Friction physics problems in junior high school!! A few problems of physical friction in the second

The direction of friction is opposite to the direction of relative motion.

In the beginning, the conveyor belt moves to the right, it does not move, and the object moves to the left relative to the conveyor belt, so the friction of the conveyor belt to its right is to the right.

Later, if the conveyor belt is horizontal, it moves at a uniform speed with the conveyor belt, there is no relative motion, so it is not subject to friction.

When two objects in contact with each other are in relative motion, a force is generated on the contact surface that hinders the relative motion, and this force is the frictional force. So, the direction of friction is opposite to the relative direction of motion of the object.

At the beginning, the object moves to the left relative to the conveyor belt, so the direction of the friction force of the conveyor belt on it is to the right.

Later, when it moves at a uniform speed with the conveyor belt, the object is stationary relative to the conveyor belt and has no relative motion or tendency to move relatively, so there is no friction between the conveyor belt and the object. (Ignoring air resistance, and the belt is horizontal).

The object moves to the left with respect to the conveyor belt, so the frictional force received is to the right. When the two are at the same velocity, there is no relative motion between the two, so there is no frictional force.

False is determined by the frictional force of the two objects.

B True C False A straight line with a constant velocity is subject to a balanced force or is not subject to a force. Thrust can be balanced with friction and friction with air.

2a needs to be otherwise, the object can be decomposed by gravity into components along the inclined plane, which will form a three-force balance and ** the problem does not match.

b needs to be otherwise, the pulling force is not equal to the indicator.

c. Otherwise, the force is unbalanced. Friction is not equal to the indication.

dNo, you don't. Because the theme is between planks and planks.

3 The object makes a straight line with a uniform velocity.

The frictional force is equal to the thrust force.

Thrust = 200n when the object is moving at a constant speed

The frictional force of the object after motion is constant 200n

Friction is reduced by sliding friction.

b also reduces friction by sliding friction.

c. Increase friction.

dReduce friction.

I'm a first-year science student.

To overcome gravity and do work is to look at the height of the object rising vertically and calculate the work. w=gh=600n*3m=1800j。

The work done by thrust is the distance of the object extending the inclined plane under the action of thrust, and the thrust is calculated first. Since it is a constant upward thrust, the thrust is equal to the gravity extending the inclined plane component, and from the height and length of the inclined plane, we can calculate that the thrust is 600n*(3 5)=360n.

Thrust work w=360n*5m=1800Because frictional resistance is not considered, we can see from the final result that work done by overcoming gravity = work done by thrust! If you understand, please adopt, if you don't understand, please ask!

The first null, w useful = gh = 600n * 3m = 1800j

Third, W total = fs, f = w s = 1800 5 = 360n

Second, W total = FS = 360 * 5 = 1800J

The work done to overcome the gravity of the heavy object is converted into gravitational potential energy, mgh = 600 * 3 = 1800 due to the uniform speed pushed to the top, the kinetic energy is considered to be unchanged, the thrust workmanship is completely converted into potential energy, the thrust work is 1800, the force of thrust and gravity along the inclined plane are balanced, the triangle of the decomposition force is similar to the triangle formed by the inclined plane, and the thrust = 600 * 3 5 = 360N

Work done to overcome gravity of heavy objects w=gh=600*3=1800j

Thrust to do work w push = w = 1800j

The thrust added is f=1800 5=360n

Overcoming gravity w = gh = 600 * 3 = 1800j

Thrust does work w=1800j, because it can only be calculated without considering friction, otherwise w=fs should be used

The thrust f=ws=1800 5=360n

Work done by overcoming gravity of heavy objects: w1 = gh = 600n 3 m = 1800j

The work done by thrust is equal to the work done by gravity: w2 = w1 = 1800j (because the object moves at a uniform speed).

by w2=fxs, f=w2, s=1800j, 5m=360n

The work done by gravity is w=-mgh=-600*3=-1200j; The work done to overcome the gravity of a heavy object is 1200j. The thrust work done is 1200j.

fs=1200, f=1200 s=1200 5=240n, and the thrust added is 240 n.

Personally, I don't think the conditions of the question are clear.

1。Traction.

2。Rolling friction is on the surface of an object.

3。The saw edge can be widened, and the friction of the effective material on the saw When the saw is sawed, both sides will be close to the saw, and the pressure will definitely be larger: and when the sawing is small, this pressure is much smaller or even none, so that the sliding friction is smaller.

PS: Point A will have, point B will not [short circuit state].

It can widen the saw mouth and see the friction of the effective material on the saw.

1.Rear-wheel drive, ground hinders the wheels from turning backwards, and ground friction against the car forward. The front wheels hinder the ground backwards, and the ground friction against the car is backwards.

2.Sliding friction: The friction of two contact surfaces with different speeds or directions.

Rolling friction: The friction of the velocity of two touching objects at the same magnitude and direction.

I sharpen a pencil and turn it, and let's not move the blade inside.

