P=F 1v 1
According to the uniform motion of two objects, the pulling force is equal to the friction force, that is, f1= f.
The ground friction is f = μ n = μ (m+m) g.
Alternative data: μ=0.2
(2) At the moment of 2) 1s, the object moves at a uniform speed, and the tension of the rope is equal to the static friction between the plate and the object. According to the uniform motion of the plate, it is known that the static friction between the object and the plate is equal to the sliding friction relative to the plate:
f 1=μ(M+m)g=6N
As can be seen from the figure, starting from 2s, the flat plate hits the steps and stops, so the slider starts to decelerate at a uniform speed on the flat plate, so the friction force on the slider is sliding friction force, so the friction force remains unchanged before stopping.
3s, the slider slides on the board, so the friction on the slider is sliding friction.
Because when the last piece is uniform again, the speed is:
V2 = 0.3m/s
P=F2v2
Therefore, F 1V 1=F2V2.
F2=f2= 10N
(3) The deceleration time of the stop on the flat plate is t=6-2=4s,
In the whole process, the work done by the motor is W=Pt, and the friction force is always sliding. The friction force is f2= 10N.
From the kinetic energy theorem:
pt-f2l 1 = 12mv 22- 12mv 12
L 1 = 1.84m is obtained.
The time for the slider to move uniformly on the flat plate is: t ′ =10-6 = 4 s.
Displacement of uniform motion: L2 = V2T ′ = 0.3× 4 =1.2m.
Total length of flat plate: l = l1+L2 =1.84m+1.2m = 3.04m.
Answer: (1) The dynamic friction coefficient μ between the flat plate and the ground is 0.2.
(2) At the end of1s and 3s, the friction forces acting on the stopper are 6N and 10N, respectively.
(3) The length L of the flat plate is 3.04m.