I helped solve the physics problem in the first year of high school, thank you 5

Pick B. It's very simple, because of uniform acceleration, the car does positive work on people, so people do negative work on cars.

To be more detailed, the person is subjected to forward friction (car-to-foot) and backward support (car-to-opponent), and the friction is greater than the support force because of uniform acceleration. On the contrary, according to Newton's third law, the backward friction force of man on the car is less than the forward thrust, so it is obvious that the direction of the resultant force of the person on the car is backward, and the negative work is obvious.

According to Newton's second law of motion, it can be known that the direction of the combined external force should be horizontal to the left, and the external force can only be ** in the carriage, so according to Newton's third law of motion, it can be known that the direction of the resultant force of the human force on the carriage should be horizontal to the right, and the carriage should move to the left, so people should do negative work on the carriage.

c, to the left at a constant speed, you can imagine him as being in a stationary carriage with a person pushing the carriage inside the carriage. w=fs s=0

So reactive!

a, but it is offset by the work done by the elasticity of the carriage.

There is still friction on your feet.

s is equal to force f=f one positive attack and one negative work.

A person standing on a car can't do much work no matter how hard he tries to do it, and it's impossible to move the car in one direction without an external effect.

I just can't figure out why I do the work, but the teacher is just talking about doing the right work, so it's helpless.

The thrust of the person is a positive work, and the person is a negative work on the friction of the car.

1. Analyze the state when not affected by f first:

The spring is compressed. A is supported by the elastic force of the spring FK1 inclined plane Na gravity mg from which FK1=mg*sin30°=1 2mg, at this time, the spring is compressed x1=fk1 k=1 2mg k, when b outline leaves c, f=2mg>2mg*sin30° So the spring is stretched, and AB is regarded as a whole, at this time, the whole is only supported by the inclined plane, f, gravity 2mg, so the overall acceleration a=(2mg-2mg*sin30°) 2m=1 2g is analyzed for a.

At this time, the tensile force of the spring to a is FK2=2mg-mg*sin30°-1 2mg=mg (where 1 2mg is ma).

The spring is pulled up x2=fk2 k=mg k

The intersection of the spring and a is displaced by s=1 2mg k+mg k =3 2mg k, and the object a is also displaced by s=3 2mg k

Because a does a uniform acceleration motion with an initial velocity of 0.

s=1/2at^2

3/2mg/k=1/2 * 1/2g *t^2t=√6m/k

2. Consider from the perspective of conservation of energy.

fThe work done causes the elastic potential energy of the spring to change ek=k*s 2 2 =k 2*(3 2mg k) 2=9(mg) 2 8k

aThe velocity at this time v=at=1 2g* 6m kf does the work wf= ek + 1 2mv 2 = 9(mg) 2 8k + 3gm 2 k

That's the way of thinking, because it's all mental arithmetic, so the answer may be wrong, so it's better to do the math yourself.

This is not a simple problem, first of all, you have to analyze the whole process, because there is no friction, and it is an inclined plane, the spring is compressed at first, and when b leaves, the spring is stretched again, but the question tells you that the object is moving at a uniform acceleration, and the force of the spring is a variable force, so the f-pulling force is also a variable force.

f-fk=m*a where a is the acceleration This is the analysis of a by the row of columns known from the last sentence in the question.

f=2mg and fk=k*mgsin30 k m can be approximated, so we know that the acceleration is 15 meters per second squared.

According to the kinematic formula, the displacement is the amount of compression plus the amount of tension, which is equal to the power of 1 2at, and t can be found here.

This is the first question, you understand the first question, and I will write the second question.

The stiffness coefficient is k, c is Fig. 3 3 9 A fixed baffle, the system is in a stationary state. Now use a force f along the inclined plane to pull block A to accelerate it upwards evenly, and when block B is about to leave C, the size of f is exactly 2mg. Seeking:

1) From the time F starts to the time when block B is about to leave C.

2) The work done by force f when block b is just about to leave c. Please.

Although the questions are different, the ideas can be used for reference.

The idea downstairs is correct, but there is a mistake, the length change of the spring is only mb*g*sin0 k, and the two cannot be added, the real one is wrong, the elastic deformation is only for one force, not the addition of two forces.

Do you have a membrane for this baffle?

Solution: 360km h=100m s

In the first 10s is a flat throwing motion, the 10s object descends from a distance of h=1 2gt=500m, and the speed after opening the parachute is v=gt=100m

At this time, the time to land is t1=1000 100=10s, so the object is t=10+10=20s when it is thrown to the ground, and because the speed in the horizontal direction is unchanged in the whole process, it should be at a horizontal distance somewhere, l=20s, 100m, s=2000m

This proves that the release was started at a distance of 2000m from this level.

