Integral (final) form. Theorem on the change in kinetic energy of a material point: the change in the kinetic energy of a material point at some of its displacement is equal to the algebraic sum of the work of all forces acting on this point at the same displacement.

The theorem on the change in kinetic energy of a mechanical system is formulated: the change in the kinetic energy of a mechanical system when it moves from one position to another is equal to the sum of the work of all external and internal forces applied to the system during this movement:

In the case of an unchangeable system, the sum of the work done by internal forces on any displacement is equal to zero (), then

Law of conservation of mechanical energy. When a mechanical system moves under the influence of forces that have potential, changes in the kinetic energy of the system are determined by the dependencies:

Where ,

The sum of the kinetic and potential energies of a system is called total mechanical energy systems.

Thus, When a mechanical system moves in a stationary potential field, the total mechanical energy of the system during movement remains unchanged.

Task. A mechanical system, under the influence of gravity, comes into motion from a state of rest. Taking into account the sliding friction of body 3, neglecting other resistance forces and the masses of the threads assumed to be inextensible, determine the speed and acceleration of body 1 at the moment when the path traveled by it becomes equal s(Fig. 3.70).

In the task, accept:

Solution. The mechanical system is acted upon by active forces , , . Applying the principle of freeing the system from constraints, we will show the reactions of the hinged-fixed support 2 and the rough inclined surface. We will depict the directions of velocities of the bodies of the system taking into account the fact that body 1 is descending.

Let's solve the problem by applying the theorem on the change in kinetic energy of a mechanical system:

Where T and is the kinetic energy of the system in the initial and final positions; - the algebraic sum of the work done by external forces applied to the system to move the system from the initial position to the final position; - the sum of the work done by the internal forces of the system at the same displacement.

For the system under consideration, consisting of absolutely rigid bodies connected by inextensible threads:

Since the system was at rest in the initial position, then . Hence:

The kinetic energy of the system is the sum of the kinetic energies of bodies 1, 2, 3:

The kinetic energy of load 1 moving forward is equal to:

Kinetic energy of block 2 rotating around an axis Oz, perpendicular to the drawing plane:

Kinetic energy of body 3 in its forward motion:

Thus,

The expression for kinetic energy contains the unknown velocities of all bodies in the system. The definition must begin with . Let's get rid of unnecessary unknowns by creating equations of connections.

Constraint equations are nothing more than kinematic relationships between the velocities and movements of points in the system. When composing the constraint equations, we will express all the unknown speeds and movements of the bodies of the system through the speed and movement of load 1.

The speed of any point on the rim of small radius is equal to the speed of body 1, as well as the product of the angular speed of body 2 and the radius of rotation r:

From here we express the angular velocity of body 2:

The rotational speed of any point on the rim of a block of large radius, on the one hand, is equal to the product of the angular speed of the block and the radius of rotation, and on the other hand, the speed of body 3:

Substituting the value of the angular velocity, we get:

Having integrated expressions (a) and (b) under the initial conditions, we write the ratio of the displacements of the points of the system:

Knowing the basic dependences of the velocities of the points of the system, we return to the expression of kinetic energy and substitute equations (a) and (b) into it:

The moment of inertia of body 2 is equal to:

Substituting the values ​​of the body masses and the moment of inertia of body 2, we write:

Determination of the sum of work of all external forces of the system at a given displacement.

Now, according to the theorem on the change in kinetic energy of a mechanical system, we equate the values T And

The speed of body 1 is obtained from the expression (g)

The acceleration of body 1 can be determined by differentiating the equality (g) with respect to time.

Let us introduce the concept of another basic dynamic characteristic of motion - kinetic energy. The kinetic energy of a material point is a scalar quantity equal to half the product of the mass of the point and the square of its speed.

The unit of measurement for kinetic energy is the same as work (in SI - 1 J). Let's find the relationship that connects these two quantities.

Let's consider a material point with mass moving from a position where it has speed to a position where its speed

To obtain the desired dependence, let us turn to the equation expressing the basic law of dynamics. Projecting both of its parts onto the tangent to the trajectory of point M, directed in the direction of movement, we obtain

Let us represent the tangential acceleration of the point included here in the form

As a result, we find that

Let's multiply both sides of this equality by and enter it under the differential sign. Then, noting that where is the elementary work of force, we obtain the expression of the theorem on the change in the kinetic energy of a point in differential form:

Having now integrated both sides of this equality within the limits corresponding to the values ​​of the variables at the points, we will finally find

Equation (52) expresses the theorem about the change in the kinetic energy of a point in final form: the change in the kinetic energy of a point during some displacement is equal to the algebraic sum of the work of all forces acting on the point at the same displacement.

