小华用如图所示的滑轮组将重900N的物体在30s内升高9m,所用的拉力为400N,求:(1)小华拉力做功的功率 (2)该
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(1)W总=FS=400N*27m=10800J

P=W总/t=10800J/30s=360W

(2)W有=Gh=900N*9m=8100J

η=W有/W总=8100J/10800J=75%

(3)G动=G物-(1/3F拉)=400N-300N=100N ∴提升1200N的物体的拉力为(1200N/3)+100N=500N

η=W有/W总=Gh/Fs=(1200N*9m)/(500N*27m)=80%

我估计最后一问lz想到拉力还是400N