I have some questions about the problem of mechanics, and the master comes in

Nope. Don't be cranky.

Because before reaching a constant velocity.

Because of friction, the force experienced by the object is not the same as the constant velocity, assuming a constant force.

f1-f）s=

When the velocity at a constant velocity is reached.

As in the above question, 10 meters per second and 20 meters per second.

The force becomes the same as the frictional force, and you are studying the state after that.

Just think about it that way. Because it is necessary to move in a straight line at a uniform speed, the force of pulling b (i.e., the pulling force) and the frictional force (generated by the friction of the ground) must be as large as the friction force on bThe sliding friction experienced by b sliding on the ground is fixed, which is equal to its own gravity multiplied by the friction coefficient.

Therefore, if the tensile force is small, less than the fixed friction force, B cannot be pulled, if the tensile force is large, pull B, after B slides, the tensile force and the frictional force on B are equal, then B will do uniform linear motion. If the tensile force is greater, it will either move in a straight line at a uniform speed, or it will accelerate in a straight line, and the speed of b will be faster and faster. As the question, if the ground is smooth and the friction coefficient is small, then B can be pulled with a relatively small tensile force, and if the friction coefficient is large, the tensile force will be large.

As for the speed, you are talking about the speed of the pulling force, that is, the speed of 20m s and the speed of 10m s to pull, of course, in the process of pulling, the magnitude of the pulling force is changed, that is, it is not a constant force. One more step, you don't care about this, you won't take the test... Does that make it clear?

At the moment of release, object b is only subjected to gravity and tension, and acceleration ab is in the vertical direction.

The rope is not extensible, and the acceleration of object A The component of the acceleration of object A in the vertical direction should be equal to the acceleration of object B ab.

Let the tension of the rope be t, then there is.

m2g-t=m2ab

m1aa=(m1g+t)sinα

ab=aasinα

The solution yields t=m1m2g cos m1+m2sin )

Since it slides at a constant speed along the diagonal AC, the direction of the sliding friction force is diagonally upwards along the AC, and the gravitational force can be decomposed into the component n on the perpendicular inclined plane upwards and the component f parallel to the downward slope surface'=GSIN37=15N,F=UN=UGCOs37=20N (the direction is diagonally upwards along AC).

On an inclined plane, the object is subjected to a component f of gravity parallel to the downward gravity of the inclined plane', the frictional force f diagonally upwards along AC, find f and f'The resultant force f'=9n (direction is perpendicular to ac down).

The object slides at a constant speed and is subjected to a balanced force, so f=f', f in the opposite direction.

f=9n (direction is perpendicular to ac upwards).

Under the premise of constant motion:

p = f*v for the first time, p has = 12*, i.e. p total = 6 80% = so f total =

Because it is a pulley block, the two ropes are equally stressed in the movable pulley.

So f1 = known f1:f2=3:2

This gives f2=5

In the second time: the useful tensile force of object c f=10

Object C has a constant velocity of 4V, that is, the velocity is 2m, and the mechanical rate is 60%.

So the total power of c is p=20 and 1 3

The distance of the moving pulley is doubled, and half of the force is saved.

Then the power of object a at this time p = p (c total power) 4 = 8 and 1 3 (w).

p=f*v

In the first time: p has = 12*, i.e. p total = 6 80% =

So f1 = known f1:f2=3:2

This gives f2 = 10 3

In the second time, object c has a constant velocity of 4 v, i.e., the velocity is 2 m s

Then the velocity of a is 3*4V=6m s

p=f2*v=20w

It's not that complicated at all, and that 60% is just a distraction, as long as you know at all times which object you are seeking to get or power, you won't have a problem.

According to the formula of doing work w=fs, the force you use when lifting water f=mg=450n, and the lifting distance is 2 meters, then the work w=450*2=900joules=

1) For object a, pull a with a horizontal tensile force f, a still remains at rest, according to the two forces are not equal, object a must be affected by another horizontal withering force in order to balance with force f, and continue to remain at rest, and this force is equal to f, the direction is opposite to f, from the question, it can be seen that this force is the friction force of rough ground against a, so there is f=fthen a is correct.

2) Since A and B are both at rest and there is no relative motion between them, the friction between objects A and B is f=0, then D is correct.

So the answer is AD

It is important to choose the right object to be forced:

When analyzing the friction between ABs, select B above, and if the friction of A is applied horizontally, it will not be balanced. So D is right.

When analyzing the friction between A and the ground, if AB is the whole, then the horizontal tension f of the closed answer is balanced by the frictional pre-friction force of the ground in the horizontal direction. It's right to bury so A.

Considering that if the horizontal direction of b and b receives friction, it will not be able to balance the world, and it will not be stationary.

The friction between A and B is 0

In the same way, the friction force between A and the ground is equal to the search segment-F

So choose D

This is the first session of high school physics!

First of all, a and b are both stationary, so they must be balanced by force, i.e., a must be subject to the friction of the ground.

If B is also stationary, it must be balanced by force, i.e., it is not affected by the force.

The answer is, AD

The ada option is very good to explain, and AB is considered as an object, so the friction is f

Option d looks at it this way, all the forces experienced by b physics in the direction of the horizontal chain are only frictional force, if there is friction, the net force is equal to the frictional force, and it will move, so the frictional force is 0

a,d；First of all, the preparation hall regards A and B as a whole, so the friction force between A and B is f=f (Newton's first law), and then A and B have no tendency to move relative to the sails, so the frictional rolling force is zero. Thank you.

ad.First of all, if there is friction between B, then B is not in contact with any other object except this force, so B is only affected by this one force, so B will not remain in equilibrium. Therefore, there is no friction between the blades of AB.

Re-analyzing A, AB is regarded as a whole affected by an external force F, and in order to maintain balance, the ground must bury an equivalent reverse friction force F to A.