The board with length of L = 1.6m and mass of M = 3kg is placed on a smooth horizontal plane, and the small block with mass of M = 1kg is placed on the right end of the board The friction coefficient between the block and the object is u = 0.1, and a constant force F is applied to the board horizontally to the right. If the speed of the block leaving the board is v = 2m / S2, what is the force F? (g = 10m / S2)

The small block has been accelerating, and the acceleration to the right is A0 = UMG / M = 1m / S ^ 2
Let the acceleration of the board be a
The relative acceleration is A-1
Using x = 1 / 2at ^ 2, we can get l = 1 / 2 (A-1) T ^ 2
Then t = V / A0 = 2S
Then the solution is a = 1.8m/s ^ 2
F=Ma+ma0=6.4N

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