Abstract. Law of energy conservation. Theorem on kinetic energy Theorem on kinetic energy derivation of the formula

Let's start with a definition. Job A strength F when moving X of the body to which it is applied is defined as the scalar product of vectors F And X .

A= F x= Fxcosα. (2.9.1)

Where α – the angle between the directions of force and displacement.

Now we will need expression (1.6 a), which was obtained for uniformly accelerated motion. But we will draw a universal conclusion, which is called the theorem on kinetic energy. So, let's rewrite equality (1.6 a)

a· x=(V 2 –V 0 2)/2.

Let's multiply both sides of the equation by the mass of the particle, we get

Fx=m(V 2 –V 0 2)/2.

Finally

A= m V 2 /2 – m V 0 2 /2.

(2.9.1) Size= m E

V 2 /2 is called the kinetic energy of the particle.

You are used to the fact that in geometry theorems have their own oral formulation. To keep up with this tradition, let us present the theorem on kinetic energy in text form.

The change in the kinetic energy of a body is equal to the work done by all the forces acting on it.

This theorem is universal, i.e. it is valid for any type of movement. However, its exact proof involves the use of integral calculus. Therefore we omit it.

Let's consider an example of the movement of a body in a gravitational field. The work of gravity does not depend on the type of trajectory connecting the start and end points, but is determined only by the difference in heights in the start and end positions: A=mg( 1 –A=mg( 2). (2.9.2)

h Let's take some point in the gravitational field as the origin and consider the work done by gravity when moving a particle to this point from another arbitrary point R A=mg(, located at a height . This work is equal to mgh Size and is called potential energy Let's take some point in the gravitational field as the origin and consider the work done by gravity when moving a particle to this point from another arbitrary point:

Size n particles at a point . This work is equal to (2.9.3)

n =

A= m V 2 /2 – m Now we transform equality (2.9.1), the mechanical theorem about kinetic energy takes the form V 0 2 /2= E Size p1 –

m p2. V 0 2 /2=(2.9.4) m V 2 /2+ V 0 2 /2= n2 =

V 0 2 /2+

p1. Size.

Size=V 0 2 /2= In this equality, on the left side is the sum of kinetic and potential energy at the final point of the trajectory, and on the right - at the initial point. This amount is called total mechanical energy. We will denote it k +

P. We have arrived at the law of conservation of total energy: in a closed system, total energy is conserved.. These forces depend only on the position in space. And the work done by such forces when moving a body from one position to another depends only on these two positions and does not depend on the path. The work done by a conservative force is mechanically reversible, that is, it changes its sign when the body returns to its original position. Gravity is a conservative force. In the future, we will get acquainted with other types of conservative forces, for example, with the force of electrostatic interaction.

But in nature there are also non-conservative forces. For example, sliding friction force. The longer the path of a particle, the more work is done by the sliding friction force acting on this particle. In addition, the work of the sliding friction force is always negative, i.e., such a force cannot “return” energy.

For closed systems, the total energy is, of course, conserved. But for most problems in mechanics, a special case of the law of conservation of energy, namely the law of conservation of total mechanical energy, is more important. Here is his wording.

If only conservative forces act on a body, then its total mechanical energy, defined as the sum of kinetic and potential energies, is conserved.

In what follows we will need two more important equalities. As always, we will replace the conclusion with a simple demonstration of a special case of the gravity field. But the form of these equalities will be valid for any conservative forces.

Let us reduce equality (2.9.4) to the form

A=F∆x= Size E Size n2 = –( This amount is called total mechanical energy. We will denote it p.kon – Size n.beg)= – ∆U.

Here we looked at the work A when moving a body a distance ∆ x. The value ∆U, equal to the difference between the final and initial potential energy, is called the change in potential energy. And the resulting equality deserves a separate line and a special number. Let us hasten to assign it to him:

A=– ∆U (2.9.5)

From here follows the mathematical relationship between force and potential energy:

F= – ∆U/∆ x (2.9.6)

In the general case, not related to the gravitational field, equality (2.9.6) is the simplest differential equation

F= – dU/ dx.

Let's consider the last example without proof. The gravitational force is described by the law of universal gravitation F(r)= GmM/ r 2 and is conservative. The expression for the potential energy of the gravitational field has the form:

U(r)= – GmM/ r.

