As shown in the figure, a wooden board is still on a smooth and long horizontal plane, with mass m = 4kg and length L = 1.4m; at the right end of the board, there is a small slider n with mass m = 1kg, whose size is far less than l, and the dynamic friction coefficient between the small slider and the wooden block μ = 0.4 & nbsp; (g = 10m / S2). (1) in order to make n slide down from m, we use constant force F to act on M? (2) Other conditions remain unchanged, if constant force F = 22.8n, and is always used on M, n can finally slide down from M. Q: how long does n slide on M?

As shown in the figure, a wooden board is still on a smooth and long horizontal plane, with mass m = 4kg and length L = 1.4m; at the right end of the board, there is a small slider n with mass m = 1kg, whose size is far less than l, and the dynamic friction coefficient between the small slider and the wooden block μ = 0.4 & nbsp; (g = 10m / S2). (1) in order to make n slide down from m, we use constant force F to act on M? (2) Other conditions remain unchanged, if constant force F = 22.8n, and is always used on M, n can finally slide down from M. Q: how long does n slide on M?

(1) The sliding friction force between the small slider and the board is f = μ n = μ mg, and the acceleration of the right uniformly accelerating motion of the board under the action of F is A1 = FM = μ g = 4 & nbsp; m / S2. The acceleration of the right uniformly accelerating motion of the board under the action of tension F and F is A2 = f − FM, and the condition for n to slide down from M is A2 > A1, so f > μ (M + m) g = 20 & nbsp; N (2) let the sliding time of N on M be t, when the constant force F = 22.8 & nbsp; N, the acceleration of the board is: A2 = f − FM = 22.8 − 0.4 × 104  4.7 & nbsp; m / S2, the displacement of the small slider in time t is: S1 = 12a1t2, the displacement of the board in time t is: S2 = 12a2t2, because s2-s1 = L, the solution is: 12 × 4.7 & nbsp; t2-12 × 4T2 = 1.4, the solution is: T = 2 & nbsp; s. A: (1) the size range of F is f > 20n; (2) the sliding time of N on M is 2S