Lecture 5. The theorem on the change in kinetic energy

5. 1. Work of power

Let the force - 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 dot product

Elementary work is a scalar. 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 the force on the direction of elementary displacement (or the direction of the point's velocity).

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

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

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

.

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

. (5.3)

Integration result is complete work A strength F on the considered final displacement along the path.

5. 1. 1. Work of gravity

Let be m - point mass, g- acceleration free fall... Then

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

where is the height of the point lowering.

When lifting a point, therefore,.

5. 1. 2. Work of the 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 at small deviations of the point, it can be assumed 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 by the difference between the squares of the initial and final elongation (or compressions) of the spring.

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

Consider the motion of a body in a plane. Let be O- an arbitrarily selected point on a solid (Figure 5.4). Let's call it a pole. Then the movement of the body in the plane can be represented as the sum of the simplest: translational movement together with the pole and the rotation of the body around the pole. Then, the speed of a point relative to a fixed coordinate system is defined 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 speed of Euler, that is, the speed of a point as it rotates around the pole.

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

We calculate the displacement of a point under the action of a force:

Then .

An elementary work, according to (5.1), is written as follows

Using the properties of the mixed product of vectors , we rewrite the last expression as

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

.

Then (a) will be written like this

According to (3.1 and 3.2), the principal vector and main point internal forces of the system are equal to zero, we obtain

here: Is the main vector, - the main moment of external forces relative to the point O.

Special cases

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

. (5.7)

B. Rotation of a rigid body around a fixed axis... Let the axis z passes through the pole O(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 point A(fig. 5.6). The coil rolls to the right without slipping on a rough surface.

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

Solution. The coil moves flat. Since rolling occurs without sliding, the instantaneous center of velocities is at the point where the coil touches the plane, i.e. at the point R(Figure 5.6). Let us direct the S axis horizontally to the right. In accordance with the direction of movement, we take the positive direction of the rotation angle counterclockwise.

Let the center of the coil WITH will move to. This will rotate the coil through an angle. Then, whence

Taking the point R for the instantaneous axis of rotation, we calculate the elementary work by the formula (5.8):

(a)

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

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

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

Suppose that a point moves in some space and a force acts on it from the side of space, which depends on the position of the point in this space, but does not depend on the speed of the point. In this case, they say that the space is given force field and also that the point is moving in the force field. The corresponding concepts for the 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 the force field in nature is the 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 gravitation.

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

The function is called power function.

Consider the properties of the strength function.

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

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

Full work of force on the site from the point to the point (Figure 5.1)

those. ... (5.10)

It follows from the expressions obtained that

1. the work of force in a potential force field along any closed path is equal to zero;

2.the work of the 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 NS at the considered point of the force field R call the work done by the field forces acting on a material point when it moves from a point R to start point 1, i.e.

NS= or NS=

Link the strength function U with potential energy. We have

Examples of calculating potential energy

1. Homogeneous gravity field... Let be m- point mass; g - acceleration of gravity. 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, setting 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 call half the product of the mass of a point by the square of its velocity, i.e. . Kinetic energy is a scalar positive value. In the SI system, the unit of measurement for kinetic energy is 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 the system (5.1) are determined relative to the stationary frame of reference.

Let's align the origin with the center of mass of the system. Suppose that the mechanical system, together with the coordinate system, moves translationally relative to the stationary coordinate system (Figure 5.7). Point is the point of the system.

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

, (a)

where is the speed of the origin of the moving coordinate system (portable speed, i.e. the speed of the center of mass of the system); - point speed Rk relative to the moving coordinate system Oohz (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 is O 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 the system is equal to the sum of the kinetic energy that would have a material point located in the center of mass of the system and having a mass, equal to the mass system, and the kinetic energy of motion of the system relative to the center of mass.

5. 3. 2. Kinetic energy of a solid

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 over into integrals. So, for a rigid body, formula (5.11) takes the form

. (5.14)

1. Kinetic energy of a rigid body moving forward.

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

. (5.15)

The kinetic energy of a rigid body moving translationally is equal to half the product of the body's massMby the square of its speed.

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

Speed ​​module V 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(Figure 5.9). Substituting into formula (5.14), we obtain

here - moment of inertia of a rigid body about 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 by the square of the angular velocity of the body.

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

In a plane-parallel motion, the speed of any point of the body consists of the geometric sum of the speed of the pole and the speed of the point as it rotates around the pole. Let the body move flat in the plane Oxy, then

|| ... For the pole we choose the center of mass of the body, then in the formula (5.13), the speed is the speed of the point k body as it rotates about the pole (center of mass) and is equal to where the distance k- th point to the pole. Then (5.13) can be rewritten

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

, (5.17)

with plane-parallel motion of a body, kinetic energy is made up 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 of motion.

