Physics questions in the first year of high school are urgent

Force f to do work. But friction also does work.

The kinetic energy theorem refers to the work done by the combined external force.

Answer: b, c

By the definition of work w fs, since f and s are the same, the work of force is the same, by the kinetic energy theorem: to a: w ek(a).

To B: w wf ek (B).

So the kinetic energy of A is greater than the kinetic energy of B.

In w fs, f is the resultant force! For B, there is friction.

By the definition of work w fs, since f and s are the same, the work of force is the same, by the kinetic energy theorem: to a: w ek(a).

To B: w wf ek (B).

So the kinetic energy of A is greater than the kinetic energy of B.

Answer: b, c

You ignore the problem of fB overcoming friction to do work. The answer can be obtained by using energy to keep the balance. BC is correct.

There is also friction in B to do negative work, so the total work is smaller than A, so the kinetic energy is smaller.

Choose BC, B doesn't need to explain it, C option you can calculate it with the kinetic energy theorem, w = 1 2mv -1 2mv'The initial velocity of A and B is the same, both are zero, and the acceleration of B is a=(f-f) m m, and the acceleration of A is a=f m, so the acceleration of A is greater than B, so the final velocity of A is greater than B, so the kinetic energy obtained by A is greater than that of B.

Hello! Solution: Force analysis, on the horizontal slide by the horizontal left pull force, the horizontal right friction force, the support force and gravity.

The support force perpendicular to AB, the vertical downward gravitational force, the upward pull force and the downward friction force horizontal to AB on the inclined slide.

1) From Newton's second law, derived.

f=ma acceleration on a horizontal slide is a=fm m

a=(f-pull-f) m=( meters per square second.

The acceleration on the inclined chute ab is a=f, m=(f-pull-f-mgx), m= -4 meters per quadratic second.

2) From (1), the acceleration of BC is 1 meter per square second, and the acceleration of AB is -4 meters per square second.

In the BC segment. From vt square - v0 square = 2as, obtained.

vb=10m/s

In the ab segment, it is obtained by the vt square - the v0 square = 2as.

sab=3) is obtained from (2), vb=10m s

From vt=v0+at, obtained.

tbc=10s

tab=t=tab+tbc=

(1) (2) There should be no problem, I write the answer directly and don't know how to ask 1m s. 4m s in the direction of the inclined downward.

The third question is preliminarily judged that it cannot be written according to the information of the question stem, and there is not even a distance of a paragraph, but if it is calculated as the distance given in the second question, t=

Solution: The thrust f can be orthogonally decomposed into the vertical downward component fy and the horizontal rightward component fx.

From the figure, we can see that fx cos a *f, fy sin a *f sin a 2+cos a 2 =1

Pressure fn=g+fy

The maximum static friction fmax fslip 0*fn 0*mg+ 0*fy

When fx < fmax, the wooden block does not slide.

Then cos a*f < 0*mg+ 0*sin a *fcos a*f 0*sin a *f < 0*mg (less than or equal to the sign is fine, because fmax is a critical value).

Let the angle between the force f and the horizontal plane be a

If the block does not move, the friction force is f=f*cos(a), and the downward component is f1=f*sin(a).

If the wooden block moves, the friction force is: f1=(mg+f1)*0, when f is greater than or equal to f1. No matter how big f is, the block never moves: f*cos(a) >mg+f*sina)*0

It can be calculated that a is of different sizes, and > changed to a greater than or equal sign.

Decompose f into a horizontal force f1 and a vertically downward force f2

Then f1=fcos(a),f2=fsin(a)as long as f1<=(f2+mg)*0 is satisfied, and the substitution gets: fcos(a)<=(fsin(a)+mg) 0, the judgment condition upstairs is wrong.

At the beginning of exercise fm=kx1=200*

ff=kx2=200* at constant velocity

So the kinetic friction factor =ff mg=4 (2*10)=

The maximum friction is fmax=3*10 -2*200=6n. The frictional force is equal to the tensile force when doing uniform motion, i.e., mg=kxSubstituting data yields =

Let the spring stiffness coefficient be k

Yes: f1=k

f2=k divides by the equation to get :

Find v1:v2.

1: It can not be used to measure gravity, only to measure the tension of two hands, because f=kx has no requirements for the environment in which the spring is located. 2：

Let the speed of the car be V, the supporting force of the bridge is F, and the gravity is G, then G-F=MV 2 R, bring the data into the calculation R, and then find the corresponding speed of the vehicle when F=0. 3: A is obviously wrong, from mv 2 r=m(an) to know b pair, the distance that the ball passes in time t should be vt, that is, (t root number under ran), the period of the ball in a circular motion t=2 r v=2 r under the root number ran=2 multiplied by the root number under r an [see if there is a miscalculation, hehe, do it a little fast].

1: Spring dynamometer has nothing to do with gravitational acceleration.

2: f=mv*v r when 10, 1 4mg=mv*v r when mg=mv1*v1 r compare it yourself.

First of all, you don't specify the direction of the force f, otherwise it can't be solved, you can analyze the force first, and then list the equations to solve with equilibrium conditions.

Decomposes gravity into those that are down the slope and perpendicular to the incline.

The applied force is downward along the inclined plane.

First of all, force analysis: the object is supported by gravity and friction. The gravity force is split into a downward force in the direction of the inclined plane and a downward force perpendicular to the downward force of the inclined plane by orthogonal decomposition.

Friction is equal to the friction factor multiplied by the pressure. Therefore, the friction force along the inclined plane, the sliding component of gravity and the f-trivial balance are balanced. After the force analysis is completed, the equation can be listed.

Solution: Establish a coordinate system along the center of gravity of the inclined plane, and decompose the vector g orthogonally along the x-axis of the inclined plane (decompose it into components along the direction of the coordinate axis), then the equilibrium equation in the x-axis direction has:

cos30‘g=fmin+fn 1 fmax=fn+cos30’g 2

In the y-axis direction there is:

fn=usin30'g(u is the coefficient of dynamic friction) 3 is solved by 1 and 2 and 3.

fmin，fmax

So fmin =< f=< fmax to do it yourself. Hands are tired...

The force f can be upward along the inclined plane, f f = GSIN30°

n=gcos30°

f=f size is: f=g(

f maximum is the minimum is.