A long board with mass of M and length of L is placed on a smooth horizontal plane, and a small block with mass of M is placed at the rightmost end of the board Now add a horizontal constant force F to the right end of the long board, so that the long board is pulled out from under the small block, and the dynamic friction coefficient between the small block and the long board is μ, so as to find the work done by the horizontal constant force F to pull out the long board

A long board with mass of M and length of L is placed on a smooth horizontal plane, and a small block with mass of M is placed at the rightmost end of the board Now add a horizontal constant force F to the right end of the long board, so that the long board is pulled out from under the small block, and the dynamic friction coefficient between the small block and the long board is μ, so as to find the work done by the horizontal constant force F to pull out the long board

Small block acceleration A1 = μ G
Long board acceleration A2 = (F - μ mg) / M
Board displacement S = relative displacement L + small block displacement = L + 1 / 2a1t ^ 2
Board displacement S = 1 / 2a2t ^ 2
F work done = FS
From the solutions of the above equations, it can be concluded that
W=FL（F-μmg）/（（F-μ（m+M）g）