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
(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
练习册系列答案
相关题目

违法和不良信息举报电话:027-86699610 举报邮箱:58377363@163.com

精英家教网