(20 13? The bridge construction team uses the pulley block as shown in the figure to salvage the submerged workpiece. When the workpiece is suspended in water, the pulling force f is 180N.

(1) The worker's pulling force when the workpiece is in water: F= 12(G object +G moving -F floating) = 180N,

After the workpiece leaves the water, the worker's pulling force: F 1= 13(G object +G motion) =480N,

The above formula is simultaneous, and ffloat = 900 n is obtained;

When the workpiece is suspended in water: η=W, there is W total =G object +G floating object +G moving +G floating,

80%=G+900N480X 3+900N，

You can get: g object = 1332N,

Mass m = gg =1332n10n/kg =133.2n,

From F-float = ρ water gV, row V = 900 n1.0×103kg/m3×10n/kg = 9×10-2? m3；

Workpiece density ρ = mvobject =133.2kg9×10? 2 m3 = 1.48× 103?/m3。

(2) In the process of rising from the water to the surface, the buoyancy gradually decreases, the resistance overcome by the pulley block increases, and the pulling force of people also increases. According to the interaction between objects, the tension of the rope gradually increases, so the pressure of people on the ground decreases. According to P=FS, the worker pressure on the ground is reduced.

Answer: (1) The buoyancy of the workpiece suspended in water is 900N, and the density of the workpiece ρ =1.48×103/m3;

(2) In the process of rising from the water to the surface, the buoyancy gradually decreases, the resistance overcome by the pulley block increases, and the pulling force of people also increases. According to the interaction between objects, the tension of the rope gradually increases, so the pressure of people on the ground decreases. According to P=FS, the worker pressure on the ground is reduced.