Author: – Let's look at a simple case. A body of mass m located on a horizontal plane is acted upon by T horizontal force F. There is no friction. What is the work done by force? F?

Student: – During T the body will move a distance S= AT 2 /2, where A=F/m. Therefore, the required job is A=F S= F 2 T 2 /(2m).

Author: Everything is correct if we assume that the body was at rest before the force began to act on it. Let's complicate the task a little. Let the body move rectilinearly and uniformly before the onset of the force with a certain speed V 0, co-directed with the external force. What is the work done in time now? T?

Student: – To calculate the displacement, I will take a more general formula S= V 0 T+AT 2/2, I get it for work A=F(V 0 T+AT 2 /2). Comparing with the previous result, I see that the same force produces different work over the same periods of time.

A body of mass m slides down an inclined plane with an angle of inclination α. Coefficient of sliding friction of a body on a plane k. A horizontal force acts on the body all the time F. What is the work done by this force when moving the body a distance S?

Student: – Let's arrange the forces and find their resultant. The body is acted upon by an external force F, as well as the forces of gravity, support reaction and friction.

Student: – It turns out that work A = F S cosα and that's it. I was really let down by the habit of looking for all the forces every time, especially since the problem indicated the mass and coefficient of friction.

Student: – Work of force F I already calculated: A 1 = F S cosα. The work done by gravity is A 2 =mgS sinα. The work of the friction force ... is negative, since the vectors of force and displacement are oppositely directed: A 3 = – kmgS cosα. Reaction force work N is equal to zero, because the force and displacement are perpendicular. Is it true that I don’t really understand the meaning of negative work?

Author: – This means that the work of a given force reduces the kinetic energy of the body. By the way. Let's discuss the motion of the body shown in Fig. 2.9.1 from the point of view of the law of conservation of energy. First, find the total work done by all forces.

Student: - A= A 1 + A 2 + A 3 = FS cosα+ mgS sinα– kmgS cosα.

According to the kinetic energy theorem, the difference between the kinetic energies in the final and initial states is equal to the work done on the body:

Size To - Size n = A.

Student: – Maybe these were other equations not related to this problem?

Author: – But all equations should give the same result. The point is that potential energy is contained latent in the expression for total work. Indeed, remember A 2 = mgS sinα=mgh, where h is the height of descent of the body. Now obtain from the kinetic energy theorem an expression for the law of conservation of energy.

Student: – Since mgh=U n – U k, where U n and U k are respectively the initial and final potential energies of the body, we have:

m V n 2 /2 + U n + A 1 + A 3 = m V to 2 /2+ U To.

Student: – This, in my opinion, is easy. The work done by the friction force is exactly equal in magnitude to the amount of heat Q. That's why Q= kmgS cosα.

Student: m V n 2 /2 + U n + A 1 – Q= m V to 2 /2+ U To.

Author: – Now let's somewhat generalize the definition of work. The fact is that relation (2.9.1) is true only for the case of a constant force. Although there are many cases when the force itself depends on the movement of the particle. Give an example.

Student: – The first thing that comes to mind is spring stretching. As the loose end of the spring moves, the force increases. The second example is related to a pendulum, which, as we know, is more difficult to hold with large deviations from the equilibrium position.

Author: – Fine. Let's look at the spring example. The elastic force of an ideal spring is described by Hooke's law, according to which when the spring is compressed (or stretched) by an amount X a force arises opposite to the displacement, linearly depending on X. Let's write Hooke's law as an equality:

F= – k x (2.9.2)

Here k is the spring stiffness coefficient, x– the amount of spring deformation. Draw a graph of the relationship F(x).

Student: My drawing is shown in the picture.

Fig.2.9.2

The left half of the graph corresponds to compression of the spring, and the right half corresponds to tension.

Author: – Now let's calculate the work done by force F when moving from X=0 to X= S. There is a general rule for this. If we know the general dependence of force on displacement, then the work on the section depends on x 1 up to x 2 is the area under the curveF(x) on this segment.

Student: – This means that the work done by the elastic force when moving a body from X=0 to X=S is negative, and its modulus is equal to the area of a right triangle: A= kS 2 /2.

A= k X 2 /2. (2.9.3)

This work is converted into potential energy of the deformed spring.

Story.

Rutherford demonstrated to listeners the decay of radium. The screen alternately glowed and darkened.

- Now you see – said Rutherford, – that nothing is visible. And why nothing is visible, you will now see.

