1. **Problem statement:** A bead moves on a frictionless track starting at point A with speed 20 m/s at height 200 m. We want to find its speed at point C, which is at height 160 m. 2. **Relevant principle:** Since the track is frictionless and air resistance is negligible, mechanical energy is conserved. This means the total energy (kinetic + potential) at point A equals that at point C. 3. **Formulas:** - Kinetic energy: $KE = \frac{1}{2}mv^2$ - Potential energy: $PE = mgh$ - Conservation of mechanical energy: $$KE_A + PE_A = KE_C + PE_C$$ 4. **Apply conservation of energy:** Let $m$ be the mass of the bead, $g = 9.8$ m/s$^2$ acceleration due to gravity. At point A: $$KE_A = \frac{1}{2}m(20)^2 = 200m$$ $$PE_A = mg(200)$$ At point C: $$KE_C = \frac{1}{2}mv_C^2$$ $$PE_C = mg(160)$$ Using conservation: $$200m + mg(200) = \frac{1}{2}mv_C^2 + mg(160)$$ 5. **Simplify equation:** Divide entire equation by $m$ (mass cancels out): $$200 + g(200) = \frac{1}{2}v_C^2 + g(160)$$ 6. **Isolate $v_C^2$:** $$\frac{1}{2}v_C^2 = 200 + 9.8 \times 200 - 9.8 \times 160$$ Calculate the terms: $$9.8 \times 200 = 1960$$ $$9.8 \times 160 = 1568$$ So: $$\frac{1}{2}v_C^2 = 200 + 1960 - 1568 = 592$$ 7. **Solve for $v_C$:** $$v_C^2 = 2 \times 592 = 1184$$ $$v_C = \sqrt{1184} \approx 34.4 \text{ m/s}$$ **Final answer:** The speed of the bead at point C is approximately **34.4 m/s**.