High School Physics Problems... Overweight and weightlessness ... Seek the solution process, find t

Calculate the height of the free fall first:

1/2gt2=3125m

Calculate the speed at this time:

gt=250m/s

The acceleration at the time of deceleration is calculated according to the maximum pressure the student is exerted on:

2mg-mg=ma,a=g；

So slow down to walk:

v2/(2*a)=3125m

So the total height is:

h=3125+3125+500=6750m

Some formulas are needed, so give more points.

Let's analyze the process first: the plane is free falling, the initial velocity is 0, and it should start to decelerate after 25 seconds. In order to achieve the minimum height, it is of course necessary to ensure that the deceleration motion is carried out with maximum acceleration.

According to the title, the maximum acceleration that a college student can withstand should be g (upward). When decelerating to v=0, ensure that the height is not less than 500m.

So the idea is that you can work backwards from 500m (of course, you can set the initial height h, the column, and solve the equation).

According to the characteristics of the problem, after a little analysis, it is known that the process time of acceleration and deceleration is equal (because the initial and final velocities are 0, and the magnitude of acceleration is also equal). At 500m, the acceleration motion is accelerated at 10m s 2 for 25s, and then the free throw is just stationary for 25s, and the altitude at this time is the desired flight altitude.

It is not difficult to find that the height is 6750m.

The main problem is this. 1. Weightlessness of 25s. From the initial velocity of 0 to start free fall.

2. Overweight x seconds, x unknown. The main function is to slow it down.。。 Landed safely.

That is, the speed of reaching the ground is just reduced to 0. Overweight. The support force can be twice as strong as gravity.

That is, the overweight acceleration is also g, and this is the whole idea. So according to symmetry. The weightlessness distance is equal to the deceleration distance.

So at least 2*1 2*g*t 2=10*25*25=6250m

Overweight is the apparent weight is greater than the actual weight, weightlessness is the apparent weight is less than the actual weight, the apparent weight here refers to the weight seen on the instrument, the actual weight refers to the static situation, the weight of the instrument, is the actual weight, is this meaning, if the word is a bit of a problem, please ignore.

When you take the elevator, the elevator runs upwards to accelerate or downward to decelerate, you are in a state of overweight (there is a clear sense of weight on your legs), at this time, if you have a scale under your feet, it will show that your weight is larger than usual, and when the elevator runs downward, you are in a state of weightlessness when accelerating or running upwards to decelerate (feeling the blood rushing to the top of your head), if there is a scale under your feet at this time, it shows that your weight is smaller than usual. If you still can't understand it, it is recommended to sit in the elevator to experience it, practice helps to understand, and it is estimated that you will not forget it in the future. Hope it helps.

Being overweight is to give you an upward force greater than your weight, and you will feel very heavy, and weightlessness is to give you an upward force less than your weight, just like when you are on a roller coaster, you have a feeling of panic.

It's very simple, the elevator goes up, and when you go up at the beginning, you are overweight; The elevator goes down, and when it goes down at first, it is weightless. Note: It was at the beginning.

Overweight – The gravitational force of an object is greater than the gravitational force of an object when it is static on the earth's surface – such as the gravitational force experienced by an object in an accelerating rocket (or vertical elevator) is greater than the gravitational force of an object standing on the earth's surface.

Weightlessness – the gravitational value of an object is less than the value of its gravity when it is static on the Earth's surface – such as an object in a spacecraft (capsule) orbiting in space, which experiences zero gravitational force. (The object in a vertical elevator is less gravitational than the resting force during the deceleration of the ascent).

If you find that the force is greater than gravity, you are overweight.

To put it simply, first of all, you need to understand that this weight means the gravity of oneself without exercise, not the body weight.

Second, let's take the example of an elevator (where gravity is almost always based on what is exerted on the Earth's surface).

Overweight, which means that the gravity you bear is greater than your own gravity, in the elevator ascending process, your gravity is downward, and at this time, if the elevator speed is up and appropriate, you and the elevator as a whole will be subject to downward gravity, and the upward process will also be subject to downward resistance, so that the gravity + resistance "gravity" gravity, this is overweight.

Weightlessness, which means that the force you bear is less than the personal gravity, in the process of elevator descent, you and the elevator as a whole are subjected to a downward gravity, followed by a direction of upward resistance, then the direction of gravity and resistance is opposite to subtract, so it is weightless.