Questions and tasks

1. List situations encountered in everyday life in which non-conservative forces are involved.

2. You slowly lift the book from the table onto a high shelf. List the forces acting on the book and determine which ones are conservative and which are not.

3. The resulting force acting on the particle is conservative and increases its kinetic energy by 300 J. What is the change in a) the potential energy of the particle, b) its total energy?

4. Does the following statement make physical sense: the use of poles made of flexible plastic in high jumps has led to an increase in results due to the fact that its greater flexibility provides additional elastic energy, converted into potential energy of the gravitational field?

5. There is an inclined plane, one end of which is raised to a height N. Body mass M rolls down (without initial speed) from the top point. Does the speed of this body at the base of the inclined plane depend on the angle it makes with the horizon, if a) there is no friction, b) there is friction?

6. Why do we still get tired when we first climb a mountain and then descend it? After all, the total work done in a gravitational field is zero.

7. This example is even tougher. Imagine that you are holding a dumbbell at arm's length. Don't worry, it's not very heavy. But still the hand gets tired. But there is no mechanical work, because there is no movement. Where does the energy of your muscles go?

8. Spring mass m rests in a vertical position on the table. Will the spring be able to jump and come off the table after you compress it, pressing from above, and then release it? Explain your answer using the law of conservation of energy.

9. What happens to the potential energy that the water had at the top of the waterfall when the water reaches its base? What happens to the kinetic and total energy?

10. Experienced tourists prefer to step over a fallen log, rather than step on it and jump off from the opposite side. Explain the phenomenon.

11. Two people are on different platforms that move relative to each other with speed V. They observe a log that is being pulled along a rough horizontal surface. Do the values obtained by these people coincide: a) kinetic energy of the log; b) total work done on the body; c) mechanical energy converted into thermal energy due to the presence of friction? Does the answer to question c) contradict the answers to questions a) and b)?

12. Where does the kinetic energy of a car come from when it accelerates uniformly from a state of rest? How can we relate the increase in kinetic energy to the presence of friction between the tires and the highway?

13. In winter, the Earth approaches the Sun at the shortest distance. When is the Earth's potential energy greatest?

14 Can total mechanical energy be negative? Give examples.

15. At what point is the force greatest? For each numbered point, indicate in which direction the force is acting. Which point corresponds to the equilibrium position?

Tasks

16. A bullet penetrates a fixed board at a minimum speed of 200 m/s. How fast must the bullet travel to pierce this board suspended by a long thread? Bullet weight 15 G, board weight 90 G, the bullet hits exactly the center of the board perpendicular to its surface.

17. Wooden ball of mass M =1 kg hangs on a cord so that the distance from the point of suspension of the cord to the center of the ball is equal to L= 1 m. The ball is hit by a plane flying horizontally at speed V 1 =400 m/s bullet mass m= 10 G, which pierces the ball exactly along its diameter and flies out of it at a speed V 2 =230 m/s. Define Angle  maximum deviation of the suspension from the vertical. Neglect air resistance and the time it takes for a bullet to penetrate the ball.

18. On a plane inclined to the horizon at an angle α, two bodies of mass m. Friction coefficient between bodies and plane k>tgα. The bodies are given the same counter velocities V. At what maximum initial distance L will they collide between bodies?

19. The cart rolls down smooth rails forming a vertical loop of radius R. From what minimum height H min should the cart roll so that it does not leave the rails along their entire length? What will be the motion of the cart if it rolls down from a height? A=mg(, smaller H min?

20. Determine the force acting on the vertical wall from the falling dumbbell at the moment when the axis of the dumbbell makes an angle  with the horizontal. The dumbbell begins its movement from a vertical position without an initial speed. The mass of each dumbbell ball is m.

21. On a thread length 2 A=mg( suspended weight m. On distance A=mg( a nail is driven under the suspension point. The thread was deflected from the equilibrium position by an angle of /2 and released. To what maximum height will the weight rise after passing through the equilibrium position?

22. Mass stand M with hemispherical recess radius R stands on a smooth horizontal plane. Small body of mass m Place it on the edge of the notch and release it. Find the velocities of the body and the stand, the force acting on the body at the moment of passing the lowest point

23. Weight mass m, suspended on a stiffening spring k, is held by the stand so that the spring is in an undeformed state. The stand is suddenly removed. Find the maximum elongation of the spring and the maximum speed of the load.

