A physics question in high school compulsory 2 and a topic in senior 1 physics compulsory 2

Diagonally pushed up. Resultant force fcosa- (mg-fsina) frictional force f= (mg-fsina).

Acceleration a=[fcosa- (mg-fsina)] m displacement x=1 2at 2=[fcosa- (mg-fsina)]t 2 2m

The thrust force is in time t and does work on the object.

w=fcosa[fcosa-μ(mg-fsina)]t^2/2m

There are some problems with your title, but I can barely understand it.

The first question is the most difficult, so start with the second question.

2. Friction is equal to the pressure of the object on the horizontal plane multiplied by the dynamic friction factor. Thus it is equal to the gravitational force plus or minus (I don't know if f is inclined upwards or downwards) the vertical component of f, multiplied by the friction factor.

3. In the vertical direction, it is balanced, and the net force is 0, which is equal to the horizontal component of f minus the friction force in the horizontal direction.

1. The first two questions are made which is simple, there are many ways, such as ft=mv, find v, you know the kinetic energy of the object, and the kinetic energy increased by the object is equal to the thrust work. Or according to the resultant force, find the acceleration, then find the displacement in time t, and then w=fs, find the work done.

To avoid confusion between the included angle and the acceleration, set the included angle to

1) If f is diagonally upward.

fcosθ-u(mg-fsinθ)=ma

A is obtained by s=1 2at2 to obtain s

So w=fscos gives w=[fcos -u(mg-fsin )t2]fcos 2m

2) If f is oblique downwards.

fcosθ-μmg+fsinθ)=ma

In the same way, w=[fcos - mg+fsin)t2]fcos 2m friction (1) is f= (mg-fsin) in 2) f= (mg+fsin) in the case of f

The resultant force is fcos -f

Why is it so complicated ...

The above is based on the movement of the object in the horizontal plane, I don't know whether the description in your question is moving on an inclined plane or moving on a horizontal plane and the thrust is not horizontal, I answered w according to the latter

Isn't this topic: an object with a mass m, on a rough horizontal plane, is subjected to an oblique downward thrust f at an angle to the horizontal direction to start moving from rest, and the dynamic friction factor u between the object and the ground is known, if so, it is more clear.

Because the pulling force affects the frictional force of the formula fcosa=fcosa-(fsina-mg)u

The questions are incomplete, so you can't search for them accurately.

The conditions are too unclear, whether the movement is constant or accelerating, and the distance traveled.

1. Finding the relation:

1) After the ball is launched, it will move horizontally at a constant speed, and the initial velocity should be v:

Get v=s t

2) Vertical free-fall movement:

From h=(1 2)g*t*t we get t, and then substitute t into the above equation to get v

3) When the ball just leaves the spring, the kinetic energy e=(1 2)*m*v*v(4) The desktop is smooth, and the potential energy of the spring is obtained from the conservation of energy e=e, which can be obtained from the sum up e=(mgs*s) (4h)2, from the data in the table, it can be seen that the image of x and s is a straight line, and the slope can be known, and the relationship between x and s can be obtained by drawing a straight line (ignoring individual special points), and then the result in (1) can be solved.

Input is inconvenient, good luck.

1、g=gm/r^2

g’=gm/r’^2

r'=r*root number (g g').

The central angle of any two satellites of the three geostationary satellites is 120 degrees, and the distance between any two satellites is s=2*r'sin60=2*r*root(g g')*sin60

2. From "the velocity size when it passes through the lowest point of the ring (n-1) times is 7m s, and the velocity size when it passes through the lowest point of the ring for the nth time is 5m s", it can be seen that it goes around a circle and overcomes the friction force to do work w is (49-25) m 2 = 24m 2. After another circle, because the speed has decreased, the physical and mental strength has decreased, and the pressure of the ball on the track has also decreased, and the friction force has decreased. The work done to overcome friction is w'Also decreased, it will be less than 24m2.

Thus mv 2 2=25m 2-w'

v>1

1. Let the orbital radius of the geostationary satellite be r, the mass of the satellite be m, and the mass of the earth be m

At ground level, gmm r = mg

At the satellite gmm r = mg'

R = gr g'

The orbital circumference of the satellite motion c=2 r

Since the three satellites should be equally spaced, s=c 3

Lianli can be solved, but it is not good to say, so solve it yourself.