5. 4. Theorem on the change in kinetic energy

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

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

We multiply both sides of this equation by, and, taking into account that , we get

. (5.19)

On the right-hand side of this equality, 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 value changes point kinetic energy characterizing the motion of a point and, in particular, the modulus of its velocity. If a point shifts from position to, and its velocity 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 of 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. The theorem on the change in the kinetic energy of the system

Let the points of the system with mass moved so that their radius vectors in the inertial frame of reference received an increment. Let us find how the kinetic energy has changed in this case T systems.

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

.

We 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.

A special case ... For an absolutely rigid body, the sum of the work of 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 rigid 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, respectively, the kinetic energy and, we obtain

. (5.25)

Example 1... Disc mass m= 5 kg and the radius is set in motion by a constant force applied at the point A(fig. 5.6). The disc 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 disc

Solution. The disc is in a flat motion. Let us write down the theorem on the change in the kinetic energy for a rigid 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 end position of the disc

Let us introduce the concept of one more basic dynamic characteristic of motion - kinetic energy. The kinetic energy of a material point is a scalar value equal to half the product of the point's mass by the square of its velocity.

The unit of measurement for kinetic energy is the same as for work (in SI - 1 J). Let us find the dependence by which these two quantities are related.

Consider a material point with mass moving from a position where it has velocity to a position where its velocity is

To obtain the required 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 towards the direction of motion, we obtain

The tangential acceleration of a point entering here is represented in the form

As a result, we find that

We multiply both sides of this equality by and add them under the differential sign. Then, noticing that where is the elementary work of the 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 finally find

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

A case of non-free movement. In the case of non-free motion of a point, the work of the given (active) forces and the work of the coupling reaction will enter the right-hand side of equality (52). We restrict ourselves to considering the motion of a point along a fixed smooth (frictionless) surface or curve. In this case, the reaction N (see Fig. 233) will be directed along the normal to the trajectory of the point and. Then, according to formula (44), the reaction work of a fixed smooth surface (or curve) for any displacement of a point will be equal to zero, and from equation (52) we obtain

Consequently, when moving along a fixed smooth surface (or curve), the change in the kinetic energy of a point is equal to the sum of the work on this movement of 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 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 work of the reaction N will not be equal to zero (for example, the work of the reaction of the lift platform).

Solving problems. The theorem on the change in kinetic energy [formula (52)] allows, knowing how when a point moves, its velocity changes, 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 the point changes when moving (the second problem of dynamics ). When solving the second problem, when 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, for example, the forces of elasticity or gravity (see § 88).

Thus, formula (52) can be directly used to solve the second problem of dynamics, when in the problem the data and sought quantities include: acting forces, displacement of a point and its initial and final velocities (i.e., quantities), and the forces must 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 at a speed from point A, located at a height (Fig. 235), has a speed at the point of fall C Determine what the work of the air resistance force acting on the load is equal to

Solution. During its movement, the load is acted upon by the force of gravity P and the force of air resistance R. By the theorem on the change in kinetic energy, considering the load as 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 way the load will travel to a stop (see Fig, 223, where is the initial position of the load, and is the final one).

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

In this case, 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 where we find

According to the results of task 96, the deceleration time increases proportionally initial speed, and the braking distance, as we found, is proportional to the square of the initial speed. In the case of land 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 initial speed. When moving, a resistance force R acts on the load, which is approximately replaced by its average value.Find the speed of the load at that moment in time when the thread forms 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 thread of resistance, represented by its average value R. For the force P according to the formula (47) for the force N, since we finally obtain for the force, since according to the formula (45) it will be (the length s of the arc is equal to the product radius l by the 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 this the well-known formula of Galileo, which is obviously valid for the speed of a freely falling weight (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. The valve spring has a length of cm in an undeformed state. When the valve is fully open, its length is cm, and the valve lift height is cm (Fig. 237). Spring rate valve weight kg. Ignoring the action 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, the work is performed only by the elastic force of the spring. Then, by 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 if the load falls on the beam from a height N.

Solution. As in the previous problem, we will use Eq. (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 force of gravity P on displacement and the elastic force of the beam F on displacement.

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

Solving it quadratic equation and taking into account that, according to the conditions of the problem, we must find

It is interesting to note that at is obtained.Consequently, if the 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. In the future, the load will begin to vibrate together with the beam about the equilibrium position. Under the influence of resistances, these vibrations will damp and the system will balance in a position at which the deflection of the beam is equal to

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

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

The work here is performed 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, at the found value of the work, equation (a) gives

Let's consider special cases:

a) let H be very small in comparison with 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) we find at what initial speed the thrown body will go to infinity, dividing the numerator and denominator by A, we get

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

,

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

,

. (2)

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

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

Differential forms of the kinetic energy change theorem can be used to compose differential equations movement, 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 displacement of the system from a certain initial position, where the kinetic energy is equal, to the position where the value of the kinetic energy becomes equal , will have

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

Unlike the previous theorems, internal forces are not excluded in the equations. Indeed, if and are the forces of interaction between the points and the system (see Fig. 51), then. But in this case, the point can move towards, and the point towards. The work of each of the forces will then be positive and the amount of work will not be zero. An example is the rollback phenomenon. Internal forces (pressure forces) acting both on the projectile and on the rolling back 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 the points changes, the elastic forces applied to the points will do the work. But if the system consists of absolutely rigid bodies and the connections between them are unchangeable, not 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.

