Lecture 5. Theorem on the change of kinetic energy

5. 1. Work of force

May the force – the resultant of all forces of the system, applied to point P, and ( dx, dy, dz) – elementary movement of point P along its trajectory P 1 P 2 (Fig. 5.1). Elementary work dA forces are called scalar product

Elementary work is a scalar quantity. If is the angle between the force and the direction of displacement , then expression (5.1) can be represented as

where is the projection of force onto the direction of elementary displacement (or the direction of point velocity).

The sign of the elementary work depends on the sign of the function. If - acute angle, then , if is an obtuse angle, then , if , then .

Let the point R makes a final movement from position to position, describing an arc. Let's split the arc into n arbitrary small sections, indicating the length of the section with the number k through . Then the elementary work of force on k- section will be equal to , and all the way from to - the amount of work in individual sections

We obtain the exact value of the work by moving to the limit, provided that the number of sections n increases indefinitely, and the length of each section decreases:

.

Such a limit is called a curvilinear integral of the first kind along an arc and is written as follows

. (5.3)

The result of integration is the complete work A strength F on the considered finite displacement along the path.

5. 1. 1. Work of gravity

Let m – point mass, g– acceleration free fall. Then

Calculating the work using formulas (5.1) and (5.3), we have

where is the height of the point's descent.

When the point rises, therefore, .

5. 1. 2. Work of linear elastic force

Let the material point R moves along the axis Oh(Fig. 5.3) under the action of the spring to which it is attached. If at , , then the spring is deformed and for small deviations of the point we can assume that an elastic force is applied to it from the side of the spring. Then the work of the elastic force on displacement x 0 x 1 will be equal

. (5.5)

The work of the elastic force is equal to half the product of the stiffness coefficient and the difference between the squares of the initial and final elongation (or compression) of the spring.

5. 1. 3. Elementary work of forces applied to a solid body

Let's consider the motion of a body in a plane. Let ABOUT– an arbitrarily selected point on a solid body (Fig. 5.4). Let's call it a pole. Then the motion of a body in a plane can be represented as the sum of the simplest: translational motion together with the pole and rotation of the body around the pole. Then, the speed of the point relative to the fixed coordinate system will be determined as the geometric sum of two speeds

where is the speed of the pole, is the vector of the angular velocity of the rigid body, is the Euler speed, i.e. the speed of the point as it rotates around the pole.

We will represent a solid body as a mechanical system consisting of N individual points, the mutual distance between which does not change.

Let's calculate the displacement of a point under the influence of force:

Then .

The elementary work, according to (5.1), will be written as follows

Using the properties of a mixed product of vectors , we rewrite the last expression in the form

Let be the resultant of all forces, external and internal (Fig. 5.4), applied at a point on the body, i.e.

.

Then (a) will be written like this

According to (3.1 and 3.2), the main vector and main point internal forces systems are equal to zero, we get

Here: – main vector, – the main moment of external forces relative to the point ABOUT.

Special cases

A. Translational motion of a rigid body. All points of the body have the same displacements (Fig. 5.5, a) both in magnitude and in direction, then, from (5.6), we obtain (here):

. (5.7)

B. Rotating a rigid body around fixed axis . Let the axis z passes through the pole ABOUT(Fig. 5.5b). Then , ; from (5.6) we obtain

. (5.8)

Example. Coil mass m and radius R driven by constant force F, applied at the point A(Fig. 5.6). The reel rolls to the right without slipping on the rough surface.

Calculate the work of all external forces if the center of the coil has moved by a distance, - rolling friction coefficient, - friction force, r - radius of the coil core to which the force is applied.

Solution. The coil moves in a plane motion. Since rolling occurs without sliding, the instantaneous center of speed is located at the point of contact of the coil with the plane, i.e. at the point R(Fig. 5.6). Let's direct the S axis horizontally to the right. In accordance with the direction of movement we take positive direction rotation angle counterclockwise.

Let the center of the coil WITH will move to . In this case, the coil will rotate at an angle. Then where from

Having accepted the point R for the instantaneous axis of rotation, we calculate the elementary work using formula (5.8):

(A)

Here: lines of action of forces and mg intersect the axis of rotation, therefore; further, where N– strength of normal reaction.

