1. **State the problem:** A spring with spring constant $k = 75$ N/m is compressed by $x = 10$ cm. We need to find the force acting on the spring. 2. **Formula used:** The force exerted by a spring is given by Hooke's Law: $$F = kx$$ where $F$ is the force, $k$ is the spring constant, and $x$ is the displacement from the equilibrium position. 3. **Important notes:** - The displacement $x$ must be in meters to match the units of $k$. - The force is restorative, meaning it acts opposite to the direction of displacement. 4. **Convert units:** $$x = 10\text{ cm} = 0.10\text{ m}$$ 5. **Calculate the force:** $$F = 75 \times 0.10 = 7.5\text{ N}$$ 6. **Interpretation:** The force acting on the spring is 7.5 N, which corresponds to option d. **Final answer:** $7.5$ N