There is a physics question about doing work, and there are 2 questions about doing work in physics

w=f*s, it can be seen that the work in different reference systems is different, and the reference system must also be specified when calculating the work.

Taking the ground as the reference frame, the car is displaced in the direction of the human force, and the person is doing positive work on the car.

By the way, there is no question of useful and useless work here, and it is only in the case of mechanical efficiency that the distinction between useful and useless work is necessary, and there is no such concept here.

People do negative work on the car.

The force analysis of the person in the horizontal direction shows that the friction f on the person is opposite to the thrust f of the carriage wall against him.

Because people accelerate evenly with the car, so f f

According to Newton's third law, the force of a person on the wall of a carriage is equal to the force of the wall of a carriage on a person, and in opposite directions.

Therefore, the work done by a person on a car is less than the work done by a car on a person, that is, a person does negative work on a car.

Positive work. The person in the car moves backward relative to the car, and the direction of movement of the car is consistent with the direction of the person's thrust.

There are two cases:

1 is useful for a separate carriage. Positive work.

2. It is useless to do it for the whole car

Because in a carriage with a uniform acceleration, when he pushes the wall of the carriage with his hand in the direction of speed, his feet are pushing the carriage in the opposite direction of speed

1 is useful for a separate carriage. Positive work. It's an external force.

2. For the whole car, it is useless to do it, and it is the internal force.

What he did for the carriage was useless

Because in a carriage with a uniform acceleration, when he pushes the wall of the carriage with his hand in the direction of speed, his feet are pushing the carriage in the opposite direction of speed

1 A car with a mass of 2 tons, driving at a constant speed of 72km h on a straight road for 30min, it is known that the resistance of the car is twice the weight of the car (g takes 10n kg), find: 1What is the work done by the resistance of the car?

The gravity of the car.

g = mg = 2000kg * 10n kg = 20000n drag on the car.

f = the distance traveled by the car.

s=vt=72km/h*

The work done by the resistance of the car owner.

w=fs=1200n*36000m=43200000j2.What is the power of the engine when the car is moving?

p=w/t=43200000j/1800s=24000w

First, in the stress analysis, there are three forces experienced by an object: gravity, support, and friction.

Since it is "slow" rotation, the motion of the object is considered to be moving at a uniform rate, and the kinetic energy is constant. According to the kinetic energy theorem, the sum of the work done by the external forces should be zero. Let gravity do work w1, friction do work w2, and support force do work w3, then w1+w2+w3=0.

Investigate the work done by gravity w1, which is derived from the title, w1=-2j;

Investigate the work done by friction w2, because the direction of friction is always perpendicular to the instantaneous velocity direction of the object, so w2=0;

Therefore, the support force does work w3=-w1-w2=2j.

ps: This question has changed the usual "convention" of not doing work with support. The reason why the supporting force does the work in this problem is because the direction of the supporting force is always the same as the direction of the instantaneous velocity of the object, and not the "perpendicular" that is usually encountered. ）

The force analysis shows that the block is only supported by force and gravity.

If the block goes from rest to rest, then the work done by the supporting force and the work done by gravity are equal to the gravitational potential energy increases by 2j, then the support force does the work by 2j

Gravity does negative work, and support does positive work.

The blocks on the board are always stationary relative to the plank, indicating that the direction of the support force is always in the vertical direction

The object on the wooden board can be regarded as always in equilibrium during the motion, that is, it is only supported by gravity and the wooden plank, and according to the conservation of energy, it can be known how much work the object increases to support the potential energy (in fact, the object also collects friction but does not work).

So how do you say gravitational potential energy increases?

w=p=w / 120=

You can calculate the specific data yourself, and the formula should be this.

C is also right!

When a force overcomes other forces to do work, the potential energy of this force increases, for example: just as an object overcomes gravity to do work, and the gravitational potential energy increases, which is the same reason.

You can think of option c like this: the spring overcomes the tensile force to do "negative work", (the elastic force of the spring is in the opposite direction of motion, i.e., negative work), so the elastic potential energy increases.

Regardless of whether the tensile force does positive work on the spring, or the spring overcomes the tensile force to do the work, the spring is elongated or shortened, the length of the spring changes to x, and the spring elastic potential energy = 1 2kx 2, do you know? x Regardless of positive or negative, the elastic potential energy of one square always increases.

A external force does positive work, and it is certain that the elastic potential energy of the spring increases.

The external force does negative work on the object and the object overcomes the external force to do the same work, here the external force is to do positive work, should not be said to be the spring overcoming the external force work.

The work done by the spring is negative, and the direction of the force is opposite to the direction of displacement, so it is negative work, but it does do work. c said that the spring overcame the tensile force to do work, but did not say that it was positive work.

Whether a force does positive or negative work depends on the direction of the force and the displacement (action distance), if the direction is the same, it does positive work, and if it is in the opposite direction, it does negative work. Stretch a spring of natural length, and the direction of displacement is along the direction of tension, so do positive work. In this process, the elastic potential energy of the spring increases, but the elastic force does negative work.

The work done by the spring to overcome the tensile force means that the spring does negative work, not the tensile force does negative work on the spring.

I guess only A is correct. The work of the outside makes the spring potential energy increase.

The force is always in the direction of the hypotenuse, and the magnitude remains the same, the work w = f * the distance of the rope end movement, the distance of the rope end = the hypotenuse - the height = under the root number (h 2 + s 2) - h, w = f * [under the root number (h 2 + s 2) - h].

2. The person did not work on the bucket. The force of the person on the bucket is vertically upward, and the distance traveled is horizontal, that is, the force of the person on the bucket has no effect.

According to w=f*s where s in the f direction is 0, so the work done is 0.

The other options have a distance in the direction of the force.

Do work, 2 do no work. The condition for doing work is that there is a moving distance in the direction of the force + force. 2 There is no distance of movement in the direction of the force (the direction of the force is perpendicular to the direction of movement).

1 Yes. 2 No, the direction of force is perpendicular to the direction of motion, and it is zero after multiplication.

3 Yes. 4 Yes.

The acceleration of an object is obtained from Newton's second law.

g-f=ma a=9m/s^2

2s The height at which the object falls.

h=at^2/2=18m

The work done by gravity wg=gh=180j

Work done to overcome resistance wf=fh=18j

Fall acceleration a=(

The displacement in 2s s=1 2gt 2=18m

The work done by gravity w1=mgs=180j

The resistance does work w2=-fs=-18j, and the object overcomes the resistance to do work 18j

Choose D, remember: your purpose is to have useful work, what you have to do is extra work, and the total amount you pay is the total work.

The purpose of this problem is to fish for the bucket, so it is useful to overcome the gravity of the bucket to do work; It is inevitable to carry water, so overcoming the gravity of water to do work is additional work; The sum of the work of lifting water and the work of carrying a bucket is the total work.