题目内容
如图1-3-10所示,劲度系数为k2的轻质弹簧,竖直放在桌面上,上面压一质量为m的物块,劲度系数为k1的轻质弹簧竖直地放在物块上面,其下端与物块上表面连接在一起,而使物块处于静止状态,现用力拉弹簧上端A使物体缓慢运动,使下面弹簧承受物重的2/3,应该将上面弹簧的上端A竖直向上提高多大的距离?
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241159171901665.jpg)
图1-3-10
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241159171901665.jpg)
图1-3-10
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824115917221219.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824115917236252.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824115917252258.gif)
如右图,弹簧2开始时产生的弹力为F2,形变量为x2,后来产生的弹力为F2′,形变量为x2′,依题意及受力分析:
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241159172682465.jpg)
F2=mg=k2x2 ①
F2′=
mg=k2x2′ ②
由①②式得:
mg=k2(x2-x2′),即
x2-x2′=
.
弹簧1开始时处于原长,后来产生的弹力为F1′,形变量为x1′.
由F1′+F2′=mg得:
F1′=
mg=k1x1′
x1′=
.
因此A端提高
x1′+(x2-x2′)=
mg·(
+
).
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241159172682465.jpg)
F2=mg=k2x2 ①
F2′=
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824115917283224.gif)
由①②式得:
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824115917221219.gif)
x2-x2′=
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824115917299448.gif)
弹簧1开始时处于原长,后来产生的弹力为F1′,形变量为x1′.
由F1′+F2′=mg得:
F1′=
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824115917221219.gif)
x1′=
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824115917330428.gif)
因此A端提高
x1′+(x2-x2′)=
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824115917221219.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824115917236252.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824115917252258.gif)
![](http://thumb2018.1010pic.com/images/loading.gif)
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