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 a scalar quantity equal to half the product of the mass of a point and the square of its speed
.
(43)
Kinetic energy characterizes the mechanical action of force that can be converted into other types of energy, for example, thermal.
Work of force at a given displacement is the characteristic of the action of force that leads to a change in the velocity module.
Elementary work of force is defined as the scalar product of the force vector and 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 elementary displacement vector; 
- projection of the force vector onto the tangent.
The total work on 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 get 
.
When solving problems, it is often convenient to use the analytical method of calculating force
Where F x , F y , F z– projections of force onto 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 the material point M of mass m moves under the influence of 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 theorem formulated above.
The theorem is convenient to use when the given and sought parameters include the mass of a point, its initial and final speed, forces and displacement.
Calculation of the work of characteristic forces.
1. Work of gravity is calculated as the product of the force modulus and the vertical displacement of the point of its application
.
(50)
When moving up, the work is positive; when moving down, the work is negative.
2. Work of elastic force of a spring F=-cx equal to
,
(51)
Where x 0 – initial elongation (compression) of the spring;
x 1 – final elongation (compression) of the spring.
The work 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. Work of friction force.
Since the friction force is always directed in the direction opposite to the direction of movement, its work is equal to
The work done by the friction force is always negative. Forces whose work is always negative are called dissipative.
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
2.4.1. Kinetic energy of a mechanical system. The kinetic energy of a material point of mass moving with speed is called the quantity
The kinetic energy of a mechanical system is the sum of the kinetic energies of the material points included in this system:
In cases where the mass of the system is distributed continuously, the summation in expression (7) is replaced by integration over the distribution area.
The relationship between the values of the kinetic energy of a mechanical system in two reference systems, one of which is stationary, and the other moves translationally at speed , where point C is the center of mass of the mechanical system, is given by Koenig’s theorem:
. (8)
Here 
- kinetic energy of a mechanical system in a moving coordinate system.
Using expressions (6, 7, 8) allows you to write formulas for calculating the kinetic energy of a solid body:
When a body of mass moves forward with speed
When rotating with angular velocity around a fixed axis of a body with a moment of inertia
in plane-parallel motion of a rigid body with angular velocity at a value of the central moment of inertia relative to the axis perpendicular to the plane of motion, and a value of the moment of inertia relative to the instantaneous axis of rotation
. (11)
2.4.2. Energy characteristics. The energy characteristics of a force include its power, work and potential energy.
Power force, the point of application of which moves with speed, is called the magnitude
Job strength on an elementary interval time and the elementary displacement of the point of application corresponding to this period of time is determined by the rule
Work strength on a finite interval time and the corresponding change in the radius - the vector of the point of application of this force from to - is called the magnitude
. (14)
The work done by the moment of a pair of forces is calculated in a similar way.
Potential energy is defined only in cases where expression (13) is a total differential:
When condition (15) is met, the force is said to be potential. Relations connecting the projections of force on the axis of the selected coordinate system with the function:
If the point of application of the force has moved from position to position , then by integrating (15) we can obtain
. (17)
Note: potential energy is determined up to a constant term; The noted feature allows us to assume that the potential energy is equal to zero at a point we choose (for example, at the origin of coordinates).
In the case when for the set of forces acting on a mechanical system, it is possible to write down the expression for potential energy, the mechanical system is called conservative. Such mechanical systems have important features - the work of the acting forces does not depend on the type of trajectory and the law of motion along it; work when moving along a closed loop is zero.
Conditions under which a function exists:
2.4.3. Theorem on the change in kinetic energy. Writing the theorem on the change in kinetic energy of a mechanical system in differential form:
The time derivative of the kinetic energy of a mechanical system is equal to the power of external and internal forces.
Integral form of writing the theorem on the change in kinetic energy
, (20)
Where ; ; ; .
In the particular case when the expression for potential energy can be written for the totality of external and internal forces of the system, the law of conservation of total mechanical energy is satisfied
and the system itself turns out to be conservative.
EXAMPLE 3. For the mechanical system shown in Fig. 2, obtain a differential equation for the motion of the load.
SOLUTION. Let us use the theorem on the change in kinetic energy in differential form (19). Let's mentally free ourselves from connections by applying appropriate reactions to the bodies of the mechanical system (see Fig. 2). Note: the forces applied at the stationary center of mass of the coaxial block are not depicted, since their power is zero.
Let's create an expression for the kinetic energy of a mechanical system.
Energy is a scalar physical quantity that is a unified measure of various forms of motion of matter and a measure of the transition of the motion of matter from one form to another.
To characterize various forms of motion of matter, the corresponding types of energy are introduced, for example: mechanical, internal, energy of electrostatic, intranuclear interactions, etc.