The case of unfree movement. When the point moves in a non-free manner, the right side of equality (52) will include the work of the given (active) forces and the work of the coupling reaction. Let us limit ourselves to considering the motion of a point along a stationary smooth (frictionless) surface or curve. In this case, the reaction N (see Fig. 233) will be directed normal to the trajectory of the point and. Then, according to formula (44), the reaction work of a stationary smooth surface (or curve) for any movement of the point will be equal to zero, and from equation (52) we obtain

Consequently, when moving along a stationary smooth surface (or curve), the change in the kinetic energy of a point is equal to the sum of the work done on this movement of the active forces applied to the point.

If the surface (curve) is not smooth, then the work of the friction force will be added to the work of the active forces (see § 88). If the surface (curve) is moving, then the absolute displacement of point M may not be perpendicular to N and then the reaction work N will not be equal to zero (for example, the reaction work of the elevator platform).

Problem solving. The theorem on the change in kinetic energy [formula (52)] allows, knowing how the speed of a point changes when a point moves, to determine the work of the acting forces (the first problem of dynamics) or, knowing the work of the acting forces, to determine how the speed of a point changes when moving (the second problem of dynamics ). When solving the second problem, when the forces are given, it is necessary to calculate their work. As can be seen from formulas (44), (44), this can be done only when the forces are constant or depend only on the position (coordinates) of the moving point, such as the force of elasticity or gravity (see § 88).

Thus, formula (52) can be directly used to solve the second problem of dynamics, when the data and required quantities in the problem include: acting forces, the displacement of a point and its initial and final velocities (i.e., quantities ), and the forces should be constant or depending only on the position (coordinates) of the point.

The theorem in differential form [formula (51)] can, of course, be applied for any acting forces.

Problem 98. A load weighing kg, thrown with speed from point A, located at a height (Fig. 235), has a speed at the point of fall C. Determine what is the work done by the air resistance force acting on the load during its movement

Solution. As the load moves, the force of gravity P and the force of air resistance R act on the load. According to the theorem on the change in kinetic energy, considering the load to be a material point, we have

From this equality, since according to the formula we find

Problem 99. Under the conditions of problem 96 (see [§ 84), determine which path the load will travel before stopping (see Fig. 223, where is the initial position of the load, and is the final position).

Solution. The load, as in problem 96, is acted upon by forces P, N, F. To determine the braking distance, taking into account that the conditions of this problem also include a constant force F, we will use the theorem on the change in kinetic energy

In the case under consideration - the speed of the load at the moment of stopping). In addition, since the forces P and N are perpendicular to the displacement, As a result, we get from where we find

According to the results of problem 96, the braking time increases in proportion to the initial speed, and the braking distance, as we found, is proportional to the square of the initial speed. When applied to ground transport, this shows how the danger increases with increasing speed.

Problem 100. A load of weight P is suspended on a thread of length l The thread together with the load is deflected from the vertical at an angle (Fig. 236, a) and released without an initial speed. When moving, a resistance force R acts on the load, which we approximately replace with its average value. Find the speed of the load at the moment in time when the thread makes an angle with the vertical

Solution. Taking into account the conditions of the problem, we again use Theorem (52):

The load is acted upon by the force of gravity P, the reaction of the resistance thread, represented by its average value R. For the force P, according to formula (47) for the force N, since we finally obtain, for the force since, according to formula (45) it will be (the length s of the arc is equal to the product radius l per central angle). In addition, according to the conditions of the problem As a result, equality (a) gives:

In the absence of resistance, we obtain from here the well-known Galileo formula, which is obviously also valid for the speed of a freely falling load (Fig. 236, b).

In the problem under consideration Then, introducing another notation - the average resistance force per unit weight of the load), we finally obtain

Problem 101. In an undeformed state, the valve spring has a length of cm. When the valve is fully open, its length is cm, and the height of the valve lift is cm (Fig. 237). Spring stiffness valve weight kg. Neglecting the effects of gravity and resistance forces, determine the speed of the valve at the moment it is closed.

Solution, Let's use the equation

According to the conditions of the problem, work is performed only by the elastic force of the spring. Then, according to formula (48) it will be

In this case

In addition, Substituting all these values ​​into equation (a), we finally obtain

Problem 102. A load lying in the middle of an elastic beam (Fig. 238) deflects it by an amount (statistical deflection of the beam). Neglecting the weight of the beam, determine what its maximum deflection will be equal to if the load falls on the beam from a height H.

