A solid sphere and a ring, both of mass 1 kg, roll without slipping down a 30° incline. Which reaches the bottom first, and what are their accelerations? (g = 10 m/s²) Step 1: Use a = g sinθ / (1 + k²/R²). For the solid sphere k²/R² = 2/5, so a = 10 × sin30° / (1 + 0.4) = 5 / 1.4 ≈ 3.57 m/s². Step 2: For the ring k²/R² = 1, so a = 10 × 0.5 / 2 = 2.5 m/s². Step 3: Compare: 3.57 > 2.5, so the solid sphere accelerates faster and reaches the bottom first. Answer: The solid sphere wins (3.57 m/s² vs 2.5 m/s²). The 1 kg mass was irrelevant — only the mass distribution (k²/R²) matters.