To determine the required work, it remains to take a definite integral from (a) in the range from 0 to SA. We get

5. 2. Force field. Power function. Potential energy

Let us assume that a point is moving in some space and is acted upon by a force from space that depends on the position of the point in this space, but does not depend on the speed of the point’s movement. In this case, they say that the space is given force field, and also that the point moves in a force field. The corresponding concepts for a system of material points are similar.

Forces depending on the position of the points of their application are often encountered in mechanics. For example, an elastic force applied to a material point that moves along a horizontal line under the action of a spring. The most important example force field in nature there is a gravitational field: the action of the Sun on a planet of a given mass is determined at each point in space by the law universal gravity.

The force field is called potential, if there is a scalar function U, depending only on the coordinates , , point - point material system(possibly from time to time), such that

The function is called power function.

Let's consider the properties of the force function.

Elementary work (5.1) is related to the force function as follows

Thus, the elementary work of force in a potential force field is equal to full differential from power function ii.

Total work of force in the area from the point to the point (Fig.5.1)

those. . (5.10)

From the obtained expressions it follows that

1. the work done by a force in a potential force field along any closed path is zero;

2. the work of force in a potential force field depends only on the position of the final and initial points, but the path of movement itself does not matter.

Potential energy. Potential energy P at the considered point of the force field R is the work done by field forces acting on a material point when it moves from a point R V starting point 1, i.e.

P= or P=

Let's connect the force function U with potential energy. We have

Potential Energy Calculation Examples

1. Uniform gravity field. Let m– point mass; g – free fall acceleration. Then (Fig. 5.2)

2. Elastic spring force field. Let the material point move along the axis Oh(Fig. 5.3) under the action of the spring to which it is attached. If at the spring is not deformed, then, assuming in formula (5.5), we obtain

.

5. 3. Kinetic energy

5. 3. 1. Kinetic energy of the system. Koenig's theorem

Kinetic energy material point they call half the product of the mass of a point and the square of its velocity, i.e. . Kinetic energy is a positive scalar quantity. In the SI system, the unit of kinetic energy is the joule: .

Kinetic energy mechanical system is the sum of the kinetic energies of all points included in the system:

(5.11)

The velocities of the points of system (5.1) are determined relative to a fixed reference frame.

Let us align the origin of coordinates with the center of mass of the system. Let us assume that the mechanical system, together with the coordinate system, moves translationally relative to the fixed coordinate system (Fig. 5.7). Point – point of the system.

Then, based on the theorem on the addition of velocities, the absolute speed of the point Rk. the system will be written like this vector sum portable and relative speeds:

, (A)

where is the speed of the origin of the moving coordinate system (transferable speed, i.e. the speed of the center of mass of the system); – point speed Rk relative to the moving coordinate system Ohooz (relative speed).

Substituting (a) into formula (5.11), we obtain

(5.12)

Here is the mass of the entire system.

The radius vector of the center of mass of the system in the moving coordinate system is determined, according to (2.1), – , where , i.e. . Since the origin ABOUT is the center of mass of the system, then , then, i.e. the second sum in expression (5.12) is equal to zero.

Thus, the kinetic energy of system (5.12) has the form

(5.13)

This equality determines Koenig's theorem.

Theorem. The kinetic energy of a system is equal to the sum of the kinetic energy that a material point located at the center of mass of the system and having a mass equal to the mass of the system would have, and the kinetic energy of motion of the system relative to the center of mass.

5. 3. 2. Kinetic energy of a solid body

A rigid body is a special case of a mechanical system and is considered as a continuously distributed mass, then all the sums included in the expression for the kinetic energy of the system go into integrals. Thus, for a solid body, formula (5.11) will take the form

. (5.14)

1. Kinetic energy of a rigid body moving forward.

With this type of movement, the velocities of all points of the body are the same (Fig. 5.8). Taking out the integral sign in formula (5.14), we obtain

. (5.15)

The kinetic energy of a rigid body moving translationally is equal to half the product of the mass of the bodyMby the square of its speed.

