如图是某工人用滑轮组提升重物，工人用360N拉力将重为600N的重物匀速提升10m，需要时间10s．请根据提供的信息，计算五个相关的物理量．（g取10N/kg）_刷题宝

题干

如图是某工人用滑轮组提升重物，工人用360N拉力将重为600N的重物匀速提升10m，需要时间10s．请根据提供的信息，计算五个相关的物理量．（g取10N/kg）

上一题下一题 0.0难度选择题更新时间:2017-11-22 02:32:53

答案(点此获取答案解析）

解：根据题意和已知条件，可以求出绳端移动的距离、有用功、总功、额外功、机械效率．（1）由题目图中可以看出有两段绳子在拉重物，故绳端移动的距离：s=2h=2×10m=20m．（2）有用功为：W_有用=Gh=600N×10m=6000J；（3）总功为：W_总=FS=F×2h=360N×20m=7200J；（4）额外功为W_额外=W_总﹣W_有用=7200J﹣6000J=1200J；（5）机械效率为：η= {#m

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