The work done by gravity is equal to. The work of gravity. Potential energy of a body raised above the ground. Control questions and tasks

02.11.2019

The work of gravity. gravity R material point mass t near the Earth's surface can be considered a constant, equal to mg

directed vertically down.

Work BUT strength R on the move from the point M 0 to the point M

where h = z 0 - z x - point lowering height.

The work of gravity is equal to the product of this force and the height of lowering (work is positive) or the height of lifting (work is negative). The work of gravity does not depend on the shape of the trajectory between points M 0 and M|, and if these points coincide, then the work of gravity is equal to zero (the case of a closed path). It is also equal to zero if the points M 0 and M lie in the same horizontal plane.

The work of the linear force of elasticity. The linear elastic force (or linear restoring force) is the force acting according to Hooke's law (Fig. 63):

F = - withr,

where r- distance from the point of static equilibrium, where the force is zero, to the considered point M; with- constant coefficient - stiffness coefficient.

A=--().

According to this formula, the work of the linear elastic force is calculated. If point M 0 coincides with the point of static equilibrium O, so then r 0 \u003d 0 and for the work of the force on displacement from the point O to the point M we have

Value r- the shortest distance between the considered point and the point of static equilibrium. We denote it by λ and call it a deformation. Then

The work of the linear elastic force on displacement from the state of static equilibrium is always negative and equal to half the product of the stiffness coefficient and the square of the deformation. The work of the linear elastic force does not depend on the form of displacement and the work on any closed displacement is zero. It is also equal to zero if the points Mo and M lie on the same sphere circumscribed from the point of static equilibrium.

The work of a variable force in curvilinear motion.

The work of a force on a curved section

Consider the general case of finding the work of a variable force, the point of application of which moves along a curvilinear trajectory. Let the point M of application of the variable force F move along an arbitrary continuous curve. Denote by the vector of infinitely small displacement of the point M. This vector is directed tangentially to the curve in the same direction as the velocity vector.

Elementary work of a variable force F on an infinitesimal displacement

ds is called the scalar product of the vectors F and ds:

where a- angle between vectors F and ds

That is, the elementary work of the force is equal to the product of the modules of the vectors of force and an infinitesimal displacement, multiplied by the cosine of the angle between these vectors.

We decompose the force vector F into two components: - directed along the tangent to the trajectory - and - directed along the normal. line of force

is perpendicular to the tangent to the path along which the point moves, and its work is zero. Then:

dA= Ftds.

In order to calculate the work of the variable force F on the final section of the curve from a to b, one should calculate the integral of elementary work:

Potential and kinetic energy.

Potential energy P matserial point in considerationmy force field point M call work, performed by the forcesla acting on a material point when moving it from a pointMto the starting pointM 0 , i.e.

P = Umm 0

P = =-U=- U

The constant С 0 is the same for all points of the field, depending on which point of the field is chosen as the initial one. It is obvious that potential energy can be introduced only for a potential force field in which the work does not depend on the form of movement between points M and M 0 . A nonpotential force field has no potential energy, and there is no force function for it.

dA = dU= -dP; BUT = U - U 0 = P 0 - P

From the above formulas it follows that P is determined up to an arbitrary constant, which depends on the choice of the starting point, but this arbitrary constant does not affect the forces calculated through the potential energy and the work of these forces. Considering this:

P= - U+ const or P =- U.

The potential energy at any point of the field, up to an arbitrary constant, can be defined as the value of the force function at the same point, taken with a minus sign.

Kinetic energy system is called a scalar value T, equal to the sum of the kinetic energies of all points of the system:

Kinetic energy is a characteristic of both translational and rotational motions of the system. Kinetic energy is a scalar quantity and, moreover, essentially positive. Therefore, it does not depend on the directions of movement of the parts of the system and does not characterize changes in these directions.

Let us also note the following important circumstance. Internal forces act on parts of the system in mutually opposite directions. Changes in kinetic energy are influenced by the action of both external and internal forces.

Uniform motion of a point.

Uniform motion of a point- movement, with Krom kasat. acceleration ω t point (in the case of rectilinear motion, the total acceleration ω )constantly. The law of uniform motion of a point and the law of change in its speed υ during this motion are given by the equalities:

where s is the distance of the point measured along the trajectory arc from the reference point chosen on the trajectory, t- time, s 0 - value of s at the beginning. moment of time t = = 0. - beg. point speed. When the signs υ and ω identical, uniform motion. is accelerated, and when different - slowed down.

