Kinetic energy questions in senior one, physical energy and momentum in senior one solution question

When the object slides along the inclined plane to the bottom of the return inclined plane, the work done by gravity is zero, and the work done by friction is 2fl=2 mglcos45°. According to the kinetic energy theorem, there are:

2µmglcos45°=1/2 mv²-1/2 m（vo）²

v²=（vo）²-4µglcos45°

Because cos45°=sin45°, LCOS45°=lsin45°=H. Therefore, the above equation can become:

v²=（vo）²-4µgh ……1）

Then apply the kinetic energy theorem to the up-sliding process:

mglcos45°+mgh =1/2 m（vo）² lcos45°=lsin45°=h）

i.e.: gh( +1) = 1 2(vo).

That is: gh = (vo) 2 ( +1) and substitute it into (1) to get:

v²=（vo）²-2µ（vo）²/（µ+1)=（1-µ）vo）²/（1+µ）=（2/3）（vo）²

v=√6vo/3≈

Set the maximum height h

1 2MV0 2=MGH+W resistance.

W resistance = FSF = mgcos 45°

h=s×sin45°

1 2mv 2 = 1 2mv 0 2-2w resistance.

The solution is v=......

1. The conveyor belt problem must be clear about the nature of the movement of the object from one end to the other. When the nature of motion is clear, all problems can be solved using the corresponding laws. The motion of the object may be fully accelerated, or it may be the same speed after the acceleration wheel loses cavity, or other possibilities, the specific problem can be analyzed by the given conditions.

2. When recoiling, the initial momentum of the system is 0, and the final momentum is the sum of the momentum of the two objects, but the empty let is the conservation of momentum is the vector equation, so we must pay attention to the selection of the positive direction, and the positive and negative of the initial and final momentum.

If pavement refers to do, it should be like this:

According to the kinetic energy theorem:

According to the kinetic energy theorem:

Final kinetic energy – initial kinetic energy = work done by the combined external force.

In question: the first time:

The kinetic energy at point A - the kinetic energy at point D = negative work done by gravity + negative work done by friction.

The kinetic energy of point A - the kinetic energy of point D1 = mghoa + fs1 = e1 It can be seen from the above formula: E1 is less than 0

The kinetic energy at point A remains unchanged, and the work done by gravity remains unchanged.

The second time: because the displacement of the horizontal plane decreases, the work done by friction decreases.

So there is e2 greater than e1

Kinetic energy at point A – kinetic energy at point D1 = E1

Kinetic energy at point A – kinetic energy at point D2 = E2

The kinetic energy at point d2 is greater than the kinetic energy at point d1.

The kinetic energy at point d2 is less than the kinetic energy at point d1.

So the square of the velocity of d2 is less than the square of the velocity of d1 (100 m s) so the velocity of d2 is less than 10 m s

Select C (back upstairs: the pavement is OBA, it is written on it).

Draw a diagram to decompose the force into F1 of the vertical inclined plane, and F2F1 along the inclined plane = mg*cos angle abo

Friction = U*mg*COS angle abo

ab=ob*1 cos angle abo

Multiply the above two to obtain the work done by friction on AB for u*mg*ob cos angle ABO dissipation.

The work done by the frictional force in the whole process is equal to u*mg*ob + u*mg*bc=u*mg*oc Note that this is an invariant.

The initial kinetic energy minus the work done by friction is all converted into gravitational potential energy, since the negative work done by friction is constant, the gravitational potential energy m*g*oa is also constant The initial kinetic energy required is naturally the same, that is, the muzzle velocity is the same.

The answer is a 20j

The process is as follows: Since the angle of the inclined plane does not change, it can be assumed that the work done to overcome friction (the energy expended) and the potential energy converted from kinetic energy are both proportional to the distance traveled by the object.

When it reaches point m, the total energy (kinetic energy + potential energy) possessed by the object is 100-32=68j, where the kinetic energy is 20j and the potential energy is 48j.

At this time, the object continues to move upwards until the kinetic energy is 0, and the distance that continues to travel in proportion to the distance from the bottom end to the point m is 1 4 (that is, the distance that the object needs to walk from 20J kinetic energy through frictional consumption and conversion potential energy to 0 kinetic energy is 1 4 of the distance it travels from 100J kinetic energy through frictional consumption and conversion potential energy to 20J kinetic energy).

So the energy consumed by overcoming friction in this section is 32*1 4=8j, the kinetic energy is 0, and the potential energy is 68-8=60j, which is the total energy possessed by the object.

Then the object starts to descend, and the energy consumed to overcome the friction on the way down is the same as 32 + 8 = 40 J as it goes up, so when the object reaches the bottom at the end, the potential energy is 0 and the kinetic energy is 60-40 = 20J

In addition, if it is an exam, you can simply judge a as the correct answer by the elimination method, because the object has to overcome friction up and down to do work, according to the question, even if the object slides directly after m, the remaining energy of the bottom is 100-32-32=36j, and in fact it should be smaller than 36j, and it is directly judged that a is correct.

It's a very simple topic, and now let's analyze it.

