题目内容
如图所示,传送带与地面倾角θ=37°,从A到B长度为16m,传送带以10m/s的恒定速率针转动,在传送带上端A处无初速度放一个质量为0.5kg的小物体,它与传送带的摩擦系数为0.5,其它摩擦不计.,已知sin37°=0.6,cos37°=0.8,g取10m/s2。求:
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/201408250026226838714.png)
(1)若传送带顺时针转动,物体由A滑到B的时间?(2)若传送带逆时针转动,物体由A滑到B的时间?
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/201408250026226838714.png)
(1)若传送带顺时针转动,物体由A滑到B的时间?(2)若传送带逆时针转动,物体由A滑到B的时间?
(1)4s (2)2s
试题分析:(1)若传送带顺时针转动,则物体沿传送带加速下滑
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/201408250026226981282.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/201408250026227141123.png)
(2)若传送带逆时针转动,物体放上传送带后,开始一段时间t1内做初速度为0的匀加速直线运动,小物体受到沿斜面向下的摩擦力,可知,物体所受合力F合=mgsinθ+f,又因为f=μN=μmgcosθ
所以根据牛顿第二定律可得:
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825002622730934.png)
当物体速度增加到10m/s时产生的位移
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825002622745804.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825002622761594.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/201408250026227761296.png)
匀速运动的位移为16-x,设所用时间为t′,则16-x=11=vt′+
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825002622792339.png)
解得:t′=1s或-11s(舍去) t总=1s+1s=2s
![](http://thumb2018.1010pic.com/images/loading.gif)
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