1. **Problem statement:** A spring stretches by 0.5 m when a mass of 0.51 kg is attached. Find the spring constant $k$. 2. **Relevant formula:** The spring force $F_s$ balances the gravitational force $F_g$ at equilibrium, so: $$F_g = F_s \Rightarrow mg = kx$$ where $m$ is mass, $g$ is acceleration due to gravity (9.8 m/s$^2$), $k$ is spring constant, and $x$ is stretch distance. 3. **Rearranging for $k$:** $$k = \frac{mg}{x}$$ 4. **Substitute values:** $$k = \frac{0.51 \times 9.8}{0.5}$$ 5. **Calculate numerator:** $$0.51 \times 9.8 = 4.998$$ 6. **Calculate $k$:** $$k = \frac{4.998}{0.5} = 9.996$$ 7. **Final answer:** $$k \approx 10 \text{ N/m}$$ This means the spring constant is approximately 10 newtons per meter, indicating the spring stretches 1 meter for every 10 newtons of force applied.