When acting. uniform motion of a rigid body, all of the above applies to each point of the body; with uniform rotation around a fixed axis of angle. acceleration e of the body is constant, and the law of rotation and the law of change of angle. the velocities ω of the body are given by the equalities

where φ is the angle of rotation of the body, φ 0 is the value of φ in the beginning. moment of time t= 0, ω 0 - beg. ang. body speed. When the signs of ω and ε match, the rotation is accelerated, and when they do not match, it is slow.

The work of a constant force in rectilinear motion.

Let us define the work for the case when the acting force is constant in magnitude and direction, and the point of its application moves along a rectilinear trajectory. Consider a material point C, to which a force constant in value and direction is applied (Fig. 134, a).

For a certain period of time t, point C has moved to position C1 along a rectilinear trajectory at a distance s.

The work W of a constant force during rectilinear motion of the point of its application is equal to the product of the modulus of force F times the distance s and the cosine of the angle between the direction of the force and the direction of movement, i.e.

The angle α between the direction of the force and the direction of motion can vary from 0 to 180°. For α< 90° работа положительна, при α >90° is negative, at α = 90° the work is zero.

If the force makes an acute angle with the direction of motion, it is called the driving force, the work of the force is always positive. If the angle between the directions of the force and the movement is obtuse, the force resists the movement, performs negative work and is called the resistance force. Examples of resistance forces are the forces of cutting, friction, air resistance, and others, which are always directed in the direction opposite to the movement.

When α = 0°, i.e., when the direction of the force coincides with the direction of the velocity, then W = F s, since cos 0° = 1. The product F cos α is the projection of the force onto the direction of motion of the material point. Therefore, the work of a force can be defined as the product of the displacement s and the projection of the force and the direction of movement of the point.

33. Forces of inertia of a rigid body

In classical mechanics, the representations of forces and their properties are based on Newton's laws and are inextricably linked with the concept of the inertial frame of reference.

Indeed, the physical quantity called force is introduced into consideration by Newton's second law, while the law itself is formulated only for inertial frames of reference. Accordingly, the concept of force initially turns out to be defined only for such frames of reference.

The equation of Newton's second law, which relates the acceleration and mass of a material point with the force acting on it, is written as

It directly follows from the equation that only forces are the cause of acceleration of bodies, and vice versa: the action of uncompensated forces on a body necessarily causes its acceleration.

Newton's third law complements and develops what was said about forces in the second law.

force is a measure of the mechanical action on a given material body of other bodies

in accordance with Newton's third law, forces can only exist in pairs, and the nature of the forces in each such pair is the same.

any force acting on a body has a source of origin in the form of another body. In other words, forces are necessarily the result interactions tel.

No other forces in mechanics are brought into consideration or used. The possibility of the existence of forces that have arisen independently, without interacting bodies, is not allowed by mechanics.

Although the names of the Euler and d'Alembert forces of inertia contain the word force, these physical quantities are not forces in the sense accepted in mechanics.

34. The concept of plane-parallel motion of a rigid body

The motion of a rigid body is called plane-parallel if all points of the body move in planes parallel to some fixed plane (the main plane). Let some body V make a plane motion, π - the main plane. From the definition of plane-parallel motion and the properties of an absolutely rigid body, it follows that any segment of the straight line AB, perpendicular to the plane π, will perform translational motion. That is, the trajectories, speeds and accelerations of all points of the segment AB will be the same. Thus, the movement of each point of the section s parallel to the plane π determines the movement of all points of the body V lying on the segment perpendicular to the section at this point. Examples of plane-parallel motion are: wheel rolling along a straight segment, since all its points move in planes parallel to the plane perpendicular to the wheel axis; a special case of such a movement is the rotation of a rigid body around a fixed axis, in fact, all points of a rotating body move in planes parallel to some fixed plane perpendicular to the axis of rotation.

35. Forces of inertia in the rectilinear and curvilinear motion of a material point

The force with which a point resists a change in motion is called the force of inertia of a material point. The force of inertia is directed opposite to the acceleration of the point and is equal to the mass times the acceleration.

