题目内容
如图所示,在倾角为θ的光滑斜面上有两个用轻质弹簧相连接的物块A、B,它们的质量分别为mA、mB,弹簧的劲度系数为k,C为一固定挡板.系统处于静止状态,现开始用一恒力F沿斜面方向拉物块A使之向上运动,重力加速度为g,问:
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/201408250002078876304.png)
(1)物块B刚要离开C时,弹簧形变量为多少?
(2)物块B刚要离开C时,物块A的加速度多大?
(3)从开始到物块B刚要离开C时的过程中,物块A的位移多大?
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/201408250002078876304.png)
(1)物块B刚要离开C时,弹簧形变量为多少?
(2)物块B刚要离开C时,物块A的加速度多大?
(3)从开始到物块B刚要离开C时的过程中,物块A的位移多大?
(1)
(2)
(3)![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825000207949961.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825000207918751.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/201408250002079331026.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825000207949961.png)
试题分析:(1)当B刚要离开C时,弹簧处于伸长状态,对B根据物体的平衡条件得:
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825000207965823.png)
所以,
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825000208089903.png)
设此时A的加速度为a,对A应用牛顿第二定律有:
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825000208105999.png)
解得:
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/201408250002082141090.png)
系统静止时,弹簧处于压缩状态,对A根据物体的平衡条件,弹簧的弹力大小为:
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825000208230812.png)
则弹簧的压缩量为:
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825000208245897.png)
物块A的位移即为弹簧长度的改变量:
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/201408250002082611283.png)
![](http://thumb2018.1010pic.com/images/loading.gif)
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