Energy obeys the law of conservation, which is one of the most important laws of nature.
Mechanical energy E characterizes the movement and interaction of bodies and is a function of the speeds and relative positions of bodies. It is equal to the sum of kinetic and potential energies.
Kinetic energy
Let us consider the case when a body of mass m there is a constant force \(~\vec F\) (it can be the resultant of several forces) and the vectors of force \(~\vec F\) and displacement \(~\vec s\) are directed along one straight line in one direction. In this case, the work done by the force can be defined as A = F∙s. The modulus of force according to Newton's second law is equal to F = m∙a, and the displacement module s in uniformly accelerated rectilinear motion is associated with the modules of the initial υ 1 and final υ 2 speeds and accelerations A expression \(~s = \frac(\upsilon^2_2 - \upsilon^2_1)(2a)\) .
From here we get to work
\(~A = F \cdot s = m \cdot a \cdot \frac(\upsilon^2_2 - \upsilon^2_1)(2a) = \frac(m \cdot \upsilon^2_2)(2) - \frac (m \cdot \upsilon^2_1)(2)\) . (1)
A physical quantity equal to half the product of a body’s mass and the square of its speed is called kinetic energy of the body.
Kinetic energy is represented by the letter E k.
\(~E_k = \frac(m \cdot \upsilon^2)(2)\) . (2)
Then equality (1) can be written as follows:
\(~A = E_(k2) - E_(k1)\) . (3)
Kinetic energy theorem
the work of the resultant forces applied to the body is equal to the change in the kinetic energy of the body.
Since the change in kinetic energy is equal to the work of force (3), the kinetic energy of the body is expressed in the same units as the work, i.e. in joules.
If the initial speed of movement of a body of mass m is zero and the body increases its speed to the value υ , then the work done by the force is equal to the final value of the kinetic energy of the body:
\(~A = E_(k2) - E_(k1)= \frac(m \cdot \upsilon^2)(2) - 0 = \frac(m \cdot \upsilon^2)(2)\) . (4)
Physical meaning of kinetic energy
The kinetic energy of a body moving with a speed v shows how much work must be done by a force acting on a body at rest in order to impart this speed to it.
Potential energy
Potential energy is the energy of interaction between bodies.
The potential energy of a body raised above the Earth is the energy of interaction between the body and the Earth by gravitational forces. The potential energy of an elastically deformed body is the energy of interaction of individual parts of the body with each other by elastic forces.
Potential are called strength, the work of which depends only on the initial and final position of a moving material point or body and does not depend on the shape of the trajectory.
In a closed trajectory, the work done by the potential force is always zero. Potential forces include gravitational forces, elastic forces, electrostatic forces and some others.
Powers, the work of which depends on the shape of the trajectory, are called non-potential. When a material point or body moves along a closed trajectory, the work done by the nonpotential force is not equal to zero.
Potential energy of interaction of a body with the Earth
Let's find the work done by gravity F t when moving a body of mass m vertically down from a height h 1 above the Earth's surface to a height h 2 (Fig. 1). If the difference h 1 – h 2 is negligible compared to the distance to the center of the Earth, then the force of gravity F t during body movement can be considered constant and equal mg.
Since the displacement coincides in direction with the gravity vector, the work done by gravity is equal to
\(~A = F \cdot s = m \cdot g \cdot (h_1 - h_2)\) . (5)
Let us now consider the movement of a body along an inclined plane. When moving a body down an inclined plane (Fig. 2), the force of gravity F t = m∙g does work
\(~A = m \cdot g \cdot s \cdot \cos \alpha = m \cdot g \cdot h\) , (6)
Where h– height of the inclined plane, s– displacement module equal to the length of the inclined plane.
Movement of a body from a point IN exactly WITH along any trajectory (Fig. 3) can be mentally imagined as consisting of movements along sections of inclined planes with different heights h’, h'' etc. Work A gravity all the way from IN V WITH equal to the sum of work on individual sections of the route:
\(~A = m \cdot g \cdot h" + m \cdot g \cdot h"" + \ldots + m \cdot g \cdot h^n = m \cdot g \cdot (h" + h"" + \ldots + h^n) = m \cdot g \cdot (h_1 - h_2)\), (7)
Where h 1 and h 2 – heights from the Earth’s surface at which the points are located, respectively IN And WITH.
Equality (7) shows that the work of gravity does not depend on the trajectory of the body and is always equal to the product of the gravity modulus and the difference in heights in the initial and final positions.
When moving downward, the work of gravity is positive, when moving up it is negative. The work done by gravity on a closed trajectory is zero.
