(1)由图可知:n=2, 绳子自由端移动的距离: s绳=ns物=2×6m=12m,
拉力F1做的功:WF=F拉s绳=200N×12m=2400J
P= WF/ t= 2400/30S=80W
(2) η=W有/W总=
G1h/F1S绳=G1/nF=360N/2×200N=90%
(3) 不计绳重和摩擦
F1=1(G1+G动)/2,
∴动滑轮重: G动=2F1-G1=2×200N-360N=40N,
F2=1(G2+G动)/2=(600N+40N)/2=320N
S2=2h2=2×10m=20m
W2=F2s2=320N×20m=6400J