题目内容

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点水平抛出后,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运动。BC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,轨道最低点为OA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,物块与斜面间的滑动摩擦因数为=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0.3,传送带的速度为5m/s,则PA间的距离是多少?

20090505

 
 


解:(1)对小物块,由AB

                                                                                                     (1分)

B                                                                                       (2分)

所以                                                                                            (1分)

   (2)对小物块,由BO

                                                              (2分)

其中                                                                        (1分)

O                                                                                (1分)

所以N=43N

由牛顿第三定律知对轨道的压力为                                              (1分)

   (3)物块沿斜面上滑:                              (1分)

所以

物块沿斜面下滑:

a2=6m/s                                                                                                        (1分)

由机械能守恒知

小物块由C上升到最高点历时                                                (1分)

小物块由最高点回到D点历时                                  (1分)

                                                                                  (1分)

                                                                                           (1分)

   (4)由题意可知小物块在传送带上加速过程:                         (1分)

PA间的距离是                                                                 (1分)

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