Equality (7) can be represented as follows:
\(~A = - (m \cdot g \cdot h_2 - m \cdot g \cdot h_1)\) . (8)
A physical quantity equal to the product of the mass of a body by the acceleration modulus of free fall and the height to which the body is raised above the surface of the Earth is called potential energy interaction between the body and the Earth.
Work done by gravity when moving a body of mass m from a point located at a height h 2, to a point located at a height h 1 from the Earth's surface, along any trajectory, is equal to the change in the potential energy of interaction between the body and the Earth, taken with the opposite sign.
\(~A = - (E_(p2) - E_(p1))\) . (9)
Potential energy is indicated by the letter E p.
The value of the potential energy of a body raised above the Earth depends on the choice of the zero level, i.e., the height at which the potential energy is assumed to be zero. It is usually assumed that the potential energy of a body on the Earth's surface is zero.
With this choice of the zero level, the potential energy E p of a body located at a height h above the Earth's surface, equal to the product of the mass m of the body by the absolute acceleration of free fall g and distance h it from the surface of the Earth:
\(~E_p = m \cdot g \cdot h\) . (10)
The physical meaning of the potential energy of interaction of a body with the Earth
the potential energy of a body on which gravity acts is equal to the work done by gravity when moving the body to the zero level.
Unlike the kinetic energy of translational motion, which can only have positive values, the potential energy of a body can be both positive and negative. Body mass m, located at a height h, Where h < h 0 (h 0 – zero height), has negative potential energy:
\(~E_p = -m \cdot g \cdot h\) .
Potential energy of gravitational interaction
Potential energy of gravitational interaction of a system of two material points with masses m And M, located at a distance r one from the other is equal
\(~E_p = G \cdot \frac(M \cdot m)(r)\) . (eleven)
Where G is the gravitational constant, and the zero of the potential energy reference ( E p = 0) accepted at r = ∞.
Potential energy of gravitational interaction of a body with mass m with the Earth, where h– height of the body above the Earth’s surface, M e – mass of the Earth, R e is the radius of the Earth, and the zero of the potential energy reading is chosen at h = 0.
\(~E_e = G \cdot \frac(M_e \cdot m \cdot h)(R_e \cdot (R_e +h))\) . (12)
Under the same condition of choosing zero reference, the potential energy of gravitational interaction of a body with mass m with Earth for low altitudes h (h « R e) equal
\(~E_p = m \cdot g \cdot h\) ,
where \(~g = G \cdot \frac(M_e)(R^2_e)\) is the module of gravity acceleration near the Earth's surface.
Potential energy of an elastically deformed body
Let us calculate the work done by the elastic force when the deformation (elongation) of the spring changes from a certain initial value x 1 to final value x 2 (Fig. 4, b, c).
The elastic force changes as the spring deforms. To find the work done by the elastic force, you can take the average value of the force modulus (since the elastic force depends linearly on x) and multiply by the displacement module:
\(~A = F_(upr-cp) \cdot (x_1 - x_2)\) , (13)
where \(~F_(upr-cp) = k \cdot \frac(x_1 - x_2)(2)\) . From here
\(~A = k \cdot \frac(x_1 - x_2)(2) \cdot (x_1 - x_2) = k \cdot \frac(x^2_1 - x^2_2)(2)\) or \(~A = -\left(\frac(k \cdot x^2_2)(2) - \frac(k \cdot x^2_1)(2) \right)\) . (14)
A physical quantity equal to half the product of the rigidity of a body by the square of its deformation is called potential energy elastically deformed body:
\(~E_p = \frac(k \cdot x^2)(2)\) . (15)
From formulas (14) and (15) it follows that the work of the elastic force is equal to the change in the potential energy of an elastically deformed body, taken with the opposite sign:
\(~A = -(E_(p2) - E_(p1))\) . (16)
If x 2 = 0 and x 1 = X, then, as can be seen from formulas (14) and (15),
\(~E_p = A\) .
Physical meaning of the potential energy of a deformed body
the potential energy of an elastically deformed body is equal to the work done by the elastic force when the body transitions to a state in which the deformation is zero.
Potential energy characterizes interacting bodies, and kinetic energy characterizes moving bodies. Both potential and kinetic energy change only as a result of such interaction of bodies in which the forces acting on the bodies do work other than zero. Let us consider the question of energy changes during the interactions of bodies forming a closed system.