2. Kinetic energy of a rigid body rotating around a fixed axis

Speed ​​module V of any point of a rigid body rotating around a fixed axis is equal to , where is the modulus of the angular velocity of the rigid body, is the distance from the point to the axis of rotation z(Fig. 5.9). Substituting into formula (5.14), we get

Here – moment of inertia of a rigid body relative to the axis z.

The kinetic energy of a rigid body rotating around a fixed axis is equal to half the product of the moment of inertia of the body relative to the axis of rotation and the square of the angular velocity of the body.

3. Kinetic energy of a rigid body during plane-parallel motion

In plane-parallel motion, the speed of any point on the body consists of geometric sum the speed of the pole and the speed of the point when rotating around the pole. Let the body move flat in a plane Oxy, Then

|| . We choose the center of mass of the body as the pole, then in formula (5.13), the speed is the speed of the point k body during its rotation relative to the pole (center of mass) and is equal to , where is the distance k- oh point to the pole. Then (5.13) will be rewritten

Bearing in mind that – moment of inertia of the body relative to the axis z passing through the pole WITH, the last expression can be rewritten as

, (5.17)

in plane-parallel motion of a body, kinetic energy consists of the kinetic energy of translational motion together with the center of mass and kinetic energy from rotation around an axis passing through the center of mass and perpendicular to the plane movements.

5. 4. Theorem on the change in kinetic energy

5. 4. 1. Theorem on the change in kinetic energy of a point

Let's find the connection between work and change in speed. Let a material point with mass m moves along the axis Oh under the influence of a force, for example a compressed or decompressed spring, fixed at the origin of coordinates - a point ABOUT(Fig. 5.10). The equation of motion of a point has the form

Let's multiply both sides of this equation by , and, taking into account that , we get

. (5.19)

On the right side of this equality we replace V x by and multiply by dt right and left sides. Then

. (5.20)

In this form, equality has a very clear meaning: when the point is shifted by dx, the force does work, as a result of which the quantity changes kinetic energy of a point, characterizing the motion of a point and, in particular, the modulus of its velocity. If a point moves from a position to , and its speed changes from to , then, integrating (5.20), we have

. (5.21)

Considering that , we finally find

. (5.22)

The change in the kinetic energy of a material point during any movement is equal to the work done by the force acting on the point at the same movement.

Carrying out all the previous procedures, we get

,

here is the arc along which the point moves (Fig. 5.11).

5. 4. 2. Theorem on the change in the kinetic energy of the system

Let the points of the mass system move so that their radius vectors in the inertial reference system receive an increment. Let's find how the kinetic energy changed T systems.

According to (5.11), the kinetic energy of the system

.

Let us calculate the differential of the kinetic energy of the system and transform the resulting expression

Here

Taking into account that , where is the acceleration of point a and are the resultant external and internal forces applied to the point, we rewrite the last equality in the form

Thus,

. (5.23)

The last equality expresses the theorem on the change in the kinetic energy of a mechanical system in differential form: the differential of the kinetic energy of the system is equal to the elementary work of all forces of the system.

Special case . For an absolutely rigid body, the sum of the work done by all internal forces of the system is equal to zero:

.

Consequently, the theorem on the change in kinetic energy (5.23) for a rigid body can be written in the form

The change in the kinetic energy of a solid body during any elementary displacement is equal to the elementary work of external forces acting on the body.

If both sides of (5.24) are integrated between two positions – initial and final, in which the kinetic energy and are respectively, we obtain

. (5.25)

Example 1. Disk mass m=5 kg and the radius is set in motion by a constant force applied at the point A(Fig. 5.6). The disk rolls on a rough surface to the right without slipping. Determine the speed of the center of mass WITH coil at the moment when it moves a distance , sliding friction coefficient , , radius of gyration of the disk

Solution. The disk moves in a plane motion. Let us write down the theorem on the change in kinetic energy for a solid body

Let's calculate the kinetic energy of the disk. At the initial moment of time, the disk was at rest, i.e. . Kinetic energy at the final position of the disk

The kinetic energy of a mechanical system consists of the kinetic energies of all its points:

Differentiating each part of this equality with respect to time, we obtain

Using the basic law of dynamics to To th point of the system m k 2i k= Fj., we arrive at the equality

The scalar product of force F and velocity v at the point of its application is called force power and denote R:

Using this new notation, we represent (11.6) in the following form:

The resulting equality expresses the differential form of the theorem on the change in kinetic energy: the rate of change of kinetic energy of a mechanical system is equal to the sum of jpowers of all cm acting on the system.

