Question

工人有滑轮组提升重物（滑轮组由一动一定两个滑轮组成绳子段数n=3）,在10s内将240N的物体匀速提升2m,已知工人拉力

Answer 1

(1)P=W总/t=Fs/t=100N*2m*3/10s=60W

(2)机械效率=W有用/W总=Gh/Fs=240N*2m/(100N*2m*3)=80%

(3)动滑轮重力=W额外/s=(W总-W有用)/s=(600J-480J)/2m=60N

F=G总/n=(300N+60N)/3=120N

机械效率=W有用/W总=Gh/Fs=300N*2m/(120N*2m*3)=83.3%