Senior 3 Physics Questions Mechanics Everyone help 100

Answer: Perpendicular to b

Reason: If you simply ask this question (the direction of B's support force for A), you don't need to think too much about the force of the whole system, otherwise the problem will become complicated, and you will make mistakes if you don't get it right!

Since the B side is smooth and does not generate friction, the force at the point of application is decomposed to 0 in the parallel direction of BSo the force is acting in the vertical direction.

By grasping this, the problem can be simplified and the correct answer can be reached quickly. This is also a trick to solve similar problems.

This problem does not need to analyze the force situation at all, it must be perpendicular to B, because in this case, A touches a surface of B with a point (there is no friction between them), then the force between them must be perpendicular to this surface, and you can go to your teacher to verify this!

The force situation is as follows: B is subjected to a total of 5 forces: gravity, tension, pressure from A to B, support force from the ground, and friction from the ground to B (direction horizontal to right), and the 5 forces are balanced.

A is balanced by 4 forces, gravity, ground support force, B support force on A, and friction force (direction horizontal to right).

As for the direction of the frictional force, it is determined in this way: since A and B have no relative motion, then A and B can be regarded as a whole, and the tensile force as an external force has a horizontal leftward component of the whole, and this component is balanced by the frictional force, so the frictional force on A and B is to the right!

This question is made by the isolation method, isolating A and B, and analyzing the force on B, it can be seen that [B is supported by the force upward, gravity is vertically downward, the friction force is horizontal to the left and F is perpendicular B as shown in the figure, and there is the elastic force of A to B] Because B is stationary, the effect of the remaining four forces is vertically upward and balanced with gravity, and the classification is discussed as follows1, [If the component of F vertically upward is greater than the gravity of B, then the above situation is not true, because there is no friction, according to the magnitude of F, The direction of the elastic force of A to B can be from horizontal to right to vertical downward in this area, emphasizing that the principle of lever can know that F is magnified twice at the beginning of the center of gravity] 2, [if the vertical upward component of F is less than the gravity, then the elastic force of A to B is always horizontal to the right, and the magnitude is the sum of the friction force on the ground to B and the component force in the horizontal direction of F].

First of all, perpendicular to b, is certain. Because since b is smooth, a smooth object can only produce a force perpendicular to the object on the lower surface of the physical environment in which we live.

Second question:

You ask about the force on b, and you can get the final force by analyzing the gravitational force that is greater than the ground support force, and the resulting resultant force is downward.

The ground gives friction to B, horizontally to the right.

The tensile force f is greater than the pressure of a on b, resulting in a resultant force perpendicular to b obliquely to the upper left.

Perpendicular to b, (no friction at the intersection of ab!) It's moment balance, like a lever, a balance.

Due to the friction with the ground (rough ground), that is to say, AB is a lever with the grounding point as the fulcrum.

The mass of ab is uniform, assuming that the length is a and b respectively, the position of the intersection of a and b from the fulcrum of b is c, this c is changed with time, and as time continues, the angle between b and the ground becomes larger, and c also gradually increases.

d+ccosθ)²a²=c²sin²θ

Solve this equation, or differentiation, to know the specific change relationship......

The direction of the support force of B to A must be perpendicular to B.

The reason is that according to the knowledge of elastic force, the contact surface between A and B is the straight plane where B is located, and the support force must be perpendicular to the contact surface and pointed towards the supported object.

Perpendicular to b!

The reason is that the B-side is smooth and friction is impossible!

So some people say that the A-side is also smooth? Note that the point of action of the A stick is at the end of the stick, not the face!

Regarding the force on B, note that the force exerted by the ground on the B-stick can be in any direction! The ground is rough! And the b stick is good and the ground touches the end of the stick, not the face!

I will analyze the force balance of B, B is subjected to 5 forces, gravity G, tensile force F, ground support force G, ground friction force F1, A pressure on B F2. The friction of the ground is to the left, and an object can be analyzed as a particle. The force can be thought of as a common point force, and you break it down horizontally and vertically to find equilibrium.

Perpendicular to BB is subjected to 5 forces. Gravity g, pull force f, ground support force g, ground friction force f1, a pressure on b f2. Although these five forces are not at one point of action, the lines of action of all forces intersect at one point. Able to maintain b-balance.

Perpendicular to the stick B.

