Determine the minimum area of a flat ice floe 1.0 meter thick if it is to support a 2 000-kg car above seawater. (ρ_ice = 920 kg/m³, ρ_sea = 1 020 kg/m³.) A) 20 m² B) 40 m² C) 60 m² D) 80 m² E) 100 m²

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To support the car and the ice floe, the weight of the displaced seawater must equal the total weight of the car and the ice. The volume of seawater displaced is equal to the volume of the ice floe submerged. Let $ V $ be the volume of the ice floe, $ A $ its area, and $ h = 1.0 \, \text{m} $ its thickness. Thus, $ V = A \times h $.The mass of the car is $ 2000 \, \text{kg} $, and the density of ice ($ \rho_{\text{ice}} $) is $ 920 \, \text{kg/m}^3 $. The mass of the ice floe is $ \rho_{\text{ice}} \times V = 920 \times A \times 1.0 $.The total weight to be supported is the weight of the car plus the weight of the ice, which is $ 2000 \, \text{kg} + 920A \, \text{kg} $.The density of seawater ($ \rho_{\text{sea}} $) is $ 1020 \, \text{kg/m}^3 $. The weight of the displaced seawater must equal the total weight, so:$$ \rho_{\text{sea}} \times V_{\text{displaced}} = 2000 + 920A $$Since $ V_{\text{displaced}} = A \times h $ (the volume of the ice submerged equals the volume of the ice because it needs to be completely submerged to support the car and itself), and $ h = 1.0 \, \text{m} $, we have:$$ 1020 \times A = 2000 + 920A $$Solving for $ A $:$$ 100A = 2000 $$$$ A = 20 \, \text{m}^2 $$Therefore, the minimum area of the ice floe needed to support the car above seawater is $ 20 \, \text{m}^2 $.

Determine the Minimum Area of a Flat Ice Floe 1