As shown in the figure, the wood block P with mass m slides on the long board AB with mass m, and the long board is still on the horizontal ground. If the dynamic friction coefficient between the long board AB and the ground is μ 1, and the dynamic friction coefficient between the wood block P and the long board AB is μ 2, the friction force of the long board AB on the ground is () A. μ1MgB. μ1（m+M）gC. μ2mgD. μ1Mg+μ2mg

As shown in the figure, the wood block P with mass m slides on the long board AB with mass m, and the long board is still on the horizontal ground. If the dynamic friction coefficient between the long board AB and the ground is μ 1, and the dynamic friction coefficient between the wood block P and the long board AB is μ 2, the friction force of the long board AB on the ground is () A. μ1MgB. μ1（m+M）gC. μ2mgD. μ1Mg+μ2mg

The object m slides to the right relative to m and receives the sliding friction to the left, the size of which is: F1 = μ 2n = μ 2mg; the action of the forces is mutual, so p has the sliding friction to the right on the long board AB, so the long board AB has the tendency to slide to the right and receives the static friction from the ground to the left. According to the condition of the common point force balance, there is F2 = F1, so F2 = μ 2mg The friction force can not be calculated by F = μ 1 (M + m) g, so B is wrong, so C