Solution. As in the previous problem, we will use equation (52) to solve. In this case, the initial speed of the load and its final speed (At the moment of maximum deflection of the beam) are equal to zero and equation (52) takes the form

The work here is performed by the gravitational force P on the displacement and the elastic force of the beam F on the displacement. Moreover, since for the beam Substituting these quantities into equality (a), we obtain

But when the load is in equilibrium on the beam, the force of gravity is balanced by the force of elasticity, therefore, the previous equality can be represented in the form

Solving this quadratic equation and taking into account that according to the conditions of the problem we should find

It is interesting to note that when it turns out Therefore, if a load is placed in the middle of a horizontal beam, then its maximum deflection when lowering the load will be equal to twice the static one. Subsequently, the load will begin to oscillate together with the beam around the equilibrium position. Under the influence of resistance, these oscillations will dampen and the system will be balanced in a position in which the deflection of the beam is equal to

Problem 103. Determine the minimum vertically directed initial velocity that must be imparted to the body so that it rises from the Earth's surface to a given height H (Fig. 239). The force of attraction is considered to vary inversely with the square of the distance from the center of the Earth. Neglect air resistance.

Solution. Considering the body as a material point with mass, we use the equation

The work here is done by the gravitational force F. Then, using formula (50), taking into account that in this case where R is the radius of the Earth, we obtain

Since at the highest point, with the found value of work, equation (a) gives

Let's consider special cases:

a) let H be very small compared to R. Then - a value close to zero. Dividing the numerator and denominator we get

Thus, for small H we arrive at Galileo’s formula;

b) let’s find at what initial speed the thrown body will go to infinity. Dividing the numerator and denominator by A, we get

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

1. mechanical motion is transferred from one mechanical system to another as mechanical motion;
2. 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 done by 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.

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An example of solving a problem using the theorem on the change in kinetic energy of a system with rigid bodies, blocks, pulleys and a spring.

Content

The task

The mechanical system consists of weights 1 and 2, a stepped pulley 3 with step radii R 3 = 0.3 m, r 3 = 0.1 m and radius of gyration relative to the axis of rotation ρ 3 = 0.2 m, block 4 radius R 4 = 0.2 m and movable block 5. Block 5 is considered a solid homogeneous cylinder. Coefficient of friction of load 2 on plane f = 0,1 . The bodies of the system are connected to each other by threads thrown over blocks and wound on pulley 3. Sections of the threads are parallel to the corresponding planes. A spring with stiffness coefficient c = is attached to the movable block 5 280 N/m.

Under the influence of force F = f (s) = 80(6 + 7 s) N, depending on the displacement s of the point of its application, the system begins to move from a state of rest. The deformation of the spring at the moment the movement begins is zero. When moving, pulley 3 is acted upon by a constant moment M = 1.6 Nm resistance forces (from friction in bearings). Body masses: m 1 = 0 , m 2 = 5 kg, m 3 = 6 kg, m 4 = 0 , m 5 = 4 kg.

Determine the value of the center of mass of the body 5 V C 5 at the moment in time when the displacement s of load 1 becomes equal to s 1 = 0.2 m.

Note. When solving a problem, use kinetic energy change theorem.

The solution of the problem

Given: R 3 = 0.3 m, r 3 = 0.1 m, ρ 3 = 0.2 m, R 4 = 0.2 m, f = 0,1 , s = 280 N/m, m 1 = 0 , m 2 = 5 kg, m 3 = 6 kg, m 4 = 0 , m 5 = 4 kg, F = f (s) = 80(6 + 7 s) N, s 1 = 0.2 m.

Find: V C 5 .

Variable designations

R 3 , r 3- radii of pulley steps 3;
ρ 3 - radius of inertia of pulley 3 relative to the axis of rotation;
R 5 - block radius 5;
V 1 , V 2 - speeds of bodies 1 and 2;
ω 3 - angular speed of rotation of pulley 3;
V C 5 - speed of the center of mass C 5 block 5;
ω 5 - angular speed of rotation of block 5;
s 1 , s 2 - movement of bodies 1 and 2;
φ 3 - pulley rotation angle 3;
s C 5 - movement of the center of mass C 5 block 5;
s A, s B - moving points A and B.

Establishing kinematic relationships

Let us establish kinematic relations. Since loads 1 and 2 are connected by one thread, their speeds are equal:
V 2 = V 1.
Since the thread connecting loads 1 and 2 is wound on the outer stage of pulley 3, the points of the outer stage of pulley 3 move with speed V 2 = V 1. Then the angular speed of rotation of the pulley is:
.
Velocity of the center of mass V C 5 block 5 is equal to the speed of the points of the internal stage of pulley 3:
.
The speed of point K is zero. Therefore, it is the instantaneous speed center of block 5. Angular velocity of rotation of block 5:
.
The speed of point B - the free end of the spring - is equal to the speed of point A:
.

Let's express the speeds in terms of V C 5 .
;
;
.

