Question

(08年苏州中学期中) (12分)质量为m=1kg的小物块轻轻放在水平匀速运动的传送带上的P点，随传送带运动到A点后水平抛出，小物块恰好无碰撞的沿圆弧切线从B点进入竖直光滑圆孤轨道下滑。B、C为圆弧的两端点，其连线水平。已知圆弧半径R=1.0m圆弧对应圆心角θ=106°，轨道最低点为O，A点距水平面的高度h=0.8m。小物块离开C点后恰能无碰撞的沿固定斜面向上运动，0.8s后经过D点，物块与斜面间的动摩擦因数为μ₁=0.33（g=10m/s²，sin37°=0.6，cos37°=0.8）试求：

（1）小物块离开A点的水平初速度v₁。

（2）小物块经过O点时对轨道的压力。

（3）斜面上CD间的距离。

（4）假设小物块与传送带间的动摩擦因数为μ₂=0.3，传送带的速度为5m/s，则PA间的距离？

Answer 1

解析：（1）对小物块，由A到B有

 

                                                                                                      

在B点

                                                                                                      

所以        

                                                                      

（2）对小物块，由B到O有

 

                                                                                                      

其中

                                                                                                        

在O点

                                                                                                        

所以N=43N

 

由牛顿第三定律知对轨道的压力为

                                                                                                        

（3）物块沿斜面上滑：

                                                                                                        

所以

 

物块沿斜面下滑：

                                                                                                        

由机械能守恒知

 

小物块由C上升到最高点历时

                                                                                                        

小物块由最高点回到D点历时

                                                                                                        

故

                                                                                                        

即

                                                                                                        

（4）小物块在传送带上加速过程：

                                                                                                      

       PA间的距离是