Question

如圖，AB 為斜軌道，與水平方向成45°角，BC 為水平軌道，兩軌道在 B 處通過(guò)一段小圓弧相連接，一個(gè)質(zhì)量為 m 的小物塊，自軌道 AB 的 A 處從靜止開(kāi)始沿軌道下滑，最后停在軌道上的 C 點(diǎn)，已知 A 點(diǎn)高 h，物塊與軌道間的動(dòng)摩擦因數(shù)為 μ = 0.5，求：物塊沿軌道 AB 段與軌道 BC 段滑動(dòng)的時(shí)間之比值 t₁ ∶t₂ ．

 

Answer 1

【答案】

【解析】設(shè)物塊沿軌道AB滑動(dòng)的加速度為a₁，由牛頓第二定律，有

mgsin45°－μmgcos45°＝ma₁．　                                                                                                                                                                                                                                

                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                         

 設(shè)物體達(dá)B點(diǎn)速率為 V_B

物體沿BC加速度a₂

從B勻減速達(dá)C，速度為零