题目内容
(10分)如图甲所示,足够长的光滑平行金属导轨MN、PQ所在平面与水平面成30°角,两导轨的间距l=0.50m,一端接有阻值R=1.0Ω的电阻。质量m=0.10kg的金属棒ab置于导轨上,与轨道垂直,电阻r=0.25Ω。整个装置处于磁感应强度B=1.0T的匀强磁场中,磁场方向垂直于导轨平面向下。t=0时刻,对金属棒施加一平行于导轨向上的外力F,使之由静止开始沿斜面向上运动,运动过程中电路中的电流随时间t变化的关系如图乙所示。电路中其他部分电阻忽略不计,g取10m/s2,求:
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/201408250001220544297.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/201408250001220693344.png)
(1)4.0s末金属棒ab瞬时速度的大小;
(2)4.0s末力F的瞬时功率。
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/201408250001220544297.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/201408250001220693344.png)
(1)4.0s末金属棒ab瞬时速度的大小;
(2)4.0s末力F的瞬时功率。
(1)2m/s (2)1.9W
试题分析:(1)由图乙可得:
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825000122085783.png)
根据
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825000122101616.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825000122116555.png)
联立(1)-(3)得:
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/201408250001221321059.png)
(2)由
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825000122163708.png)
由运动规律:
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825000122179420.png)
得:金属棒加速度
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825000122179679.png)
对金属棒受力分析,并由牛顿运动定律:
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825000122194915.png)
其中:
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825000122210593.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825000122225458.png)
联立(1)(6)(7)(8)(9)得:
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825000122257525.png)
![](http://thumb2018.1010pic.com/images/loading.gif)
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