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 and an unchangeable system (Fig. 51) acting on each other with forces and () have in this moment speed and. Then for a period of time dt these points will make elementary movements and , directed along the vectors and. But since the segment is unchangeable, then, according to the well-known theorem of kinematics, the projections 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 and will be the same in magnitude and opposite in sign, and in total they will give zero. This result is valid for all internal forces for any displacement of the system.

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

2) A system with perfect connections... Consider a system overlaid with connections that do not change over time. Let us divide all external and internal forces acting on points of the system into active and bond reactions. Then

,

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

As you can 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 bonds does not affect the change in the kinetic energy of the system during its motion. For such connections, the condition must obviously be satisfied:

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

Thus, the change in the kinetic energy of a system with ideal, time-invariant bonds for any of its displacement is equal to the sum of the work on this displacement applied to the system of external and internal active forces.

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

or (3)

It is the law of conservation of mechanical energy: when the system moves in a potential field, its mechanical energy (the sum of potential and kinetic) remains unchanged, 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, using the energy integral, it is possible to check the correctness of the compilation of differential equations of motion. If the system is conservative, and condition (3) is not met, then an error was made in the formulation of the equations of motion.

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

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

This theorem establishes a quantitative relationship between the work of a force (cause) and the kinetic energy of a material point (effect).

Kinetic energy of a material point is called a scalar value equal to half the product of the mass of a point by the square of its velocity

. (43)

Kinetic energy characterizes the mechanical action of the force, which can be converted into other types of energy, for example, into heat.

A work of strength on a given displacement is called the characteristic of that action of force, which leads to a change in the speed module.

Elementary work of strength is defined as the scalar product of the force vector by the elementary displacement vector at the point of its application

, (44)

where
- elementary movement.

The module of elementary work is determined by the formula

where  - the angle between the force vector and the vector of elementary displacement; is the projection of the force vector onto the tangent line.

The total work at some finite displacement is determined by the integral

. (46)

From (46) it follows that the total work can be calculated in two cases when the force is constant or depends on the displacement.

At F= const we obtain
.

When solving problems, it is often convenient to use the analytical method for calculating the force

where F x , F y , F z- the projection of the force on the coordinate axes.

Let us prove the following theorem.

Theorem: The change in the kinetic energy of a material point at some of its displacement is equal to the work of the force acting on the point at the same displacement.

Let a material point M of mass m moves by force F from position M 0 to position M 1.

OUD:
. (47)

Let's introduce the substitution
and project (47) onto the tangent

. (48)

We separate the variables in (48) and integrate

As a result, we get

. (49)

Equation (49) proves the above theorem.

It is convenient to use the theorem when the mass of a point, its initial and final velocity, forces and displacement are present among the given and sought parameters.

Calculation of the work of characteristic forces.

1. Work of gravity calculated as the product of the modulus of the force and the vertical displacement of the point of its application

. (50)

When moving up, the work is positive; when moving down, it is negative.

2. The work of the elastic force of the spring F=-cx is equal to

, (51)

where x 0 - initial elongation (compression) of the spring;

x 1 - final elongation (compression) of the spring.

The work of the force of gravity and elastic force does not depend on the trajectory of movement of their points of application. Such forces, the work of which does not depend on the trajectory, are called potential forces.

3. Frictional force work.

Since the friction force is always directed in the direction opposite to the direction of movement, its work is equal to

Frictional work is always negative... Forces whose work is always negative are called dissipative.

Integral (final) form... The theorem on the change in the 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 the 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 cules applied to the system on this movement:

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

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

Where ,

The sum of the kinetic and potential energies of the system is called full 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 action of gravity starts to move 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 to 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 the constraints, we show the reactions of the hinged-fixed support 2 and a rough inclined surface. The directions of the velocities of the bodies of the system are depicted taking into account the fact that body 1 is descending.

We solve the problem using the theorem on the change in the kinetic energy of a mechanical system:

where T and - kinetic energy of the system in the initial and final positions; - the algebraic sum of the work of external forces applied to the system, on the movement of the system from the initial position to the final one; - the sum of the work of the internal forces of the system on the same displacement.

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

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

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

The kinetic energy of the load 1, moving translationally, is equal to:

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

Kinetic energy of body 3 in its translational motion:

Thus,

The expression for kinetic energy contains the unknown velocities of all bodies in the system. It is necessary to start the definition with. Let's get rid of unnecessary unknowns by making up the equations of connections.

The constraint equations are nothing more than the kinematic relationships between the velocities and displacements of points in the system. When drawing up the equations of constraints, we will express all the unknown speeds and displacements of the bodies of the system in terms of the speed and displacement of the load 1.

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

From here we express the angular velocity of the body 2:

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

Substituting the value of the angular velocity, we get:

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

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

The moment of inertia of body 2 is:

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

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

Now, according to the theorem on the change in the 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 equality (r) in time.