题目内容

两个完全相同的物体AB,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vt图象,求:

(1)物体A所受拉力F的大小;

(2)12s末物体AB之间的距离S

F = 0.8N;⑵60m


解析:

(1)设AB两物块的加速度为a1a2,由vt图得

                                                                                             1分

                                                                                             1分

分别以AB为研究对象,摩擦力大小为f,由牛顿第二定律

Ff = ma1                                                                                                                            1分

f = ma2                                                                                                                              1分

联立解得 F = 0.8N                                                                                                               1分

(2)设AB在12s内的位移分别为S1S2,由vt图得

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

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

S = S1S2 = 60m                                                                                                              1分

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