In a straight line the direction of acceleration coincides with the trajectory. The force of inertia is directed in the direction opposite to acceleration, and its numerical value is determined by the formula:

With accelerated movement, the directions of acceleration and speed coincide and the force of inertia is directed in the direction opposite to the movement. In slow motion, when the acceleration is directed in the direction opposite to the speed, the force of inertia acts in the direction of motion.

Atcurvilinear and unevenmovement acceleration can be decomposed into normal an and tangent at components. Similarly, the force of inertia of a point also consists of two components: normal and tangential.

Normal the component of the inertial force is equal to the product of the mass of the point and the normal acceleration and is directed opposite to this acceleration:

Tangent the component of the inertial force is equal to the product of the mass of the point and the tangential acceleration and is directed opposite to this acceleration:

Obviously, the total force of inertia of the point M is equal to the geometric sum of the normal and tangent components, i.e.

Considering that the tangential and normal components are mutually perpendicular, the total inertia force is:

36. Theorems on the addition of velocities and accelerations of a point in complex motion

Velocity addition theorem:

In mechanics, the absolute velocity of a point is equal to the vector sum of its relative and translational velocities:

The speed of the body relative to the fixed frame of reference is equal to the vector sum of the speed of this body relative to the moving frame of reference and the speed (relative to the fixed frame) of the point of the moving frame where the body is located.

in a complex motion, the absolute speed of a point is equal to the geometric sum of the translational and relative speeds. The magnitude of the absolute velocity is determined where α is the angle between the vectors and .

Acceleration addition theorem ( THEOREM OF CORIOLIS)

acor = aper + afrom + acor

The formula expresses the following Coriolis theorem on the addition of accelerated

rhenium: 1 for complex motion, the acceleration of a point is equal to the geometric

the sum of three accelerations: relative, translational and rotary, or

Coriolis.

acor = 2(ω × vot)

37. d'Alembert principle

d'Alembert's principle for a material point: at each moment of the movement of a material point, active forces, reactions of bonds and the force of inertia form a balanced system of forces.

d'Alembert's principle- in mechanics: one of the basic principles of dynamics, according to which, if the forces of inertia are added to the given forces acting on the points of the mechanical system and the reactions of the superimposed bonds, then a balanced system of forces will be obtained.

According to this principle, for each i-th point of the system, the equality

where is the active force acting on this point, is the reaction of the connection imposed on the point, is the force of inertia, numerically equal to the product of the mass of the point and its acceleration and directed opposite to this acceleration ().

In fact, we are talking about the transfer of the term ma from right to left in Newton's second law () performed separately for each of the considered material points and the censure of this term by the d'Alembert force of inertia.

The d'Alembert principle makes it possible to apply simpler methods of statics to solving problems of dynamics, therefore it is widely used in engineering practice, the so-called. kinetostatic method. It is especially convenient to use it to determine the reactions of constraints in cases where the law of the ongoing motion is known or found from the solution of the corresponding equations.

A strand \u003d mg (h n - h k) (14.19)

where h n and h k are the initial and final heights (Fig. 14.7) of a material point with mass m, g is the free fall acceleration module.

The work of gravity A strand is determined by the initial and final positions of the material point and does not depend on the trajectory between them.

It can be positive, negative or zero:

a) A strand > 0 - during the descent of a material point,

b) A heavy< 0 - при подъеме материальной точки,

c) A str = 0 - provided that the height does not change, or with a closed trajectory of a material point.

The work of the friction force at constant speed b.w. ( v = const) and friction forces ( F tr = const) on the time interval t:

A tr = ( F tr, v)t, (14.20)

The work of the friction force can be positive, negative or zero. For example:

a
) the work of the friction force acting on the lower bar from the side of the upper bar (Fig. 14.8), A tr.2,1\u003e 0, because the angle between the force acting on the bottom bar from the side of the top bar F tr.2.1 and speed v 2 of the lower bar (relative to the Earth's surface) is equal to zero;

b) A tr.1,2< 0 - угол между силой трения F tr.1,2 and speed v 1 of the upper bar is equal to 180 (see Fig. 14.8);

c) A tr \u003d 0 - for example, the bar is on a rotating horizontal disk (relative to the disk, the bar is stationary).

The work of the friction force depends on the trajectory between the initial and final positions of the material point.

