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.