Solution: Apply a vertical downward force F 1 to the free end of the rope of the pulley block, and the force analysis of the heavy object when it descends in a straight line at a uniform speed along the vertical slideway is shown in Figure 1.

At this time, there is 2F 1+f=G0+GA, and a vertical downward force F2 is applied to the free end of the rope of the pulley block. The force analysis of the weight A moving downward along the vertical slide is shown in Figure 2. At this time,

2f2 = G0+GA+FT1= GA+FWF = F2× 2h = 2f2hη1= w Useful W Total = T 1HF2× 2h = T 12f2? Pf2 = F2× 2v = =2F2v。

Hang a weight B under the weight A, and freely exert a vertical downward force F3 on the rope of the pulley block. The force analysis of A and B when they move upward at a uniform speed is shown in Figure 3.

2f3 = G0+GA+Gb+FT2 = GA+Gb+F η 2 = W Useful W Total = T2HF3× 2h = T22f3pf3 = F3× vobject' = 2f3× vobject'

According to the topic, when the rising height of an object h=0.2m, then WF2=82J,

F2=WF22h=82J2×0.2m=205N，

Substituting F 1F2=404 1, we get F 1=200N.

2f2-2f1-f = G0+ga2f2 = G0+ga+f can be obtained by the difference between the two formulas, so f=F2-F 1=5N.

Simultaneous GA=355N, substituted into solution, G0=50N,

According to the equation, η 1=W useful w total =T 1hF2×2h=2F2? G02F2, η2=W Useful w Total =2F3? G02F3，F2=205N，G0=50N，

Substituting η 1: η 2 = 8 1: 82, we get F3=225N.

So pf2pf3 = 2f2vobject2f3v'' object = 2× 205n× 0.25m/S2× 225n× 0.205m/s =109.

So the answer is: 10: 9.