题目内容
![](http://thumb.1010pic.com/pic6/res/gzwl/web/STSource/20131029195901237954954/SYS201310291959012379549004_ST/images0.png)
A.若剪断Ⅰ,则a=g,竖直向下
B.若剪断Ⅱ,则a=T2/m,方向垂直Ⅰ斜向左下方
C.若剪断Ⅰ,则a=T1/m,方向沿的延长线
D.若剪断Ⅱ,则a=g,竖直向下
【答案】分析:先研究原来静止的状态,由平衡条件求出弹簧和细线的拉力.刚剪短细绳时,弹簧来不及形变,故弹簧弹力不能突变;细绳的形变是微小形变,在刚剪短弹簧的瞬间,细绳弹力可突变!根据牛顿第二定律求解瞬间的加速度.
解答:解:A、C绳子未断时,受力如图,由共点力平衡条件得,T2=mgtanθ,T1=
.
![](http://thumb.1010pic.com/pic6/res/gzwl/web/STSource/20131029195901237954954/SYS201310291959012379549004_DA/images1.png)
刚剪断弹簧Ⅰ瞬间,细绳弹力突变为0,故小球只受重力,加速度为g,竖直向下,故A正确,C错误;
B、D刚剪短细线瞬间,弹簧弹力和重力不变,受力如图
![](http://thumb.1010pic.com/pic6/res/gzwl/web/STSource/20131029195901237954954/SYS201310291959012379549004_DA/images2.png)
由几何关系,F合=T1sinθ=T2=ma,因而a=
=
,方向水平向左,故BD错误;
故选A.
点评:本题为瞬时问题,关键要抓住弹簧弹力不可突变,细绳弹力可突变!
解答:解:A、C绳子未断时,受力如图,由共点力平衡条件得,T2=mgtanθ,T1=
![](http://thumb.1010pic.com/pic6/res/gzwl/web/STSource/20131029195901237954954/SYS201310291959012379549004_DA/0.png)
![](http://thumb.1010pic.com/pic6/res/gzwl/web/STSource/20131029195901237954954/SYS201310291959012379549004_DA/images1.png)
刚剪断弹簧Ⅰ瞬间,细绳弹力突变为0,故小球只受重力,加速度为g,竖直向下,故A正确,C错误;
B、D刚剪短细线瞬间,弹簧弹力和重力不变,受力如图
![](http://thumb.1010pic.com/pic6/res/gzwl/web/STSource/20131029195901237954954/SYS201310291959012379549004_DA/images2.png)
由几何关系,F合=T1sinθ=T2=ma,因而a=
![](http://thumb.1010pic.com/pic6/res/gzwl/web/STSource/20131029195901237954954/SYS201310291959012379549004_DA/1.png)
![](http://thumb.1010pic.com/pic6/res/gzwl/web/STSource/20131029195901237954954/SYS201310291959012379549004_DA/2.png)
故选A.
点评:本题为瞬时问题,关键要抓住弹簧弹力不可突变,细绳弹力可突变!
![](http://thumb2018.1010pic.com/images/loading.gif)
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