where:

• $x_1$ is the initial displacement,
• $x_2$ is the final displacement,
• $k$ is the spring constant.

We are stretching the spring from $x_1 = 0$ to $x_2 = 0.9 \, \text{feet}$. Substituting $k = 40$, the work becomes: $W = \int_{0}^{0.9} 40x \, dx.$

Step 3: Evaluate the Integral

The integral is: $W = 40 \int_{0}^{0.9} x \, dx = 40 \left[ \frac{x^2}{2} \right]_0^{0.9}.$ Simplify: $W = 40 \cdot \frac{(0.9)^2}{2} - 40 \cdot \frac{(0)^2}{2}.$ $W = 40 \cdot \frac{0.81}{2} = 40 \cdot 0.405 = 16.2 \, \text{foot-pounds}.$

Final Answer:

The work done in stretching the spring is: $\boxed{16.2 \, \text{foot-pounds}}$

Do you want further clarification or a breakdown of any specific step?

Related Questions:

1. How does Hooke's Law apply to real-world scenarios?
2. What happens if the spring constant ($k$) is larger or smaller?
3. How is the work formula derived for a spring?
4. What is the physical significance of the area under the force-displacement curve?
5. Can this method be used for compressing springs?

Tip:

Always remember that work done on a spring is proportional to the square of its displacement!