3.After the sawtooth is tilted to both sides, the saw mouth is widened, and the friction resistance of the saw blade and the saw mouth is reduced, so that the saw blade can be pulled back and forth, and the width of the sawed position is increased, which is conducive to the front and rear expansion and contraction of the hacksaw, and the pressure on the hacksaw is reduced, so as to reduce the friction.

I don't think there will be a signal, no potential difference.

For a bicycle moving at a constant speed, the rear wheel is subjected to 3 forces: gravity, support, and directionally forward friction. The front wheels are also subjected to 3 forces: gravity, support, and backward friction. The friction of the two wheels balances each other, and the two forces are balanced.

When sharpening a pencil, the blade slides on the pencil, so it is sliding friction. (Try this for yourself).

The purpose of the saw's staggering teeth is to increase the gap between the sawing and the thing, so as to reduce the pressure between the object and the friction.

Point A has, point B doesn't.

1. There is also the power given by the engine, which can just overcome the resistance of progress 2. It is sliding friction, because the point where the pencil contacts the pencil sharpener is always in contact with the pencil sharpener, which is nothing more than sliding friction on the curved surface.

3. Increase the coefficient of friction.

4. A has, B does not.

1. The front and rear wheels are indeed subject to gravity, support force, friction, but the friction of the front and rear wheels is not the same, the rear wheel is the driving wheel, it is subject to the forward friction, it is this force that gives the car the power to move forward, the front wheel is pushed forward, and the friction force is backward, so the car as a whole is balanced by two frictional forces.

2. Sliding friction is the force that hinders the relative motion of two objects when they slide relative to each other.

Rolling friction is a little more complicated: when an object rolls without sliding or has a tendency to roll on the surface of another object, the obstruction effect on rolling due to the deformation of two objects under pressure in the contact part is called "rolling friction".

Your question is at the level of understanding, and it is not easy to say, because it is impossible to roll between the blade and the pencil, and the contact surface between the two is translational, not rolling, although the pencil is rolling, but the contact surface is not rolling.

3. The effect of staggering is indeed to increase the friction, but the increased friction here is not for faster cutting objects, the scrooked is to make the cutting gap wider, so that the friction between the blade and both sides is not too large, so that it is more labor-saving, from this point of view, he is to reduce the friction, you just imagine, if it is not crooked, it is the same as the knife we use, cutting things deep is easy to be caught and can not be pulled.

Electricity test pen will only shine or not light, will not show the number ah, your picture is not clear, there are two lines at home, the live line is generally red, should be marked, if AB is connected to the live line, and the bulb is not broken, then AB will make the test pen bright, if it is the zero line, AB will not be bright, the problem is very clear!

1.The forward pull given by the chain. 2.

The side of the pencil can be understood as an infinitely long horizontal plane on which the blade "walks". 3.The staggered teeth create a wider gap in the object and prevent the friction caused by the side of the saw coming into contact with the object.

4 If a is connected to the line of fire, then there is.

Linear motion at a uniform velocity Horizontal tension = friction force 15n

Friction is increased by (pressure).

1: The object with a mass of 20 kg moves in a uniform straight line along the ground under the action of 15n horizontal tensile force, and the friction force on the object is 15N, if the speed of motion of the object on the ground is increased by 1 times, the friction force on the object is (15)N

2: In industrial production, often through the transmission of gears or belts, as shown in Figure A is a common transmission belt, which relies on the friction between the belt and the pulley transmission, if their friction is too small, the belt will slip and can not operate normally, so the worker master.

Lower the belt, twist it to the "8" (horizontal 8) type on the mount, and it will function normally, which is by increasing the friction ( increase pressure ) .

(1) Because the magnitude of sliding friction is only related to the roughness of the contact surface and the pressure of interaction, the friction force on the object is unchanged when the motion speed of the object gradually increases.

2) When the tensile force increases, the friction force is unchanged because the tensile force is horizontal and does not affect the pressure in the vertical direction.

(1) When the speed of the object gradually increases, the friction force on the object does not change: the friction force is only related to the positive pressure and the degree of coarseness, and has nothing to do with the speed of motion.

2) When the tensile force gradually increases, the frictional force on the object does not change: the friction force is only related to the positive pressure and the degree of coarseness, and the tensile force is in the horizontal direction and does not affect the friction force.

(1) The friction force remains unchanged, because the amount of friction is only related to the pressure and friction factor of the object on the ground, that is, f=un, and has nothing to do with the velocity.

2) Immutable, for the same reasons as (1).

(1) and (2) are not the same frictional force.

Because friction is only related to the roughness and positive pressure of the contact surface.

(1) No change. Because sliding friction = kinetic friction factor * pressure, independent of velocity.

2) No change. Ditto.

(1) No change. Friction is only related to the nature of the object itself, i.e., the kinetic friction factor and pressure. 2 In the same way, unchanged.

The friction of the hand against the cup does not change 2

Stationary: The force is balanced in the vertical direction, so that the friction received by the cup is equal to the gravitational force.

The frictional force of the hand against the cup is unchanged by 2n