This problem is about calculating the work required to stretch a spring, using Hooke's Law for springs. The given data allows us to apply the work formula for a spring.

Problem Details:

• The work required to stretch the spring 1 foot beyond its natural length is 12 ft-lb.
• We are asked to find the work required to stretch it 9 inches (which is 0.75 feet) beyond its natural length.

Hooke's Law and Work Formula:

For a spring, the force needed to stretch it is proportional to the displacement, and the work done to stretch it by a distance $x$ is:

$W = \int_0^x kx \, dx = \frac{1}{2} k x^2$

Where:

• $W$ is the work done.
• $k$ is the spring constant.
• $x$ is the displacement from the natural length.

Step 1: Finding the Spring Constant $k$

From the information provided, we know that stretching the spring 1 foot requires 12 ft-lb of work. So, using the work formula for a displacement of 1 foot:

$W = \frac{1}{2} k (1)^2 = 12 \quad \Rightarrow \quad \frac{1}{2} k = 12 \quad \Rightarrow \quad k = 24$

Step 2: Calculate the Work for 9 inches (0.75 feet)

Now that we know the spring constant $k = 24$, we can calculate the work required to stretch the spring 9 inches (or 0.75 feet):

$W = \frac{1}{2} k (0.75)^2 = \frac{1}{2} \times 24 \times (0.75)^2 = 12 \times 0.5625 = 6.75 \, \text{ft-lb}$

Final Answer:

The work required to stretch the spring 9 inches beyond its natural length is 6.75 ft-lb.

Would you like further clarification or more details on this solution?

Here are 5 related questions to expand your understanding:

1. How is Hooke's Law applied in calculating work for springs?
2. What happens if the spring is stretched by double the length? How does the work change?
3. What is the physical meaning of the spring constant $k$?
4. Can we use Hooke's Law for very large displacements from the natural length?
5. How does the concept of potential energy in a spring relate to the work done on it?

Tip: Always convert units to ensure consistency before applying formulas, especially when dealing with feet, inches, or other mixed units.