A question of motion and work, does a moving object necessarily be doing work externally?

The equation for doing work is f*s

f is the force and s is the displacement in the direction of the force.

The support force is always perpendicular to the direction of the ball's motion, that is, the displacement of the ball in the direction of the support force is 0f*s=f*0=0

So don't do the work!

The key is that the trolley is moving, and the cylindrical surface is moving!

If the cylindrical surface does not move, the support force and velocity are always perpendicular, so the work done is 0!

However, when the cylindrical surface moves, the combined velocity is not perpendicular to the supporting force, so the supporting force does the work!

This work can be found using the kinetic energy theorem!

This support force does not refer to the support force of the bottom to its upward, but refers to the forward or backward support force generated by its side to the ball when the car is in variable speed motion, and this support force will do work on the ball (limited to variable speed movement), if it is a uniform linear motion, this support force is 0

The work is relative, and the work may be different in different reference frames, and the ground is generally selected as the reference system.

The ball slides down while moving with the train, so it is not perpendicular to the displacement and support of the ground.

Therefore, the support force must be worked.

First of all, it is necessary to find out what are the conditions that support the work of force.

To do work is to have a displacement in the direction of the force, so do the work.

Whereas, the support force is always present but there is no displacement, so no work is done.

The question is, how does this cylindrical surface stay? Is that so?

The trolley has a displacement in the direction of the force under the action of the supporting force.

Because the cylindrical surface moves, it also changes with the support force.

Not necessarily.

If the object is moving in a straight line at a uniform velocity, then it is not subjected to force or the resultant external force is zero, and it will not exert force on other objects or the resultant force of each force is zero, then the moving object does not do work externally. Example: A wooden block moves at a constant speed on a smooth horizontal surface.

If the object is exerting a force on another object in motion, and at the same time displaces that object in the direction of the force or causes the shape of the object to change elastically, then the object does work externally. Example: A moving object is compressing a spring.

The two elements of doing work, one is force, and the other is displacement in the direction of force. Both are indispensable, so if a moving object is doing work externally, it must have a forceful effect on the outside, and this force causes other objects to be displaced along the direction of the force, and it is not necessary to happen, for example, an object moving in a straight line with uniform speed has no force and no work without contact with other objects, for example, an object in free fall, is moving, has no contact with other objects, has no external force, and has no displacement, so there is no external work. For example, there is an object at the lower end of the inclined conveyor belt, which is relatively heavy, the conveyor belt moves upward, the object does not move up, and the conveyor belt has friction on the object, but the object has no displacement, so the force of the moving conveyor belt does not do work, so it cannot be used to express this sentence, it can be had, it can be without, not necessarily.

1 Linear motion, because of uniform motion, after equal time t, the displacement (distance) of A x1 = v A t, B's, x2 = v B t, so x1: x2 = v A: v B = 4:

1, and f1:f2=3:1, so the work ratio is w1:

w2=f1x1:f2x2=(3 4):(1 1)Royal sedan =12:

12 A I read the question stem wrong, hehe, I'm sorry, the question stem asks about the work done by people"Labor saving is not labor-saving"It is said to the object, that is, whether it is used or not, the work done for the object is the same, but for the person, it is not, because 70% of the work done by the acorn for people is the useful work, that is, the work done by lifting the body of the object like a meter, then, assuming that the work done by the person is w, so w = 10mgh 7, so it is better than lifting the hand to do the work w'=mgh does more work.

Direct second question (see upstairs for the first question): In these two cases, the work done by people on the object is the same, because they all lift the same object 2 meters high, but the work done by people concisely and loosely is different in these two cases, and the mechanical efficiency of the block block is only 70%, and people still have to do work on the Huaihe oil tanker!

Question 2: The useful work done by the hand on the object is equal in the case of the two answers, but the efficiency of pulling directly by hand is 100 percent, that is, all the work done by people is useful work, such as the efficiency of the pulley used in the chain is 70 percent, that is, the work done by the person is 70 percent scum. In the two cases, although the useful work done is equal, the total amount of work is not equal, and the work done in the second case is 10/7 of the first case!

Set the height of the pulley from the ground to the stall letter mill h, Tan brigade is:

h(cot30°-cot60°)=s

In the process of walking from A to B, the height δh of the weight ascending the bucket is equal to the length of the rope on the right side of the pulley, that is: δH=-X, and the work done by the person on the rope is: W=mg·ΔH=mgs(-1)=1 000(-1) J=732 J

Analysis: When moving horizontally, the pulling force is perpendicular to the direction of movement, and no work is done at this time.

Answer: Work done when lifting w1=gh=2000 5=10000j

The whole process w=w1+w2=10000+0=10000j

w=fs=2000*5=10000J (no work perpendicular to the force).

When lifting, the wire rope tension is 10000J, and the wire rope tension in the whole process is 10000J

w = f * s

s is the distance traveled in the direction of the force.

The pulling force is in the vertical direction, and the pulling force does not do work when moving in the horizontal direction, so, when lifting and during the whole process, the pulling force does work 10000 joules.

This formulation is problematic in itself. The normal one should be that the person pushes it and retreats, and the hand is already off the wall at this time. The action of pushing the wall itself is instantaneous, and after leaving the wall, there is no strength to do work on people.

In the process of pushing the wall, the person is given a momentum, and the person obtains the initial velocity of the Zen, which should always be the same speed on smooth ground.

I don't know if it's moving, it's stopped, or whatever. If it stopped, it would definitely not be right. If it is only said that x is moved, then it should also be the work done by the dust wall to people, and the size is equal to the kinetic energy obtained by people, as for how much it is, the conditions given here cannot be calculated.

First of all, the push of man against the wall does not make the wall move, so man does not do work on the wall.

Secondly, the force of the person pushing the wall is f, and the size of the wall to the person is also f, in the process of this force contact, the hand of the force does not move, so it does not do work, and when the hand leaves, the force is also gone, so the wall does not do work on the person's thrust.

So what kind of force is doing the work? It is the muscles in the human body that do work on the human body. The calculation of this work cannot be calculated by w=fs, and the tremor sensitive cannot find this force and the distance traveled by the human body under the action of the force.

The general calculation method uses the kinetic energy theorem: the change of kinetic energy. Just ask for it.

The remaining water at 10m is: w = m (total) * g * h = (1 + 8) * 10 * 10 = 900j

Apply the kinetic energy theorem and holistic thinking. Assuming that there is no water leakage, the total work done by lifting water is w1=mgh, where m is the total weight of water and buckets, and h is 10 meters; Lift 10 meters of water and fall a total of 2kgThe work done on falling into the water is w2=mgh, where m=2, and the tremor code is early h=1.

To sum up, the actual work done by lifting water is w=w1-w2

The average value of the tensile force f = the weight of the barrel + the average gravity of the type of liquid of the water = 1 * 10 + (blind mu 2) * 10 = 100n

Work w=fh=90*10=900j

x(f+fcos)=fx(1+cos) 1+ 3 2)fx, which is the solution for the application point of the pulley.

If it's for point F, it's a bit more complicated. When the object moves x to the right, the point f moves x to the right, and at the same time elongates x along the extension direction, so the total work is: fx + (fcos )x, and the result is the same, where the left is the work done by the elongation, and the right is the work done by the horizontal component of f along the right shift.