To determine how far the spring must be compressed to store $3.22 \, \text{J}$ of energy, we use the formula for the potential energy stored in a spring:

$U = \frac{1}{2} k x^2$

Where:

• $U$ is the stored potential energy ($3.22 \, \text{J}$),
• $k$ is the spring constant ($1500 \, \text{N/m}$),
• $x$ is the compression distance from equilibrium.

Rearranging for $x$:

$x = \sqrt{\frac{2U}{k}}$

Substitute the values:

$x = \sqrt{\frac{2 \cdot 3.22}{1500}}$

Simplify the calculation:

$x = \sqrt{\frac{6.44}{1500}} = \sqrt{0.0042933}$

$x \approx 0.0655 \, \text{m}$

Thus, the spring must be compressed by approximately:

$\boxed{0.0655 \, \text{m} \, \text{(or 6.55 cm)}}$

Let me know if you'd like to see further details or examples! Here are related questions:

1. How would the compression change if the spring constant were doubled?
2. What would the energy be if the spring were compressed by 10 cm?
3. How does potential energy in a spring relate to its compression or extension?
4. What are real-world applications of the spring constant and potential energy concepts?
5. How would the result differ if friction were introduced in the system?

Tip: Always remember to check the units of the spring constant and energy to ensure proper calculations!