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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,BA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.

解析 (1)设物块B平抛运动的时间为t,由运动学知识可得

hgt2                                                                                                                                                                                                                                    

xvBt                                                                                                                      

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

μmgmaB                                                                                                                                                                                                                                

vBaBtB                                                                                                                                                                                                                                      

xBaBtB2                                                                                                                                                                                                                                          

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

Fl1Mv12                                                                                                                                                                                                                        

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

FμmgMaA                                                                                                                                                                                                                    

l2xBv1tBaAtB2                                                                                                                                                                                                            

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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