题目内容
如图所示,轻弹簧一端固定,另一端在F=5N的拉力作用下,弹簧伸长了2cm(在弹性限度内),则该弹簧的劲度系数k等于( )
![](http://thumb.1010pic.com/pic2/upload/papers/20140611/201406110959290583761.png)
A.250N/m | B.10N/m | C.250N | D.10N |
![](http://thumb.1010pic.com/pic2/upload/papers/20140611/201406110959290583761.png)
据题可得弹簧的弹力大小 F=5N,弹簧的伸长长度为 x=2cm=0.02m
根据胡克定律F=kx,得:k=
=
N/m=250N/m
故选:A.
根据胡克定律F=kx,得:k=
F |
x |
5 |
0.02 |
故选:A.
![](http://thumb2018.1010pic.com/images/loading.gif)
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