A high school physics problem, urgent, solve a high school physics problem

The tension on the same rope is equal everywhere.

The key to winning the tug-of-war is which pair gets the ground to face the players with more friction!

The force of the rope is the same for both the winning and losing sides.

Tug-of-war is a process, and the tension of the rope may vary throughout the process.

But at every moment, the amount of force on both sides of the rope is the same.

In this case, the winning side is subjected to static friction, while the losing side is subjected to sliding friction, so it is not difficult to find that the nature and magnitude of friction are different, resulting in victory and defeat.

The factors that affect the tug-of-war are: the weight of the players, the roughness of the shoes, the skills of the players, teamwork, and so on!

Don't wait. When a person is pulled by another person, the person must have been subjected to a large force, and the force is the reason for changing the motion of the object. It doesn't have much to do with acceleration.

The force of the two is equal. It's just when A is pulled over by B. The sliding friction of A is less than the static friction of B. Therefore, the total work done by A is less than the total work done by B. That's why A is pulled by B.

Equal, as the action of force is reciprocal.

The forces are definitely equal The forces on a rope are equal everywhere.

It's just that the friction of two people is not the same, the static friction of A is greater than the sliding friction of B, and B is pulled on the 1st to 3rd floors, fooling you, don't be gullible.

You go to cram school, don't explain.

That's a definite, force is what changes the state of motion of an object.

Yes, because the resultant force is responsible for the acceleration that occurs. So the way of movement has changed, that is, the resultant force has changed.

If there is acceleration on the acceleration side, there is a force, and when the force is 0 in equilibrium, acceleration is the force in the direction of acceleration.

Solution: When an object falls from a balloon, it has to move upwards for a certain distance because it still has an upward velocity, and the time for moving upwards for another distance is 4 10=

So this distance is 4* squared =

So the height of this weight from the ground to the highest point is h=217+, so the time t to return to the ground, can be made by.

1 2*10*tsquared=

The solution is t = so the velocity to the ground v = 10*

If you are already in the first year of high school, then I don't think there is any need for me to calculate the result, I only remind you: the initial velocity of the heavy object is 4m s upward, the acceleration g is not to mention, the displacement is 217m downward, you can calculate it if you know the formula of uniform acceleration linear motion.

a, you use the kinetic energy theorem to do it

c The height of the first fall is a r

The second time is five quarters r

So it's four to five.

bYou use the kinetic energy theorem to do it.

3 minutes crack quietly 3*60s = 180s

20m buried, s*180s=3600m

3600m+1000m=4600m)

a*t1=30m/s

30m/s*t2=x2

x1+x2=4600m

t1+t2=180s

t1=800/15s

t2=(180-800 basal source 15)s

a=9/16m/s^2

Option d This question can be used by elimination. First of all, it can be known that A and B must meet for the first time before t1, and a answer, when t'=t1 is, as can be seen in the figure, the displacement of A is twice the displacement of B, if A and B meet, then d must be equal to 1 2s, and a can be excluded. From b, c, d answers, t'=1/2t1。

From the kinematic formula, v A = 2s t1, B acceleration a = 2s (squared), from the first encounter formula to get the first encounter: v a*t'=d+1/2*a*t'(of the square), substitute the above formula, you can calculate d.

choose d.This question is easy to do, because if you want to catch up twice, you need at least d s. Therefore a veto.

Then we can subtract the area of A from the area of B, then it is the distance to catch up. After calculation, d is compliant.

This is a scenario question, the time in front is the condition, and the D in the back is the conclusion, ask you which combination is in line with the facts, let's look at the BCD first, they are all the same in front.

You draw a vertical line in the OQ.

The OPQ is divided into 1:3

Right, that stands for T'=1 2t1.

In this case, the area represents the distance.

B:1 4S A:S

Right, I mean the distance they've traveled.

And then didn't the two meet at this time, at the beginning, car B was in front of car A, so, d=3 4s are looking at A, and if you want to catch up twice, at least d s. If A, B has been walking more than A since T, and it is impossible to meet again.

So, choose D. It's detailed enough, plus points.

The direction of the force is parallel to the inclined plane.

In this case, the force needs to be minimal.

Because if the direction of the force is angled to the inclined plane, there will be a component force, so that the resultant force is greater.

The magnitude of the force is equal to the component of the gravity of the wooden block diagonally downward plus the frictional force generated when it is pulled diagonally upward, i.e.:

1 2 m g + 3 2mg = ( 3 ruishen 2 + 1 2 ) m g Xi Pei Hui Wang will be useful to you.

According to the kinetic energy theorem, the total work of f with friction and gravity is 0, the magnitude of friction does not change, and gravity does not change, so the work of f does not change.

If the rope does not break.

If the radius of the ball around point b is r=l-d, then the highest point of the ball motion is h=d-(l-d)=2d-l

According to the kinetic energy theorem, mgh = mv i.e. mg(2d-l) = mv v = g*(4d-2l) (1).

And mv r=mg, i.e., mv (l-d) = mg v g*(l-d) (2).

1) (2) Lianlitech: 4d-2l=l-d launched d=3l 5=

then d is the range of values.

I'm curious, the ball "moves in a circle" around the nail, then the rope will be shortened, then it will not be in a circular motion, then the question will not be possible, so how to answer?

Use the kinetic energy theorem.

From the point at rest, to the highest point, the work done by the gravitational potential energy is used to change his speed, i.e., to increase his kinetic energy.

There is a requirement for speed--- to ensure that this rope can send the ball to the highest point, that is, gravity fully provides the centripetal force.

Just follow these two columns.

Personal opinion only.

It can be seen from the area of the V-t plot as its displacement.

s=16-2=14m

by Newton the second.

ma= mgcos37°-mgsin37° solution gives =7 8

The amount of mechanical energy increased.

e=1/2mv2^2-1/2mv1^2+mgssin37°=90j

x=vt-s+2*4/2=18m

q = fx phase = mgcos37°*δx=126j For the glory of the Soviets, if you still have questions, please ask.

(2+6)x4/2-2x2/2=14.。。

The answer is no problem, boy.

The area above the x-axis minus the area below

The displacement is the trapezoid above minus the small triangle below, the top is 16, and the bottom is 2, so the displacement is 14,; The support force of the conveyor belt to the object is 8n, the component of gravity down along the inclined plane is 6n, and the acceleration is upward along the inclined plane, so the friction force is upward, and the slope in the right figure is 1m, so the acceleration of the object is 1m s2, so the magnitude of the frictional force is 7n, and the kinetic friction factor is obtained; The increment of mechanical energy is equal to the increment of kinetic energy plus the increment of gravitational potential energy, the increment of kinetic energy is 6j, and the increment of gravitational potential energy is 84j, giving 90j; The heat generated by friction is equal to the frictional force * relative displacement, which gives 126j