What is the smallest area of an ice floe 25 cm thick to hold a 50 kg person on the water?

1). Let’s calculate the mass of the ice floe by the formula m = ρ • V, where the ice density ρ = 900 kg / cubic meter, and the volume V = S • h, the thickness of the ice floe h = 25 cm = 0.25 m; m = ρ • S • h; m = 900 • S • 0.25.
2). A buoyant force acts on the ice floe from the water side, which, according to Archimedes’ law, is F = ρ • g • V, where the water density is ρ = 1000 kg / cubic meter, g = 9.8 N / kg, and the volume of the ice floe is V = S • h ; F = 1000 • 9.8 • S • 0.25.
3). To prevent the ice floe from sinking, the buoyancy force must be no less than the total gravity acting on the ice floe together with the person, we define the force of gravity: F = (m + 900 • S • 0.25) • g, where the person’s mass is m = 50 kg.
4). We get the equation: 1000 • 9.8 • S • 0.25 = (50 + 900 • S • 0.25) • 9.8; or 100 • S • 0.25 = 5 + 90 • S • 0.25; then S = 2 sq.m. Answer. Ice floe area S = 2 sq.m.