This problem deals with Hooke's Law, which states that the extension of a spring is directly proportional to the weight (or force) applied to it, as long as the elastic limit is not exceeded. Mathematically, this can be expressed as:

$L \propto W$ where:

• $L$ is the length the spring stretches,
• $W$ is the weight applied to the spring.

Given that a 4-pound weight stretches the spring 9 inches, we can set up the ratio for the given situation:

$\frac{L_1}{W_1} = \frac{L_2}{W_2}$

Where:

• $L_1 = 9$ inches (the stretch with 4 pounds),
• $W_1 = 4$ pounds,
• $L_2$ is the unknown stretch with $W_2 = 36$ pounds.

Now, substituting the known values into the equation:

$\frac{9}{4} = \frac{L_2}{36}$

To solve for $L_2$, multiply both sides by 36:

$L_2 = \frac{9}{4} \times 36 = 81 \text{ inches}.$

So, the spring will stretch 81 inches if the weight is increased to 36 pounds.

Do you have any questions or would you like more details?

Here are 5 related questions:

1. What would the stretch be for a 10-pound weight?
2. How does Hooke’s Law apply to real-world engineering problems?
3. What happens if the spring exceeds its elastic limit?
4. Can this proportionality change if the material of the spring is altered?
5. How does the stiffness of a spring (spring constant) affect its stretch?

Tip: Always check if the weight applied exceeds the spring’s elastic limit; beyond this point, the spring may not return to its original shape.