题目内容
如图8所示,一物体分别沿三个倾角不同的光滑斜面由静止开始从顶端下滑到底端C、D、E处,三个过程中重力的冲量依次为I1、I2、I3,动量变化量的大小依次为ΔP1、ΔP2、ΔP3,到达下端时重力的瞬时功率依次为P1、P2、P3,则有( )?
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241208348135111.jpg)
图8
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241208348135111.jpg)
图8
A.I1<I2<I2,ΔP1<ΔP2<ΔP3,P1=P2=P3 |
B.I1<I2<I3,ΔP1=ΔP2=ΔP3,P1>P2>P3 |
C.I1=I2=I3,ΔP1=ΔP2=ΔP3,P1>P2>P3 |
D.I1=I2=I3,ΔP1=ΔP2=ΔP3,P1=P2=P3 |
B?
由运动学公式s=
at2,可得滑到底端的时间![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241208349222483.gif)
显然t1<t2<t3,所以三个过程中重力的冲量的关系是I1<I2<I3;又根据机械能守恒定律,可得物体下滑到底端C、D、E处的速度大小相等,所以有Δp1=Δp2=Δp3;根据功率的公式P=
由于物体滑到底端重力做的功相同,所以有P1>P2>P3,故B选项正确.?
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824120834828206.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241208349222483.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241208349221545.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824120834937668.gif)
![](http://thumb2018.1010pic.com/images/loading.gif)
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