Question

兩個完全相同的物體A、B，質量均為m = 0.8kg，在同一粗糙水平面上以相同的初速度從同一位置開始運動。圖中的兩條直線分別表示A物體受到水平拉力F作用和B物體不受拉力作用的v－t圖象，求：
（1）物體A所受拉力F的大小；
（2）12s末物體A、B之間的距離S。

Answer 1

⑴F = 0.8N；⑵60m

（1）設A、B兩物塊的加速度為a₁、a₂，由v—t圖得
                                                                                             1分
                                                                                             1分
分別以A、B為研究對象，摩擦力大小為f，由牛頓第二定律
F－f = ma₁                                                                                                                            1分
－f = ma₂                                                                                                                              1分
聯立解得F =" 0.8N                                                                                                              " 1分
（2）設A、B在12s內的位移分別為S₁、S₂，由v—t圖得
S₁ =×（4 + 8）×12m =" 72m                                                                                              " 1分
S₂ =×6×4m =" 12m                                                                                                             " 1分
故S = S₁－S₂ =" 60m                                                                                                             " 1分