For physics problems, see Supplement Don t use the limit method .

First of all, the Q force analysis shows that F=MgCosa, with the decrease of A, F becomes larger. And the component of the force of q to p in the vertical direction, which you express by a itself, should be reduced, while the resultant force of q is always zero.

Q is analyzed for forces: Q is supported by nm fn, p on fp, and earth on its gravity mg

Let the angle between the center of O and Q and the ground be , and the angle between Onm and the ground is

Orthogonal decomposition is: FNCOS +FPSIN = MG, FNSIN = FPCOs

It can be obtained: fn=mg (sin tan +cos), fp=mg (cos tan +sin).

Since when is smaller, smaller, and the difference between and is very small, it can be approximately considered that the two are equal, so the formula can be reduced to fn=mgcos, fp=mgsin, when smaller, fn becomes larger, and fp becomes smaller, so a is true and c is wrong.

P was analyzed: FPCOS = F, FPSIN + MG = N

where n is the elastic force of the ground facing p, fp becomes smaller, becomes smaller, n naturally becomes smaller, so b is wrong.

As for d, q is considered to be at rest in each phase of the process, then the net force to which it is subjected is always 0

Single choice or multiple?

It's very simple, only a pair, and I thought it was complicated at first.

In this problem, the middle mercury column is in equilibrium, and the force analysis of the mercury column shows that there are three forces, the downward one is the gravity force and the pressure given by the gas above, and the upward one is the pressure given by the gas below. So we can clearly see that the pressure of the upper gas is less than the pressure of the lower gas, that is, the pressure of the upper gas is less than the pressure of the lower gas.

Only p2 needs to be satisfied in this question< p1 is sufficient, that is to say, it doesn't matter how small p2 is, then we take the minimum value of p2 0, which avoids the problem of considering the temperature change caused by the change of gas pressure above.

The limit method is a situation that conforms to the meaning of the topic, and can be used in three cases to determine the final situation, unchanged, up or down.

Solution: See the velocity time diagram, and the problem is very easy. The area is the displacement.

The horizontal plane should move in a straight line with uniform deceleration!

1. Calculate the average velocity, v average = (s1 + s2) t = (4 + 6) 10 = 1m s

2. Calculate the maximum velocity, v = v average * 2 = 1 * 2 = 2m s3, calculate the time of inclined plane and horizontal motion, t1 = s1 v mean = 4 1 = 4s t2 = s2 v mean = 6 1 = 6s

4. Calculate the acceleration, a1=(v-0) t1=2 4= a2=(0-v) t2=-2 6=-1 35m s

Your title is contradictory; ''Then it does a uniform linear motion on the horizontal plane until it stops''