24. From a load suspended on a stiffening spring k, a part of the mass comes off m. To what height will the remaining part of the load rise after this?

25. How much force should be applied to the upper mass m, so that the lower load weighs M, connected to the upper stiffening spring k, came off the floor after the force ceased?

26. Two bodies with masses lie on a horizontal plane m 1 and m 2 connected by an undeformed spring. Find what is the smallest constant force that must be applied to the left body in order for the right one to move. The coefficient of friction between bodies and a plane is .

View: this article has been read 48440 times

Pdf Select language... Russian Ukrainian English

Short review

The entire material is downloaded above, after selecting the language

Two cases of transformation of the mechanical motion of a material point or system of points:

mechanical motion is transferred from one mechanical system to another as mechanical motion;
mechanical motion turns into another form of motion of matter (into the form of potential energy, heat, electricity, etc.).

When the transformation of mechanical motion without its transition to another form of motion is considered, the measure of mechanical motion is the vector of momentum of a material point or mechanical system. The measure of the force in this case is the vector of the force impulse.

When mechanical motion turns into another form of motion of matter, the kinetic energy of a material point or mechanical system acts as a measure of mechanical motion. The measure of the action of force when transforming mechanical motion into another form of motion is the work of force

Kinetic energy

Kinetic energy is the body's ability to overcome an obstacle while moving.

Kinetic energy of a material point

The kinetic energy of a material point is a scalar quantity that is equal to half the product of the mass of the point and the square of its speed.

Kinetic energy:

characterizes both translational and rotational movements;
does not depend on the direction of movement of the points of the system and does not characterize changes in these directions;
characterizes the action of both internal and external forces.

Kinetic energy of a mechanical system

The kinetic energy of the system is equal to the sum of the kinetic energies of the bodies of the system. Kinetic energy depends on the type of motion of the bodies of the system.

Determination of the kinetic energy of a solid body for different types of motion.

Kinetic energy of translational motion
During translational motion, the kinetic energy of the body is equal to T=m V 2 /2.

The measure of the inertia of a body during translational motion is mass.

Kinetic energy of rotational motion of a body

During the rotational motion of a body, kinetic energy is equal to half the product of the moment of inertia of the body relative to the axis of rotation and the square of its angular velocity.

A measure of the inertia of a body during rotational motion is the moment of inertia.

The kinetic energy of a body does not depend on the direction of rotation of the body.

Kinetic energy of plane-parallel motion of a body

With plane-parallel motion of a body, the kinetic energy is equal to

Work of force

The work of force characterizes the action of a force on a body during some movement and determines the change in the velocity modulus of the moving point.

Elementary work of force

The elementary work of a force is defined as a scalar quantity equal to the product of the projection of the force onto the tangent to the trajectory, directed in the direction of motion of the point, and the infinitesimal displacement of the point, directed along this tangent.

Work done by force on final displacement

The work done by a force on a final displacement is equal to the sum of its work on elementary sections.

The work of a force on a final displacement M 1 M 0 is equal to the integral of the elementary work along this displacement.

The work of a force on displacement M 1 M 2 is depicted by the area of the figure, limited by the abscissa axis, the curve and the ordinates corresponding to the points M 1 and M 0.

The unit of measurement for the work of force and kinetic energy in the SI system is 1 (J).

Theorems about the work of force

Theorem 1. The work done by the resultant force on a certain displacement is equal to the algebraic sum of the work done by the component forces on the same displacement.

Theorem 2. The work done by a constant force on the resulting displacement is equal to the algebraic sum of the work done by this force on the component displacements.

Power

Power is a quantity that determines the work of a force per unit of time.

The unit of power measurement is 1W = 1 J/s.

Cases of determining the work of forces

Work of internal forces

The sum of the work done by the internal forces of a rigid body during any movement is zero.

Work of gravity

Work of elastic force

Work of friction force

Work of forces applied to a rotating body

The elementary work of forces applied to a rigid body rotating around a fixed axis is equal to the product of the main moment of external forces relative to the axis of rotation and the increment in the angle of rotation.

