题目内容
斜面AB与水平面夹角α=30°,B点距水平面的高度h="1" m,如图5-7-7所示.一个质量为m的物体,从斜面底端以初速度v0="10" m/s沿斜面向上射出,物体与斜面间的动摩擦因数μ=0.2,且物体脱离斜面以后的运动过程中空气阻力不计,求物体落到水平面时的速度vC.
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241204535732644.jpg)
图5-7-7
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241204535732644.jpg)
图5-7-7
9.65 m/s
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241204535892677.jpg)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241204535892677.jpg)
如图所示,物体落于C点时速度大小为vC,由动能定理:
①
而Wf=-μmgcos30°
②
解①、②式及VA=V0得VC=
="9.65" m/s
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824120453589696.gif)
而Wf=-μmgcos30°
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824120453604490.gif)
解①、②式及VA=V0得VC=
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824120453620664.gif)
![](http://thumb2018.1010pic.com/images/loading.gif)
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