The first 10s are a uniform acceleration in the vertical direction 1 2*10*10 2=500m speed 10*10=100m s so there is.

1500-500) 100=10s time landing A total of 20s

360km/h=100m/s 100*20=2000m

h=1 2 gt squared, free fall height h=500m, horizontal moving distance: s1=vt=100m s*10s=1000m

The next step is to do a uniform motion, landing time (1500-500) 100=10s at this time to move horizontally s2100m s*10s=1000m

s1+s2=1000+1000=2000m

The answer is: (1) 3 10 5w

The practice is as follows: (1) When the constant power of the car is started, it moves in a straight line at a uniform speed when the maximum speed is reached after 20s, at this time f = f, and the speed is v = 15m s, it can be calculated that p = fv = fv = 3 10 5w at this time

2) From the kinetic energy theorem: pt-fs=mv 2 2-0, bring in the data to obtain s=

g: lunar gravitational constant;

mgh+1/2mv^2=1/2mv^2

m*v^2/r=mg

gmm/r^2=mg

m(month)v(month) 2 r=m(month)g

Weakly ask the landlord.

The answer is v multiplied by the root number under r 5r no?

It's simple! Use good force analysis! Judge the state of exercise! Just follow the formula again!

Let the initial velocity of the object be vo

The displacement in 1s s1=vot1+1 2at 2 2=vo*1+1 2*2*1 2 vo=1m s

The displacement in 2 s2 = vot2 + 1 2at2 2 = 1 * 2 + 1 2 * 2 * 2 = 6m

The displacement in 3 s3 = vot3 + 1 2at3 2 = 1 * 3 + 1 2 * 2 * 3 2 = 12m

Displacement in the 3rd = 12m-6m = 6m

The first second: s=v0*t+ the initial velocity v0=1m s at the end of the second second v1=v0+a*t1=1+2*2=5m s, the velocity at the end of the third second v2=7m s

The displacement in the third second is s=(v1+v2)t 2=6m

No, because it is not said that the acceleration starts from a standstill, so there is an initial velocity, the average velocity of x1 = 2m s, the average velocity of vo = v - at 2 = 2-2*, the average velocity of the second 2 is 4m s, the average velocity of the 3rd second is 5m s, so the displacement of the 3rd second is 6mYour formula is that the displacement difference of each segment is 2m, so 2+2+2=6m is the same.

If n is a finite value, there is no need to integrate, and the operation after integration is no different.

I think it should be simpler to use the kinetic energy theorem, which is seen as a tensile drag to do a uniform acceleration motion, the acceleration of each section can be found, the work of each section can be found (set a mass), and finally the final kinetic energy can be found and the velocity can be calculated.

This is my calculation, if you make a mistake, please bear with me, I hope it will help you.

I directly act as n-> infinity.

a(x)=a(1+x/s)

It is obvious that it is written in this way, the simplest second-order differential equation.

Analysis: The maximum opening angle taken by the measuring instrument is 2, then the length of the corresponding transverse line segment on the ground is.

l1＝2* h * tanα

So the distance of the airplane moving in time t is l2 s l1 s (2* h * tan ).

The photographed area is rectangular).

The average velocity of the aircraft is V-flat l2 t s (2* h t * tan ).

It's actually a math problem.

The area is like a runway.

First, find the radius: r h=tan angle a, r=h*tan angle a, then s=t*v*2*h*tan angle a+pai*(h*tan angle a) square to obtain v=[s-pai*(h*tan angle a) square] (2t*h*tan angle a).

Correct answer b c

By the law of flat throwing motion.

Vertical lsin = 1 2gt 2

Horizontal LCOS = V0T

tan = gt; 2v0 t = 2v0 tan g d error c is correct.

l=2v0*v0tan gcos a error b is correct.

Horizontal distance: l=v0t, t= 2ltan g, then there is: v0=l 2ltan g, then: b, l is proportional to the square of v0, correct.

Similarly, there is: 2ltan = gt 2, l = gt 2 2tan , then there is: gt 2 2tan =, then: c, t is proportional to v0, correct: answer: b, c

Here set the height of the descent to h, the horizontal distance is s, so h=lsin, s=lcos, and h=1 2gt 2, s=vt,, the last two formulas eliminate t, there is a combination of the first two formulas have lsin = (gl 2*cos 2) (2v0), and then simplify there is l v0 2=(2sin) (gcos 2), this is a constant, that 2 means squared, it is difficult to type, hey.