返回 满分5 > 高中物理试题 首页  

如图,AB 为斜轨道,与水平方向成45°角,BC 为水平轨道,两轨道在 B 处通过一段小圆弧相连接,一个质量为 m 的小物块,自轨道 AB 的 A 处从静止开始沿轨道下滑,最后停在轨道上的 C 点,已知 A 点高 h,物块与轨道间的动摩擦因数为 μ = 0.5,求:物块沿轨道 AB 段与轨道 BC 段滑动的时间之比值 t1 ∶t2

说明: 6ec8aac122bd4f6e

 

答案:
【解析】设物块沿轨道AB滑动的加速度为a1,由牛顿第二定律,有 mgsin45°-μmgcos45°=ma1.                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                             设物体达B点速率为 VB 物体沿BC加速度a2 从B匀减速达C,速度为零
推荐试题