Rolling resistance

In the contact zone of the stationary cylinder and the plane, local deformation of contact compression occurs, the stress is distributed according to an elliptical law, and the line of action of the resultant N of these stresses coincides with the line of action of the load force on the cylinder Q. When the cylinder rolls, the load distribution becomes asymmetrical with a maximum shifted towards movement. The resultant N is displaced by the amount k - the arm of the rolling friction force, which is also called the rolling friction coefficient and has the dimension of length (cm)

Theorem on the change in kinetic energy of a material point

The change in the kinetic energy of a material point at a certain displacement is equal to the algebraic sum of all forces acting on the point at the same displacement.

Theorem on the change in kinetic energy of a mechanical system

The change in the kinetic energy of a mechanical system at a certain displacement is equal to the algebraic sum of the internal and external forces acting on the material points of the system at the same displacement.

Theorem on the change in kinetic energy of a solid body

The change in the kinetic energy of a rigid body (unchanged system) at a certain displacement is equal to the sum of the external forces acting on points of the system at the same displacement.

Efficiency

Forces acting in mechanisms

Forces and pairs of forces (moments) that are applied to a mechanism or machine can be divided into groups:

1. Driving forces and moments that perform positive work (applied to the driving links, for example, gas pressure on the piston in an internal combustion engine).

2. Forces and moments of resistance that perform negative work:

useful resistance (they perform the work required from the machine and are applied to the driven links, for example, the resistance of the load lifted by the machine),
resistance forces (for example, friction forces, air resistance, etc.).

3. Gravity forces and elastic forces of springs (both positive and negative work, while the work for a full cycle is zero).

4. Forces and moments applied to the body or stand from the outside (reaction of the foundation, etc.), which do not do work.

5. Interaction forces between links acting in kinematic pairs.

6. The inertial forces of the links, caused by the mass and movement of the links with acceleration, can perform positive, negative work and do not perform work.

Work of forces in mechanisms

When the machine operates at a steady state, its kinetic energy does not change and the sum of the work of the driving forces and resistance forces applied to it is zero.

The work expended in setting the machine in motion is expended in overcoming useful and harmful resistances.

Mechanism efficiency

The mechanical efficiency during steady motion is equal to the ratio of the useful work of the machine to the work expended on setting the machine in motion:

Machine elements can be connected in series, parallel and mixed.

Efficiency in series connection

When mechanisms are connected in series, the overall efficiency is less than the lowest efficiency of an individual mechanism.

Efficiency in parallel connection

When mechanisms are connected in parallel, the overall efficiency is greater than the lowest and less than the highest efficiency of an individual mechanism.

Format: pdf

Language: Russian, Ukrainian

Calculation example of a spur gear
An example of calculating a spur gear. The choice of material, calculation of permissible stresses, calculation of contact and bending strength have been carried out.

An example of solving a beam bending problem
In the example, diagrams of transverse forces and bending moments were constructed, a dangerous section was found and an I-beam was selected. The problem analyzed the construction of diagrams using differential dependencies and carried out a comparative analysis of various cross sections of the beam.

An example of solving a shaft torsion problem
The task is to test the strength of a steel shaft at a given diameter, material and allowable stress. During the solution, diagrams of torques, shear stresses and twist angles are constructed. The shaft's own weight is not taken into account

An example of solving a problem of tension-compression of a rod
The task is to test the strength of a steel bar at specified permissible stresses. During the solution, diagrams of longitudinal forces, normal stresses and displacements are constructed. The rod's own weight is not taken into account

Application of the theorem on conservation of kinetic energy
An example of solving a problem using the theorem on the conservation of kinetic energy of a mechanical system

Theorem on the kinetic energy of a point in differential form

Multiplying scalarly both sides of the equation of motion of a material point by the elementary displacement of the point we obtain

or, since , then

A scalar quantity or half the product of the mass of a point and the square of its speed is called the kinetic energy of a point or the living force of a point.

The last equality constitutes the content of the theorem on the kinetic energy of a point in differential form, which states: the differential of the kinetic energy of a point is equal to the elementary work acting on the point of force.

The physical meaning of the theorem on kinetic energy is that the work done by a force acting on a point accumulates in it as kinetic energy of motion.

Theorem on the kinetic energy of a point in integral form

Let the point move from position A to position B, passing along its trajectory the final arc AB (Fig. 113). Integrating the equality from A to B:

where are the velocities of the point in positions A and B, respectively.

The last equality constitutes the content of the theorem on the kinetic energy of a point in integral form, which states: the change in the kinetic energy of a point over a certain period of time is equal to the work done during the same time by the force acting on it.