Presenting the derivative f in (8.5) in fraction form -- and performing

then separating the variables, we get:

Where dT- kinetic energy differential, i.e. its change over an infinitesimal period of time dr, dr k = k dt - elementary movement To- th points of the system, i.e. movement in time dt.

Scalar product of force F and elementary displacement dr its points of application are called basic work forces and denote dA:

Using Properties dot product the elementary work of force can also be represented in the form

Here ds = dr - arc length of the trajectory of the force application point, corresponding to its elementary displacement s/g; A - the angle between the directions of the force vector F and the elementary displacement vector c/r; F„ F y , F,- projections of the force vector F onto the Cartesian axes; dx, dy, dz - projections onto the Cartesian axes of the vector of elementary displacement s/g.

Taking into account notation (11.9), equality (11.8) can be represented in the following form:

those. the differential of the kinetic energy of the system is equal to the sum basic work all forces acting on the system. This equality, like (11.7), expresses the differential form of the theorem on the change in kinetic energy, but differs from (11.7) in that it uses not derivatives, but infinitesimal increments - differentials.

Performing term-by-term integration of equality (11.12), we obtain

where the following are used as integration limits: 7 0 - kinetic energy of the system at a moment in time? 0 ; 7) - kinetic energy of the system at the moment of time tx.

Definite integrals by time period or A(F):

Note 1. To calculate work, it is sometimes more convenient to use a non-arc parameterization of the trajectory M(s), and coordinate M(x(t), y(/), z(f)). In this case, for elementary work it is natural to take representation (11.11), and line integral present in the form:

Taking into account the notation (11.14) of work on a finite displacement, equality (11.13) takes the form

and represents the final form of the theorem on the change in kinetic energy of a mechanical system.

Theorem 3. The change in the kinetic energy of a mechanical system when it moves from the initial position to the final position is equal to the sum of the work of all forces acting on points of the system during this movement.

Comment 2. The right side of equality (11.16) takes into account the work with all our might, acting on the system, both external and internal. Nevertheless, there are mechanical systems for which the total work done by all internal forces is zero. Egos so called immutable systems, in which the distances between interacting material points do not change. For example, the system solids, connected by frictionless hinges or flexible inextensible threads. For such systems, in equality (11.16) it is sufficient to take into account only the work of external forces, i.e. Theorem (11.16) takes the form:

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 times angular velocity bodies.

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 final movement is equal to the sum of the work done 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)

If we consider any point of the system with mass , having speed , then for this point it will be

,

where and - elementary work of external and internal forces acting on a point. Compiling such equations for each of the points of the system and adding them term by term, we obtain

,

. (2)

Equality expresses the theorem about the change in the kinetic energy of the system in differential form.

If the resulting expression is related to the elementary time period during which the movement in question occurred, we can obtain a second formulation for the differential form of the theorem: the time derivative of the kinetic energy of the mechanical system is equal to the sum of the powers of all external () and internal () forces, i.e.

Differential forms of the theorem on the change in kinetic energy can be used to compile differential equations movements, but this is done quite rarely, because there are more convenient techniques.

Having integrated both sides of equality (2) within the limits corresponding to the movement of the system from some initial position, where the kinetic energy is equal to , to a position where the value of the kinetic energy becomes equal , we will have

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

Unlike previous theorems, internal forces are not excluded from the equations. In fact, if and are the forces of interaction between points and the system (see Fig. 51), then . But at the same time, the point , can move towards , and the point - towards . The work done by each force will then be positive and the sum of the work will not be zero. An example is the phenomenon of rollback. Internal forces (pressure forces), acting on both the projectile and the rolling parts, do positive work here. The sum of these works, which is not equal to zero, changes the kinetic energy of the system from the value at the beginning of the shot to the value at the end.

Another example: two points connected by a spring. When the distance between points changes, the elastic forces applied to the points will do work. But if the system consists of absolutely rigid bodies and the connections between them are unchangeable, non-elastic, ideal, then the work of internal forces will be equal to zero and they can be ignored and not shown at all on the design diagram.

