Easy High School Physics, Easy High School Physics

The key to this problem is to understand the movement of the blocks once they are placed on the conveyor belt.

The friction force of the conveyor belt on the block is to the right, and the block accelerates evenly under the action of this friction until the speed is the same as the size of the conveyor belt, and the acceleration stops. The other case is that a constant acceleration motion is maintained throughout the process, until B, the speed is not as great as the speed of the conveyor belt.

First of all, it is required to determine the acceleration generated by the friction force of the conveyor belt on the object, and let the acceleration be a, assuming that at this stage, the block has been accelerating, then there is.

a=5 9, then the final velocity of the block is >> 2m s, then the assumption is not true. Blocks should move at a uniform acceleration first, and then move at a constant speed.

Then the acceleration time t1 = 2m s a

The distance covered in the acceleration phase is.

s= a t^2=2/a

It is time to walk the remaining distance.

t2=(10-2/a)/2=5-1/a

Composed. t=t1+t2=6s=5+1/a

a=1m/s^2

That is, when the friction force is maximum, the object produces an acceleration of 1m s 2, in order to make the distance to the B end the shortest, during this time, the object must go with the maximum acceleration, and the speed of the object must always be less than the conveyor belt, otherwise it will move at a uniform speed.

Then, the time it takes for the object to walk 10m with a is t'

t'= 2 root number (5)s

i.e., the velocity of the object at b is.

v'=at'=2 root number (5)m s

In order to keep it accelerating, the minimum speed of the conveyor belt is v'

It's not an easy question. First of all, you need to understand the meaning, when is the shortest time to transmit? It should be that the object has been on it and has been accelerated by the friction of the conveyor belt until it reaches point B!

If the conveyor belt velocity is 2 and it takes 6 seconds, it means that the object is first accelerated to 2 and then goes to point B with the conveyor belt. Strictly speaking, this needs to be proven, and 78 words are omitted here.

Suppose S1 is the displacement of the acceleration segment and T1 is the acceleration segment time.

2^2=2gus1

2=gut1

Learns that s1=t1

10-s1)/2=6-t1

It is learned that t1 = 2 and u = 1 g

Well, the ideal state is v 2 = 2 gu * 10

Arrive at the answer v = written in red.

Acceleration requires the combined external force, the spacecraft is affected by gravity mg and the upward magnitude of the air resistance, the resultant force is downward, so the acceleration is equal to the combined external force divided by the amount of qualitative defeat, the magnitude is.

Then we use the formula of uniform acceleration, s=(at 2) 2=(to fall three kilometers.)

If the buried void is positive for 3s and returns to the throwing point, it is: vo 3 + g 3 2 = 0, and vo = 15m s

2S displacement before bending is VO 2 + G 2 2 = 10m

symmetrical. Let's draw the picture yourself, it's not a solution anyway.

The answer is a

The ball moves from the inclined plane in an inclined plane, gradually accelerates, reaches the lower end of the inclined plane, and lands on the table, making a short horizontal movement, and then leaves the edge of the table and flies out. That is, the horizontal flat throwing motion.

When the ball leaves the edge of the table, its horizontal velocity multiplied by the flight time produces a horizontal distance, i.e. x= v0 * t, so v0= x t.

The flight time is determined by gravity, i.e., the time of landing.

This question should assume that the table friction is 0 and the air resistance is zero.

Therefore, the horizontal position is removed to the landing time, and the velocity of the ball in a horizontal table motion is obtained.

Solution: Measure the horizontal distance x from the edge of the table where the ball lands.

Measure the height of the table surface to the ground.

It is very good to solve t= (2y g) from y=1 2 g t 2 (time set to t seconds).

Then the horizontal velocity is:

v0= x / t = x/√(2y/g) =x*√(g/2y）

This is actually a problem of calculating the initial velocity of the flat throwing motion, which is usually decomposed into a uniform linear motion in the horizontal direction and a free fall motion in the vertical direction. The movement of the ball on the horizontal table that we require is the speed of the ball in the horizontal direction during the flat throwing movement.

1.Measure the height h of the table top from the ground. From the formula h=, the time t. is calculated

2.Measure the horizontal distance between the landing point of the ball and the edge of the table.

3.From the formula v=s t, calculate the velocity of the ball on the table horizontally.

I remember one of my high school physics classes doing just that. You're really.