In addition, the most obvious weightlessness is in space, this time, in theory, is considered to be g=0, then your personal gravity is 0, in a state of weightlessness.

I'm bored, I'll give you a call in person, tomorrow's exam (physics) is the same as you, a freshman in high school!

To understand overweight and weightlessness, you must first understand these!

Real weight: that is, the actual gravity of the object, which does not change with the change of the motion state of the object.

Apparent weight: refers to the pressure of the object on the horizontal support surface or the tension force on the vertical suspension line, which changes with the motion state of the object.

Overweight: The phenomenon of seeing a greater weight than a real weight.

Weightlessness: The phenomenon that the apparent weight is less than the actual weight.

Total weightlessness: A phenomenon in which the apparent weight is equal to zero.

Let me give you some examples:

When you take the elevator:

When accelerating to rise or decelerating down, it is in overweight; When decelerating up or accelerating down, it is in weightlessness.

Total weightlessness case:

When you throw two blocks upwards (one up and one down), the pressure between the blocks is zero because it is completely weightless.

When weighing: When a person squats.

At the beginning, the acceleration is downward, in weightlessness, and the indicator becomes smaller;

Later, when I squatted halfway, I had to slow down, so I was overweight, and the number increased;

Later, the person is in equilibrium, and the number is restored.

Important: When the acceleration is upward, it is in overweight, and when the acceleration is downward, it is in weightlessness.

For example, bridges are designed to arch upward, because the car is on the bridge, there is a downward centripetal acceleration, so it is weightless, and the pressure on the bridge becomes smaller.

The design is concave downward, the car is on the bridge, there is an upward centripetal acceleration, so it is in overweight, and the pressure on the bridge becomes greater, therefore, the bridge is designed to arch upward. , 1. Overweight, do acceleration exercises upwards. It is conceivable that the sum of the upward forces is greater than the sum of the downward forces.

When weighing, squat and stand up again. The display will be greater than your actual weight. Because you are doing an upward acceleration motion, the upward force (support force) is greater than the downward force (gravity), and the weighing machine shows the support force, so it is more important than the strength body.

Weightlessness, on the contrary...2. Overweight, the ground support force received by the object is greater than its own gravity, weightless, the ground support force received by the object is less than its own gravity, when weighing on the scale, the squat is weightless, and vice versa. 2. Overweight, the ground support force received by the object is greater than the gravity, such as when the elevator is just started upward, it feels like the person is pressed to the middle.

Weightlessness, the ground support force received by the object is less than the gravity, for example, when the elevator just starts downward, it feels like the person is pulling to both ends, 2, about weightlessness, you can refer to my previous one.

When an object accelerates, it needs a corresponding resultant force in its direction to support the accelerated motion.

Overweight, similar, just in the opposite direction. Accelerating upwards, the resultant force should be upward, at this time, greater support is required, so compared to stationary, the support force is greater, and the pressure felt is of course greater, so it will feel heavier. But, no matter what the circumstances, human gravity still does not change.

0,

Solution: (1) Object: f (2) Object: known by Newton's second law.

mg-f=ma，f=40 n

The solution yields a=2 m s 2

That is, the elevator acceleration is downward, and the elevator may accelerate downward or decelerate upward.

I can't see it clearly. I guess it's because the spring ball system hangs on the roof of the elevator.

The idea is this: the spring indicator becomes larger, overweight; The indication is small and weightless.

When overweight, the elevator accelerates up or decelerates, and weightlessness is the opposite: deceleration and fire acceleration.

It is found by the formula: f=t-mg=ma, where t is the spring tension.

If the mass is greater than the normal state, it is overweight.

The mass is less than that of the general state, and the 5kg glass ball in the question has become 4kg, which is obviously weightless. The acceleration is 2, and the elevator is in a state of acceleration and descent or deceleration and ascent.

[The direction of acceleration is the same as the direction of force], when the object has an upward acceleration, it is in overweight, and vice versa.

A may be deceleration down, just imagine, when you want to decelerate when descending, there must be an upward force, so the acceleration is also upward, in overweight, the indicator increases.

A spring scale is equivalent to a definite amount of upward force.