A b is of uniform quality.

b is subject to gravity, the support force of the ground, the elastic force between it and a, and the friction force of the ground.

Add a little friction that is colloquially flexible.

It must be perpendicular to the stick B, and the stick is subjected to five forces.

Perpendicular to b, the support force is named after the effect

The opposite side of the vertical stick point B is vertical.

Of course it's b.

Reasons they all said.

I only know that it is vertical b and I don't know why.

3g/2；√3g/6.

The force situation is briefly drawn.

It's strange that there is no direct contact between the ground and C, where does the elastic force come from, let alone the friction force, only if there is elasticity in direct contact, there can be friction if there is elasticity.

The elastic force and friction force of the ground on C are both zero, because elastic force and friction are both contact forces, and C is not in contact with the ground, so there is no elastic force and friction between them!

The elasticity of the B ball is 3 2g The friction of the B ball is 1 2g I am level, however, I can't send **, please forgive me. Pay attention to the force.

1) First find the acceleration of train B 2a*s=v2-o algebra to get 2a*200m=20*20 to get a=1m s2

The time required for car B to reach a speed of 4m s from braking to 4m s is obtained by the formula v=at

At this time, the distance traveled by car A is s=vt=4m, s*16s=64m

64m+125m=185m The distance traveled by car B: s=vt-1 2at2=192m

So be able to catch up!

Note: The first question takes into account that car B must not be hit when it reaches the speed of car A! Because car A does a constant speed, car B does slow down! It's not about slowing down to zero, it's about slowing down to the same speed! In the later time will not bump in!

2) If it does not collide, when the speed of car B is the same as the speed of car A, the distance traveled by car A +125m is equal to the distance of car B, which is equal to the critical state, and S A = 125 + V A * t

s B = s A = (20 * 20 - 4 * 4) 2 * a

a=(20-4)/t

Combining the above three formulas yields a=128 125

Since the friction force of braking remains unchanged, it is a linear motion with uniform speed, so according to V square = 2ax, a = 1m s is solved

So the distance between the two cars is x=4t+125-(square), so that x=0, so t is equal to 16 + root number 6, or 16-root number 6, and when car B is at a speed of 0, t is equal to 20s, so they collide when they are 16-root number 6.

The distance formula of the two cars remains unchanged, x=4t+125-(square), if x is greater than or equal to 0, it will not collide, so there is no solution according to the 2-yuan one-time equation, a is greater than or equal to 128 125.

Late at night, I hope to give points.

1) Will collide.

Proof: When the braking acceleration of car B a1=v 2 (2x)=1 and the speed of the two cars are equal, t=16s, x(a)=125+4*16=189mx(b)=192 x(a) will collide.

2) If they do not collide, when the speed of the two cars is equal, x(a) x(b) is equal to the critical state, x(a)=125+v a*txb=(20 2-4 2) (2a).

t=（20-4）/a

Combining the above three formulas yields a=128 125

a f2+mgsina>=fmax+f1cosa, which is obliquely split into the air, and the direction of the friction force on the ground is to the left, and the acceleration direction of f2cosafmax b must be along the inclined downward direction.

b Only f1 f2 + mgsina-fmax=ma a is removed from ground friction.

CD only removes F2 F1 still partially provides acceleration along the diagonal split, and some of the bucket marks are offset by the elastic slow collapse.

The ground friction f=f1(sina) 2 experienced by a does not change.

a If the force is removed, the object is on an inclined plane, the acceleration is downwards along the inclined plane, b when only f2 is subjected, the force is analyzed by the integral method, and the external force of the branch is to the left, then the frictional force of a is like the left.

c From the meaning of the title, it can be seen that f1 is greater than f2, and f2 is still sliding, so the friction force of a may be to the right.

d In the same way, the friction force with c will change.

Due to the smooth bevel, there is no friction between the xy and the bevel.

Taking Y as the object of study, Y is subjected to only three forces: the pressure n of the inclined plane, the force of gravity g, and the force of x on it f.

These 3 forces form a right triangle.

f=gcosθ

Just substitute it.

1 All personal opinion: the direction you refer to along the od is the direction of the resultant force of the ball, because according to the momentum theorem, the impulse of the resultant force on the object is equal to the change of the momentum of the object, and the od direction you use the velocity vector operation should be the direction of the impulse of the external force, and the billiard ball in the process of colliding with the table, the resultant force should not refer to the elastic force alone, but also the frictional force, and the direction of the resultant force of the elastic force and the frictional force is along the od direction.

