(2013?昌平区二模)如图甲所示是一个装置示意图,正方体A作为配重使用...

发布网友发布时间：2024-10-22 06:05

共1个回答

热心网友时间：2024-11-22 03:29

（1）由图象可得hD=3m-2m=1m
VD=SDhD=1.2×10-2m2×1m=1.2×10-2m3，
F浮=ρ水gVD=1×103kg/m3×10N/kg×1.2×10-2m3=120N；
（2）由图象可得未被提出水面匀速上升和完全提出水面后匀速上升的功率分别为P1=6W，P2=12W；
∵P=Fv
∴P1P2=F13vF23v=F1F2=6W12W=12，
当D在水中匀速上升时，动滑轮及物体B和D整体、杠杆、定滑轮的受力分析

即3F1+F浮=GB+GD+G0-----①
FF1=4F1+2G0-----②
FE1OE=FF1OF------③
当D在空气中匀速上升时，动滑轮及物体B和D整体、杠杆、定滑轮的受力分析

3F2=GB+GD+G0---------④
FF2=4F2+2G0---------⑤
FE2OE=FF2OF--------⑥
物体A单独在地面、D在水中、D在空气中的受力分析：

GA=N0=p0SA--------⑦
GA=N1+FE1=p1SA+FE1--------⑧
GA=N2+FE2=p2SA+FE2------⑨
由①④式得3F1+F浮=3F2
再由F1F2=12，F浮=120N及上式可得：F1=40N，F2=80N；
由③⑥式可得FE1FE2=FF1FF2，
再由②⑤⑦⑧⑨及上式可得
GA?p1SAGA?p2SA=4F1+2G04F2+2G0
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