A force of 6 pounds is required to stretch a spring 0.4 feet beyond its equilibrium position.

find the force function (in lbs) for the spring. f(x) = _____ pounds
what is the work (in ft/lb) required to stretch the spring 0.8 feet beyond its equilibrium position? round to 5 decimal places. W = _____ ft-lb


This problem is about finding the force function for a spring using **Hooke's Law**, which states that the force \( F(x) \) needed to stretch or compress a spring is proportional to the displacement \( x \) from its equilibrium position.

### Step 1: Find the Force Function

Hooke's Law is mathematically represented as:

\[
F(x) = kx
\]

where:
- \( F(x) \) is the force (in pounds),
- \( k \) is the spring constant (to be determined),
- \( x \) is the displacement (in feet) from the equilibrium position.

We are given that a force of 6 pounds is required to stretch the spring by 0.4 feet. So, we use the given information to find \( k \):

\[
6 = k(0.4)
\]

Solving for \( k \):

\[
k = \frac{6}{0.4} = 15
\]

Thus, the force function is:

\[
F(x) = 15x
\]

### Step 2: Calculate the Work Required to Stretch the Spring

The work \( W \) required to stretch the spring from 0 to a distance \( x \) is given by:

\[
W = \int_0^x F(x) \, dx
\]

Since \( F(x) = 15x \), the work to stretch the spring to 0.8 feet is:

\[
W = \int_0^{0.8} 15x \, dx
\]

First, calculate the integral:

\[
W = 15 \int_0^{0.8} x \, dx = 15 \left[ \frac{x^2}{2} \right]_0^{0.8}
\]

\[
W = 15 \left( \frac{(0.8)^2}{2} \right) = 15 \left( \frac{0.64}{2} \right) = 15 \times 0.32 = 4.8 \, \text{ft-lb}
\]

Thus, the work required to stretch the spring 0.8 feet beyond its equilibrium position is:

\[
W = 4.8 \, \text{ft-lb}
\]

### Summary:
- Force function: \( F(x) = 15x \, \text{pounds} \)
- Work required: \( W = 4.8 \, \text{ft-lb} \)

Do you want further details or have any questions?

Here are 5 related questions:
1. What is the physical meaning of the spring constant \( k \)?
2. How would the force function change if a different force was applied?
3. How can you calculate the work if the spring is stretched by a variable force?
4. What happens to the work if the spring is compressed instead of stretched?
5. How would Hooke’s Law change if the spring behaved non-linearly?

**Tip:** When dealing with spring problems, always use Hooke's Law to find the force function before calculating the work or energy involved.

A force of 6 pounds is required to stretch a spring 0.4 feet beyond its equilibrium position. Find the force function (in lbs) for the spring. f(x) = _____ pounds. What is the work (in ft-lb) required to stretch the spring 0.8 feet beyond its equilibrium position? Round to 5 decimal places. W = _____ ft-lb.

Solve a spring problem using Hooke's Law: A force of 6 pounds stretches a spring 0.4 feet. Find the force function and calculate the work required to stretch the spring by 0.8 feet. Detailed step-by-step solution provided, suitable for high school calculus students.

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