Physics questions in the first year of high school, physics questions in the first year of high scho

For the object, the displacement is the same, the sliding friction is the same, so the work done is the same, and the final velocity remains the same.

For conveyor belts, the frictional work increases when moving counterclockwise, and the increased part of the work leads to an increase in the amount of heat generated.

So the answer is correct.

Both times are subject to sliding friction. The size is the same, the direction is the same, and the displacement is the same. The work is the same. The last kinetic energy is the same. The speed is the same.

The work done by the above sliding friction force is equal to the product of the frictional force and the displacement (to the ground). This displacement is not the sliding distance relative to the pulley.

If you have any questions, feel free to ask. Please also be prompt.

Analysis: 1) The object has the property of keeping the original state of motion unchanged, which is called inertia;

2) the object that was originally moving, due to inertia, is still moving; The object that was originally at rest, due to inertia, remains at rest

Solution: When the conveyor belt is stationary, it will be subjected to friction in the horizontal direction opposite to the direction of movement, and the block will do a uniform deceleration motion, and do a flat throwing motion when leaving the conveyor belt; When the conveyor belt rotates counterclockwise, the object moves forward relative to the conveyor belt, and the direction of sliding friction is opposite to the direction of motion, and the object moves uniformly and decelerates, and when it leaves the conveyor belt, it also does a flat throwing motion, and the throwing speed is the same as that when the conveyor belt is not moving, it falls at the Q point

Therefore, please accept B, thank you.

When the object slides down from the inclined plane, when the conveyor belt is stationary, it will be subjected to friction in the horizontal direction opposite to the horizontal direction, and the object will do a uniform deceleration motion, and when it leaves the conveyor belt, it will do a flat throwing motion; When the conveyor belt is driven counterclockwise, the block moves forward relative to the conveyor belt, and the block moves evenly and decelerates in the direction of sliding friction and the direction of movement, and when it leaves the conveyor belt, it also does a flat throwing motion, and the throwing speed is the same as that when the conveyor belt is not moving, it falls at the q point. Therefore, choose B - Peishu Peiyou Education will answer for you.

From the formula w=fs, it is clear that the force and displacement in the formula must be in a straight line, and remember to decompose the direction of force and displacement into a straight line when using it.

The topic says that f A = 30, f B = 40, the two forces are perpendicular, then the displacement in the direction of the force needs to be decomposed to find out, orthogonal decomposition displacement (10 meters) to the two directions of A and B, the movement of the A direction is 6 meters, and the direction of Party B is 8 meters, and then the work of A and B is calculated to be 180j and 320j respectively.

The amount of work done by the box to overcome the friction force is numerically equal to the negative work done by the friction force on the box. In the future, when we see what force to overcome, we can directly understand that it is to find the value of the negative work done by a certain force. Here, the friction is 10n, and the displacement is still 10 meters, so there is no need to decompose, because the friction and displacement direction are the same.

Therefore, if you directly find out that the work done by friction is -100j, then the work done by overcoming friction is 100j.

To do the work of the resultant force of the box, you need to find the resultant force first, the parallelogram rule, and find that the resultant force is equal to 50n, at this time, the resultant force is in the same direction as the displacement, so you can multiply it directly to get the final result of 500j

First of all, after the resultant force of A and B is determined, the product of the component force and the distance of the thrust force exerted by A and B in the direction of motion is the work done by the force. F = 50N, F A = 18N, F B = 32N, F Mo = 10N, then W A = 18x10 = 180J, W B = 320J, W Mo = 100J, F = 50X10 = 500J

This question should be answered in reverse.

4、w4=1/2*m*v^2=2

3. The work done according to and external forces is equal to the change in the kinetic energy of the object, and w3=w4=2 is obtained

2、w2=fs=10*1=10

1. The work done by man on the apple plus the work done by gravity on the apple (negative) is equal to the change in the kinetic energy of the apple w1 = w2 + w4 = 12

It is better to teach a man to fish than to teach him to fish.

If you want to count work, you have to know what work is.

Work is the accumulation of force over distance.

i.e. f*s=w

According to this formula, it can be seen that work is only related to the force and the distance traveled in the direction of reason, and nothing else, and secondly, kinetic energy and work.

Kinetic energy is energy.

Whereas, work is the amount of change in energy.

That is, how much work you have done.

It means how much energy has increased or decreased.

So when he asks you about kinetic energy, he is actually asking you how much work you have done, you just need to write w=ek, if you only want a standard answer to this question.

I'm not going to be verbose.

If you don't understand, you can ask.

The object must move at a uniform speed, so the force is balanced.

By f(1+cosa)=mgsin + fsina-mgcos )f=mg(sin - cos) (1+cosa- sina), only the maximum value of (1+cosa- sina) is required, and the tip is 1+cosa- sina=1-root(1+ 2)sin(a-b) tanb=1

It is the hand that does the work on the object as 12j

It can be calculated in terms of functional relationships.

The amount of work done by the combined external force corresponds to the increase in kinetic energy The amount of kinetic energy increase is 1 2mv 2-0=2j, and the work done by the hand on the object is used to overcome gravity (increase gravitational potential energy) and increase kinetic energy, so w=1 2mv 2+mgh=12j

The work done by the combined external force on the object is 12j

w=mgh+1/2mv^2=10+2=12(j)

The hand does positive work on the object, and gravity does negative work on the object.

Because of the rise of 1 m, the gravitational potential energy increases, and there is kinetic energy.

Right? So where does this energy come from? It's all hand-made, so it's solved.

It is the hand that does the work on the object.

It's not calculated, it's judged according to the title.

The work done by the total external force = the amount of change in the energy of the object = the increased potential energy + the increased kinetic energy is greater than 12j

The direction of gravity is vertically downward, and the direction of elastic force (the tension of the rope) is always perpendicular to the velocity direction of the ball, and the ball is in motion, and there is a displacement along the vertical direction, therefore: gravity has work on the ball.

Whereas, the pulling force is perpendicular to the direction of velocity at all times, therefore, no work is done. Pick B

Choose b gravity to do the work.

Nothing else works.

Hee-hee......

The answer should be that the direction of movement of the b ball is always perpendicular to the elastic force! It would be different if the thin wire was replaced with a stick that was glued to the ball, and the stick could give force to the ball in any direction.

1. When kicking the ball, the player obtains kinetic energy from the ball by doing work on the football The direction of the speed of the ball kicked by the player is positive, then w =

1/2mv1&2-(-1/2mv0&2)

m=,v0=v1=10m/s

w=50j2, the analysis process is the same as above, w=1 2mv1&2

Happy 50j!

Both use the kinetic energy theorem.

Calculated as w=50j, mv0 squared - (minus mv0 squared) is calculated as w=50j, mv0 squared-0 w=50j

In addition, the formula of workmanship w=fs, s should be displaced under the action of f, the title is not reflected, so it is useless.