题目

如图3－2－16所示，在高出水平地面h＝1.8 m的光滑平台上放置一质量M＝2 kg、由两种不同材料连接成一体的薄板A，其右段长度l1＝0.2 m且表面光滑，左段表面粗糙．在A最右端放有可视为质点的物块B，其质量m＝1 kg，B与A左段间动摩擦因数μ＝0.4.开始时二者均静止，现对A施加F＝20 N水平向右的恒力，待B脱离A(A尚未露出平台)后，将A取走．B离开平台后的落地点与平台右边缘的水平距离x＝1.2 m．(取g＝10 m/s2)求： 图3－2－16 (1)B离开平台时的速度vB. (2)B从开始运动到刚脱离A时，B运动的时间tB和位移xB. (3)A左段的长度l2. 答案：解析　(1)设物块B平抛运动的时间为t，由运动学知识可得 h＝gt2                                                                                                                                                                                                                                     ① x＝vBt                                                                                                                       ② 联立①②式，代入数据得 vB＝2 m/s                                                                                                                 ③ (2)设B的加速度为aB，由牛顿第二定律和运动学的知识得 μmg＝maB                                                                                                                                                                                                                                 ④ vB＝aBtB                                                                                                                                                                                                                                       ⑤ xB＝aBtB2                                                                                                                                                                                                                                           ⑥ 联立③④⑤⑥式，代入数据得 tB＝0.5 s                                                                                                                   ⑦ xB＝0.5 m                                                                                                                 ⑧ (3)设B刚开始运动时A的速度为v1，由动能定理得 Fl1＝Mv12                                                                                                                                                                                                                         ⑨ 设B运动后A的加速度为aA，由牛顿第二定律和运动学的知识得 F－μmg＝MaA                                                                                                                                                                                                                     ⑩ l2＋xB＝v1tB＋aAtB2                                                                                                                                                                                                             ⑪ 联立⑦⑧⑨⑩⑪式，代入数据得 l2＝1.5 m                                                                                                                  ⑫ 答案　(1)2 m/s　(2)0.5 s　0.5 m　(3)1.5 m

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