1. **State the problem:** We need to find the distance $x$ by which an ideal spring is stretched when 4 joules of work is done on it, given the spring constant $k = 2500$ N/m. 2. **Formula used:** The work done in stretching or compressing a spring is given by the elastic potential energy formula: $$ W = \frac{1}{2} k x^2 $$ where $W$ is the work done, $k$ is the spring constant, and $x$ is the displacement. 3. **Rearrange the formula to solve for $x$:** $$ x = \sqrt{\frac{2W}{k}} $$ 4. **Substitute the known values:** $$ x = \sqrt{\frac{2 \times 4}{2500}} $$ 5. **Calculate the value inside the square root:** $$ x = \sqrt{\frac{8}{2500}} $$ 6. **Simplify the fraction:** $$ x = \sqrt{\cancel{\frac{8}{\cancel{2500}}}} = \sqrt{0.0032} $$ 7. **Calculate the square root:** $$ x \approx 0.0566 \text{ meters} $$ **Final answer:** The spring is stretched approximately 0.0566 meters (or 5.66 centimeters).