(1)打点计时器打下B点时木块的速度为:vB= sAC 10T = (85.73?70.18)×10?2 10×0.02 m/s=0.78m/s由△x=aT2得:a= sBC?sAB T2 ,其中T=5×0.02=0.1s带入数据解得:a=0.39m/s2.(2)木块所受摩擦力为:f=Mgμ对整体有:mg-f=(M+m)a带入数据得:μ=0.20故答案为:(1)0.78,0.39.(2)0.20.