题目内容
如图所示的装置中,轻绳将A、B相连,B置于光滑水平面上,人用拉力
使B以
匀速的由P运动到Q,P、Q处绳与竖直方向的夹角分别为α1=37°,α2=53°.滑轮离光滑水平面高度
=2.4m,已知
=10
,
=20
,不计滑轮质量和摩擦,求在此过程中拉力F做的功(取sin37°=0.6,sin53°=0.8.
取
)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408242323125163695.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824232301658304.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824232302485478.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824232303967312.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824232305090405.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824232306073382.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824232306697394.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824232306073382.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824232309177273.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824232310737575.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408242323125163695.png)
101.4J
试题分析:A的速度等于绳子收缩的速度,则由运动的合成与分解可得:物体在P点时,A上升的速度V1=vsinα1=0.6m/s;物体拉到Q点时,A上升的速度V1′=vsinα2=
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824232313670432.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824232315183631.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824232316681627.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824232317991368.png)
解得:WF=mAgh′+
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824232317991368.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824232317991368.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408242323217047512.png)
本题中由于拉力是变力,故无法直接用功的公式计算,可以使用动能定理求出拉力的功;
首先由速度的合成与分解求得A上升的速度,由几何关系可求得物体A上升的高度,则由动能定理可求得拉力的功.
![](http://thumb2018.1010pic.com/images/loading.gif)
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