The resulting theorem is valid when a point moves under the influence of any force. However, as indicated, to calculate the total work of a force, in the general case it is necessary to know the equations of motion of a point.

Therefore, the theorem on kinetic energy, generally speaking, does not give the first integral of the equations of motion.

Energy integral

The kinetic energy theorem gives the first integral of the equations of motion of a point if the total work done by a force can be determined without resorting to the equations of motion. The latter is possible, as previously indicated, if the force acting on the point belongs to the force field. In this case, it is enough to know only the trajectory of the point. Let the trajectory of a point be some kind of curve, then the coordinates of its points can be expressed through the arc of the trajectory, and, therefore, the force depending on the coordinates of the point can be expressed through

and the kinetic energy theorem gives the first integral of the form

where are the arcs of the trajectory corresponding to points A and is the projection of the force on the tangent to the trajectory (Fig. 113).

Potential energy and the law of conservation of mechanical energy of a point

Of particular interest is the motion of a point in a potential field, since the theorem on kinetic energy gives a very important integral of the equations of motion.

In a potential field, the total work done by a force is equal to the difference between the values of the force function at the end and at the beginning of the path:

Therefore, the kinetic energy theorem in this case is written as:

The force function taken with the opposite sign is called the potential energy of a point and is denoted by the letter P:

The potential energy, as well as the force function, is specified up to an arbitrary constant, the value of which is determined by the choice of the zero level surface. The sum of the kinetic and potential energy of a point is called the total mechanical energy of the point.

The theorem on the kinetic energy of a point, if the force belongs to the potential field, is written as:

where are the values of potential energy corresponding to points A and B. The resulting equation constitutes the content of the law of conservation of mechanical energy for a point, which states: when moving in a potential field, the sum of the kinetic and potential energy of the point remains constant.

Since the law of conservation of mechanical energy is valid only for forces belonging to potential fields, the forces of such a field are called conservative (from the Latin verb conservare - to preserve), which emphasizes the fulfillment of the formulated law in this case. Note that if the concept of kinetic energy has known physical foundations in its definition, then the concept of potential energy does not have this. The concept of potential energy in a certain sense is a fictitious quantity, which is defined so that changes in its value exactly correspond to changes in kinetic energy. The introduction of this quantity associated with motion helps the description of motion and, due to this, plays a significant role in the so-called energy description of motion, developed by analytical mechanics. The latter is the meaning of introducing this value.

Let's start with a definition. Job A strength F when moving X of the body to which it is applied is defined as the scalar product of vectors F And X .

A=F x= Fxcosα.(2.9.1)

Where α – the angle between the directions of force and displacement.

a x=(V 2 –V 0 2)/2.

Let's multiply both sides of the equation by the mass of the particle, we get

Fx=m(V 2 –V 0 2)/2.

Finally

A= m V 2 /2 – m V 0 2 /2. (2.9.1)

Size Size=m V 2 /2 is called the kinetic energy of the particle.

You are used to the fact that in geometry theorems have their own oral formulation. To keep up with this tradition, let us present the theorem on kinetic energy in text form.

The change in the kinetic energy of a body is equal to the work done by all the forces acting on it.

This theorem is universal, i.e. it is valid for any type of movement. However, its exact proof involves the use of integral calculus. Therefore we omit it.

Let's take some point in the gravitational field as the origin and consider the work done by gravity when moving a particle to this point from another arbitrary point Let's take some point in the gravitational field as the origin and consider the work done by gravity when moving a particle to this point from another arbitrary point, located at a height A=mg(. This work is equal to . This work is equal to and is called potential energy Size n particles at a point Let's take some point in the gravitational field as the origin and consider the work done by gravity when moving a particle to this point from another arbitrary point:

Size n = mgh(2.9.3)

Now we transform equality (2.9.1), the mechanical theorem about kinetic energy takes the form

A= m V 2 /2 – m Now we transform equality (2.9.1), the mechanical theorem about kinetic energy takes the form This amount is called total mechanical energy. We will denote it p1 – Size p2. (2.9.4)

m p2. This amount is called total mechanical energy. We will denote it n2 = m V 2 /2+ This amount is called total mechanical energy. We will denote it p1.

In this equality, on the left side is the sum of kinetic and potential energy at the final point of the trajectory, and on the right - at the initial point.

This amount is called total mechanical energy. We will denote it Size.

