High school physics is a compulsory subject in mechanics

When no plate supports are used, both springs are elongated.

At this time, the elastic force of the upper spring is greater than that of the lower spring.

The elastic force of the upper spring is (m1+m2)g

The elastic force of the lower spring is m2*g

When the plate slowly lifts up M2, the length of the two springs decreases at the same time, because the elastic force of the original lower spring is smaller, so the length of the lower spring must return to the original length first (at this time, the upper spring is still stretched).

Continue to lift M2 upwards, then the lower spring begins to be compressed, and when the lower spring is compressed at an appropriate distance, the sum of the lengths of the two springs is equal to the sum of the original lengths of the two springs (at this time, the upper spring is still in a stretched state).

We now analyze an intermediate situation:

First of all, we mark two springs: the upper spring is spring 1 with length L1, and the lower spring is spring 2 with length L2,.

When spring 2 returns to its original length, spring 2 has no force on both objects, and the object with mass m1 is subjected to gravity and the upward pulling force of spring 1, so the total length of spring is still elongated at this time, so to continue to support the plate upward, spring 2 will be compressed, so when the total length is equal to the sum of the original lengths of the two springs, spring 2 is in a compressive state, and spring 1 is in an elongated state.

For spring k1, there is a downward tension force of m1m2, so it has been kept in a stretched state, and for k2, when the gravity of m2 is greater than the tensile force, it is still in a stretched state.

The object glides to the right and the friction force is to the left, f=20 n

There is also a thrust f=20n

The gravitational and supporting forces are equal in magnitude and opposite in direction,。。 So the resultant force is f+f=40n to the left.

The initial velocity v0 of the object slides horizontally to the right, and the frictional force always hinders the relative motion, and the direction follows the relative motion.

The direction is opposite, so the direction of friction is to the left, and to the left is also subject to the tensile force f The direction of the resultant force is to the left: f = f + f = f + g = 20 + that is, the object does a uniform deceleration motion to the right.

It must be stopped in the middle, analyzed the person, and received the two pulling forces given to him by the upward rope, and the rent-type hail guess that the force is equal to its own gravity, and the tensile force of the rope is f, there are 2fsin37°=mg, and the angle is 37° is obtained by the triangular relationship, and f=500n

1) For A, when the object is stationary relative to the trolley, it should have the same acceleration a as the trolley, and the provider of this acceleration is the rightward elastic force of the spring d plus the tensile force of the spring c to the right, (m will only have the acceleration to the right when the car moves to the left, so d is compressed, and c is elongated) kd*x+kc*x=ma can get x=ma (kd+kc);

2) For B, when the object is stationary relative to the trolley, it should have the same acceleration a as the trolley, and the provider of this acceleration is the rightward tensile force of the spring d, and Niu San knows that the force is equal to the rightward tensile force of the spring c, kd*x1=kc*x2=ma, so x=ma kc+ma kd;

d, because I don't know how much sand it contains, it is possible that at first b has a downward movement trend along the inclined plane, and later the sand leaks out, and there is a upward movement tendency on the oblique side of the eye. There are also multiple possibilities, so there is no way to determine that there is only option D

Choose d to come, my sister will take you to analyze, I like to do this kind of thing the most

Before the sand leaks, the force analysis of B is as follows (no picture for you, you need a little patience).

The gravity mg supports the force f of the rope and the tension t of the rope, and t is equal to the gravity of the wooden block a (because it is a fixed pulley).

Let the resultant force of mg and f be f, case 1: if f=t, there is no friction. Case 2: If f is greater than t, the friction f is upward along the inclined plane. Case 3: If the f is less than t, the friction f is downward along the inclined plane.

When the sand leaks, the gravity mg and the support force f will continue to decrease (why does the support force f also become smaller?). If you have any doubts, you can add questions), then the resultant force of the two must also be constantly decreasing.

The point is that if case 1, then the friction f is increasing at this time, and the direction is downward along the slope.

In case 2, the friction is decreasing and the direction is upward along the inclined plane (the analysis of this situation is not detailed, if you have any doubts, you can add questions).

If case three, then the friction f is also increasing, and the direction is downward along the inclined plane.

Obviously d, if at the beginning the friction may be downward, the sand from the figure.

Small and medium-sized holes begin to leak out, and friction may decrease. If the friction starts to rise, the sand starts to leak out of the small hole in the figure, and the friction force may increase.

First of all, I don't know the original movement trend of A, so the friction of A may be up or down the inclined plane. If it is downward, then it decreases first and then increases, and finally the friction force goes up. If it starts to go up, then it keeps getting bigger.

So strictly speaking, there is no option, if you want to choose, choose D.

Force analysis is the lifeline, don't be afraid of trouble, draw a force analysis diagram, although it may be a little slower than others, but your question is 100% full score, and others may have zero points....

s = the square of one-half at.

S1 T1 square = half a = S2 T2 square = S3 T3 square.

T1 is greater than T2 is greater than T3

Therefore, if b is increased, the value of a (s1 t1 square) increases.

Untie; Let s be the distance.

t1 = s a1 t2 = s a2 out of 2, so v = s t

v = s (s a1 per 2 + s a2 per 2) = (a1 + a2) 2 a1a2

v= s sail t s is not a distance, it is a displacement state key, draw a triangle, the three angles are respectively to find the displacement is 3x2 r 3 t=(1 3)*t s t= bright sail 3x2 r