Closed system- this is a system that is not acted upon by external forces or the action of these forces is compensated. If several bodies interact with each other only by gravitational and elastic forces and no external forces act on them, then for any interactions of bodies, the work of the elastic or gravitational forces is equal to the change in the potential energy of the bodies, taken with the opposite sign:
\(~A = -(E_(p2) - E_(p1))\) . (17)
According to the kinetic energy theorem, the work done by the same forces is equal to the change in kinetic energy:
\(~A = E_(k2) - E_(k1)\) . (18)
From a comparison of equalities (17) and (18) it is clear that the change in the kinetic energy of bodies in a closed system is equal in absolute value to the change in the potential energy of the system of bodies and opposite in sign:
\(~E_(k2) - E_(k1) = -(E_(p2) - E_(p1))\) or \(~E_(k1) + E_(p1) = E_(k2) + E_(p2) \) . (19)
Law of conservation of energy in mechanical processes:
the sum of the kinetic and potential energy of the bodies that make up a closed system and interact with each other by gravitational and elastic forces remains constant.
The sum of the kinetic and potential energy of bodies is called total mechanical energy.
Let's give a simple experiment. Let's throw a steel ball up. By giving the initial speed υ inch, we will give it kinetic energy, which is why it will begin to rise upward. The action of gravity leads to a decrease in the speed of the ball, and hence its kinetic energy. But the ball rises higher and higher and acquires more and more potential energy ( E p = m∙g∙h). Thus, kinetic energy does not disappear without a trace, but is converted into potential energy.
At the moment of reaching the top point of the trajectory ( υ = 0) the ball is completely deprived of kinetic energy ( E k = 0), but at the same time its potential energy becomes maximum. Then the ball changes direction and moves downward with increasing speed. Now the potential energy is converted back into kinetic energy.
The law of conservation of energy reveals physical meaning concepts work:
the work of gravitational and elastic forces, on the one hand, is equal to an increase in kinetic energy, and on the other hand, to a decrease in the potential energy of bodies. Therefore, work is equal to energy converted from one type to another.
Mechanical Energy Change Law
If a system of interacting bodies is not closed, then its mechanical energy is not conserved. The change in mechanical energy of such a system is equal to the work of external forces:
\(~A_(vn) = \Delta E = E - E_0\) . (20)
Where E And E 0 – total mechanical energies of the system in the final and initial states, respectively.
An example of such a system is a system in which, along with potential forces, non-potential forces act. Non-potential forces include friction forces. In most cases, when the angle between the friction force F r body is π radians, the work done by the friction force is negative and equal to
\(~A_(tr) = -F_(tr) \cdot s_(12)\) ,
Where s 12 – body path between points 1 and 2.
Frictional forces during the movement of a system reduce its kinetic energy. As a result of this, the mechanical energy of a closed non-conservative system always decreases, turning into the energy of non-mechanical forms of motion.
For example, a car moving along a horizontal section of the road, after turning off the engine, travels some distance and stops under the influence of friction forces. The kinetic energy of the forward motion of the car became equal to zero, and the potential energy did not increase. When the car was braking, the brake pads, car tires and asphalt heated up. Consequently, as a result of the action of friction forces, the kinetic energy of the car did not disappear, but turned into the internal energy of thermal motion of molecules.
Law of conservation and transformation of energy
In any physical interaction, energy is transformed from one form to another.
Sometimes the angle between the friction force F tr and elementary displacement Δ r is equal to zero and the work of the friction force is positive:
\(~A_(tr) = F_(tr) \cdot s_(12)\) ,
Example 1. Let the external force F acts on the block IN, which can slide on the cart D(Fig. 5). If the cart moves to the right, then the work done by the sliding friction force F tr2 acting on the cart from the side of the block is positive:
Example 2. When a wheel rolls, its rolling friction force is directed along the movement, since the point of contact of the wheel with the horizontal surface moves in the direction opposite to the direction of movement of the wheel, and the work of the friction force is positive (Fig. 6):
Literature
- Kabardin O.F. Physics: Reference. materials: Textbook. manual for students. – M.: Education, 1991. – 367 p.
- Kikoin I.K., Kikoin A.K. Physics: Textbook. for 9th grade. avg. school – M.: Prosveshchenie, 1992. – 191 p.
- Elementary physics textbook: Proc. allowance. In 3 volumes / Ed. G.S. Landsberg: vol. 1. Mechanics. Heat. Molecular physics. – M.: Fizmatlit, 2004. – 608 p.
- Yavorsky B.M., Seleznev Yu.A. A reference guide to physics for those entering universities and self-education. – M.: Nauka, 1983. – 383 p.
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 we determine the power from the relation (14.3.5)
as the product of torque and angular velocity of the body.
The theorem on the change in kinetic energy proven for a point (§ 14.4) will be valid for any point in the system
By composing such equations for all points of the system and adding them term by term, we obtain:
or, according to (19.1.1):
which is an expression of the theorem on the kinetic energy of a system in differential form.
Integrating (19.2.2) we get:
The theorem on the change in kinetic energy in its final form: the change in the kinetic energy of a system during some final displacement is equal to the sum of the work done on this displacement 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)
