Question

(08年蘇州中學期中) (12分)質量為m=1kg的小物塊輕輕放在水平勻速運動的傳送帶上的P點，隨傳送帶運動到A點后水平拋出，小物塊恰好無碰撞的沿圓弧切線從B點進入豎直光滑圓孤軌道下滑。B、C為圓弧的兩端點，其連線水平。已知圓弧半徑R=1.0m圓弧對應圓心角θ=106°，軌道最低點為O，A點距水平面的高度h=0.8m。小物塊離開C點后恰能無碰撞的沿固定斜面向上運動，0.8s后經過D點，物塊與斜面間的動摩擦因數為μ₁=0.33（g=10m/s²，sin37°=0.6，cos37°=0.8）試求：

（1）小物塊離開A點的水平初速度v₁。

（2）小物塊經過O點時對軌道的壓力。

（3）斜面上CD間的距離。

（4）假設小物塊與傳送帶間的動摩擦因數為μ₂=0.3，傳送帶的速度為5m/s，則PA間的距離？

Answer 1

解析：（1）對小物塊，由A到B有

 

                                                                                                      

在B點

                                                                                                      

所以        

                                                                      

（2）對小物塊，由B到O有

 

                                                                                                      

其中

                                                                                                        

在O點

                                                                                                        

所以N=43N

 

由牛頓第三定律知對軌道的壓力為

                                                                                                        

（3）物塊沿斜面上滑：

                                                                                                        

所以

 

物塊沿斜面下滑：

                                                                                                        

由機械能守恒知

 

小物塊由C上升到最高點歷時

                                                                                                        

小物塊由最高點回到D點歷時

                                                                                                        

故

                                                                                                        

即

                                                                                                        

（4）小物塊在傳送帶上加速過程：

                                                                                                      

       PA間的距離是