A spring has a natural length of 16 cm. If a 25 N force is required to keep it stretched to a length of 30 cm, how much work *W* (in J) is required to stretch it from 16 cm to 23 cm? (Round your answer to two decimal places.)

*W* =  J

To calculate the work \( W \) required to stretch a spring, we use **Hooke's Law** and the formula for work done in stretching a spring:

\[
W = \int_{x_1}^{x_2} F(x) \, dx
\]

Where:
- \( F(x) \) is the restoring force required to stretch the spring to a displacement \( x \) from its natural length.
- \( x_1 \) and \( x_2 \) are the initial and final displacements from the natural length.
- The force \( F(x) \) is proportional to the displacement \( x \) by the spring constant \( k \): \( F(x) = kx \).

### Step 1: Find the spring constant \( k \)
We are given that a force of \( 25 \, \text{N} \) stretches the spring to \( 30 \, \text{cm} \) (or 0.30 m). The natural length of the spring is \( 16 \, \text{cm} \) (or 0.16 m), so the displacement \( x \) is:
\[
x = 0.30 - 0.16 = 0.14 \, \text{m}
\]
Using Hooke's Law \( F = kx \), we can solve for \( k \):
\[
25 = k \times 0.14
\]
\[
k = \frac{25}{0.14} \approx 178.57 \, \text{N/m}
\]

### Step 2: Calculate the work \( W \)
The work done to stretch the spring from 16 cm to 23 cm (or from 0.16 m to 0.23 m) is given by:

\[
W = \int_{x_1}^{x_2} kx \, dx = \frac{1}{2} k (x_2^2 - x_1^2)
\]

Where:
- \( x_1 = 0.23 - 0.16 = 0.07 \, \text{m} \)
- \( x_2 = 0.23 \, \text{m} \)

Substitute the values into the formula:
\[
W = \frac{1}{2} \times 178.57 \times ((0.07)^2 - (0)^2)
\]
Now let's calculate the work.

\[
W = \frac{1}{2} \times 178.57 \times (0.07^2)
\]
\[
W = 0.5 \times 178.57 \times 0.0049 \approx 0.437 \, \text{J}
\]

So, the work required to stretch the spring from 16 cm to 23 cm is approximately **0.44 J**.

Would you like further details or have any questions?

### Related Questions:
1. How is the spring constant \( k \) calculated in other contexts?
2. What happens if the force applied exceeds the spring's elastic limit?
3. Can Hooke's Law be applied to all materials?
4. How do damping forces affect work done on springs?
5. What is the difference between elastic potential energy and work done on a spring?

### Tip:
Always ensure units are consistent when solving physics problems! Converting lengths to meters and forces to newtons is essential in SI unit-based calculations.

A spring has a natural length of 16 cm. If a 25 N force is required to keep it stretched to a length of 30 cm, how much work W (in J) is required to stretch it from 16 cm to 23 cm? (Round your answer to two decimal places.) W = J

Learn how to calculate the work required to stretch a spring from its natural length of 16 cm to 23 cm, given a force of 25 N stretches it to 30 cm. This detailed solution covers Hooke's Law, spring constant calculations, and the work-energy theorem. Suitable for Grades 10-12.

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