题目内容
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824115555856278.gif)
(1)小球的加速度多大?
(2)若2秒后撤去外力F,求小球向上离出发点的最远距离。
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241155558872941.jpg)
(1)2.5m/s2(2)
m
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824115555887263.gif)
(1)受力分析如图:f=μ (F-mg ) cosθ(2分)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241155559023053.jpg)
a=
=2.5m/s2 (4分)
( 2 ) 2s 后v=at=2.5×2=5 m/s s=
at2=5m(2分)
撤去外力a /=
=mg sinθ+μmg cos θ=7.5m /s2 (2分)
由v2=2as s2=
=5m (2分)
最大位移s=s1+s2=
m (2分)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241155559023053.jpg)
a=
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824115555918682.gif)
( 2 ) 2s 后v=at=2.5×2=5 m/s s=
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824115555934225.gif)
撤去外力a /=
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824115555949771.gif)
由v2=2as s2=
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824115555965372.gif)
最大位移s=s1+s2=
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824115555887263.gif)
![](http://thumb2018.1010pic.com/images/loading.gif)
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