Size=This amount is called total mechanical energy. We will denote it k + This amount is called total mechanical energy. We will denote it k +

We have arrived at the law of conservation of total energy: in a closed system, total energy is conserved.

However, one note should be made. While we were looking at an example of the so-called conservative forces. These forces depend only on the position in space. And the work done by such forces when moving a body from one position to another depends only on these two positions and does not depend on the path. The work done by a conservative force is mechanically reversible, that is, it changes its sign when the body returns to its original position. Gravity is a conservative force. In the future, we will get acquainted with other types of conservative forces, for example, with the force of electrostatic interaction.

If only conservative forces act on a body, then its total mechanical energy, defined as the sum of kinetic and potential energies, is conserved.

Let us reduce equality (2.9.4) to the form

A=F∆x= E p1 – Size n2 = –( This amount is called total mechanical energy. We will denote it p.kon – Size n.beg)= – ∆U.

A=– ∆U (2.9.5)

From here follows the mathematical relationship between force and potential energy:

F= – ∆U/∆ x(2.9.6)

In the general case, not related to the gravitational field, equality (2.9.6) is the simplest differential equation

F= – dU/dx.

Let's consider the last example without proof. The gravitational force is described by the law of universal gravitation F(r)=GmM/r 2 and is conservative. The expression for the potential energy of the gravitational field has the form:

U(r)= –GmM/r.

Author: – Let's look at a simple case. A body of mass m located on a horizontal plane is acted upon by T horizontal force F. There is no friction. What is the work done by force? F?

Student: – During T the body will move a distance S= aT 2 /2, where A=F/m. Therefore, the required job is A=F S= F 2 T 2 /(2m).

Student: – To calculate the displacement, I will take a more general formula S= V 0 T+aT 2/2, I get it for work A=F(V 0 T+aT 2 /2). Comparing with the previous result, I see that the same force produces different work over the same periods of time.

Student: – Let's arrange the forces and find their resultant. The body is acted upon by an external force F, as well as the forces of gravity, support reaction and friction.

Student: - A= A 1 + A 2 + A 3 = FS cosα+ mgS sinα– kmgS cosα.

According to the kinetic energy theorem, the difference between the kinetic energies in the final and initial states is equal to the work done on the body:

Size To - Size n = A.

Student: – Maybe these were other equations not related to this problem?

Student: – Since mgh=U n – U k, where U n and U k are respectively the initial and final potential energies of the body, we have:

m V n 2 /2 + U n + A 1 + A 3 = m V to 2 /2+ U To.

Student: – This, in my opinion, is easy. The work done by the friction force is exactly equal in magnitude to the amount of heat Q. That's why Q= kmgS cosα.

Student: m V n 2 /2 + U n + A 1 – Q= m V to 2 /2+ U To.

F= – k x (2.9.2)

Here k is the spring stiffness coefficient, x– the amount of spring deformation. Draw a graph of the relationship F(x).

Student: My drawing is shown in the picture.

Fig.2.9.2

The left half of the graph corresponds to compression of the spring, and the right half corresponds to tension.

Author: – Now let's calculate the work done by force F when moving from X=0 to X= S. There is a general rule for this. If we know the general dependence of force on displacement, then the work on the section from x 1 to x 2 is the area under the curve F (x) on this segment.

Student: – This means that the work done by the elastic force when moving a body from X=0 to X=S is negative, and its modulus is equal to the area of a right triangle: A= kS 2 /2.

A= k X 2 /2. (2.9.3)

This work is converted into potential energy of the deformed spring.

Story.

Rutherford demonstrated to listeners the decay of radium. The screen alternately glowed and darkened.

– Now you see – said Rutherford, – that nothing is visible. And why nothing is visible, you will now see.

The scalar quantity T, equal to the sum of the kinetic energies of all points of the system, is called the kinetic energy of the system.

Kinetic energy is a characteristic of the translational and rotational motion of a system. Its change is influenced by the action of external forces and since it is a scalar, it does not depend on the direction of movement of the parts of the system.

Let's find the kinetic energy for various cases of motion:

1.Forward movement

The velocities of all points of the system are equal to the velocity of the center of mass. Then

The kinetic energy of the system during translational motion is equal to half the product of the mass of the system and the square of the velocity of the center of mass.

