答案
解:(1)拉力的功率P=FV=25N×3×0.2m/s=15W;
(2)f=0.1G=0.1×mg=0.1×60kg×10N/Kg=60N;
W
有用=fS
1=60N×0.5m=30J;
滑轮组绳子段数n=3
W
总=F×S
2=F×3S
1=25N×3×0.5m=37.5J;
η=
=
×100%=80%.
答:(1)拉力F做功的功率是15W.
(2)该滑轮组的机械效率是80%.
解:(1)拉力的功率P=FV=25N×3×0.2m/s=15W;
(2)f=0.1G=0.1×mg=0.1×60kg×10N/Kg=60N;
W
有用=fS
1=60N×0.5m=30J;
滑轮组绳子段数n=3
W
总=F×S
2=F×3S
1=25N×3×0.5m=37.5J;
η=
=
×100%=80%.
答:(1)拉力F做功的功率是15W.
(2)该滑轮组的机械效率是80%.