Let's first see if your so-called maximum value of a is a question of g When an object falls from A and just touches the spring, the object is subjected to gravity and elastic force, then their resultant force is.

f=mg-kx, the corresponding acceleration is a=g-kx mWhen mg-kx >0, the acceleration direction of the object has been downward, so the object has been doing downward acceleration with the acceleration gradually decreasing, so the downward velocity has been increasing.

When mg=kx, at this time, the acceleration of the object is 0, and the downward velocity reaches the maximum, and thereafter, due to the action of inertia, the object still has to move downwards and continues to compress the spring, at this time kx>mg

As for how much the object continues to move downward to make the velocity 0, quantitative analysis is needed here, rather than a general qualitative analysis that requires the kinetic energy theorem.

Suppose the kinetic energy of the object at the moment when it just comes into contact with the spring is e', apparently e'>0 then when the object moves from touching the spring to the velocity to 0, the deformation of the spring becomes x, and it is clear that the work done by gravity in this process is mgx

The spring elastic force does negative work, and the magnitude is.

It can be obtained according to the kinetic energy theorem.

e'+mgx=

Due to e'>0, so "mgx, about x, can be obtained.

kx>2mg

Therefore, when the object is stationary, the elastic force of the spring is greater than 2 times the gravitational force, then the resultant external force of the object is greater than mg, and the corresponding acceleration is greater than g, when the object falls from b, let the kinetic energy of the moment the object contact the spring be e'', apparently e''>e'

Let e''=e'+δe

E is satisfied when the object falling from B reaches X'+mgx=At this time, the object still has part of the kinetic energy δe, so the object will continue to move downward, and the elastic force of the spring will continue to increase, so the acceleration at this time a2>a1

First of all, it should be explained that the maximum acceleration is not g, because when the object moves to the position where it just touches the spring, the acceleration a=g has a velocity v at this time, according to the symmetry principle of simple harmonic motion, when it reaches its symmetry point, the velocity is also v, and the acceleration is also g, and the direction is opposite, but it is not the lowest point at this time, so if it moves downward, the spring deformation increases, and the acceleration will also increase, so the maximum acceleration is related to the initial energy, the higher the initial position, the greater the final variable, and the greater the maximum acceleration a1

The acceleration is maximum when the spring is compressed to the shortest time, and it can be seen from the energy conversion that the kinetic energy and potential energy of the object are converted into the elastic potential energy of the spring.

The maximum acceleration should be the acceleration in the direction upwards, and the acceleration is greatest when the spring is compressed for the shortest time and the object is at rest. The energy possessed is converted by the neutral potential energy, so the higher the position, the greater the gravitational potential energy, and the greater the elastic potential energy of the transformation. The greater the acceleration.

a=f/m=8÷2=4m/s²

v=at=8m/

It doesn't matter if it's north or south, it's perpendicular to west. I choose B

Divide the north-south and east-west directions, and make a uniform acceleration linear motion with an initial velocity of 0 and an acceleration of f m=4 (m s squared). Stuff:

Do a uniform linear motion with a speed of 6m s. v (north and south) = at = 8 (m s), v (east and west) = 6 (m s), isn't the synthesis 10 m s? ek (end) = 1 2 * 2 * 10 * 10 = 100 (j).

ek (initial) = 1 2 * 2 * 6 * 6 = 36 (j). Kinetic energy increment = 64 (j). I choose B

Question 1: (c).The cannonball and the artillery carriage are regarded as a whole, the force between the artillery shell and the artillery carriage is the internal force, the two wholes are in a state of overweight, the external force in the vertical direction is not zero, and the horizontal direction and the external force are zero, and the momentum in the horizontal direction is conserved.

Question 2: Obviously two balls collide and both are conserved in momentum. And A catches up with B, which means that A's speed is fast, while the momentum of A's ball is small, which means that A's ball has a small mass, and after AB touches, there are two possibilities:

A has a velocity smaller than b (the sum of the momentum of the two equals 25), and there is a kind of a**, which is the opposite motion (the sum of the momentum of the two equals 25).

First, if a has a small velocity, and it has been proved earlier that a's mass is small, this proves that a's momentum is definitely smaller than b's, and the sum of the two momentum equals 25, then b is compliant.

The second type: D obviously agrees on the surface, but according to the kinetic energy of the two balls A B, the total kinetic energy of the two balls hitting each other cannot become larger, so D is excluded. Therefore, choose B

c Because the car is in an overweight state when the shell is fired, the force unbalanced momentum in the vertical direction is not constant, but the horizontal direction is constant. Question B,,, You should pay attention to the speed of the A ball after the two balls hit it must be less than the speed of B (otherwise the two balls will collide), from the question, it can be seen that when the momentum of A is 10, you can catch up with B with a momentum of 15, so it can be excluded In addition, the total momentum of the two balls after hitting must be less than or equal to the total momentum of the two balls before hitting it, so D is also excluded,

Let's give you another old idea: the resistance of the force of the pure ridge in addition to gravity and elastic force is equal to the amount of change in mechanical energy, and the ground of the inclined plane is selected as the zero potential energy surface: the total amount of mechanical energy at the beginning of E = the initial kinetic energy of E + the potential energy of the initial gravity at the end of E = the total kinetic energy at the end of E = the total amount of mechanical energy at the beginning of E - the total amount of mechanical energy at the end of E, and then substitute the number of appreciation of the pants to solve the solution WF = 3800J