Question

质量为M的卡车,载一质量为m的木箱,以速率v沿平直路面行驶

Answer 1

刹车后,木箱相对地面滑行距离为L+l,加速度为a1

a1=μ1mg/m=μ1g

L+l=½a1t²

t²=2(L+l)/μ1g

车受合外力F=μ2(M+m)g-μ1mg

车的加速度为a2=F/M=[μ2(M+m)g-μ1mg]/M

L=½a2t²=½[μ2(M+m)g-μ1mg]/M ×2(L+l)/μ1g=[μ2(M+m)-μ1m](L+l)/μ1M

μ1ML=[μ2(M+m)-μ1m](L+l)

μ1ML=[μ2(M+m)-μ1m]L+[μ2(M+m)-μ1m]l

μ1ML-[μ2(M+m)-μ1m]L=[μ2(M+m)-μ1m]l

(μ1-μ2)(M+m)L=[μ2(M+m)-μ1m]l

L=μ2(M+m)-μ1m]l/(μ1-μ2)(M+m)