The energy characteristics of motion are introduced on the basis of the concept of mechanical work or work of force.

Work A performed by a constant force F → is a physical quantity equal to the product of the force and displacement modules multiplied by the cosine of the angle α , located between the force vectors F → and the displacement s →.

This definition discussed in Figure 1. 18 . 1 .

The work formula is written as,

A = F s cos α .

Work is a scalar quantity. This makes it possible to be positive at (0° ≤ α< 90 °) , отрицательной при (90 ° < α ≤ 180 °) . Когда задается прямой угол α , тогда совершаемая сила равняется нулю. Единицы измерения работы по системе СИ - джоули (Д ж) .

A joule is equal to the work done by a force of 1 N to move 1 m in the direction of the force.

Picture 1 . 18 . 1 . Work of force F →: A = F s cos α = F s s

When projecting F s → force F → onto the direction of movement s → the force does not remain constant, and the calculation of work for small movements Δ s i is summed up and produced according to the formula:

A = ∑ ∆ A i = ∑ F s i ∆ s i .

This amount of work is calculated from the limit (Δ s i → 0) and then goes into the integral.

The graphical representation of the work is determined from the area of ​​the curvilinear figure located under the graph F s (x) of Figure 1. 18 . 2.

Picture 1 . 18 . 2. Graphic definition of work Δ A i = F s i Δ s i .

An example of a force that depends on the coordinate is the elastic force of a spring, which obeys Hooke's law. To stretch a spring, it is necessary to apply a force F →, the modulus of which is proportional to the elongation of the spring. This can be seen in Figure 1. 18 . 3.

Picture 1 . 18 . 3. Stretched spring. The direction of the external force F → coincides with the direction of movement s →. F s = k x , where k denotes the spring stiffness.

F → y p r = - F →

The dependence of the external force modulus on the x coordinates can be plotted using a straight line.

Picture 1 . 18 . 4 . Dependence of the external force modulus on the coordinate when the spring is stretched.

From the above figure, it is possible to find the work done on the external force of the right free end of the spring, using the area of ​​the triangle. The formula will take the form

This formula is applicable to express the work done by an external force when compressing a spring. Both cases show that the elastic force F → y p is equal to the work of the external force F → , but with the opposite sign.

Definition 2

If several forces act on a body, then the formula for the total work will look like the sum of all the work done on it. When a body moves translationally, the points of application of forces move equally, that is, the total work of all forces will be equal to the work of the resultant of the applied forces.

Picture 1 . 18 . 5 . Model of mechanical work.

Power determination

Power is called the work done by a force per unit time.

Recording the physical quantity of power, denoted N, takes the form of the ratio of work A to the time period t of the work performed, that is:

Definition 4

The SI system uses the watt (W t) as a unit of power, equal to the power of the force that does 1 J of work in 1 s.

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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 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 located on the table, the modulus of which is 10 N. Why angle is equal between this force and the movement of the block, if when moving the block 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. 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 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. B starting moment the stretch 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 with a mass of 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 from snowy mountain with an inclination angle α = 30º and travel 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 combustion 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.

Mechanical work This is an energy characteristic of the movement of physical bodies, which has a scalar form. It is equal to the modulus of the force acting on the body, multiplied by the modulus of the displacement caused by this force and by the cosine of the angle between them.

Formula 1 - Mechanical work.

F - Force acting on the body.

s - Body movement.

cosa - Cosine of the angle between force and displacement.

This formula has general form. If the angle between the applied force and the displacement is zero, then the cosine is equal to 1. Accordingly, the work will be equal only to the product of the force and the displacement. Simply put, if a body moves in the direction of application of force, then mechanical work is equal to the product of force and displacement.

Second special case, when the angle between the force acting on the body and its displacement is 90 degrees. In this case, the cosine of 90 degrees is equal to zero, so the work will be equal to zero. And indeed, what happens is that we apply force in one direction, and the body moves perpendicular to it. That is, the body clearly does not move under the influence of our force. Thus, the work done by our force to move the body is zero.

Figure 1 - Work of forces when moving a body.

If more than one force acts on a body, then the total force acting on the body is calculated. And then it is substituted into the formula as the only force. A body under the influence of force can move not only rectilinearly, but also along an arbitrary trajectory. In this case, the work is calculated for a small section of movement, which can be considered rectilinear, and then summed up along the entire path.

Work can be both positive and negative. That is, if the displacement and force coincide in direction, then the work is positive. And if a force is applied in one direction, and the body moves in another, then the work will be negative. An example of negative work is the work of a frictional force. Since the friction force is directed counter to the movement. Imagine a body moving along a plane. A force applied to a body pushes it in a certain direction. This force does positive work to move the body. But at the same time, the friction force does negative work. It slows down the movement of the body and is directed towards its movement.

Figure 2 - Force of motion and friction.

Mechanical work is measured in Joules. One Joule is the work done by a force of one Newton when moving a body one meter. In addition to the direction of movement of the body, the magnitude of the applied force can also change. For example, when a spring is compressed, the force applied to it will increase in proportion to the distance traveled. In this case, the work is calculated using the formula.

Formula 2 - Work of compression of a spring.

k is the spring stiffness.

x - moving coordinate.

« Physics - 10th grade"

The law of conservation of energy is a fundamental law of nature that allows us to describe most occurring phenomena.

Description of the movement of bodies is also possible using such concepts of dynamics as work and energy.

Remember what work and power are in physics.

Do these concepts coincide with everyday ideas about them?

All our daily actions come down to the fact that we, with the help of muscles, either set the surrounding bodies in motion and maintain this movement, or stop the moving bodies.

These bodies are tools (hammer, pen, saw), in games - balls, pucks, chess pieces. In production and agriculture people also set tools in motion.

The use of machines increases labor productivity many times due to the use of engines in them.

The purpose of any engine is to set bodies in motion and maintain this movement, despite braking by both ordinary friction and “working” resistance (the cutter should not just slide along the metal, but, cutting into it, remove chips; the plow should loosen land, etc.). In this case, a force must act on the moving body from the side of the engine.

Work is performed in nature whenever a force (or several forces) from another body (other bodies) acts on a body in the direction of its movement or against it.

The force of gravity does work when raindrops or stones fall from a cliff. At the same time, work is also done by the resistance force acting on the falling drops or on the stone from the air. The elastic force also performs work when a tree bent by the wind straightens.

Definition of work.

Newton's second law in impulse form Δ = Δt allows you to determine how the speed of a body changes in magnitude and direction if a force acts on it during a time Δt.

The influence of forces on bodies that lead to a change in the modulus of their velocity is characterized by a value that depends on both the forces and the movements of the bodies. In mechanics this quantity is called work of force.

A change in speed in absolute value is possible only in the case when the projection of the force F r on the direction of movement of the body is different from zero. It is this projection that determines the action of the force that changes the velocity of the body modulo. She does the work. Therefore, work can be considered as the product of the projection of force F r by the displacement modulus |Δ| (Fig. 5.1):

A = F r |Δ|. (5.1)

If the angle between force and displacement is denoted by α, then Fr = Fcosα.

Therefore, the work is equal to:

A = |Δ|cosα. (5.2)

Our everyday idea of ​​work differs from the definition of work in physics. You are holding a heavy suitcase, and it seems to you that you are doing work. However, from a physical point of view, your work is zero.

The work of a constant force is equal to the product of the moduli of the force and the displacement of the point of application of the force and the cosine of the angle between them.

In the general case, when a rigid body moves, the displacements of its different points are different, but when determining the work of a force, we are under Δ we understand the movement of its point of application. During the translational motion of a rigid body, the movement of all its points coincides with the movement of the point of application of the force.

Work, unlike force and displacement, is not a vector, but a scalar quantity. It can be positive, negative or zero.

The sign of the work is determined by the sign of the cosine of the angle between force and displacement. If α< 90°, то А >0, since the cosine of acute angles is positive. For α > 90°, the work is negative, since the cosine of obtuse angles is negative. At α = 90° (force perpendicular to displacement) no work is done.

If several forces act on a body, then the projection of the resultant force on the displacement is equal to the sum of the projections of the individual forces:

F r = F 1r + F 2r + ... .

Therefore, for the work of the resultant force we obtain

A = F 1r |Δ| + F 2r |Δ| + ... = A 1 + A 2 + .... (5.3)

If several forces act on a body, then the total work (the algebraic sum of the work of all forces) is equal to the work of the resultant force.

The work done by a force can be represented graphically. Let us explain this by depicting in the figure the dependence of the projection of force on the coordinates of the body when it moves in a straight line.

Let the body move along the OX axis (Fig. 5.2), then

Fcosα = F x , |Δ| = Δ x.

For the work of force we get

A = F|Δ|cosα = F x Δx.

Obviously, the area of ​​the rectangle shaded in Figure (5.3, a) is numerically equal to the work done when moving a body from a point with coordinate x1 to a point with coordinate x2.

Formula (5.1) is valid in the case when the projection of the force onto the displacement is constant. In the case of a curvilinear trajectory, constant or variable force, we divide the trajectory into small segments, which can be considered rectilinear, and the projection of the force at a small displacement Δ - constant.

Unit of work.

The unit of work can be established using the basic formula (5.2). If, when moving a body by a unit length, a force acts on it, the modulus of which is equal to one, and the direction of the force coincides with the direction of movement of its point of application (α = 0), then the work will be equal to one. The International System (SI) unit of work is the joule (denoted by J):

1 J = 1 N 1 m = 1 N m.

Joule- this is the work done by a force of 1 N on displacement 1 if the directions of force and displacement coincide.

Multiple units of work are often used: kilojoule and megajoule:

1 kJ = 1000 J,
1 MJ = 1000000 J.

Work can be completed either in a large period of time or in a very short one. In practice, however, it is far from indifferent whether work can be done quickly or slowly. The time during which work is performed determines the performance of any engine. A tiny electric motor can do a lot of work, but it will take a lot of time. Therefore, along with work, a quantity is introduced that characterizes the speed with which it is produced - power.

Power is the ratio of work A to the time interval Δt during which this work is done, i.e. power is the speed of work:

Substituting into formula (5.4) instead of work A its expression (5.2), we obtain

Thus, if the force and speed of a body are constant, then the power is equal to the product of the magnitude of the force vector by the magnitude of the velocity vector and the cosine of the angle between the directions of these vectors. If these quantities are variable, then using formula (5.4) one can determine the average power in a similar way to determining the average speed of a body.

The concept of power is introduced to evaluate the work per unit of time performed by any mechanism (pump, crane, machine motor, etc.). Therefore, in formulas (5.4) and (5.5), traction force is always meant.

In SI, power is expressed in watt (W).

Power is equal to 1 W if work equal to 1 J is performed in 1 s.

Along with the watt, larger (multiple) units of power are used:

1 kW (kilowatt) = 1000 W,
1 MW (megawatt) = 1,000,000 W.