High School Physics There are 6 small questions about conservation of energy, which are not difficul

For question 1, can you know that whether the mechanical energy increased by the object or the internal energy generated by friction comes from the conveyor belt, right, you must know that there is a force to work on the system? The internal energy generated is relative displacement with fs,s, there is no doubt about it! For question 2, the increase in mechanical energy, the change in potential energy plus the change in kinetic energy, you must know that the change is the comparison between the last moment and the initial moment, because small objects initially have velocity, this must be noted!

Question 3, the velocity v relative to the ground is the actual velocity, why do you need to change, I don't understand this, can you tell me? I think the change is wrong. I don't know if I can help you.

First of all, this physics question, the first question is to look at the graph to find the area, pay attention to the positive and negative.

The second question is that a=1m s2 is known from the figure, and mgcos -mgsin =ma is obtained by force analysis

The third question: q= x*f from the graph, v pass=4m s, x x, x, x, x, x, x, f The second question can also be calculated, q is fine Note that 6-8s is relatively stationary.

h=ssin , ep=mgh, and ek remains

Actually, v is not that complicated, it is v

That's fine, right? (You do the math and see if it's right.)

Let's talk about your question again.

The answer is to use fs to find the heat production of friction directly, and the answer should be relative displacement, if it is relative displacement, fs= This is not right, do you think S is displacement?

fs (relative displacement seems to be miscalculated = =) s = area (i.e. distance) *tan is a bit irresponsible = =

a, I don't really understand the meaning of this sentence = =

b, if you take s as a displacement, look at the third question.

c, if you take s as the relative displacement, then the relative displacement algorithm is not right, both of them are calculated as a large decrease is the relative displacement, pay attention to the 6-8s is relatively stationary.

I won't, let me know if you know.

If the lowest point of the arc is selected as the zero point of the potential energy in this question, then the gravitational potential energy at point A is mgr (you may not need the solution process, you can say it if you need to).

The ball slides down 1 4 arc stage: take the lowest point of the arc as the zero point of potential energy, and obtain mgr mv 2 2 from the conservation of mechanical energy

The velocity of the ball before touching is v root number (2gr) root number (2*10*1) 2 root number 5 m s

Collision phase between ball and block: momentum is conserved, mv 2m*v1

Their common velocity is v1=v 2 root number 5 m s

Flat throwing stage: there is a shed hidden S v1*t in the horizontal direction

Vertically, there is h g*t 2 2

The horizontal distance is sv1*root(2hg) (root:5)*root(2*16 10) 4m.

Note: A completely inelastic collision is a chain of morning halls that stick together after the rocks are touched.

Because there is no friction, the mechanical energy is conserved during the motion of the object. After cutting off part of the inclined plane, the object does an oblique throwing motion, and the velocity at the highest point is not zero, and there is a horizontal velocity. Therefore, the gravitational potential energy to the highest point is smaller than the original mgh, that is, the rising height is less than h.

When the inclined plane is bent into an arc, the problem of the centripetal force of circular motion should be considered, and the object in the vertical plane is in circular motion, and the velocity at the highest point is limited, if it is a rope model, the highest point velocity is limited to (gr)*, and if it is a rod model, the highest point velocity is limited to 0In this problem, the rope model is used, so there is a limit to the speed at the highest point, which is greater than 0If the arc bent by the inclined plane is greater than 1 4 arc, the speed can not be 0 when leaving, and when the arc is less than or equal to 1 4, the highest point velocity can be 0, so the rising height is still h

Yes. The reason why a is wrong is that there is still velocity in the horizontal direction at the highest point, and when the inclined plane is bent into an arc equal to 1 4, the horizontal velocity at the highest point is also 0, so it can.

You don't understand the answers in the counseling book?

