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The question asks: 

**What is the spring constant of a spring that stores 38 J of elastic potential energy when compressed by 6.5 cm from its relaxed length?**

To solve this, we use the formula for elastic potential energy stored in a spring:

\[
U = \frac{1}{2} k x^2
\]

Where:
- \( U \) is the elastic potential energy (38 J),
- \( k \) is the spring constant (what we are solving for),
- \( x \) is the displacement from the relaxed length (6.5 cm, which is 0.065 m).

Rearranging the formula to solve for \( k \):

\[
k = \frac{2U}{x^2}
\]

Substituting the given values:

\[
k = \frac{2 \times 38}{(0.065)^2}
\]

\[
k = \frac{76}{0.004225} \approx 17983.63 \, \text{N/m}
\]

Thus, the spring constant \( k \) is approximately **17984 N/m**.

Would you like more details on this, or do you have any questions?

Here are 5 follow-up questions you could consider:
1. How does the spring constant affect the stiffness of a spring?
2. What is the relationship between potential energy and displacement in a spring?
3. Can this formula be used for springs with non-linear properties?
4. How would the spring constant change if the compression was doubled?
5. How do real-world factors, like material properties, affect the spring constant?

**Tip:** Always convert units to SI (meters, kilograms, etc.) when solving physics problems to avoid calculation errors.

What is the spring constant of a spring that stores 38 J of elastic potential energy when compressed by 6.5 cm from its relaxed length?

Learn how to calculate the spring constant of a spring that stores 38 J of elastic potential energy when compressed by 6.5 cm. This problem involves concepts such as elastic potential energy and Hooke's Law, suitable for high school physics students.

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