(20 13? As the picture shows, there is water in the container, and there is a piece of wood tied with a thin thread in the water. It is known that the water weighs 200N and the water depth is 0.5m m.

It is known that: h=0.5m, V=4dm3=0.004m3, g= 10N/kg, ρ water = 1.0× 103kg/m3.

Find: (1) the pressure p of water on the bottom of the container, and the buoyancy f of the wood block floats;

(2) the density of wood ρ wood.

Solution: (1) Water pressure at the bottom of the container:

P = rho water GH =1.0×103kg/m3×10n/kg× 0.5m = 5000pa,

At this point, the wood block is completely immersed in the water.

∴V row =V=0.004m3,

Buoyancy of wood block:

F float = ρ water gV row =1.0×103kg/m3×10n/kg× 0.004m3 = 40n;

(2) If the rope breaks, the pulley finally floats on the water.

∵ Wood blocks float on the water,

∴ f floating' = Cowood,

∫F float =ρgV row, and ρ=mV, G=mg,

ρ water gV row' = ρ wood Vg, that is, ρ water g(V-25V)=ρ wood Vg,

Solution: ρ wood =35ρ water = 35×1.0×103kg/m3 = 0.6×103kg/m3.

Answer: (1) The pressure of water on the bottom of the container is 5000Pa, and the buoyancy of the wood block is 40 N;

(2) The density of wood block is 0.6× 103kg/m3.