Work force of dry friction formula. Mechanical work. Power

You are already familiar with mechanical work (work of force) from the basic school physics course. Recall the definition of mechanical work given there for the following cases.

If the force is directed in the same direction as the displacement of the body, then the work done by the force

In this case, the work done by the force is positive.

If the force is directed opposite to the movement of the body, then the work done by the force is

In this case, the work done by the force is negative.

If the force f_vec is directed perpendicular to the displacement s_vec of the body, then the work of the force is zero:

Work is a scalar quantity. The unit of work is called the joule (denoted: J) in honor of the English scientist James Joule, who played an important role in the discovery of the law of conservation of energy. From formula (1) it follows:

1 J = 1 N * m.

1. A bar weighing 0.5 kg was moved along the table by 2 m, applying an elastic force equal to 4 N to it (Fig. 28.1). The coefficient of friction between the bar and the table is 0.2. What is the work done on the bar:
a) gravity m?
b) normal reaction forces ?
c) elastic force?
d) forces of sliding friction tr?

The total work of several forces acting on a body can be found in two ways:
1. Find the work of each force and add these works, taking into account the signs.
2. Find the resultant of all forces applied to the body and calculate the work of the resultant.

Both methods lead to the same result. To verify this, return to the previous task and answer the questions of task 2.

2. What is equal to:
a) the sum of the work of all the forces acting on the block?
b) the resultant of all forces acting on the bar?
c) the work of the resultant? In the general case (when the force f_vec is directed at an arbitrary angle to the displacement s_vec), the definition of the work of the force is as follows.

Job A constant force is equal to the product of the modulus of force F times the modulus of displacement s and the cosine of the angle α between the direction of the force and the direction of displacement:

A = Fs cos α (4)

3. Show that the general definition of work leads to the conclusions shown in the following diagram. Formulate them verbally and write them down in your notebook.

4. A force is applied to the bar on the table, the module of which is 10 N. What is the angle between this force and the movement of the bar, if, when moving the bar along the table by 60 cm, this force did the work: a) 3 J; b) –3 J; c) –3 J; d) -6 J? Make explanatory drawings.

2. The work of gravity

Let a body of mass m move vertically from the initial height h n to the final height h k.

If the body moves down (h n > h k, Fig. 28.2, a), the direction of movement coincides with the direction of gravity, so the work of gravity is positive. If the body moves up (h n< h к, рис. 28.2, б), то работа силы тяжести отрицательна.

In both cases, the work done by gravity

A \u003d mg (h n - h k). (5)

Let us now find the work done by gravity when moving at an angle to the vertical.

5. A small block of mass m slid along an inclined plane of length s and height h (Fig. 28.3). The inclined plane makes an angle α with the vertical.

a) What is the angle between the direction of gravity and the direction of movement of the bar? Make an explanatory drawing.
b) Express the work of gravity in terms of m, g, s, α.
c) Express s in terms of h and α.
d) Express the work of gravity in terms of m, g, h.
e) What is the work of gravity when the bar moves up along the entire same plane?

Having completed this task, you made sure that the work of gravity is expressed by formula (5) even when the body moves at an angle to the vertical - both up and down.

But then formula (5) for the work of gravity is valid when the body moves along any trajectory, because any trajectory (Fig. 28.4, a) can be represented as a set of small "inclined planes" (Fig. 28.4, b).

Thus,
the work of gravity during movement but any trajectory is expressed by the formula

A t \u003d mg (h n - h k),

where h n - the initial height of the body, h to - its final height.
The work of gravity does not depend on the shape of the trajectory.

For example, the work of gravity when moving a body from point A to point B (Fig. 28.5) along path 1, 2 or 3 is the same. From here, in particular, it follows that the work of gravity when moving along a closed trajectory (when the body returns to the starting point) is equal to zero.

6. A ball of mass m, hanging on a thread of length l, is deflected by 90º, keeping the thread taut, and released without a push.
a) What is the work of gravity during the time during which the ball moves to the equilibrium position (Fig. 28.6)?
b) What is the work of the elastic force of the thread in the same time?
c) What is the work of the resultant forces applied to the ball in the same time?

3. The work of the force of elasticity

When the spring returns to its undeformed state, the elastic force always does positive work: its direction coincides with the direction of movement (Fig. 28.7).

Find the work of the elastic force.
The modulus of this force is related to the modulus of deformation x by the relation (see § 15)

The work of such a force can be found graphically.

Note first that the work of a constant force is numerically equal to the area of the rectangle under the graph of force versus displacement (Fig. 28.8).

Figure 28.9 shows a plot of F(x) for the elastic force. Let us mentally divide the entire displacement of the body into such small intervals that the force on each of them can be considered constant.

Then the work on each of these intervals is numerically equal to the area of the figure under the corresponding section of the graph. All the work is equal to the sum of the work in these areas.

Consequently, in this case, the work is also numerically equal to the area of the figure under the F(x) dependence graph.

7. Using Figure 28.10, prove that

the work of the elastic force when the spring returns to the undeformed state is expressed by the formula

A = (kx 2)/2. (7)

8. Using the graph in Figure 28.11, prove that when the deformation of the spring changes from x n to x k, the work of the elastic force is expressed by the formula

From formula (8) we see that the work of the elastic force depends only on the initial and final deformation of the spring, Therefore, if the body is first deformed, and then it returns to its initial state, then the work of the elastic force is zero. Recall that the work of gravity has the same property.

9. At the initial moment, the tension of the spring with a stiffness of 400 N / m is 3 cm. The spring is stretched another 2 cm.
a) What is the final deformation of the spring?
b) What is the work done by the elastic force of the spring?

10. At the initial moment, a spring with a stiffness of 200 N / m is stretched by 2 cm, and at the final moment it is compressed by 1 cm. What is the work of the elastic force of the spring?

4. The work of the friction force

Let the body slide on a fixed support. The sliding friction force acting on the body is always directed opposite to the movement and, therefore, the work of the sliding friction force is negative for any direction of movement (Fig. 28.12).

Therefore, if the bar is moved to the right, and with a peg the same distance to the left, then, although it returns to its initial position, the total work of the sliding friction force will not be equal to zero. This is the most important difference between the work of the sliding friction force and the work of the force of gravity and the force of elasticity. Recall that the work of these forces when moving the body along a closed trajectory is equal to zero.

11. A bar with a mass of 1 kg was moved along the table so that its trajectory turned out to be a square with a side of 50 cm.
a) Did the block return to its starting point?
b) What is the total work of the friction force acting on the bar? The coefficient of friction between the bar and the table is 0.3.

5. Power

Often, not only the work done is important, but also the speed of the work. It is characterized by power.

The power P is the ratio of the work done A to the time interval t during which this work is done:

(Sometimes power in mechanics is denoted by the letter N, and in electrodynamics by the letter P. We find it more convenient to use the same designation of power.)

The unit of power is the watt (denoted: W), named after English inventor James Watt. From formula (9) it follows that

1 W = 1 J/s.

12. What power does a person develop by uniformly lifting a bucket of water weighing 10 kg to a height of 1 m for 2 s?

It is often convenient to express power not in terms of work and time, but in terms of force and speed.

Consider the case when the force is directed along the displacement. Then the work of the force A = Fs. Substituting this expression into formula (9) for power, we obtain:

P = (Fs)/t = F(s/t) = Fv. (10)

13. A car is driving along a horizontal road at a speed of 72 km/h. At the same time, its engine develops a power of 20 kW. What is the force of resistance to the movement of the car?

Clue. When a car is moving along a horizontal road at a constant speed, the traction force is equal in absolute value to the drag force of the car.

14. How long will it take to evenly lift a concrete block weighing 4 tons to a height of 30 m, if the power of the crane motor is 20 kW, and the efficiency of the crane motor is 75%?

Clue. The efficiency of the electric motor is equal to the ratio of the work of lifting the load to the work of the engine.

Additional questions and tasks

15. A ball of mass 200 g is thrown from a balcony 10 high and at an angle of 45º to the horizon. Having reached a maximum height of 15 m in flight, the ball fell to the ground.
a) What is the work done by gravity in lifting the ball?
b) What is the work done by gravity when the ball is lowered?
c) What is the work done by gravity during the entire flight of the ball?
d) Is there extra data in the condition?

16. A ball weighing 0.5 kg is suspended from a spring with a stiffness of 250 N/m and is in equilibrium. The ball is lifted so that the spring becomes undeformed and released without a push.
a) To what height was the ball raised?
b) What is the work of gravity during the time during which the ball moves to the equilibrium position?
c) What is the work of the elastic force during the time during which the ball moves to the equilibrium position?
d) What is the work of the resultant of all forces applied to the ball during the time during which the ball moves to the equilibrium position?

17. A sledge weighing 10 kg slides down a snowy mountain without initial speed with an inclination angle α = 30º and travels some distance along a horizontal surface (Fig. 28.13). The coefficient of friction between the sled and snow is 0.1. The length of the base of the mountain l = 15 m.

a) What is the modulus of the friction force when the sled moves on a horizontal surface?
b) What is the work of the friction force when the sled moves along a horizontal surface on a path of 20 m?
c) What is the modulus of the friction force when the sled moves up the mountain?
d) What is the work done by the friction force during the descent of the sled?
e) What is the work done by gravity during the descent of the sled?
f) What is the work of the resultant forces acting on the sled as it descends from the mountain?

18. A car weighing 1 ton moves at a speed of 50 km/h. The engine develops a power of 10 kW. Gasoline consumption is 8 liters per 100 km. The density of gasoline is 750 kg/m 3 and its specific heat of combustion is 45 MJ/kg. What is the engine efficiency? Is there extra data in the condition?
Clue. The efficiency of a heat engine is equal to the ratio of the work done by the engine to the amount of heat released during the combustion of fuel.

Instruction

Example of problem 3: a block of mass 1 kg slid off the top of an inclined plane in 5 seconds, the path is 10 meters. Determine the force of friction if the angle of inclination of the plane is 45o. Consider also the case where the block was subjected to an additional force of 2 N applied along the angle of inclination in the direction of motion.

Find the acceleration of the body in the same way as in examples 1 and 2: a = 2*10/5^2 = 0.8 m/s2. Calculate the friction force in the first case: Ftr \u003d 1 * 9.8 * sin (45o) -1 * 0.8 \u003d 7.53 N. Determine the friction force in the second case: Ftr \u003d 1 * 9.8 * sin (45o) +2-1*0.8= 9.53 N.

Case 6. A body moves uniformly along an inclined surface. So, according to Newton's second law, the system is in equilibrium. If the sliding is spontaneous, the motion of the body obeys the equation: mg*sinα = Ftr.

If an additional force (F) is applied to the body, which prevents uniformly accelerated movement, the expression for motion has the form: mg*sinα–Ftr-F = 0. From here, find the friction force: Ftr = mg*sinα-F.

Sources:

slip formula

According to the mechanical law of Coulomb, the sliding force is F = kN, where k is the coefficient of friction, and N is the reaction force of the support. Since the reaction force of the support is directed strictly vertically, then N \u003d Fthrust \u003d mg, where m is the body mass, g is the acceleration free fall. This condition follows from the immobility of the body relative to the vertical direction.

Thus, the coefficient of friction can be found by the formula k = Ffr/N = Ffr/mg. To do this, you need to know the power. If the body is uniformly accelerated, then the friction force can be found by knowing the acceleration a. Let the driving force F act on the body and Ftr directed oppositely to it. Then according to Newton's second law (F-Ftr)/m = a. Expressing Ffr from here and substituting into the formula for the friction coefficient, we get: k = (F-ma)/N.

It can be seen from these formulas that the coefficient of friction is a dimensionless quantity.

Consider a more general case, when from an inclined plane, for example, from a fixed block. Such tasks are very common in school course in the Mechanics section.

Let the angle of inclination of the plane be equal to φ. The support reaction force N will be directed perpendicular to the inclined plane. The force of gravity and friction will also act on the body. We direct the axes along and perpendicular to the inclined plane.

According to Newton's second law, the body equations can be written: N = mg*cosφ, mg*sinφ-Ftr = mg*sinφ-kN = ma.

Substituting the first equation into the second and reducing the mass m, we get: g*sinφ-kg*cosφ = a. Hence, k = (g*sinφ-a)/(g*cosφ).

Consider an important special case sliding down an inclined plane, when a = 0, that is, the body moves uniformly. Then the equation of motion has the form g*sinφ-kg*cosφ = 0. Hence, k = tgφ, that is, to determine the slip coefficient, it is enough to know the tangent of the plane inclination angle.

Related videos

Friction is the process of interaction between two bodies, causing a slowdown in movement when displaced relative to each other. Find strength friction- means to determine the magnitude of the impact directed in the direction opposite to the movement, due to which the body loses energy and, in the end, stops.

Instruction

Force friction- a vector quantity that depends on many factors: bodies on top of each other, the materials from which they were made, speed. The surface area does not matter in this case, since the larger it is, the greater the mutual pressure (support force N), which is already involved in finding the force friction.

Coefficient friction rolling is generally a known quantity for common materials. For example, for iron, it is 0.51 mm, for iron for wood - 5.6, wood for wood - 0.8-1.5, etc. You can find it by the moment ratio formula friction to the pressing force.

Force friction rest appears with minimal displacement of bodies or deformation. This force is always present in dry sliding. Its maximum value is μ N. There is also internal friction, inside one body between its layers or.

note

The uniform motion of a body is characterized by a balance between the external force and the force of friction.

In school problems in physics, to determine the force of sliding friction, it is mainly considered a rectilinear uniform or rectilinear uniformly accelerated motion of a body. See how you can find the friction force in different cases, depending on the conditions of the problem. To correctly evaluate the effects of forces and to formulate the equation of motion, always draw a drawing.

If on a body of mass m, located on a smooth horizontal surface, acts
constant force F directed at some angle α to the horizon while the body moves a certain distance S, then they say that the force F did the job A. The amount of work is determined by the formula:

A= F× S cos α (1)

However, ideally smooth surfaces do not exist in nature, and friction forces always arise on the contact surface of two bodies. Here is how it is written in the textbook: “The work of the static friction force is zero, since there is no displacement. When sliding solid surfaces, the friction force is directed against the movement. Her work is negative. As a result, the kinetic energy of the rubbing bodies is converted into internal energy - the rubbing surfaces are heated.

A TR = FTP ×S = μNS (2)

Where μ - coefficient of sliding friction.

Only in the textbook O.D. Khvolson considered the case of ACCELERATED MOVEMENT in the presence of friction forces: “So, two cases of work production should be distinguished: in the first, the essence of work is to overcome external resistance to movement, which is performed without increasing the speed of the body; in the second, work is revealed by an increase in the speed of movement, to which the outside world is indifferent.

In fact, we usually have a CONNECTION OF BOTH CASES: power f overcomes any resistance and at the same time changes the speed of the body.

Let's assume that f" not equal f, namely that f"< f. In this case, the force acting on the body
f- f", Job ρ which causes an increase in the speed of the body. We have ρ =(f- f")S,
where

fS= f"S+ ρ (*)

Job r= fS consists of two parts: f"S spent on overcoming external resistance, ρ to increase the speed of the body.

Let's imagine it in a modern interpretation (Fig. 1). On a body of mass m traction force works F T , which is greater than the force of friction FTP = μN = μmg. The work of the traction force in accordance with the formula (*) can be written as follows

A=F T S=F TP S+F a S= ATP+ A a(3)

Where Fa=F-T-F-TP- force causing accelerated motion of the body in accordance with Newton's II law: Fa= ma. The work of the friction force is negative, but hereinafter we will use the friction force and the work of friction modulo. Further reasoning requires a numerical analysis. Let's take the following data: m=10 kg; g\u003d 10 m / s 2; F T=100 N; μ = 0,5; t=10 s. We carry out the following calculations: FTP= mmg= 50 N; Fa= 50 N; a=Fa/m\u003d 5 m / s 2; V= at= 50 m/s; K= mV 2 /2 \u003d 12.5 kJ; S= at 2/2 = 250 m; A a= F a S=12.5 kJ; ATP=F TP S=12.5 kJ. So the total work A= ATP+ A a=12.5 +12.5 = 25 kJ

And now we calculate the work of the traction force F T for the case when there is no friction ( μ =0).

Carrying out similar calculations, we get: a \u003d 10 m / s 2; V=100m/s; K = 50 kJ; S = 500 m; A = 50 kJ. In the latter case, in the same 10 s, we got twice as much work. It may be objected that the path is twice as long. However, no matter what they say, it turns out a paradoxical situation: the powers developed by the same force are two times different, although the impulses of the forces are the same I =F T t =1 kN.s. As M.V. wrote Lomonosov in 1748: "... but all the changes that take place in nature occur in such a way that how much is added to something, the same amount will be taken away from the other ...". Therefore, let's try to get another expression for defining work.

We write Newton's II law in differential form:

F. dt = d(mV ) (4)

and consider the problem of acceleration of an initially immobile body (no friction). Integrating (4), we get: F × t = mV . Squaring and dividing by 2 m both sides of the equation, we get:

F 2 t 2/2m= mV 2 / 2 A= K (5)

Thus, we got another expression for calculating the work

A=F 2 t 2 / 2m = I 2/2m (6)

Where I = F × t - force impulse. This expression is not related to the path S passed by the body during the time t, i.e. it can be used to calculate the work done by an impulse of force even if the body remains motionless, although, as they say in all physics courses, no work is done in this case.

Turning to our problem of accelerated motion with friction, we write the sum of the force impulses: I T = I a + I TP, Where I T = F T t; I a= F a t; I TP = F TP t. Squaring the sum of the impulses, we get:

F T 2 t 2= Fa 2 t2+ 2F a F TP t 2 + F TP 2 t 2

Dividing all terms of equality by 2m, we get:

or A= A a + A UT + A TP

Where A a=Fa 2 t 2 / 2 m- work expended acceleration; ATP = FTP 2 t 2 /2 m - the work expended on overcoming the friction force in uniform motion, and AUT =F a F TP t 2 / m- work expended on overcoming the friction force during accelerated motion. Numerical calculation gives the following result:

A=A a + AUt + ATP = 12.5 + 25 +12.5 = 50 kJ,

those. we got the same amount of work done by the force F T in the absence of friction.

Consider a more general case of motion of a body with friction, when a force acts on the body F directed at an angle α to the horizon (Fig. 2). Now the traction force F T = F cosα, but force F L= F sinα - let's call it the levitation force, it reduces the force of gravity P=mg, and in the case F L = mg the body will not exert pressure on the support, it will be in a quasi-weightless state (a state of levitation). Friction force FTP = μN = μ (P - F L) . The pull force can be written as F T= Fa+ FTP, and from a right-angled triangle (Fig. 2) we get: F 2 =F T 2 + F L 2 . Multiplying the last relation by t2 , we obtain the balance of force impulses, and dividing by 2m, we get the balance of energies (ra-bot):

Let us give a numerical calculation for the force F = 100 N and α = 30o under the same conditions (m = 10 kg; μ = 0,5; t = 10 With). Force work F will be equal to A=F 2 t 2 /2m= 50 , and formula (8) gives the following result (up to the third decimal place):

50=15.625+18.974-15.4-12.5+30.8+12.5 kJ.

Calculations show that the force F = 100 N, acting on a body of mass m = 10 kg at any angle α 50 kJ does the same work in 10 s.

The last term in formula (8) is the work of the friction force during uniform motion of the body on a horizontal surface with a speed V

Thus, no matter at what angle this force acts F on a given body of mass m, with or without friction, over time t the same work will be done (even if the body is motionless):

Fig.1

Fig.2

BIBLIOGRAPHY

Matveev A.N. mechanics and theory of relativity. Textbook for physical.special.universities. -M.: Higher school, 1986.
Strelkov SP. Mechanics. General course of physics. T. 1. - M.: GITTL, 1956.
Khvolson O.D. Physics course. T. 1. RSFSR State Publishing House, Berlin, 1923.

Bibliographic link

IVANOV E.M. WORK DURING THE MOVEMENT OF BODIES WITH FRICTION // Modern problems of science and education. - 2005. - No. 2.;
URL: http://science-education.ru/ru/article/view?id=1468 (date of access: 20.04.2019). We bring to your attention the journals published by the publishing house "Academy of Natural History"

With the relative motion of one body on the surface of another, friction forces arise, that is, the bodies interact with each other. However, this type of interaction is fundamentally different from those considered earlier. The most significant difference is the fact that the force of interaction is determined not by the relative position of the bodies, but by their relative speed. Consequently, the work of these forces depends not only on the initial and final positions of the bodies, but also on the shape of the trajectory, on the speed of movement. In other words, friction forces are not potential.
Let us consider in more detail the work of various types of friction.
The simplest case is static friction. Suffice it to say that in the absence of displacement, the work is equal to zero, so the static friction does no work.
When one body moves over the surface of another, a force of dry friction arises. According to the Coulomb-Amonton law, the magnitude of the friction force is constant and directed in the direction opposite to the speed of movement. Therefore, at any moment of time, at any point of the trajectory, the velocity and friction force vectors are directed in opposite directions, the angle between them is equal to 180°(remember cos180° = -1). Thus, the work of the friction force is equal to the product of the friction force and the length of the trajectory S:
A mp = −F mp S. (1)
Between two points, you can lay as many trajectories as you like, the lengths of which can vary over a wide range, while moving along each of these trajectories, the friction force will perform different work.
Using the concept of work is also useful in the presence of friction forces. Let's consider a simple example. Let there be a bar on a horizontal surface, to which the speed was given by a push v o. Let us find the path the bar will take to stop in the presence of dry friction, the coefficient of which is equal to μ . Since when stopping, the kinetic energy vanishes, the change in the kinetic energy of the body is equal to:

According to the kinetic energy theorem, the change in the latter is equal to the work of external forces. The only force that does work is the friction force, which in this case is equal to:
A = -μmgS.
By equating these expressions, we can easily find the path to the stop:
S = v o 2 /(2μg).
In order for the considered bar to move along a horizontal surface at a constant speed, it is necessary to apply a constant, horizontally directed force to it F equal in modulus to the force of friction. This external force will do positive work A, equal in modulus to the work of the friction force. The kinetic energy of the bar will not increase during such a movement. Note that there is no contradiction with the kinetic energy theorem in this statement - so, the total external force acting on the bar is equal to zero. Nevertheless, it is necessary to firmly understand that the work of any force is a measure of the transition of energy from one form to another, therefore, it is necessary to determine what changes with the system (bar and surface) occurred as a result of the work done. The answer is known: both the surface and the bar were heated. In other words, the work of the external force went to increase the internal, thermal energy. Similarly, when braking, the initial kinetic energy of the bar was converted into internal energy. In any case, the work of the friction force leads to an increase in thermal energy.
When moving in a viscous medium, a resistance force acts on the body, which depends on the speed and is directed in the direction opposite to the speed vector, so the work of these forces is always negative, and depends on the trajectory of the body. Consequently, the forces of viscous friction are not potential. Energy transformations occurring in the presence of viscous friction are similar to those considered earlier, however, their calculation is complicated by the dependence of forces on velocity. Non-potential forces leading to an increase in internal energy are called dissipative 1 . Friction forces are examples of such forces.

1 Here is how O.D. Khvolson “Force does work when its point of application moves ... ... two cases of work production should be distinguished: in the first, the essence of work is to overcome external resistance to movement, which is performed without increasing the speed of the body; in the second, work is revealed by an increase in the speed of movement, to which the outside world is indifferent. In fact, we usually have a combination of both cases: the force overcomes any resistance and at the same time changes the speed of the body.

To calculate the work of a constant force, the following formula is proposed:

Where S- movement of a body under the action of a force F, a- the angle between the directions of force and displacement. At the same time, they say that “if the force is perpendicular to the displacement, then the work of the force is zero. If, despite the action of the force, the point of application of the force does not move, then the force does not perform any work. For example, if a load is hanging motionless on a suspension, then the force of gravity acting on it does not do work.

It also says: “The concept of work as a physical quantity, introduced in mechanics, only to a certain extent agrees with the idea of work in the everyday sense. Indeed, for example, the work of a loader in lifting weights is considered the greater, the greater the load being lifted and the greater the height it must be lifted. However, from the same worldly point of view, we are inclined to call "physical work" any activity of a person in which he performs certain physical efforts. But, according to the definition given in mechanics, this activity may not be accompanied by work. In the well-known myth of Atlanta, supporting the firmament on its shoulders, people meant the effort required to support the enormous weight, and regarded this effort as a colossal work. There is no work for mechanics here, and Atlas's muscles could simply be replaced by a strong column.

These arguments are reminiscent of the well-known statement by I.V. Stalin: "There is a person - there is a problem, no person - there is no problem."

The physics textbook for grade 10 suggests the following way out of this situation: “When a person holds a load motionless in the Earth’s gravity field, work is done and the arm experiences fatigue, although the apparent movement of the load is zero. The reason for this is that human muscles are constantly contracting and stretching, resulting in microscopic movements of the load. Everything is fine, but how to calculate these contractions-stretchings?

It turns out this situation: a person tries to move the cabinet at a distance S for which he acts by force F for a time t, i.e. conveys the momentum of the force. If the cabinet has a small mass and there are no friction forces, then the cabinet moves and, therefore, work is done. But if the cabinet has a large mass and high friction forces, then a person, acting on the same impulse of force, does not move the cabinet, i.e. work is not done. Something here does not fit with the so-called conservation laws. Or take the example shown in Fig. 1. If power F a, That . Since , then, naturally, the question arises, where did the energy equal to the difference in work () disappear?

Picture 1. Force F directed horizontally (), then work, and if at an angle a, That

Let us give an example showing that work is done if the body remains motionless. Let's take an electrical circuit consisting of a current source, a rheostat and an ammeter of a magnetoelectric system. With the rheostat fully inserted, the current strength is infinitely small and the ammeter needle is at zero. We begin to gradually move the rheostat rheochord. The ammeter needle begins to deviate, twisting the spiral springs of the device. This is done by the Ampere force: the force of the interaction of the frame with the current with the magnetic field. If you stop the reochord, then a constant current will be established and the arrow will stop moving. They say that if the body is motionless, then the force does no work. But the ammeter, holding the needle in the same position, still consumes energy, where U- the voltage supplied to the ammeter frame, - the current strength in the frame. Those. Ampere's force, holding the arrow, still does work to keep the springs in a twisted state.

Let us show why such paradoxes arise. First, we get the generally accepted expression for work. Consider the work of acceleration on a horizontal smooth surface of an initially resting body of mass m due to the influence of a horizontal force on it F for a time t. This case corresponds to the angle in Fig.1. We write Newton's II law in the form . Multiply both sides of the equation by the distance traveled S: . Since , we get or . Note that multiplying both sides of the equation by S, we thereby deny the work to those forces that do not produce a displacement of the body (). Moreover, if the power F acting at an angle a to the horizon, we thereby deny the work of all the power F, "allowing" the work of only its horizontal component: .

Let's carry out another derivation of the formula for work. We write Newton's II law in differential form

The left side of the equation is the elementary momentum of the force, and the right side is the elementary momentum of the body (momentum). Note that the right side of the equation can be equal to zero if the body remains stationary () or moves uniformly (), while the left side is not equal to zero. The last case corresponds to the case of uniform motion, when the force balances the friction force .

However, let us return to our problem of acceleration of an immobile body. After integrating equation (2), we obtain , i.e. the momentum of the force is equal to the momentum (momentum) received by the body. We square and divide by both sides of the equality, we get

Thus we get another expression for calculating the work

(4)

where is the momentum of the force. This expression is not related to the path S passed by the body during the time t, so it can be used to calculate the work done by the impulse of the force even if the body remains motionless.

In case the strength F acting at an angle a(Fig. 1), then we decompose it into two components: the traction force and the force, which we will call the levitation force, it seeks to reduce the force of gravity. If it is equal to , then the body will be in a quasi-weightless state (a state of levitation). Using the Pythagorean Theorem: , we find the work of the force F

or (5)

Since , and , then the work of the thrust force can be represented in the generally accepted form: .

If the levitation force , then the work of levitation will be equal to

(6)

This is exactly the work that Atlas did, holding the vault of heaven on his shoulders.

Now consider the work of friction forces. If the friction force is the only force acting along the line of motion (for example, a car moving along a horizontal road at a speed turned off the engine and began to slow down), then the work of the friction force will be equal to the difference in kinetic energies and can be calculated using the generally accepted formula:

(7)

However, if the body moves along a rough horizontal surface with a certain constant speed, then the work of the friction force cannot be calculated using the generally accepted formula, since in this case the movements must be considered as the movement of a free body (), i.e. as a motion by inertia, and the speed v creates not the force it was acquired earlier. For example, a body moved along a perfectly smooth surface at a constant speed, and at the moment when it enters a rough surface, the traction force is turned on. In this case, the path S is not associated with the action of the force. If we take the path m, then at a speed of m / s the time of the force will be s, at m / s time s, at m / s time s. Since the friction force is considered independent of speed, then, obviously, on the same segment of the path m, the force will do much more work in 200 s than in 10 s, because in the first case, the momentum of the force, and in the latter -. Those. in this case, the work of the friction force must be calculated by the formula:

(8)

Denoting the "ordinary" work of friction through and taking into account that , formula (8), omitting the minus sign, can be represented as