§fifteen. mechanical energy

Kinetic energy of a material point K - SFV, equal to half the product of the mass of b.w. to the square of the modulus of its speed:

(15.1)

The kinetic energy due to the motion of the body depends on the frame of reference and is a non-negative quantity:

Unit of kinetic energy-joule: [K] = J.

Kinetic energy theorem- increment of kinetic energy b.w. is equal to the work A p of the resultant force:

K = A p. (15.3)

The work of the resultant force can be found as the sum of the works A i of all forces F i (i = 1,2,…n) applied to the b.w.:

(15.4)

Velocity modulus of a material point: at A p > 0 - increases; at A p< 0 - уменьшается; при A р = 0 - не изменяется.

Kinetic energy of a system of material points K c is equal to the sum of the kinetic energies K i of all n b.w. belonging to this system:

(15.5)

where m i and v i are the mass and velocity modulus of the i-th m.t. this system.

The increment of the kinetic energy of the system b.t.K с is equal to the sum of works А рi of all n resultant forces applied to the i-th material points of the system:

(15.6)

Force field- a region of space, at each point of which forces act on the body.

Stationary force field- a field whose forces do not change over time.

Uniform field of forces- a field, the forces of which are the same at all its points.

Central Force Field- a field, the directions of action of all forces of which pass through one point, called the center of the field, and the modulus of forces depends only on the distance to this center.

Non-conservative forces (nx.sl)- forces whose work depends on the trajectory between the initial and final positions of the body .

An example of non-conservative forces is friction forces. The work of friction forces along a closed trajectory in the general case is not equal to zero.

Conservative Forces (ks.sl)- forces, the work of which is determined by the initial and final positions of the m.t. and does not depend on the trajectory between them. With a closed trajectory, the work of conservative forces is zero. The field of conservative forces is called potential.

An example of conservative forces is gravity and elasticity.

Potential energy P - SPV, which is a function of the relative position of the parts of the system (body).

Unit of potential energy-joule: [P] = J.

Potential energy theorem

Loss of potential energy of a system of material points is equal to the work of conservative forces:

–P s = P n – P c = A ks.sl (15.7 )

Potential energy is determined up to a constant value and can be positive, negative or equal to zero.

Potential energy of a material point P at any point of the force field - SPV, equal to the work of conservative forces when moving the b.w. from a given point of the field to a point where the potential energy is assumed to be zero:

P \u003d A ks.sl. (15.8)

Potential energy of elastically deformed spring

(15.9)

G de x - displacement of the loose end of the spring; k is the stiffness of the spring, C is an arbitrary constant (selected from the condition of convenience in solving the problem).

P(x) graphs for various constants: a) C > 0, b) C = 0, c) C< 0  параболы (рис.15.1).

Under the condition P (0) = 0, the constant C = 0 and

(15.10)

In this lesson, we will consider the various motion of a body under the influence of gravity and learn how to find the work of this force. We will also introduce the concept of the potential energy of a body, find out how this energy is related to the work of gravity, and derive the formula by which this energy is found. Using this formula, we will solve the problem taken from the collection for preparing for the unified state exam.

In the previous lessons, we studied the varieties of forces in nature. For each force, it is necessary to correctly calculate the work. This lesson is devoted to the study of the work of gravity.

At small distances from the Earth's surface, gravity is constant and modulo equal to , where m- body mass, g- acceleration of gravity.

Let the body mass m falls freely from a height above any level from which the count is taken to a height above the same level (see Fig. 1).

Rice. 1. Free fall of the body from height to height

In this case, the modulus of displacement of the body is equal to the difference between these heights:

Since the direction of movement and gravity are the same, the work done by gravity is:

The height value in this formula can be calculated from any level (sea level, bottom level of a hole dug in the ground, table surface, floor surface, etc.). In any case, the height of this surface is chosen equal to zero, so the level of this height is called zero level.

If a body falls from a height h to zero, then the work done by gravity will be:

If a body thrown upwards from the zero level reaches a height h above this level, then the work done by gravity will be equal to:

Let the body mass m moving on an inclined plane h and at the same time makes a movement, the module of which is equal to the length of the inclined plane (see Fig. 2).

Rice. 2. Movement of a body along an inclined plane

The work of the force is equal to the scalar product of the force vector and the displacement vector of the body made under the action of this force, that is, the work of gravity in this case will be equal to:

where is the angle between the gravity and displacement vectors.

