Question

两个完全相同的物体A、B，质量均为m = 0.8kg，在同一粗糙水平面上以相同的初速度从同一位置开始运动。图中的两条直线分别表示A物体受到水平拉力F作用和B物体不受拉力作用的v－t图象，求：

（1）物体A所受拉力F的大小；

（2）12s末物体A、B之间的距离S。

Answer 1

⑴F = 0.8N；⑵60m

解析:

（1）设A、B两物块的加速度为a₁、a₂，由v—t图得

                                                                                             1分

                                                                                             1分

分别以A、B为研究对象，摩擦力大小为f，由牛顿第二定律

F－f = ma₁                                                                                                                            1分

－f = ma₂                                                                                                                              1分

联立解得 F = 0.8N                                                                                                               1分

（2）设A、B在12s内的位移分别为S₁、S₂，由v—t图得

S₁ =×（4 + 8）×12m = 72m                                                                                               1分

S₂ =×6×4m = 12m                                                                                                              1分

故S = S₁－S₂ = 60m                                                                                                              1分