2. Rotational movement(Fig. 77)

Speed of any point on the body: . Then

or using formula (15.3.1):

The kinetic energy of a body during rotation is equal to half the product of the moment of inertia of the body relative to the axis of rotation and the square of its angular velocity.

3. Plane-parallel motion

For a given movement, kinetic energy consists of the energy of translational and rotational movements

The general case of motion gives a formula for calculating kinetic energy similar to the last one.

We made the definition of work and power in paragraph 3 of Chapter 14. Here we will look at examples of calculating the work and power of forces acting on a mechanical system.

1.Work of gravity forces. Let , coordinates of the initial and final positions of point k of the body. The work done by the force of gravity acting on this particle of weight will be . Then the complete work:

where P is the weight of the system of material points, is the vertical displacement of the center of gravity C.

2. Work of forces applied to a rotating body.

According to relation (14.3.1), we can write , but ds according to Figure 74, due to its infinite smallness, can be represented in the form - an infinitesimal angle of rotation of the body. Then

Magnitude called torque.

We rewrite formula (19.1.6) as

Elementary work is equal to the product of torque times elementary rotation.

When rotating through the final angle we have:

If the torque is constant, then

and the power is determined from the relation (14.3.5)

as the product of torque and angular velocity of the body.

The theorem on the change in kinetic energy proven for a point (§ 14.4) will be valid for any point in the system

By composing such equations for all points of the system and adding them term by term, we obtain:

or, according to (19.1.1):

which is an expression of the theorem on the kinetic energy of a system in differential form.

Integrating (19.2.2) we get:

The theorem on the change in kinetic energy in final form: the change in the kinetic energy of a system during some finite movement is equal to the sum of the work on this movement of all external and internal forces applied to the system.

We emphasize that internal forces are not excluded. For an unchangeable system, the sum of the work done by all internal forces is zero and

If the constraints imposed on the system do not change over time, then the forces, both external and internal, can be divided into active and reaction constraints, and equation (19.2.2) can now be written:

In dynamics, the concept of an “ideal” mechanical system is introduced. This is a system in which the presence of connections does not affect the change in kinetic energy, that is

Such connections, which do not change with time and whose sum of work on an elementary displacement is zero, are called ideal, and equation (19.2.5) will be written:

The potential energy of a material point in a given position M is the scalar quantity P, equal to the work that the field forces will produce when moving the point from position M to zero

P = A (mo) (19.3.1)

Potential energy depends on the position of point M, that is, on its coordinates

P = P(x,y,z) (19.3.2)

Let us explain here that a force field is a part of a spatial volume, at each point of which a force of a certain magnitude and direction acts on a particle, depending on the position of the particle, that is, on the coordinates x, y, z. For example, the Earth's gravitational field.

A function U of coordinates whose differential is equal to work is called power function. A force field for which a force function exists is called potential force field, and the forces acting in this field are potential forces.

Let the zero points for two force functions P(x,y,z) and U(x,y,z) coincide.

Using formula (14.3.5) we obtain, i.e. dA = dU(x,y,z) and

where U is the value of the force function at point M. Hence

П(x,y,z) = -U(x,y,z) (19.3.5)

Potential energy at any point of the force field is equal to the value of the force function at this point, taken with the opposite sign.

That is, when considering the properties of the force field, instead of the force function, we can consider potential energy and, in particular, equation (19.3.3) will be rewritten as

The work done by a potential force is equal to the difference between the potential energy values of a moving point in the initial and final positions.

In particular, the work of gravity:

Let all forces acting on the system be potential. Then for each point k of the system the work is equal to

Then for all forces, both external and internal, there will be

where is the potential energy of the entire system.

We substitute these sums into the expression for kinetic energy (19.2.3):

or finally:

When moving under the influence of potential forces, the sum of the kinetic and potential energy of the system in each of its positions remains constant. This is the law of conservation of mechanical energy.

A load weighing 1 kg oscillates freely according to the law x = 0.1sinl0t. Spring stiffness coefficient c = 100 N/m. Determine the total mechanical energy of the load at x = 0.05 m, if at x = 0 the potential energy is zero . (0,5)

A load of mass m = 4 kg, falling down, causes a cylinder of radius R = 0.4 m to rotate with the help of a thread. The moment of inertia of the cylinder relative to the axis of rotation is I = 0.2. Determine the kinetic energy of the system of bodies at the moment of time when the speed of the load v = 2m/s . (10,5)