用定滑轮匀速提升重物,绳子自由端5秒内上升了3米,拉力F做功294焦,若滑轮重力和摩擦不计求1)拉力F为多少牛2

1个回答

  • 1)W=Pt=FS

    F=Pt/S=1000W*2s/4m=500N

    2)G动=nF-G =3*500N-1200N=300N

    3)若用此滑轮组匀速提起1800N的重物,滑轮组的机械效率为=G'h/(G'h+G动h)=G'/(G'+G动)=1800N/(1800N+300N)=85.7%