From P=FS, P=G box s = 400n0.4m2 = 1000pa.
(2)v=st=It=4m 10s=0.4m/s。
(3)W has =G box h = 400 n× 0.5m = 200 j .
(4) Work done by F: W 1 = FLL = 75N× 4m = 300J.
The work done by a person walking on an inclined plane is: W2=G, h = 600 n× 0.5h = 300 J. 。
W has =Gh=400N×0.5m=200J，
Mechanical efficiency of inclined plane: η'=W has w total,
In this case, the mechanical efficiency is: η = w+0+w2×100% = 200J300J+300J×100% = 33.3%.
Answer: (1) The pressure of the wooden box on the horizontal ground is1000 Pa; ; (2) The wooden box speed is 0.4m/s;
(3) the useful work is 200 j; (4) The mechanical efficiency in this case is about 33.3%.