What is the work done by friction forces? Modern problems of science and education
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 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 what general definition The work follows 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 done by 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 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.
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 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 the force 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 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 of 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
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?
where is the path traveled by the body during the action of the force.
After substituting the numerical values we get:
Example 3. A ball with a mass of =100 g fell from a height of =2.5 m onto a horizontal plate and bounced off it due to an elastic impact without loss of speed. Determine average speed
Solution. According to Newton’s second law, the product of an average force and the time of its action is equal to the change in the momentum of the body caused by this force, i.e.
where and are the velocities of the body before and after the action of the force; - the time during which the force was applied.
From (1) we get
If we take into account that the speed is numerically equal to the speed and opposite to it in direction, then formula (2) will take the form:
Since the ball fell from a height, its speed upon impact is
Taking this into account, we get
Substituting numerical values here, we find
The minus sign shows that the force is directed opposite to the speed of the ball's fall.
Example 4. To lift water from a well with a depth of =20 m, a pump with a power of =3.7 kW was installed. Determine the mass and volume of water raised in time = 7 hours, if efficiency. pump =80%.
Solution. It is known that pump power taking into account efficiency is determined by the formula
where is the work done during time; - efficiency factor.
The work done when lifting a load without acceleration to a height is equal to the potential energy that the load has at this height, i.e.
where is the acceleration of free fall.
Substituting the expression for work according to (2) into (1), we obtain
Let us express the numerical values of the quantities included in formula (3) in SI units: =3.7 kW = 3.7 103 W; =7 h = 2.52 104 s; =80%=0.8; =20 m.
kg kg m2 s2/(s3 m m), kg=kg
Let's calculate
kg=3.80 105 kg=380 t.
To determine the volume of water, you need to divide its mass by its density
Example 5. An artificial Earth satellite moves in a circular orbit at an altitude of =700 km. Determine the speed of its movement. The radius of the Earth = 6.37 106 m, its mass = 5.98 1024 kg.
Solution. A satellite, like any body moving in a circular orbit, is affected by a centripetal force
where is the mass of the satellite; V is the speed of its movement; - radius of curvature of the trajectory.
If we neglect the resistance of the environment and the gravitational forces from all celestial bodies, then we can assume that the only force is the force of attraction between the satellite and the Earth. This force plays the role of centripetal force.
According to the law of universal gravitation
where is the gravitational constant.
Equating the right-hand sides of (1) and (2), we obtain
Hence the speed of the satellite
Let's write down the numerical values of the quantities in SI: = 6.67*10-11 m3/(kg s2); =5.98 1024 kg; = 6.37 106 m; = 700 km = 7,105 m.
Let's check the units of the right and left sides of the calculation formula (3) to make sure that these units coincide. To do this, substitute in the formula instead of quantities their dimensions in the International System:
Let's calculate
Example 6. A flywheel in the form of a solid disk with a mass m = 80 kg and a radius = 50 cm began to rotate uniformly accelerated under the influence of a torque = 20 N m. Determine: 1) angular acceleration; 2) kinetic energy acquired by the flywheel during time = 10 s from the start of rotation.
Solution. 1. From the basic equation of the dynamics of rotational motion,
where is the moment of inertia of the flywheel; - angular acceleration, we get
It is known that the moment of inertia of the disk is determined by the formula
Substituting the expression for from (2) into (1), we obtain
Let's express the values in SI units: = 20 N m; t = 80 kg; = 50 cm = 0.5 m.
Let's check the units of the right and left sides of the calculation formula (3):
1/s2 = kg x m2/(s2x kg x m2) = 1/s2
Let's calculate
2. The kinetic energy of a rotating body is expressed by the formula:
where is the angular velocity of the body.
With uniformly accelerated rotation, the angular velocity is related to the angular acceleration by the relation
where is the angular velocity at the moment of time; - initial angular velocity.
Since according to the conditions of the problem =0, it follows from (5)
Substituting the expression for from (6), from (2) into (4), we obtain
Let's check the units of the right and left sides of formula (7):
Let's calculate
Example 7. The equation of an oscillating point has the form (displacement in centimeters, time in seconds). Determine: 1) vibration amplitude, circular frequency, period and initial phase; 2) displacement of the point at time s; 3) maximum speed and maximum acceleration.
Solution. 1. Let us write the equation of harmonic oscillatory motion in general form
where x is the displacement of the oscillating point; A - vibration amplitude; - circular frequency; - oscillation time; - initial phase.
Comparing the given equation with equation (1), we write: A = 3 cm,
The period of oscillation is determined from the relation
Substituting the value into (2), we get
2. To determine the displacement, we substitute the time value into the given equation:
3. We find the speed of the oscillatory motion by taking the first derivative of the displacement of the oscillating point:
(The speed will have its maximum value at =1:
Acceleration is the first derivative of speed with respect to time:
Maximum acceleration value
The minus sign indicates that the acceleration is directed in the direction opposite to the displacement.
Myakishev G.Ya., Kondrasheva L., Kryukov S. Work of friction forces // Quantum. - 1991. - No. 5. - P. 37-39.
By special agreement with the editorial board and editors of the journal "Kvant"
The friction force, like any other force, does work and accordingly changes the kinetic energy of the body, provided that the point of application of the force moves in the chosen reference system. However, the friction force differs significantly from other so-called conservative forces (gravity and elasticity), since its work depends on the shape of the trajectory. That is why the work of friction forces under no circumstances can be represented in the form of a change in the potential energy of the system. In addition, additional difficulties when calculating work are created by the specifics of the static friction force. There are a number of stereotypes of physical thinking that, although meaningless, are very stable.
