题目内容
(12分)如图,原长分别为L1和L2,劲度系数分别为k1和k2的轻质弹簧竖直地悬挂在天花板上,两弹簧之间有一质量为m1的物体,最下端挂着质量为m2的另一物体,整个装置处于静止状态。现用一个质量为m的平板把下面的物体竖直地缓慢地向上托起,直到两个弹簧的总长度等于两弹簧原长之和,这时托起平板竖直向上的力是多少?m2上升的高度是多少?![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241220433653227.jpg)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241220433653227.jpg)
h=![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824122043599727.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824122043599727.gif)
当两个弹簧的总长度等于两弹簧原长之和时,下面弹簧的压缩量应等于上面弹簧的伸长量,设为x
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241220436461787.jpg)
对m1受力分析得:m1g= k1x+k2x①
对平板和m1整体受力分析得:
F=(m2+m)g+k2x②
①②联解得托起平板竖直向上的力F=![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824122043802753.gif)
未托m2时,上面弹簧伸长量为x1=
③
下面弹簧伸长量为x2=
④
托起m2时:m1上升高度为:h1=x1-x⑤
m2相对m1上升高度为:h2=x2+x⑥
m2上升高度为:h=h1+h2⑦
③④⑤⑥⑦联解得h=![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824122043599727.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241220436461787.jpg)
对m1受力分析得:m1g= k1x+k2x①
对平板和m1整体受力分析得:
F=(m2+m)g+k2x②
①②联解得托起平板竖直向上的力F=
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824122043802753.gif)
未托m2时,上面弹簧伸长量为x1=
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824122043817592.gif)
下面弹簧伸长量为x2=
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824122043833426.gif)
托起m2时:m1上升高度为:h1=x1-x⑤
m2相对m1上升高度为:h2=x2+x⑥
m2上升高度为:h=h1+h2⑦
③④⑤⑥⑦联解得h=
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824122043599727.gif)
![](http://thumb2018.1010pic.com/images/loading.gif)
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