If there is an icon that represents the title, you don't have to guess. Here's how to solve it:

The velocity of the falling object before the collision is v, and the mechanical energy is conserved according to the falling process, and we get: mv 2=mgh

The hanger collides inelastically with the falling object, the instantaneous momentum of the collision is conserved, and the hanger is denoted by m:

mv=(m+m)v

The two fall process together, the mechanical energy is conserved, (m+m)v 2=(m+m)ghh=v (2g)=m v [2(m+m) g]=m h (m+m) =

e motion = u bomb.

The velocity is zero when the lowest point is reached.

e dynamic = e potential = e bomb.

mgh=½kδx²

x= (2mgh k)= (2 in.

Ho kinetic energy theorem.

It's a matter of minutes.

You have to choose B. The momentum of the system consisting of ab and ball is conserved. The total momentum of the system is zero, and the final total momentum is also zero. m1v1=m2v2, and finally b and the ball have a large mass, so the velocity is small, while a has a small mass and a large velocity.

The advantage of conservation of momentum is that it ignores the intermediate complex process, and the net external force of the whole system is zero, so the whole process is the conservation of momentum, and there is no need to consider the intermediate process, and if you consider it, you will fall into chaos.

Since the ball is in A's hand at the beginning and finally falls in B's hand, that is to say, the total mass of A and the rook becomes smaller, and the total mass of B and the rook becomes larger, because at the beginning the total mass of A and the rook is equal to the total mass of B and the rook and due to the theorem of conservation of momentum, correspondingly, the velocity of A is larger, and the velocity of B is smaller.

The first question is an analysis of the entire system.

du, is the conversion of gravitational force potential energy into kinetic energy dao. There is a change in the mass return potential energy of 2m in the system, and this change increases the kinetic energy of the whole system. So there is 2mgH=, i.e. v= (4gh3).

As for the question of force cancellation you asked, it can be seen that the pulling force does cancel out the gravitational force of a part of the weight and does the work, but it also pulls the object on the horizontal table to do the work, which is equivalent to the problem of "cart pulling the horse" and "horse-drawn cart", when studying the whole system, there is no need to consider the role of the internal force of the system.

The second question you can calculate, w=mv -2mgh=-2mgh 3, your answer column is equivalent to a vector formula, and the work done by the pulling force on the weight is negative work.

First question. The wooden block on top is placed on a smooth flat surface, meaning straight it has a friction force of 0There is no pull.

Second question. If there is no friction, there will be no tensile workmanship, and when calculating, the direction itself is considered in the work.

1.Because according to the formula of the conservation of mass. If m is subtracted, then the right side is greater than the left side. One is the vertical force, and the other is the horizontal force. Both are the same size.

Regarding cancellation, you think a little simpler, except for friction, the force is a vector and is not in the same direction.

The first question is to choose two objects as the object of study, the decrease of gravitational potential energy is equal to the increase of kinetic energy, and the wooden block is still specialized.

On the surface of the table.

The gravitational potential energy does not change.

The second question: is to choose a weight with a mass of 2m as the research object, and apply the kinetic energy theorem 2mgh + w(t) = 2mv 2 2, but w(t) is found to be a negative value, indicating that the tensile force does negative work on the weight.

(1) The answer bai is solved with the conservation of energy, and only du is related to the part of the dao of the energy change of zhi, the back on the table.

The gravitational potential energy of the wooden block has not changed, and the answer is not to contribute to the speed, do not use force, and consider it with the idea of energy.

2) Because only gravity and tension do work on 2m, so the sum is equal to kinetic energy, kinetic energy theorem, here w(t) is negative, that is, the pulling force does negative work, you are right.

Because the bai table top is smooth, there is no friction, and the gravitational potential energy of the falling weight is all converted into dao

For the kinetic energy of the wooden blocks and weights, because the rope plate is straightened.

right, so they both have the same velocity, so the answer is 2mgh=(m+2m)v 2 2, the second question, the reason is that the downward direction is positive, and the consideration is only the weight, the work obtained is a negative value, hehe...