The object A can only move downward in a straight line at a uniform speed along the vertical slide, and the object A is subjected to upward friction F,

Tension F 1= 12(GA+G moving -f)-.

The object A can just move vertically upwards at the speed of v 1, and the object A is subjected to downward friction F,

Tension F2= 12(GA+G shift +F)-②

∫W = Fs，

∴F2=W2S2=82J2×0.2m=205N，

∫f 1:F2 = 40:4 1，

∴F 1=200N，

①+② Obtain:

f 1+F2 = 12(GA+G move-f)+ 12(GA+G move+f)= GA+G move。

That is, 200N+205N=355N+G motion,

∴G shift =50N,

Substitute GA=355N, F 1=200N, G =50N into ① to obtain:

f = 5N

(2) Objects A and B can just move vertically upwards at the speed v2, and object A is subjected to downward friction force F,

Tension F3 =12 (ga+GB+g+f) =12 (355n+GB+50n+5n) =12 (410n+GB).

According to the meaning of the question:

η 1=W Useful 1W Total1= (ga+f) hf22h = 355n+5n2× 205n = 360n410n = 3641,

η2=W Total number of useful 2W 2 = (ga+gb+f) hf32h = 355n+gb+5n12 (410n+gb) × 2 = 360n+gb 410n+gb,

∵η 1:η2=8 1:82,

Namely: 3641:360n+GB 410n+GB = 81:82,

Solution:

GB=40N。

Answer: (1) The friction on the weight A is 5N;

(2) The gravity of the weight B is 40N. ..