Question

兩個(gè)完全相同的物體A、B，質(zhì)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(yùn)動(dòng)。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

解析:

（1）設(shè)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(duì)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(lián)立解得 F = 0.8N                                                                                                               1分

（2）設(shè)A、B在12s內(nèi)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