Senior 1 physics analysis questions, seeking analysis, senior 1 physics questions

1. Placed on a horizontal tabletop weighing 100N, when subjected to a 22N horizontal force, it just starts to move - this sentence tells us that the maximum static friction force (that is, the minimum horizontal force that can make the object change from rest to motion) is 22N;

2. It can move at a uniform speed under the action of a horizontal force of 21n - this sentence tells us the sliding friction force on the object (that is, the minimum horizontal force that keeps the secondary object moving at a uniform speed);

3. Premise - when the object is at rest;

4, the force of 10N is less than the maximum static friction force experienced by the object, that is, the object exerts a horizontal force of 10N, the object is still stationary, and the friction force of the object is 10N by the principle of action and reaction force;

The force of 5,21n is still less than the maximum static friction force on the object, that is, the object exerts a horizontal force of 21n, the object is still stationary, and the frictional force of the object is 21n by the principle of action and reaction force;

The force of 5,50N is greater than the maximum static friction of the object, that is, the object exerts a horizontal force of 50N, the object begins to move, and the object is subjected to sliding friction at this time, (that is, after any object moves, the friction force is the sliding friction force), and the magnitude is 21N.

In summary, the answers are 10n, 21n, 21n;

Give it a few points. It's really not easy to type.

The frictional force between objects is divided into static friction and kinetic friction. When the object is at rest, push it hard, and the static friction should balance with this force to keep the object stationary. But static friction, there is a limit, this limit is called maximum static friction, once the force exceeds this limit, the object starts to move.

When the object starts moving, the static friction disappears and is replaced by kinetic friction, which is slightly less than the maximum static friction. The sliding friction can be calculated by the formula f= n, where f is a constant when n is constant. Usually, in high school physics questions, the maximum static friction is equal to the sliding friction (unless this knowledge point is specifically tested, such as this one).

In this problem, the force applied to 10n is less than the maximum static friction, and the object is balanced, and the friction force is equal to 10n, and the force applied to 21n is the same, which is equal to 21n

When a force of 50n is applied, the frictional force is equal to the sliding friction force 21n Note: Kinetic friction is further divided into sliding friction and rolling friction, and the sliding friction is mainly studied in high school.

When the horizontal force of 22N is applied, it just starts to move - the horizontal force of 22N at this time is just equal to the maximum static friction (the maximum static friction is slightly greater than the kinetic friction).

Then it can move at a uniform speed under the action of a horizontal force of 21N, which means that the dynamic friction force is 21N.

When 10N is applied, the object will be stationary because it is less than the maximum static friction, and its static friction (distinguished from the maximum maximum static friction) is equal to 10N, and when 21N is still less than the maximum static friction of 22N, it is still stationary, so its friction is static friction, and the magnitude is 21N.

When 50N is applied, do you say the object is moving? If the static friction is greater than the maximum, it will move. So what kind of friction does the object experience at this time?

The answer is: kinetic friction. The dynamic friction force is only related to the positive pressure (100N) and the coefficient of friction of the object on the contact surface.

It has nothing to do with the horizontal force of the object, so it is 21n. Analyzed earlier.

Give it a few points. Typing is not easy.

When the object is just moving, the force is the maximum static friction, that is, 22N, and when the object is moving at a constant speed, the frictional force and thrust are equal to 21N

When the thrust force is greater than its maximum static friction force, it is calculated as 21n.

When choosing a question like BC, you must draw a V-T diagram to dismantle the ruler and dismantle the ruler to facilitate the analysis process.

1. A [The force and acceleration of the block on the conveyor belt remain unchanged] kg 5kg [m=5kg].

n m 7cm 4cm [increase 50n elongation 2cm k = 50, original length = 11cm - 100n k].

4. mg f [stationary, force balance f=mg; Accelerated descent Sliding friction f= f].

The initial velocity is positive and the initial velocity is +2m s, and according to the title, the final velocity may be +4m s and may be -4m s

If +4m s, then v=4m s-2m s=2m s direction is the same as the initial velocity direction.

If -4m s, then v=-4m s-2m s=-6m s negative sign indicates the opposite direction to the initial velocity.

So choose ACD

Because velocity is a vector to add direction.

Velocity change u = 4-2 C

4-(-2) a d

4-2 a d

4-(-2) c

There are four situations in total.

