题目内容
(1)滑板受到的阻力大小;
(2)运动员匀加速下滑至底端时所需的时间。
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241155286185014.jpg)
(1)50N(2)
s
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824115528634233.gif)
(1)匀减速上滑,受力分析如图,选取沿斜面向上为正方向
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241155286962480.jpg)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241155286962426.jpg)
由
得a1=![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824115528790306.gif)
-6m/s2 ----------(1分)
由牛顿第二定律可得:
-------(2分)
解得:
50N ----------(1分)
(2)匀加速下滑,受力分析如图,选取沿斜面向下为正方向
由牛顿第二定律可得:
代入数据可得:a2=4m/s2 -------(2分)
运动员上滑的位移为:![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824115528868501.gif)
可得:x=12m -------(2分)
下滑的位移与上滑时的位移相同,设下滑的时间为t2
由
可得
=
s -------(2分)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241155286962480.jpg)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241155286962426.jpg)
由
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824115528727424.gif)
得a1=
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824115528790306.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824115528805332.gif)
由牛顿第二定律可得:
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824115528821674.gif)
解得:
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824115528821233.gif)
(2)匀加速下滑,受力分析如图,选取沿斜面向下为正方向
由牛顿第二定律可得:
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824115528852604.gif)
代入数据可得:a2=4m/s2 -------(2分)
运动员上滑的位移为:
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824115528868501.gif)
可得:x=12m -------(2分)
下滑的位移与上滑时的位移相同,设下滑的时间为t2
由
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824115528883490.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824115528992519.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824115528634233.gif)
![](http://thumb2018.1010pic.com/images/loading.gif)
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