工人用的滑轮组匀速提升重物为1000N的货物 ,所用的拉力为600N绳子在50S内被拉下2m

G1=1000N,F=600N,L=2m
拉力F做的功：W总 = FL = 600*2 = 1200J
.拉力F的功率P = W/t = 1200/50 = 24W
1/2＜F/G1＜1,滑轮组由一个动滑轮一个定滑轮组成
H = 1/2L = 1m
.滑轮组的机械效率 = G1H/(W总)*100% = 1000*1/1200*100% ≈ 83.33%