Answer: ACD

If the initial velocity direction of ACD is positive, then V1=2M S, V2=4M S or V2=-4M S The velocity change can be V2-V1=2M S or V2-V1=-6M S.

No, when the object is at rest, the frictional force f=n; (n is the thrust) The frictional force is related to the thrust force.

f=n=100n;

When the object is in motion, the friction is related to the coefficient of friction and the positive pressure f=un

Because previously at a constant velocity, f=un=160n

So the friction produced by 240 pushes = 160n

The answers are 100n, 160 n, respectively

100 N push, the object is still in motion, the friction is the sliding friction, for 160 N is not moving, the static friction is 100 Nm, this is the balance of the two forces of 240 N push, by the friction of 160 Nm, 160 N is the sliding friction, 200 N is the maximum static friction, the maximum static friction is a little more than the sliding friction. There's nothing wrong with that. Thank you.

The horizontal force of 100N is not enough to make the trolley move, so the trolley is stationary, the horizontal force is balanced, and the friction force on the trolley is 100N

When the horizontal thrust of 240N pushes the trolley, it is greater than the maximum static friction, the trolley accelerates, and the sliding friction on the trolley is 200N

Because the maximum static friction is greater than the sliding friction, the thrust required to push a stationary object is greater than the sliding friction. In the problem (at least 200N horizontal thrust) is known, the maximum static friction of the trolley is 200N, and it is known (with a horizontal force of 160N, the trolley can move forward at a constant speed) that the sliding friction is 160N. The answer is derived from this - the first empty is 100N, and the second empty is 160N.

1)t=

2) The position of the point p is not determined, and the time of movement cannot be determined. （3）？？

If the answer is 5 forces, then the spring should be at an angle to the vertical direction.

Typical Force Balance Problems:

1) Treat p and q as a research system, and the system is balanced by forces. The forces acting on P are: gravity, the elastic force of the inclined plane (also called the support force) and the frictional force.

2) Taking Q as the research object, it is subjected to gravity and the tensile force of the spring, and the direction of action of the two is not collinear, that is, it is unbalanced, so it should also be subjected to the elastic force of P against Q, and according to the action force and reaction force, O has pressure on P.

3) Taking springs as the research object, note that this question contains an implicit condition, that is, the gravity of light springs is not counted. Because the spring is pulled by o on it, a force cannot be balanced, and it is clear that p must have a strong effect on the spring, and in the same way as (2), the spring also acts on p and there is a force on it. Spikes.

Two states of equilibrium: stationary, uniform linear motion. This is the root of solving this type of problem.

Gravity, inclined plane support force, friction force given by inclined plane, tensile force given by spring, pressure given by slider Q.

3 forces. Kai Zhichang's gravity, the support force of the vertical bevel with silver, and the elastic force (repulsion) of the spring.

Note: There is no friction here. There are 3 conditions for the existence of friction. Positive pressure, relative movement tendency, coefficient of friction (here the upper surface is smooth and therefore non-frictional).

I have done this problem, and the force is the elastic force of the spring, such as the friction of the saki bevel, and the elastic force, gravity, and slider pressure of the oblique surface.

Finally, just give the best one, and ask me if you have anything you can't do.

Amount of charge = change in magnetic flux resistance.

The magnetic flux in the circular wire loop has an initial magnitude of b*(b2-a2) and a magnitude of 0 at the end, and a change of b*(b2-a2), so the amount of charge passing through is.

b*(πb^2-πa^2)]/r;There is a question with your answer.

【Answer】ABCD.

Analysis] Since the distance between the points is uncertain, there are three situations:

If the positive charge is at infinity to the left, it has no effect on the other positive charge, and D is chosen.

The positive charge is far away from point A but the electric charge is larger, the negative charge is closer to point A and the electric charge is smaller, and the force on the other positive charge is equal and the direction is opposite, then the net force of the charge is zero, when the charge moves a small distance in the direction of point b, the influence of the positive charge on him does not change much, and the influence of the negative charge on him changes relatively largely, so the direction of the resultant force points to the left and continues to increase. If AB is very close, it is obvious that B should be chosen, and if it is very far away, it should be A.

If point A is moved slightly to the left on the basis of 2, then the direction of the net force of the other charge is directed to the right, and when the charge moves a small distance, the net force decreases until it continues to move to the right until zero, and then it increases. And the AB distance is not too large, choose C.

From the above three situations, it is possible to choose ABCD.