A force of 320 lb is required to hold a spring stretched 5 ft beyond its natural length. How much work *W* is done in stretching it from its natural length to 3 inches beyond its natural length?

To solve this problem, we first need to understand the relationship between force, spring constant, and displacement in Hooke's Law, which states:

\[
F = kx
\]

where:
- \( F \) is the force applied to the spring,
- \( k \) is the spring constant,
- \( x \) is the displacement from the natural length.

### Step 1: Find the Spring Constant \( k \)

Given that a force of \( 320 \) lb is required to stretch the spring \( 5 \) ft (or \( 60 \) inches) beyond its natural length, we can find the spring constant \( k \) using Hooke's Law:

\[
320 = k \times 60
\]

Solving for \( k \):

\[
k = \frac{320}{60} = \frac{16}{3} \, \text{lb/in}
\]

### Step 2: Calculate the Work Done \( W \)

The work done in stretching a spring is given by the integral of the force over the displacement:

\[
W = \int_{0}^{x} F \, dx = \int_{0}^{x} kx \, dx
\]

where \( x \) is the displacement from the natural length. In this case, we need to calculate the work done in stretching the spring from its natural length to \( 3 \) inches:

\[
W = \int_{0}^{3} \frac{16}{3} x \, dx
\]

Evaluating the integral:

\[
W = \frac{16}{3} \int_{0}^{3} x \, dx = \frac{16}{3} \left[\frac{x^2}{2}\right]_{0}^{3}
\]

\[
W = \frac{16}{3} \times \frac{3^2}{2} = \frac{16}{3} \times \frac{9}{2} = \frac{16 \times 9}{6} = 24 \, \text{ft-lb}
\]

Therefore, the work done in stretching the spring from its natural length to \( 3 \) inches beyond its natural length is **24 ft-lb**.

Would you like any further details or have any questions? Here are some related problems you might find interesting:

1. How much work is done in stretching the spring from 3 inches to 5 feet beyond its natural length?
2. What would be the spring constant if the force required was 400 lb for the same stretch?
3. How does the work change if the spring is stretched only 2 inches instead of 3?
4. What is the potential energy stored in the spring when stretched 3 inches beyond its natural length?
5. How much force is required to stretch the spring 10 inches beyond its natural length?

**Tip:** Always make sure units are consistent when performing calculations in physics, especially with mixed units like feet and inches!

A force of 320 lb is required to hold a spring stretched 5 ft beyond its natural length. How much work W is done in stretching it from its natural length to 3 inches beyond its natural length?

Learn how to calculate the work done in stretching a spring using Hooke's Law. In this problem, a 320 lb force stretches a spring 5 feet, and we find the work required to stretch it 3 inches beyond its natural length. This problem involves Hooke's Law, integration, and physics principles suitable for advanced high school students.

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