And what others refer to as the elastic force should be perpendicular to the edge of the table, this statement should be correct.

b.Because the process of collision between the billiard ball and the table is not only affected by elastic force, but also by friction force, the resultant force of the two is along the OD direction.

Note: The elastic force in this topic does not include the support force, but the support force is of course also elastic force.

A cyclist starts to ride in a straight line from a standstill, and he passes a distance of 1m, 2m, 3m, and 4s in the 1st, 2nd, 3rd, and 4th s, respectively.

The instantaneous velocity of the end is.

The instantaneous rate of the end is .

c.The average speed for the first 4s is.

d.The average velocity within the 4s is.

a is false, because it is not possible to obtain the instantaneous velocity at a certain moment according to the conditions;

B false, ibid.

The average velocity of the first 4s = (1 + 2 + 3 + 4) 4 = 10 4 = so the average velocity within the 4th pair of c pairs = 4 1 = 4, so d pair.

First 4s: The period from the beginning of the first second to the end of the fourth second, which is the length of the line segment from 0 to 4 on the number axis.

Within 4s: The period from the beginning of the 4th second to the end of the 4th second, which is the length of the line segment from 3 to 4 on the number axis.

End of 4s: is the moment when the end of the 4th second, the point represented by 4 on the number line.

First of all, the title doesn't say that this is a uniform acceleration motion. So the options for "instantaneous velocity" are all wrong, because there is no instantaneous velocity at all. So the rate is also not available. But the average velocity v= x t

Therefore, v(first 4 seconds) =(1+2+3+4) 4=;

v (4th second) = 4 1 = 4m s

First 4s: refers to the period from the first second to the fourth second.

4s: refers to the beginning of the fourth second to the end of the fourth second, and the time period of 4s refers to the time of the fourth second.

Within 4s: Same as the previous 4s

Within the 4th s: Same as the 4th s

This question is variable acceleration and is not easy to find.

A is based on the average speed of the first 4s, which is not true.

b is calculated according to the uniform variable speed, so it is not correct.

c Average velocity of the total distance in the first four seconds v=s t=(1+2+3+4) 4=average velocity in d v=4 1=4

The first four seconds refer to the first 4 seconds after the start of the exercise, the 4th second refers to the fourth second, and the fourth second, 4s is the moment when the fourth second has just reached after three seconds, and the 4th second is the same as the 4s.

Draw **A:

Within the first second, within the second second, within the third second, within the fourth second.

The first four seconds refer to the overall time, i.e. 4s (1+2+3+4) 4m s= c correct.

The fourth second is the last one in the picture, just like the fourth second! ( Moment )!

The fourth second is marked in the graph, which is 1s (is the time)!! 4 1m s= d correct.

Options a and b do not tell the motion displacement in the fifth second, so they cannot be solved, and it is not yet a uniform acceleration motion!!

Explain your "first 4s, 4s, 4s, 4s", "4s".

Suppose there is a train, and the train has many sections, the first 4s: refers to a period of time, and in the case of a train, it refers to four trains, from beginning to end.

4s: also refers to a period of time, less than the first 4s, is a quarter of it, in terms of trains, it refers to the fourth train, the first three trains are not counted.

There should be no such vague concept at the end of 4s or the beginning of 4s, the end of 4s refers to that moment, and the end of the train refers to the end of the fourth train, excluding the body, and the same goes for the beginning of 4s.

Within the 4s: It is similar to the 4s, but the words are more accurate.

Now when it comes to the topic, if it is a uniform acceleration motion, and the acceleration is 2m per square second, then the distance passed in the 1s, 2s, 3s, and 4s is 1m, 3m, 5m, and 7m respectively, so it can be judged that it is not a uniform acceleration, and it is impossible to decelerate evenly, because the distance traveled is increasing, and it is also not a uniform velocity, so it cannot be judged what kind of motion it is. Look at the CD two items again, the average speed: the distance in a period of time divided by the time, so the C term: the distance in the first 4s is 1+2+3+4=10m, and then divide by 4 to get the average speed; D term: The distance in the fourth second is known, is 4m, and the time is 1s, so the speed is obtained.