Figure 2 shows that the displacement () is the hypotenuse of a right triangle, and the height h- cathet. According to the property of a right triangle:

Hence

We have obtained the expression for the work of gravity is the same as in the case of the vertical motion of the body. It can be concluded that if the trajectory of the body is not rectilinear and the body moves under the action of gravity, then the work of gravity is determined only by a change in the height of the body above a certain zero level and does not depend on the trajectory of the body.

Rice. 3. Motion of a body along a curvilinear trajectory

Let us prove the previous assertion. Let the body move along some curvilinear trajectory (see Fig. 3). We mentally divide this trajectory into a number of small sections, each of which can be considered a small inclined plane. The movement of the body along the entire trajectory can be represented as movement along a set of inclined planes. The work of gravity on each of the sections will be equal to the product of the force of gravity and the height of this section. If the height changes in individual sections are equal, then the work of gravity on them is equal:

The total work on the entire trajectory is equal to the sum of the work on individual sections:

- the total height that the body has overcome,

Thus, the work of gravity does not depend on the trajectory of the body and is always equal to the product of gravity and the difference in heights in the initial and final positions. Q.E.D.

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

Let some body move along a closed trajectory, that is, it first went down, and then returned to the starting point along some other trajectory. Since the body ended up at the same point where it was originally, the height difference between the initial and final position of the body is zero, so the work of gravity will be zero. Hence, the work done by gravity when a body moves along a closed trajectory is zero.

In the formula for the work of gravity, we take (-1) out of the bracket:

It is known from previous lessons that the work of forces applied to a body is equal to the difference between the final and initial values of the body's kinetic energy. The resulting formula also shows the relationship between the work of gravity and the difference between the values of some physical quantity equal to . Such a value is called potential energy of the body which is at the height h above some zero level.

The change in potential energy is negative in magnitude if positive work is done by gravity (it can be seen from the formula). If negative work is done, then the change in potential energy will be positive.

If a body falls from a height h to the zero level, then the work of gravity will be equal to the value of the potential energy of the body raised to a height h.

Potential energy of the body, raised to a certain height above the zero level, is equal to the work that the force of gravity will do when the given body falls from a given height to the zero level.

Unlike kinetic energy, which depends on the speed of the body, potential energy may not be zero even for bodies at rest.

Rice. 4. The body below the zero level

If the body is below the zero level, then it has a negative potential energy (see Fig. 4). That is, the sign and modulus of the potential energy depend on the choice of the zero level. The work that is done when moving the body does not depend on the choice of the zero level.

The term "potential energy" applies only to a system of bodies. In all the above reasoning, this system was "Earth - a body raised above the Earth."

Homogeneous rectangular parallelepiped with mass m with ribs are placed on a horizontal plane on each of the three faces in turn. What is the potential energy of the parallelepiped in each of these positions?

Given:m- mass of the parallelepiped; - the length of the edges of the parallelepiped.

To find:; ;

Decision

If it is necessary to determine the potential energy of a body of finite dimensions, then we can assume that the entire mass of such a body is concentrated at one point, which is called the center of mass of this body.

In the case of symmetrical geometric bodies, the center of mass coincides with the geometric center, that is (for this problem) with the point of intersection of the diagonals of the parallelepiped. Thus, it is necessary to calculate the height at which this point is located at various locations of the parallelepiped (see Fig. 5).

Rice. 5. Illustration for the problem

In order to find the potential energy, it is necessary to multiply the obtained height values by the mass of the parallelepiped and the acceleration of free fall.

Answer:; ;

In this lesson, we learned how to calculate the work of gravity. At the same time, we saw that, regardless of the trajectory of the body, the work of gravity is determined by the difference between the heights of the initial and final positions of the body above some zero level. We also introduced the concept of potential energy and showed that the work of gravity is equal to the change in the potential energy of the body, taken with the opposite sign. What work must be done to shift a bag of flour weighing 2 kg from a shelf located at a height of 0.5 m relative to the floor to a table located at a height of 0.75 m relative to the floor? What is the potential energy of the bag of flour lying on the shelf, and its potential energy when it is on the table, relative to the floor?

In the previous lessons, we studied the varieties of forces in nature. For each force, it is necessary to correctly calculate the work. This lesson is devoted to the study of the work of gravity.

At small distances from the Earth's surface, gravity is constant and modulo equal to , where m- body mass, g- acceleration of gravity.

Let the body mass m falls freely from a height above any level from which the count is taken to a height above the same level (see Fig. 1).

Rice. 1. Free fall of the body from height to height

In this case, the modulus of displacement of the body is equal to the difference between these heights:

Since the direction of movement and gravity are the same, the work done by gravity is:

If a body falls from a height h to zero, then the work done by gravity will be:

If a body thrown upwards from the zero level reaches a height h above this level, then the work done by gravity will be equal to:

Let the body mass m moving on an inclined plane h and at the same time makes a movement, the module of which is equal to the length of the inclined plane (see Fig. 2).

Rice. 2. Movement of a body along an inclined plane

where is the angle between the gravity and displacement vectors.

Figure 2 shows that the displacement () is the hypotenuse of a right triangle, and the height h- cathet. According to the property of a right triangle:

Hence

Rice. 3. Motion of a body along a curvilinear trajectory

The total work on the entire trajectory is equal to the sum of the work on individual sections:

- the total height that the body has overcome,

Thus, the work of gravity does not depend on the trajectory of the body and is always equal to the product of gravity and the difference in heights in the initial and final positions. Q.E.D.

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

In the formula for the work of gravity, we take (-1) out of the bracket:

If a body falls from a height h to the zero level, then the work of gravity will be equal to the value of the potential energy of the body raised to a height h.

Unlike kinetic energy, which depends on the speed of the body, potential energy may not be zero even for bodies at rest.

Rice. 4. The body below the zero level

The term "potential energy" applies only to a system of bodies. In all the above reasoning, this system was "Earth - a body raised above the Earth."

Given:m- mass of the parallelepiped; - the length of the edges of the parallelepiped.

To find:; ;

Decision

Rice. 5. Illustration for the problem

In order to find the potential energy, it is necessary to multiply the obtained height values by the mass of the parallelepiped and the acceleration of free fall.

Answer:; ;

The work of gravity. Problem solving

The purpose of the lesson: determine the formula for the work of gravity; determine that the work of gravity does not depend on the trajectory of the body; develop practical problem solving skills.

During the classes.

1. Organizational moment. Greeting students, checking absentees, setting the goal of the lesson.

2. Checking homework.

3. Study of new material. In the previous lesson, we defined a formula for determining work. What is the formula for the work done by a constant force? (A=FScosα)

What is A andS?

Now let's apply this formula for gravity. But first, let's remember what is the force of gravity? (F= mg)

Consider case a) a body falls vertically downwards. As you and I know, gravity is always directed straight down. In order to determine the directionSremember the definition. (Displacement is a vector connecting the start and end points. It is directed from start to end)

That. for determining , Since the direction of movement and the force of gravity are the same, thenα =0 and the work done by gravity is:

Consider case b) the body moves vertically upwards. Because direction of gravity and displacement are opposite, thenα =0 and the work done by gravity is .

That. Thus, if you compare two formulas modulo, they will be the same.

Consider case c) the body moves along an inclined plane. The work of the force is equal to the scalar product of the force vector and the displacement vector of the body made under the action of this force, that is, the work of gravity in this case will be equal to, where is the angle between the gravity and displacement vectors. The figure shows that the displacement () is the hypotenuse of a right triangle, and the heighth- cathet. According to the property of a right triangle:

.Hence

That. what conclusion can be drawn?(that the work of gravity does not depend on the trajectory of motion.)

Consider the last example, when the trajectory movement will be a closed line. Who will say what the work will be equal to and why? (A=0 because displacement is 0)

Note!: the work done by gravity when a body moves along a closed trajectory is zero.

4. Fixing the material.

Task 1. The hunter shoots from a cliff at an angle of 40° to the horizon. During the fall of the bullet, the work of gravity was 5 J. If the bullet entered the ground at a distance of 250 m from the rock, then what is its mass?

Task 2. While on Neptune, the body moved as shown in the figure. With this displacement, the work of gravity was 840 J. If the mass of this body is 5 kg, then what is the acceleration of free fall on Neptune?

5. Homework.

The work done by gravity is equal to. The work of gravity. Potential energy of a body raised above the ground. Control questions and tasks

§fifteen. mechanical energy

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