We will consider several issues related to the not entirely correct understanding of the role of the friction force in changing the energy of a system of bodies.
About the force of sliding friction
It is often said that the sliding friction force always does negative work and this leads to an increase in the internal (thermal) energy of the system.
This statement requires important clarification - it is true only if we're talking about not about the work of one individual sliding friction force, but about the total work of all such forces acting in the system. The fact is that the work of any force depends on the choice of reference system and can be negative in one system, but positive in another. The total work of all friction forces acting in the system does not depend on the choice of reference system and is always negative. Here's a concrete example.
Let's place the brick on a moving cart so that it begins to slide along it (Fig. 1). In the frame of reference associated with the earth, the friction force F 1, acting on the brick until the sliding stops, does positive work A 1 . At the same time the friction force F 2, acting on the cart (and equal in magnitude to the first force), does negative work A 2, modulo greater than work A 1, since the cart path s more brick way s - l (l- path of the brick relative to the cart). Thus, we get
\(~A_1 = \mu mg(s - l), A_2 = -\mu mgs\) ,
and the total work of friction forces
\(~A_(tr) = A_1 + A_2 = -\mu mgl< 0\) .
Therefore, the kinetic energy of the system decreases (turns into heat):
\(~\Delta E_k = -\mu mgl\) .
This conclusion has general significance. Indeed, the work of two forces (not only friction forces) that interact between bodies does not depend on the choice of reference system (prove this yourself). You can always go to a reference system relative to which one of the bodies is at rest. In it, the work of the friction force acting on a moving body is always negative, since the friction force is directed against the relative speed. But it is negative in any other frame of reference. Therefore, always, for any number of bodies in the system, A tr< 0. Эта работа и уменьшает механическую энергию системы.
About the force of static friction
When a static friction force acts between contacting bodies, neither the mechanical nor the internal (thermal) energy of these bodies changes. Does this mean that the work done by the static friction force is zero? As in the first case, this statement is correct only in relation to the total work of static friction forces on all interacting bodies. One single static friction force can do work, both negative and positive.
Consider, for example, a book lying on a table in a train picking up speed. It is the static friction force that gives the book the same speed as that of a train, that is, it increases its kinetic energy, doing a certain amount of work. Another thing is that a force of the same magnitude, but opposite in direction, acts from the book on the table, and therefore on the train as a whole. This force does exactly the same work, but only negative. As a result, it turns out that the total work of the two static friction forces is zero, and the mechanical energy of the system of bodies does not change.
About car movement without wheel slipping
The most persistent misconception is related to this issue.
Let the car first be at rest and then begin to accelerate (Fig. 2). The only external force imparting acceleration to the car is the static friction force F tr acting on the drive wheels (we neglect the force of air resistance and rolling friction force). According to the theorem on the motion of the center of mass, the impulse of the friction force is equal to the change in the momentum of the car:
\(~F_(tr) \Delta t = \Delta(M \upsilon_c) = M \upsilon_c\) ,
if the speed of the center of mass at the beginning of the movement was zero, and at the end υ c. By acquiring momentum, i.e. increasing its speed, the car simultaneously receives a certain portion of kinetic energy. And since the impulse is imparted by the friction force, it is natural to assume that the increase in kinetic energy is determined by the work of the same force. This statement turns out to be completely incorrect. The friction force accelerates the car, but does not do any work. How so?
Generally speaking, there is nothing paradoxical in this situation. As an example, it is enough to consider a very simple model - a smooth cube with a spring attached to the side (Fig. 3). The cube is moved towards the wall, squeezing the spring, and then released. “Pushing off” from the wall, our system (a cube with a spring) acquires a certain impulse and kinetic energy. The only external force acting horizontally on the system is, obviously, the wall reaction force F p. It is she who imparts acceleration to the system. However, of course, no work is done - after all, the point of application of this force is motionless (in the coordinate system associated with the earth), although the force acts for some finite time Δ t.
A similar situation occurs when accelerating a car without slipping. The point of application of the friction force acting on the driving wheel of a car, i.e., the point of contact of the wheel with the road, is at rest relative to the road at any moment (in the reference frame associated with the road). When the car moves, it disappears at one point and immediately appears at the next one.
Doesn't this contradict the law of conservation of mechanical energy? Of course not. In our case with a car, the change in the kinetic energy of the system occurs due to its internal energy released during fuel combustion.
For simplicity, consider a purely mechanical system: a spring-wound toy car. The engine of such a car does not use the internal energy of the fuel, but the potential energy of a compressed spring. Initially, the spring is wound, and its potential energy E p1 is different from zero. If the toy's motor is simply a stretched spring, then \(~E_(p1) = \frac(k (\Delta l)^2)(2)\). The kinetic energy is zero, and the total initial energy of the car is E 1 = E p1. In the final state, when the spring deformation disappears, the potential energy is zero and the kinetic energy \(~E_(k2) = \frac(M \upsilon_c^2)(2)\). Total Energy E 2 = E k2. According to the law of conservation of energy (we neglect friction),
\(~\frac(M \upsilon_c^2)(2) = \frac(k (\Delta l)^2)(2)\) .
In case of a real car
\(~\frac(M \upsilon_c^2)(2) = \Delta U\) ,
where Δ U- energy obtained from fuel combustion.
If the car wheels are slipping, then A tr<0, так как точка соприкосновения колес с дорогой движется против направления силы трения. Следовательно,
\(~\frac(M \upsilon_c^2)(2) = \frac(k (\Delta l)^2)(2) + A_(tr)\) .
It can be seen that the kinetic energy of the car in the final state is less than in the absence of slipping.