如图所示,质量为M=1kg,长度l=2.5m的木板B静止在水平面上,其右端上表面紧靠一固定斜面轨道的底端(斜面底端与木板B右端的上表面之间有一段小圆弧平滑连接),轨道与水平面的夹角θ=
![](http://math.21cnjy.com/mml2svg?mml=%3Cmath+xmlns%3D%22http%3A%2F%2Fwww.w3.org%2F1998%2FMath%2FMathML%22%3E%3Cmrow%3E%3Cmsup%3E%3Cmrow%3E%3Cmn%3E37%3C%2Fmn%3E%3C%2Fmrow%3E%3Cmo%3E%C2%B0%3C%2Fmo%3E%3C%2Fmsup%3E%3C%2Fmrow%3E%3C%2Fmath%3E)
。质量为m=1kg的物块A在斜面轨道上距轨道底端x
0=
![](http://math.21cnjy.com/mml2svg?mml=%3Cmath+xmlns%3D%22http%3A%2F%2Fwww.w3.org%2F1998%2FMath%2FMathML%22%3E%3Cmrow%3E%3Cmfrac%3E%3Cmn%3E4%3C%2Fmn%3E%3Cmn%3E3%3C%2Fmn%3E%3C%2Fmfrac%3E%3C%2Fmrow%3E%3C%2Fmath%3E)
m处静止释放,一段时间后从右端滑上木板B。已知斜面轨道光滑,物块A与木板 B 上表面间的动摩擦因数μ
1=0.3,木板B与地面间的动摩擦因数μ
2=0.1,设最大静摩擦力等于滑动摩擦力,sin
![](http://math.21cnjy.com/mml2svg?mml=%3Cmath+xmlns%3D%22http%3A%2F%2Fwww.w3.org%2F1998%2FMath%2FMathML%22%3E%3Cmrow%3E%3Cmsup%3E%3Cmrow%3E%3Cmn%3E37%3C%2Fmn%3E%3C%2Fmrow%3E%3Cmo%3E%C2%B0%3C%2Fmo%3E%3C%2Fmsup%3E%3C%2Fmrow%3E%3C%2Fmath%3E)
= 0.6,cos
![](http://math.21cnjy.com/mml2svg?mml=%3Cmath+xmlns%3D%22http%3A%2F%2Fwww.w3.org%2F1998%2FMath%2FMathML%22%3E%3Cmrow%3E%3Cmsup%3E%3Cmrow%3E%3Cmn%3E37%3C%2Fmn%3E%3C%2Fmrow%3E%3Cmo%3E%C2%B0%3C%2Fmo%3E%3C%2Fmsup%3E%3C%2Fmrow%3E%3C%2Fmath%3E)
=0.8,g取10m/s
2 , 物块A可视为质点,求: