Senior One Physics, Application of Ox Er, Senior One Physics About Ox Two and F F. Thanks for the a

Alas. s=(1/2)at^2

t=2,s=4

The solution yields a=2 m s 2

Force Analysis, Orthogonal Decomposition Method:

Vertical direction: mg + fsin30° = n (equilibrium condition) Horizontal direction: FCOS30°- f = ma (this equation is the application of Niu Er) auxiliary equation: f = un (sliding friction).

Three equations algebraic, synchronous:

2×10 + 1/2)f = n （1）

f (root number 3 2) -f = 2 2 (2) f = (3).

Three equations, three unknowns, solvable.

f = 9 (root number 3 + 1).

The problem of Newton's law should pay attention to force analysis, orthogonal decomposition!

Solution: Decompose the thrust f into horizontal and right component force f1 and vertical component force f2 easy to know f2 = f*sin30 f1 = f*cos30

The pressure of the object on the table is n=g+f2

The net force in the horizontal direction is f=f1-f=f1-un=f1-(g+f2)*, which is obtained by f=ma=ma=2*2=4n, that is, f1-(g+f2)*, and the answer can be removed.

Find the magnitude of the acceleration first.

a=2s/t^2=2m/s^2

Then find the magnitude of the combined external force.

f=ma=4n

Decompose the external force f into f1 in the horizontal direction and f2 in the vertical direction, then.

f1=fcos30=√3f/2

f2=fsin30=f/2

The magnitude of the resultant external force is.

3f/2-u(f2+mg)=4

3f can be solved with f.

According to the kinetic friction factor, the square of the acceleration = ug= is obtained.

7 meters per square second.

Displacement x = , end velocity v = 0, acceleration a = 7 meters per square second according to the formula v 2-v 0 2 = 2ax

v0^2=v^2-2ax

It is equivalent to about 30km h

So the car was speeding.

This is a typical example problem, and we can analyze it like this. The topic says to determine whether the car is speeding, then we need to know the original speed of the car when it is driving before braking. According to the topic, I will analyze the movement process of the truck, at the beginning is the original speed driving, and then emergency braking, the truck is doing deceleration movement at this time, assuming that this movement is a uniform deceleration motion (high school physics thinking is generally considered to be a uniform deceleration motion) at this time the acceleration direction of the car is opposite to the direction of motion, acceleration is the problem to be solved at present, we know the dynamic friction factor, we can find the resultant force of the truck when braking, and then it is not difficult to find the acceleration of the truck compared with the mass of the truck with the resultant force.

Finally, the original velocity is solved, the whole process of the truck is the movement of the original velocity v after uniform deceleration linear motion at the end of the velocity is 0, the acceleration has been solved above, and then the velocity formula: the square of the original velocity - the square of the end velocity = 2as.

This is the idea of solving the whole problem, this problem is a typical example problem, understand the solution method of this problem, and you can easily solve similar problems in the future

The wheels are locked to a stop, and the car does a uniform beam reduction motion, and the final velocity is zero vt 2-vo 2=2as

mg=ma, a=the direction of the initial velocity is opposite to the direction of the plus number of degrees, vo=

Newton's second law of motion, Newton's second law of motion

1. The content of the law: The acceleration of an object is directly proportional to the resultant external force f experienced by the object, inversely proportional to the mass of the object, and the direction of acceleration is the same as the direction of the resultant external force. From the point of view of physics, Newton's second law of motion can also be expressed as "the rate of change of momentum of an object with time is proportional to the sum of the external forces subjected to it".

That is, the first derivative of momentum to time is equal to the sum of external forces.

2 Formula: f = ma

Newton's original formula: f=δ(mv) δt (see Newton's Mathematical Principles of Natural Philosophy). That is, the rate of change of the force is proportional to the momentum of the object, which is also called the momentum theorem.

In the theory of relativity, f=ma is not true because the mass changes with velocity, while f=δ(mv) δt is still used.

3 A few notes:

1) Newton's second law is the law of instantaneous action of force. Force and acceleration are generated, changed, and disappeared at the same time.

2) f=ma is a vector equation, the positive direction should be specified when applying, where the force or acceleration is the same as the positive direction is taken as a positive value, and the negative value is taken otherwise, and the direction of acceleration is generally taken as the positive direction.

3) According to the principle of independent action of forces, when Newton's second law is used to deal with the problem of the motion of an object in a plane, the forces on the object can be orthogonally decomposed[1], and the component form of Newton's second law can be applied in two mutually perpendicular directions: fx=max, fy=may.

4.The six properties of Newton's second law:

1) Causality: Force is responsible for the generation of acceleration.

2) Vectorability: Force and acceleration are both vector quantities, and the direction of the acceleration of the object is determined by the direction of the external force on the nucleus of the object. In the mathematical expression of Newton's second law f = ma, the equal sign not only indicates that the left and right sides are equal in value, but also indicates that the direction is the same, that is, the direction of the acceleration of the object is the same as the direction of the combined external force.

3) Transientness: When the external force on the object (with a certain mass) changes suddenly, the magnitude and direction of the acceleration determined by the force should also change abruptly at the same time; When the resultant external force is zero, the acceleration is zero at the same time, and the acceleration and the resultant external force maintain a one-to-one correspondence. Newton's second law is an instantaneous law that indicates the instantaneous effect of force.

4) Relativity: There is a coordinate system in nature, in which the object will maintain a uniform linear motion or a stationary state when it is not subjected to force, and such a coordinate system is called an inertial reference system. The ground and objects that are stationary or moving in a straight line at a uniform speed relative to the ground can be regarded as inertial frames of reference, and Newton's laws are only true in inertial frames of reference.

5) Independence: Each force acting on the object can independently produce an acceleration, and the vector sum of the acceleration generated by each force is equal to the acceleration generated by the combined external force.

6) Identity: A and F correspond to a certain state of the same object.

Newton's second law states that the acceleration of an object is directly proportional to the resultant external force on the object and inversely proportional to its mass. The formula f=ma

This is the maximum static friction, when the tensile force is not greater than the maximum static friction, the maximum static friction is always equal to the tensile force.

The pulling force acts on A, then for them to move together, B must receive a frictional force in the direction of the pulling force. If there is no friction between the two, B is not forced, and A is pulled, then relative motion will occur.

The sliding friction of the iron block is equal to 4N, which is equal to the maximum static friction, when the wooden board is subjected to the force of 6N, the iron block is exactly subject to the force of 4N (you can draw the force analysis diagram), this is a critical point, when it is less than 6N, the two are relatively stationary, but there is static friction, when it is greater than 6N, the iron can only collect 4N friction at most, so that the iron will quickly move relatively, but in the process of movement, the sliding friction is equal to the friction coefficient * mass, so it is always equal to 4N

f = mg-f (relative movement trend downward, resistance upward) f = mg 8

a=f and m

So a=7g8= direction, vertically downward.

F = mg + f (relative movement trend upward, resistance downward, gravity downward, no other external force, net force down).

f=mg/8

a=f and m

a=9g/8=

1. Downward movement:

Let the weight be m; Then the gravity g = m * g, the air resistance f = 1 8ga = (g - 1 8g) m = 7 8g = (m s 2 ) is vertically downward (consistent with the direction of gravity).

2. Upward movement:

Let the weight be m; Then the gravity g=m*g, and the air resistance f=1 8ga=(g+1 8g) m =9 8g = (m s2 ) is vertically downward (consistent with the direction of gravity).

The direction is judged based on the resultant force of the object.

1. When falling, the direction of resistance is upward. Then a=10(1-1 8)=

2. When moving upward, the direction of gravitational acceleration is still downward, but the drag force is also downward at this time, so the two should be added up, that is, a=10(1+1 8)=

a=g-1 8g=7 8g= g1 8g vertically downward.

The object moves upwards a=g+1 8g=9 8g= vertically downward.

Under force analysis, the ball is smooth and frictionless.

It is supported by the inclined plane n and the trolley supports n'and gravity mg three forces action (1) considering the horizontal direction, the trolley to the right with acceleration a uniform acceleration linear motion then by ox two, the horizontal direction of the external force f = ma, this f by the inclined ball support force n horizontal component provides nsin = ma so by the question the pressure of the ball on the inclined plane = the inclined ball support force is ma sin (2) consider the vertical direction, no acceleration, the force balance vertical direction n'+ncosθ=mg

n'=mg-macotθ

The pressure of the ball on the trolley is equal to the support force of the trolley on the ball as n'=mg-macotθ

The ball is subjected to gravity mg, and the pressure n on it in the upper right direction is obliquely faced, and the pressure T on it vertically upwards is carried out by A, and the motion is uniformly accelerated to the right with A. The horizontal direction is obtained by the ox two: ncos = ma, and n = ma cos, so the pressure of the ball on the inclined plane n = n=ma cos, and the vertical direction:

nsin +t=mg, we get t=matan -mg, so the pressure of the ball on the trolley t=t=matan -mg

As can be seen from the question, ncos@ am (@就是斜面的角度) and nsin@ mg (you can do without this, but it is better to habitually mark it after the force analysis is drawn), you can get it from the first formula, n ma cos@ (this is the first void) Since the action of the force is mutual, the second void is also the same answer.

When the ball slides to point B, the centripetal force f=n-mg (the direction of the centripetal force is upward), why n>mg, because there is velocity, so the greater the h, the greater the velocity, and the greater the centripetal force required f=m(v 2 r), that is, the greater n. If you don't understand, you can ask.

The supporting force is greater than the gravitational force because the resultant force of the supporting force with the gravitational force at the lowest point provides the centripetal force, and the imagination points to the center of the circle f=(m v 2) r

The higher the height, the greater the speed.

The obvious answer is that for circular motion there needs to be a centripetal force for point b and the equation is n-mg=mv2 r, so n>mgThe higher the altitude, the greater the gravitational potential energy, and the greater the kinetic energy converted, that is, the greater the velocity to point b, so the greater n. It's a pleasure to help you graduate students in physics, and if you have any questions in the future, please feel free to ask questions.

The maximum pulling force of the rope is the same.

For the first there is m1g-t=m1a

For the second there is t-m2g=m2a

Synact a=

Solution: (g is the acceleration due to gravity).

3kg*(g-a)=1kg*(g+a).

The solution yields a=1 2 g=

The magnitude of acceleration a.

Force analysis of an object:

In the vertical direction, there is:

The first: g1-t=m1*a;

the second type: t-g2=m2*a;

Remove the t up and down.

You can find: a=1 2*g

Force analysis, orthogonal decomposition: mgsin30°+fcos30°-f=ma1f=f

fn=mg-fsin30°

The solution yields a1=10sin60°m s

After the car goes to F, mgsin30°-f = ma2f = mgcos30°

The solution yields a2=5(1-cos30°)m s

The block is put up and the friction is provided, so the resultant force is the friction force, the acceleration is the friction force divided by the mass is equal to 2, the block is put up to do a uniform acceleration movement, the acceleration is 2, the maximum speed is the speed of the conveyor belt, according to at=v, so t=v a=1s