High School Physics Don t understand, high school physics problems?

Let the total distance be s and the total time be t

Pair A: The first half of the time is t1=t 2, the speed is 5m s, and the distance traveled is s1=v1t1=5*t 2

The second half of the time is t2=t 2, the speed is 6m s, and the distance traveled is s2=v2t2=6*t 2

So s=s1+s2=11t 2 t=t1+t2=t

The average velocity of A is s t=11 2=

For B: The first half of the distance is s1=s 2, the speed is 5m s, and the time taken is t1=s1 v1=(s 2) 5=s 10

The second half of the distance is s2=s 2, the speed is 6m s, and the time taken is t2=s2 v2=(s 2) 6=s 12

So s=s1+s2=s t=t1+t2=s 10+ s 12=s*22 120

The average velocity of B is s t=120 22=

The average speed is x t, which is the total distance divided by the total time.

A: The total distance is 5t + 6t = 11t. Time 2t. Average.

In the same way, the total distance must be divided by the total time.

All the questions that calculate the average speed are the total distance traveled by the total time.

According to the meaning of the actual operation of A and B to draw out, and then tell you the known conditions of the title in the diagram, think about what the average speed means, how to solve it, I believe you can do it.

The fixed pulley can change the direction of the action: when the fixed pulley under the ceiling is hung on both sides, the gravity on both sides acts on each other through the rope, and when the mass of the two sides is not the same (ma>mb), the two forces are not unbalanced, so there is a relationship of "force = mass acceleration" between the force of the two objects moving together as a "whole" and the hidden change of the overall mass and acceleration, where "mag-mbg" is the "combined external force" when the whole accelerates the motion.

mag is the weight of weight a:=ma gmbg is the weight of weight b:=mb g because the two weights a and b are connected by a rope around the fixed pulley.

Both of them move together and have the same acceleration.

Without considering the friction of the pulley, consider these two punching weights as a whole. At this point, the resultant force they experience is equal to:

f = ga - gb = mag - mbg = ma - mb)g

So let's dress up: f = ma+mb) *a

i.e.: f = ma+mb) *a = m-mb)g

Mag-MBG refers to the difference in gravity between MA and MB (G is the acceleration due to gravity) and acts on a system with two weights as a whole.

It's envy that I didn't quarrel with my brother Cha Liang like this:

After deceleration together on a rough horizontal plane, the forces on the block and the board are different.

On the rough surface as shown in Figure 2, in addition to the friction of the wooden block, there is also the friction of the ground, the force of the two is different, the resultant force is different, the acceleration is also different, naturally there will be friction between the two, otherwise the wooden block can not be slowed down.

In response to such problems, the resolute approach is to first conduct a force analysis to analyze the direction of the resultant force. According to different problems, different methods are used, and sometimes the whole method is used, which requires the two forces to be combined into a whole, judge the acceleration of the whole, and solve the problem.

The first one is the overall analysis of A and B, and the second is the isolation analysis of A.

AB is in a state of force equilibrium, and the resultant force in all directions can be decomposed orthogonally for A, and F = MGCOS can be obtained from the isolation analysis of A

f is the resultant force of the elastic force of the oblique face A and B, so the elastic force of the oblique face B is f -f.

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