Question

如图,小强使用滑轮组提升重为800N的物体A时,他对地面产生的压强为3000Pa,已知小强的体重为620N,他每只脚

Answer 1

1、

F拉=G-F支

=G-F压

=G-pS

=630N-3000Pa*2*0.02m^2

=520N

2、

F总=2*F拉

=2-520N

=1040N

G轮=F总-G物

=10400N-800N

=240N

3、

η=W有/W总*100%

=G物/F总*100%

=800N/1020N*100%

=78.4%

4、

F总'=2G

=2*620N

=1240N

F物'=F总'-G轮

=1240N-240N

=1000N

1楼的答案算错了、正确是这样的.