足够长5kg木板,恒力10N的作用下向右加速运动,达3m/s时,放2kg木块于上,U=0.25.相对静止时相对滑行距离

木板M=5kg,木块m=2kg,v0 = 3m/s,μ=0.25
木板与木块之间的动摩擦力f=μmg
木块加速度a1=f1/m = μg = 0.25*10 = 2.5m/s^2,v1=a1t
木板加速度a2=(F-f)/M = (F-μmg)/M = (10-0.25*2*10)/5 = 1m/s^2,v2=v0+a2t
相对静止时v1=v2
a1t = v0+a2t
t = v0/(a1-a2) = 3/(2.5-1) = 2s
滑行距离L = S2-S1 = v0t-1/2a2t^2 - 1/2a1t^2 = 3*2-1/2*1*2^2 - 1/2*2.5*2^2 = 2m
相对静止时相对滑行距离2米