The object is subjected to two forces, one is its own gravity g (downward, the magnitude is unchanged), and the other is the support force t of the spring scale (upward, the magnitude is readable). Their net force is ** of the acceleration of the object.

If the reading is accurate, indicating g t, equilibrium state, it may move at a uniform speed or may be stationary.

If the reading is on the high side, then t' g, the resultant force is upward, and the acceleration is upward. At this point, it may be an upward acceleration or a downward deceleration.

In the same way, if the reading is small, then t" g, the resultant force is downward, and the acceleration is downward. At this point, it may be a downward acceleration or an upward deceleration.

In summary, b, c are paired.

If D wants to accelerate upwards and then decelerate upwards, the direction is always upward, and the readings are sometimes large and sometimes small, so D is wrong.

A high reading indicates that the acceleration of the device is in an upward direction, and the direction of acceleration is not necessarily the same as the direction of velocity. So the device may be doing an acceleration motion upwards or a meiotic motion downwards. In the same way, b is also wrong.

This kind of problem requires students to distinguish the relationship between velocity and acceleration, the direction of velocity is not necessarily the same as the direction of acceleration, the direction of acceleration of an accelerating object is the same as the direction of its motion, and the direction of acceleration of a decelerating object is opposite to the direction of its motion.

The size of the reading does not "indicate" that it is only one of the above states and is not comprehensive. For example, the reading may be decelerating or accelerating, or it may be tilting to accelerate up or tilting to decelerate down, as long as there is a sub-acceleration of acceleration in the vertical direction and the upward movement is eligible. The same is true for the reduction of readings.

Note the analysis of the stressed person standing on the ground n (support 1) = mg when the person squats f = mg-n (support 2) = ma Obviously, n (support 2) is less than n (support 1), so the support force decreases first, and when the person squats, it is still on the ground

If moving in the erection direction:

It turns out that A is stationary, that is, it is not pulled by the spring, which means that the tensile force of the spring is less than the maximum static friction between A and the board. When A is suddenly pulled while the spring force is unchanged, the maximum static friction between the boards of A is reduced. i.e. a pressure on the planks is reduced.

It turns out that A does a uniform linear motion, and the pressure of A on the plate is equal to the gravitational force of A. After A is pulled, the pressure of A on the plate decreases, indicating that the pressure of A on the plate is less than the gravitational force, so it should be weightless now. I choose ab

If moving in the horizontal direction:

It turns out that the uniform speed of motion, the balance of tensile force and static friction force is 0, assuming that A does not move, if it accelerates to the left, what requires the resultant force of A to the left, that is, it is necessary to increase the static friction force, and the friction force at this time exceeds the maximum static friction force, so it moves to the right. If the acceleration is to the right, the direction of the resultant force is to the right, and the static friction force needs to be reduced, if the acceleration is relatively large, the static friction force may be reduced to 0, or it may be reversed to the right, and no matter how large a is, it may move to the left.

When moving at a constant speed, there is f-f=0

Pulling to the right, it means that there is acceleration to the right, f-f=ma is greater than the frictional force, so that the friction force becomes smaller.

One is to reduce the pressure, so a accelerates the fall.

The other is to reduce the relative motion, and the friction force is analyzed to the left, then the acceleration of d is to the left, so AD is selected

Whether it is accelerating down or decelerating, the acceleration is downward, it is weightless, weightless, weightless, the pressure on the bottom of the box is smaller, and the static friction force is also smaller, so it is pulled to the right by the spring, so a b is right.

c No, because the net force is zero at a constant velocity, it will not move.

D is right, because the box suddenly accelerates to the left, and A doesn't have time to accelerate, so it moves to the right relative to the wooden box.

Originally, the uniform linear motion indicates that the force is balanced, and the spring is elongated, indicating that the object is subjected to left friction. The magnitude is equal to the pulling force. Later, it was found that A was suddenly pulled to the right by the spring, indicating that the friction decreased, whether it was accelerating down or decelerating, the acceleration was downward, and it was weightless Weightless A The pressure on the bottom of the box was smaller, and the static friction force was also reduced, so it was pulled to the right by the spring, so A B was right.

The box suddenly accelerates to the left, and A doesn't have time to accelerate, so the relative wooden box moves to the right, and D is right.