题目内容

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。

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

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