1. **State the problem:** A box slides down a frictionless incline of length 12.0 m at an angle of 30.0° starting from rest. We need to find its speed at the bottom. 2. **Relevant formula:** Use conservation of energy or kinematics. Here, we use energy conservation: potential energy at top converts to kinetic energy at bottom. 3. **Calculate height:** Height $h = L \sin(\theta) = 12.0 \times \sin(30^\circ) = 12.0 \times 0.5 = 6.0$ m. 4. **Energy conservation:** Potential energy at top $= mgh$, kinetic energy at bottom $= \frac{1}{2}mv^2$. 5. Set $mgh = \frac{1}{2}mv^2$ and cancel $m$: $$gh = \frac{1}{2}v^2$$ 6. Solve for $v$: $$v^2 = 2gh$$ $$v = \sqrt{2gh}$$ 7. Substitute $g=9.8$ m/s$^2$ and $h=6.0$ m: $$v = \sqrt{2 \times 9.8 \times 6.0} = \sqrt{117.6} \approx 10.84$$ m/s. **Final answer:** The speed of the box at the bottom is approximately $10.84$ m/s.