2. Select D, let the mass of the ball be m, and the kinetic energy can be found to be reduced by ek=12m when (n-1) times and n times pass through the lowest point of the ring

Due to the decrease of velocity, the pressure between the ball and the orbit decreases, the friction decreases, and the kinetic energy of the last circle decreases should be less than ek=12m, and the kinetic energy when passing through the lowest point of the ring for the nth time is, so the kinetic energy when passing through the lowest point for the last time is greater than, it can be seen that the velocity is greater than 1m s

Static friction provides centripetal force f= n= mg

From f=f.

So mg=mw r

260cm-220cm=40cm, that is to say, the distance passed by uniform deceleration linear motion is 40cm. The final velocity is 0, which is multiplied by the formula vt -v0 = 2as, and vt = 2 is multiplied and multiplied into m), and the solution is v = root number and my g is used.

This question is a more conventional one.

First of all, the ball reaches the inclined plane and is exactly perpendicular to the inclined plane, so the angle between the velocity direction of the ball and the vertical direction is 37°

From this, it can be solved that the velocity direction at this time is: vy=v0 tan37°=20m s, then the time t=vy g=(20 10)s=2s, and then, the height h of the throwing point from the bottom of the inclined plane is divided into two parts, and the height of the demarcation is the height of the ball falling on the inclined plane.

The vertical displacement of the ball falling on the inclined plane h1 = 1 2gt 2 = 1 2 * 10 * 4m = 20m

Horizontal displacement l=v0t=15*2m=30m

The remaining section is h1 = l * tan 37 ° = 30 * 3 4 = h = h1 + h2 =

In summary, (1)t=2s

2) H=If you don't understand something, you can ask questions.

This is a parabolic problem, and the horizontal wind speed does not affect the landing time; However, it should be noted that the vertical direction is always moving at a uniform velocity of 5m s;

1) h=vt 100=5t t=20s2) v1 2=1 2+5 2 v1= 26 m s3) s=20*1=20m horizontally, moving 20m north

The wind does not affect the movement in the vertical direction, this problem can be regarded as the horizontal direction, and the vertical direction is a uniform motion.

h=1/2gt²

t=2√5s

5m s (independent of wind).

x=vt=1x2√5=2√5m

(1) The time of movement in the air is not closed to 100 5=20s

2) There are many algorithms for landing speed: Pythagorean theorem, okay: 5, 2+1, 2, and then root number, root number 11

3) Horizontal distance Time is already known, speed * time bei 1 * 20 = 20m

The bullet came to rest after being fired into the plank.

1.The work done by ff is exactly equal to the reduction of kinetic energy by 400j according to w=fsff=400j

2.All kinetic energy is converted into heat energy: q=400j

1: F direction = MV R = mg = 5n F direction = mg + n n = The force exerted by the rod on the ball is opposite to the direction of gravity of the ball, so the rod exerts an upward force on the ball, and according to the interaction force, the force exerted by the ball on the rod in the direction of the rod, i.e., the pressure, is magnitude.

2: The rope tension provides the centripetal force, then mg f to n + mg brings in f direction = mv r then root number 2 v 4

For problems like this, we can think of a spaceship as a spaceship moving on the surface of a planet, and its trajectory is a circle with a radius r, because it is moving in a circle, and in the absence of a specific question, we can see it as a uniform circular motion.

From Newton's formula of gravitation and circular motion, we can get g (m r 2) m = m (2 t) r

Reducing m is calculated as: m=(4 2 t 2g) r 3 (2 is the exponent, m is the mass of the celestial body).

From the mass formula: m= v, the mass of the planet is known m, according to the volume formula v=4 3 r 3, (the calculation process is written by yourself) to solve:

3π/t^2g

It can be seen that 3 t 2g is a fixed value, so it is also a fixed value.

So pt 2=k, k is a constant quantity.

Or derived from the formula: p=m v= (4 3* r 3)=(3) gt 2).

So pt 2 = (3) g