Let's consider two important special cases.

1) Immutable system. Immutable we will call a system in which the distances between the points of application of internal forces do not change when the system moves. In particular, such a system is an absolutely rigid body or an inextensible thread.

Fig.51

Let two points of an unchangeable system (Fig. 51), acting on each other with forces and () have velocities and at the moment. Then over a period of time dt these points will make elementary movements and , directed along the vectors and . But since the segment is immutable, then, according to the well-known theorem of kinematics, the projection of vectors and , and, consequently, both the displacements and the direction of the segment will be equal to each other, i.e. . Then the elementary work of forces will be identical in magnitude and opposite in sign and in total will give zero. This result is valid for all internal forces for any movement of the system.

From here we conclude that for an unchangeable system, the sum of the work done by all internal forces is zero and the equations take the form

2) System with ideal connections. Let us consider a system on which connections are imposed that do not change over time. Let us divide all external and internal forces acting on points of the system into active And connection reactions. Then

,

where is the elementary work acting on k- th point of the system of external and internal active forces, a is the elementary work of reactions imposed on the same point by external and internal connections.

As we see, the change in the kinetic energy of the system depends on the work and active forces and reactions of bonds. However, it is possible to introduce the concept of such “ideal” mechanical systems in which the presence of connections does not affect the change in the kinetic energy of the system during its movement. For such connections, the following condition must obviously be satisfied:

If for connections that do not change with time, the sum of the work done by all reactions during an elementary displacement of the system is equal to zero, then such connections are called ideal. For a mechanical system on which only ideal connections that do not change with time are imposed, we will obviously have

Thus, the change in the kinetic energy of a system with ideal connections that do not change over time during any movement of it is equal to the sum of the work on this movement applied to the system external and internal active forces.

The mechanical system is called conservative(its energy is, as it were, conserved and does not change), if for it the energy integral takes place

or (3)

It's there law of conservation of mechanical energy: when a system moves in a potential field, its mechanical energy (the sum of potential and kinetic) remains unchanged and constant all the time.

A mechanical system will be conservative if the forces acting on it are potential, for example gravity, elastic forces. In conservative mechanical systems, the energy integral can be used to check the correctness of the differential equations of motion. If the system is conservative, and condition (3) is not satisfied, then an error was made in drawing up the equations of motion.

The energy integral can be used to check the correctness of equations in another way, without calculating the derivative. To do this, after carrying out numerical integration equations of motion, calculate the value of the total mechanical energy for two various moments time, for example, start and end. If the difference in values ​​turns out to be comparable to the calculation errors, this will indicate the correctness of the equations used.

All previous theorems made it possible to exclude internal forces from the equations of motion, but all external forces, including previously unknown reactions external relations, were retained in the equations. The practical value of the theorem on the change in kinetic energy is that, with ideal connections that do not change over time, it will allow one to exclude from the equations of motion All previously unknown reactions of connections.

The kinetic energy of a mechanical system is the sum of the kinetic energies of all its material points:

Let's calculate the differential from the expression of kinetic energy and perform some simple transformations:

Omitting intermediate values ​​and using the symbol previously introduced to denote elementary work, we write:

So, the differential of the kinetic energy of a mechanical system is equal to the sum of the elementary works of all external and internal forces acting on points of the system. This is the content of the theorem on the change in kinetic energy.

Note that the sum of the work done by the internal forces of the system is not equal to zero in the general case. It vanishes only in some special cases: when the system is an absolutely rigid body; a system of absolutely rigid bodies interacting with the help of non-deformable elements (ideal hinges, absolutely rigid rods, inextensible threads, etc.). For this reason, the theorem on the change in kinetic energy is the only one general theorems dynamics, which takes into account the effect of internal forces.

One can be interested in the change in kinetic energy not over an infinitesimal period of time, as is done above, but over a certain finite period of time. Then using integration we can get:

Here - the values ​​of kinetic energy, respectively, at moments of time - the sum of the total work of external and internal forces for the considered period of time.

The resulting equality expresses the theorem on the change in kinetic energy in a final (integral) form, which can be formulated as follows: the change in kinetic energy during the transition of a mechanical system from one position to another is equal to the sum of the total work of all external and internal forces.