(1) Draw a line segment with an arrow passing through the center of gravity of the lever in the vertical downward direction, which is indicated by the symbol G; The schematic diagram is as follows:

(2) When the barbell bar is just turned over:

G piece l 1=G rod l2

That is: m gl 1=m poles gl2.

Therefore: m bar =l 1l2m piece =36cm45cm×25kg=20kg.

A: The mass of the lever is at least 20kg. ..

(3)W=nG, total h=m, total GH =10× 80kg×10n/kg× 40x10-2m = 3200J.

P=Wt=3200J20s= 160W。

A: The recommended power for doing work in this exercise is about160 w. 。