Now let's install connections between body movements and rotation angles pulley and block. Since velocities and angular velocities are time derivatives of displacements and rotation angles
,
then the same connections will be between displacements and rotation angles:
s 2 = s 1;
;
;
.

Determination of the kinetic energy of the system

Let's find the kinetic energy of the system. Load 2 makes translational motion with speed V 2 . Pulley 3 performs rotational motion with angular rotation speed ω 3 . Block 5 performs plane-parallel motion. It rotates with angular velocity ω 5 and its center of mass moves with speed V C 5 . Kinetic energy of the system:
.

Since the radius of inertia of the pulley relative to the axis of rotation is given, the moment of inertia of the pulley relative to the axis of rotation is determined by the formula:
J 3 = m 3 ρ 2 3.
Since block 5 is a solid homogeneous cylinder, its moment of inertia relative to the center of mass is equal to
.

Using kinematic relations, we express all velocities through V C 5 and substitute expressions for moments of inertia into the formula for kinetic energy.
,
where we entered the constant
kg.

So, we have found the dependence of the kinetic energy of the system on the speed of the center of mass V C 5 moving block:
, where m = 75 kg.

Determination of the amount of work of external forces

Consider external forces, acting on the system.
At the same time, we do not consider the tension forces of the threads, since the threads are inextensible and, therefore, they do not produce work. For this reason, we do not consider internal stresses acting in bodies, since they are absolutely solid.
Body 1 (with zero mass) is acted upon by a given force F.
Load 2 is acted upon by gravity P 2 = m 2 g 2 and friction force F T .
Pulley 3 is acted upon by gravity P 3 = m 3 g, N axis pressure force 3 and the moment of friction forces M.
Pulley 4 (with zero mass) is affected by the pressure force of the N axis 4 .
The movable block 5 is acted upon by gravity P 5 = m 5 g, the elastic force F y of the spring and the tension force of the thread T K at point K.

The work that a force does when moving the point of its application by a small displacement is equal to the scalar product of vectors, that is, the product of the absolute values ​​of the vectors F and ds by the cosine of the angle between them. A given force applied to body 1 is parallel to the movement of body 1. Therefore, the work done by force when body 1 moves a distance s 1 is equal to:


J.

Consider load 2. It is acted upon by the force of gravity P 2 , surface pressure force N 2 , thread tension force T 23 , T 24 and friction force F T . Since the load does not move in the vertical direction, the projection of its acceleration onto the vertical axis is zero. Therefore, the sum of the projections of forces on the vertical axis is equal to zero:
N 2 - P 2 = 0;
N 2 = P 2 = m 2 g.
Friction force:
F T = f N 2 = f m 2 g.
Forces P 2 and N 2 perpendicular to the displacement s 2 , so they don't produce work.
Work of friction force:
J.

If we consider load 2 as an isolated system, then we need to take into account the work produced by the tension forces of the threads T 23 and T 24 . However, we are interested in the entire system, consisting of bodies 1, 2, 3, 4 and 5. For such a system, the tension forces of the threads are internal forces. And since the threads are inextensible, the sum of their work is zero. In the case of load 2, you also need to take into account the tension forces of the threads acting on pulley 3 and block 4. They are equal in magnitude and opposite in direction to the forces T 23 and T 24 . Therefore, the work done by the tension forces of threads 23 and 24 over load 2 is equal in magnitude and opposite in sign to the work done by the tension forces of these threads over pulley 3 and block 4. As a result, the amount of work produced by the tension forces of the threads is zero.

Consider pulley 3. Since its center of mass does not move, the work done by gravity P 3 equal to zero.
Because C axis 3 is motionless, then the pressure force of the N axis 3 does not produce work.
The work done by the torque is calculated similarly to the work done by the force:
.
In our case, the vectors of the moment of friction forces and the angle of rotation of the pulley are directed along the axis of rotation of the pulley, but opposite in direction. Therefore, the work of the moment of friction forces:
J.

Let's look at block 5.
Since the speed of point K is zero, the force T K does not produce work.
Center of mass of block C 5 moved a distance s C 5 up. Therefore, the work done by the block’s gravity is:
J.
The work done by the elastic force of the spring is equal to the change in the potential energy of the spring with a minus sign. Since the spring is not deformed at first, then
J.

The sum of the work of all forces:

J.

Application of the theorem on the change in kinetic energy of a system

Let us apply the theorem on the change in the kinetic energy of the system in integral form.
.
Since the system was at rest at the beginning, its kinetic energy at the beginning of its motion is
T 0 = 0 .
Then
.
From here
m/s.

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 the properties of the scalar product, we can represent the elementary work of force also 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 of the elementary works of 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 over time 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 represent the curvilinear integral 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, a system of solid bodies 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: