Instructions
Example problem 3: a block of mass 1 kg slid from the top of an inclined plane in 5 seconds, a distance of 10 meters. Determine the friction force if the angle of inclination of the plane is 45°. Consider also the case when the block was subjected to an additional force of 2 N applied along the angle of inclination in the direction of movement.
Find the acceleration of the body similarly to examples 1 and 2: a = 2*10/5^2 = 0.8 m/s2. Calculate the friction force in the first case: Ftr = 1*9.8*sin(45о)-1*0.8 = 7.53 N. Determine the friction force in the second case: Ftr = 1*9.8*sin(45о) +2-1*0.8= 9.53 N.
Case 6. A body moves uniformly along an inclined surface. This means that according to Newton's second law, the system is in equilibrium. If the sliding is spontaneous, the movement of the body obeys the equation: mg*sinα = Ftr.
If an additional force (F) is applied to the body, preventing 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 Coulomb's mechanical law, the sliding force is equal to F = kN, where k is the friction coefficient and N is the support reaction force. Since the support reaction force is directed strictly vertically, then N = Fweight = mg, where m is the mass of the body, g is the acceleration of free fall. This condition follows from the immobility of the body relative to the vertical direction.
Thus, the friction coefficient can be found using the formula k = Ftr/N = Ftr/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 body be acted upon by a driving force F and directed oppositely to it by Ftr. Then according to Newton's second law (F-Ftr)/m = a. Expressing Ftr from here and substituting it into the formula for the friction coefficient, we obtain: k = (F-ma)/N.
From these formulas it is clear that the friction coefficient is a dimensionless quantity.
Let's consider a more general case, when from an inclined plane, for example, from a fixed block. Such problems are very often found in the 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. Let's direct the axes along and perpendicular to the inclined plane.
According to Newton's second law, we can write the equations of the body: 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).
Let us consider an important special case of 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 = tanφ, that is, to determine the slip coefficient it is enough to know the tangent of the plane’s inclination angle.
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note
One should not confuse Coulomb's law in mechanics with Coulomb's law in electrostatics!
When two bodies move relative to each other, friction occurs between them. It can also occur when moving in a gaseous or liquid medium. Friction can either interfere with or facilitate normal movement. As a result of this phenomenon, a force acts on the interacting bodies.
Instructions
The most general case is considered force, when one of the bodies is fixed and at rest, and the other slides along its surface. From the side of the body along which the moving body slides, the support reaction force directed perpendicular to the sliding plane acts on the latter. This force is the letter N. A body can also be at rest relative to a fixed body. Then strength friction, acting on it Ftrfriction. It depends on the materials of the rubbing surfaces, the degree of their polishing and a number of other factors.
In the case of body motion relative to the surface of a fixed body, the force friction slip becomes equal to the product of the coefficient friction on force support reactions: Ftr = ?N.
Let now a constant force F>Ftr = ?N act on the body, parallel to the surface of the contacting bodies. When a body slides, the resulting component of the force in the horizontal direction will be equal to F-Ftr. Then, according to Newton’s second law, the acceleration of the body will be related to the resulting force according to the formula: a = (F-Ftr)/m. Hence, Ftr = F-ma. The acceleration of a body can be found from kinematic considerations.
A frequently considered special case of force friction when a body slips from a fixed plane. Let be? - the angle of inclination of the plane and let the body slide evenly, that is, without . Then the equations of motion of the body will look like this: N = mg*cos?, mg*sin? = Ftr = ?N. Then from the first equation of motion force friction can be expressed as Ftr = ?mg*cos?. If the body moves along an inclined plane with a, then the second equation will have the form: mg*sin?-Ftr = ma. Then Ftr = mg*sin?-ma.
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If the force directed parallel to the surface on which the body stands exceeds the static friction force, then movement will begin. It will continue as long as the driving force exceeds the sliding friction force, which depends on the friction coefficient. You can calculate this coefficient yourself.
You will need
- Dynamometer, scales, protractor or protractor
Instructions
Find the mass of the body in kilograms and place it on a flat surface. Attach a dynamometer to it and start moving your body. Do this in such a way that the dynamometer readings stabilize, maintaining a constant speed. In this case, the traction force measured by the dynamometer will be equal, on the one hand, to the traction force, which is shown by the dynamometer, and on the other hand, the force multiplied by the sliding.
The measurements taken will allow us to find this coefficient from the equation. To do this, divide the traction force by body weight and the number 9.81 (gravitational acceleration) μ=F/(m g). The resulting coefficient will be the same for all surfaces of the same type as those on which the measurement was made. For example, if a body was moving on a wooden board, then this result will be valid for all wooden bodies moving by sliding on the tree, taking into account the quality of its processing (if the surfaces are rough, the value of the sliding friction coefficient will change).
You can measure the sliding friction coefficient in another way. To do this, place the body on a plane that can change its angle relative to the horizon. It could be an ordinary board. Then start carefully at one edge. At the moment when the body begins to move, sliding down a plane like a sled down a hill, find the angle of its inclination relative to the horizon. It is important that the body does not move with acceleration. In this case, the measured angle will be extremely small, at which the body will begin to move under . The sliding friction coefficient will be equal to the tangent of this angle μ=tg(α).
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Friction is the process of interaction between two bodies, causing a slowdown in movement when moving relative to each other. Find force 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 ultimately stops.
Instructions
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 resistance is, as a rule, a known quantity for common materials. For example, for iron it is 0.51 mm, for iron on wood – 5.6, wood on wood – 0.8-1.5, etc. It can be found using the torque ratio formula friction to pressing force.
Force friction rest appears with minimal body movements or deformation. This force is always present during dry sliding. Its maximum value is μ N. There is also internal friction, inside one body between its layers or.
note
Uniform movement of a body is characterized by a balance between external force and frictional force.
In school problems in physics to determine the force of sliding friction, rectilinear uniform or rectilinear uniformly accelerated motion of a body is mainly considered. See how you can find the friction force in different cases depending on the conditions of the problem. To correctly evaluate the impact of forces and create an equation of motion, always draw a drawing.
It remains for us to consider the work of the third mechanical force - the sliding friction force. Under terrestrial conditions, the force of friction is manifested to one degree or another during all movements of bodies.
The sliding friction force differs from the force of gravity and the force of elasticity in that it does not depend on coordinates and always arises with the relative motion of contacting bodies.
Let us consider the work of the friction force when a body moves relative to a stationary surface with which it comes into contact. In this case, the friction force is directed against the movement of the body. It is clear that in relation to the direction of movement of such a body, the friction force cannot be directed at any angle other than an angle of 180°. Therefore, the work done by the friction force is negative. The work done by the friction force must be calculated using the formula
where is the friction force, is the length of the path along which the friction force acts
When a body is acted upon by gravity or an elastic force, it can move both in the direction of the force and against the direction of the force. In the first case, the work of force is positive, in the second - negative. When a body moves back and forth, the total work done is zero.
The same cannot be said about the work of the friction force. The work of the friction force is negative both when moving “there” and when moving back.” Therefore, the work done by the friction force after the body returns to the starting point (when moving along a closed path) is not equal to zero.
Task. Calculate the work done by the friction force when braking a train weighing 1200 tons to a complete stop, if the speed of the train at the moment the engine was turned off was 72 km/h. Solution. Let's use the formula
Here is the mass of the train, equal to kg, is the final speed of the train, equal to zero, and is its initial speed, equal to 72 km/h = 20 m/sec. Substituting these values, we get:
Exercise 51
1. A friction force acts on the body. Can the work done by this force be zero?
2. If a body on which a frictional force acts, after passing a certain trajectory, returns to the starting point, will the work done by the friction be equal to zero?
3. How does the kinetic energy of a body change when a friction force works?
4. A sleigh weighing 60 kg, having rolled down the mountain, drove along a horizontal section of the road for 20 m. Find the work done by the friction force on this section if the coefficient of friction of the runners of the sleigh on the snow is 0.02.
5. The part to be sharpened is pressed against a sharpening stone with a radius of 20 cm with a force of 20 N. Determine how much work is done by the engine in 2 minutes if the grindstone makes 180 rpm and the coefficient of friction of the part on the stone is 0.3.
6. The driver of the car turns off the engine and begins to brake 20 m from the traffic light. Assuming the friction force to be equal to 4,000 k, find at what maximum speed of the car it will have time to stop in front of the traffic light if the mass of the car is 1.6 tons?
You are already familiar with mechanical work (work of force) from the basic school physics course. Let us recall the definition of mechanical work given there for the following cases.
If the force is directed in the same direction as the movement 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
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 done by the force is zero:
Work is a scalar quantity. The unit of work is called the joule (symbol: 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 block weighing 0.5 kg was moved along the table 2 m, applying an elastic force of 4 N to it (Fig. 28.1). The coefficient of friction between the block and the table is 0.2. What is the work acting on the block?
a) gravity m?
b) normal reaction forces?
c) elastic forces?
d) sliding friction forces tr?
The total work done by several forces acting on a body can be found in two ways:
1. Find the work of each force and add up 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 make sure of this, go back to the previous task and answer the questions in task 2.
2. What is it equal to:
a) the sum of the work done by all forces acting on the block?
b) the resultant of all forces acting on the block?
c) work 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.
The work A of a constant force is equal to the product of the force modulus F by the displacement modulus 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 a block on the table, the modulus of which is 10 N. What is the angle between this force and the movement of the block if, when moving the block 60 cm along the table, this force does the work: a) 3 J; b) –3 J; c) –3 J; d) –6 J? Make explanatory drawings.
2. 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 downwards (h n > h k, Fig. 28.2, a), the direction of movement coincides with the direction of gravity, therefore the work of gravity is positive. If the body moves upward (h n< h к, рис. 28.2, б), то работа силы тяжести отрицательна.
In both cases, the work done by gravity
A = 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 block? 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 done by gravity when the block moves upward along the entire same plane?
Having completed this task, you are convinced that the work of gravity is expressed by formula (5) even when the body moves at an angle to the vertical - both down and up.
But then formula (5) for the work of gravity is valid when a 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 done by gravity when moving along any trajectory is expressed by the formula
A t = mg(h n – h k),
where h n is the initial height of the body, h k is its final height.
The work done by 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 trajectory 1, 2 or 3 is the same. From here, in particular, it follows that the force 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, was deflected by 90º, keeping the thread taut, and released without a push.
a) What is the work done by gravity during the time during which the ball moves to the equilibrium position (Fig. 28.6)?
b) What is the work done by the elastic force of the thread during the same time?
c) What is the work done by the resultant forces applied to the ball during the same time?
3. Work of elastic force
When the spring returns to an undeformed state, the elastic force always does positive work: its direction coincides with the direction of movement (Fig. 28.7). 
Let's find the work done by the elastic force.
The modulus of this force is related to the modulus of deformation x by the relation (see § 15)
The work done by such a force can be found graphically.
Let us first note that the work done by 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 graph of F(x) for the elastic force. Let us mentally divide the entire movement of the body into such small intervals that the force at 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 work is equal to the sum of work in these areas.
Consequently, in this case, the work is numerically equal to the area of the figure under the graph of the dependence F(x). 
7. Using Figure 28.10, prove that
the work done by the elastic force when the spring returns to its undeformed state is expressed by the formula
A = (kx 2)/2. (7)
8. Using the graph in Figure 28.11, prove that when the spring deformation 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 returns to its initial state, then the work of the elastic force is zero. Let us recall that the work of gravity has the same property.
9. At the initial moment, the tension of a spring with a stiffness of 400 N/m is 3 cm. The spring is stretched by 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 done by the elastic force of the spring?
4. Work of friction force
Let the body slide along 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 in any direction of movement (Fig. 28.12). 
Therefore, if you move the block to the right, and the peg the same distance to the left, then, although it will return to its initial position, the total work done by the sliding friction force will not be equal to zero. This is the most important difference between the work of sliding friction and the work of gravity and elasticity. Let us recall that the work done by these forces when moving a body along a closed trajectory is zero.
11. A block 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) Has the block returned to its starting point?
b) What is the total work done by the frictional force acting on the block? The coefficient of friction between the block and the table is 0.3.
5.Power
Often it is not only the work being done that is important, but also the speed at which the work is being done. It is characterized by power.
Power P is the ratio of the work done A to the time period t during which this work was 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 for power.)
The unit of power is the watt (symbol: W), named after the 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 through work and time, but through force and speed.
Let's consider the case when the force is directed along the displacement. Then the work done by 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 traveling on 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 moves along a horizontal road at a constant speed, the traction force is equal in magnitude to the resistance force to the movement of the car.
14. How long will it take to uniformly 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 electric motor of the crane is 75%?
Clue. The efficiency of an 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 weighing 200 g was thrown from a balcony with a height of 10 and an angle of 45º to the horizontal. Having reached a maximum height of 15 m in flight, the ball fell to the ground.
a) What is the work done by gravity when 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 any extra data in the condition?
16. A ball with a mass of 0.5 kg is suspended from a spring with a stiffness of 250 N/m and is in equilibrium. The ball is raised so that the spring becomes undeformed and released without a push.
a) To what height was the ball raised?
b) What is the work done by gravity during the time during which the ball moves to the equilibrium position?
c) What is the work done by the elastic force during the time during which the ball moves to the equilibrium position?
d) What is the work done by the resultant of all forces applied to the ball during the time during which the ball moves to the equilibrium position?
17. A sled weighing 10 kg slides down a snowy mountain with an inclination angle of α = 30º without initial speed and travels a certain 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 is l = 15 m. 
a) What is the magnitude of the friction force when the sled moves on a horizontal surface?
b) What is the work done by the friction force when the sled moves along a horizontal surface over a distance of 20 m?
c) What is the magnitude of the friction force when the sled moves along the mountain?
d) What is the work done by the friction force when lowering the sled?
e) What is the work done by gravity when lowering the sled?
f) What is the work done by 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 efficiency of the engine? Is there any extra data in the condition?
Clue. The efficiency of a heat engine is equal to the ratio of the work performed by the engine to the amount of heat released during fuel combustion.
With the relative movement 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 discussed 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 and the speed of movement. In other words, friction forces are not potential.
Let's take a closer look at the work of various types of friction.
The simplest case is static friction. Suffice it to say that in the absence of movement, the work is zero, so static friction does not do any 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. Consequently, at any moment of time, at any point in 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 done by 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 any number of trajectories, the lengths of which can vary within wide limits; when moving along each of these trajectories, the friction force will do different work.
The use of the concept of work is also useful in the presence of friction forces. Let's look at a simple example. Let there be a block on a horizontal surface to which a speed is imparted by a push v o. Let us find the distance the block will travel before stopping in the presence of dry friction, the coefficient of which is equal to μ
. Since when stopping the kinetic energy goes to zero, the change in the kinetic energy of the body is equal to:
According to the theorem on kinetic energy, the change in the latter is equal to the work of external forces. The only force doing work is the friction force, which in this case is equal to:
A = −μmgS.
Equating these expressions, we easily find the path to the stop:
S = v o 2 /(2μg).
In order for the block in question to move along a horizontal surface at a constant speed, a constant, horizontally directed force must be applied to it F equal in magnitude to the friction force. This external force will do positive work A, equal in magnitude to the work of the friction force. The kinetic energy of the block will not increase with such movement. Note that there is no contradiction with the theorem on kinetic energy in this statement - thus, the total external force acting on the block 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 to the system (bar and surface) occurred as a result of the work performed. The answer is known: both the surface and the block were heated. In other words, the work of the external force went to increase the internal, thermal energy. Similarly, during braking, the initial kinetic energy of the block turned 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, depending on the speed and directed in the direction opposite to the speed vector, therefore the work of these forces is always negative, and depends on the trajectory of the body. Consequently, viscous friction forces are not potential. The energy transformations that occur in the presence of viscous friction are similar to those discussed earlier, however, their calculation is complicated by the dependence of forces on speed. Non-potential forces that lead to an increase in internal energy are called dissipative 1. Examples of such forces are frictional forces.
To calculate the work of a constant force, a formula is proposed:
Where S- movement of a body under the influence of 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 done by 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 do any work. For example, if any load hangs motionless on a suspension, then the force of gravity acting on it does not do any work.”
It also says: “The concept of work as a physical quantity, introduced in mechanics, is only to a certain extent consistent with the idea of work in the everyday sense. Indeed, for example, the work of a loader in lifting weights is assessed the more, the larger the load being lifted and the greater the height it must be lifted. However, from the same everyday point of view, we are inclined to call “physical work” any human activity in which he makes 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 Atlas supporting the vault of heaven on his shoulders, people were referring to the efforts required to support enormous weight, and regarded these efforts as colossal work. There is no work for mechanics here, and Atlas’ muscles could simply be replaced by a strong column.”
These arguments are reminiscent of the famous statement of I.V. Stalin: “If there is a person, there is a problem, if there is no person, there is no problem.”
The physics textbook for grade 10 offers 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 hand experiences fatigue, although the visible movement of the load is zero. The reason for this is that human muscles experience constant contractions and stretches, leading to microscopic movements of the load.” Everything is fine, but how to calculate these contractions and stretches?
It turns out this situation: a person tries to move the cabinet at a distance S why does he act by force? F for a time t, i.e. communicates a force impulse. If the cabinet has a small mass and there are no friction forces, then the cabinet moves and that means work is done. But if the cabinet is of large mass and has large friction forces, then the person, acting with the same impulse of force, does not move the cabinet, i.e. no work is done. Something here does not fit with the so-called conservation laws. Or take the example shown in Fig. 1. If strength F a, That . Since , the question naturally arises, where did the energy equal to the difference in work () disappear?
Picture 1. Force F is directed horizontally (), then the work is , 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. When the rheostat is fully inserted, the current strength is infinitesimal and the ammeter needle is at zero. We begin to gradually move the rheostat's rheochord. The ammeter needle begins to deviate, twisting the spiral springs of the device. This is done by the Ampere force: the force of interaction between the current frame and the magnetic field. If you stop the rheochord, a constant current strength is established and the arrow stops moving. They say that if the body is motionless, then the force does not do work. But the ammeter, holding the needle in the same position, still consumes energy, where U- voltage supplied to the ammeter frame, - current strength in the frame. Those. The Ampere force, holding the arrow, still does work to keep the springs in a twisted state.
Let us show why such paradoxes arise. First, let's get a generally accepted expression for work. Let us consider the work of acceleration along a horizontal smooth surface of an initially stationary body of mass m due to the influence of horizontal force on it F for a time t. This case corresponds to the angle in Fig. 1. Let us write Newton's II law in the form. Multiply both sides of the equality by the distance traveled S: . Since , we get or . Note that multiplying both sides of the equation by S, we thereby deny work to those forces that do not move the body (). Moreover, if the force F acts 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. Let's write Newton's II law in differential form
The left side of the equation is the elementary impulse of force, and the right side is the elementary impulse of the body (quantity of motion). 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 accelerating a stationary body. After integrating equation (2), we obtain, i.e. the impulse of force is equal to the impulse (amount of motion) received by the body. Squaring and dividing by both sides of the equation, we get
This way we get another expression for calculating work
(4)
where is the impulse of force. This expression is not associated with a path S traversed by the body in time t, therefore it can be used to calculate the work done by a force impulse even if the body remains motionless.
In case the power F acts at an angle a(Fig. 1), then we decompose it into two components: the traction force and the force, which we call the force of levitation, it tends to reduce the force of gravity. If it is equal to , then the body will be in a quasi-weightless state (state of levitation). Using the Pythagorean theorem: 
, let's find the work done by force F
or (5)
Since , and , then the work of the traction force can be represented in the generally accepted form: .
If the force of levitation is , then the work of levitation will be equal to
(6)
This is exactly the work that Atlas did, holding the firmament on his shoulders.
Now let's look at 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 brake), then the work done by the friction force will be equal to the difference in kinetic energies and can be calculated using the generally accepted formula:
(7)
However, if a 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 movement by inertia, and the speed V is not created by force, it was acquired earlier. For example, a body was moving along a perfectly smooth surface at a constant speed, and at the moment when it enters a rough surface, the traction force is activated. In this case, the path S is not associated with the action of force. If we take the path m, then at a speed m/s the time of action of the force will be s, at m/s the time will be s, at m/s the time will be 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 impulse of force is , and in the latter - . Those. in this case, the work of the friction force must be calculated using the formula:
(8)
Denoting the “ordinary” work of friction through 
and taking into account that , formula (8), omitting the minus sign, can be represented in the form
