Why the concept of work in physics. Mechanical work and power

In our everyday experience, the word “work” appears very often. But one should distinguish between physiological work and work from the point of view of the science of physics. When you come home from class, you say: “Oh, I’m so tired!” This is physiological work. Or, for example, the work of a team in folk tale"Turnip".

Figure 1. Work in the everyday sense of the word

We will talk here about work from the point of view of physics.

Mechanical work is performed if a body moves under the influence of a force. Work is designated by the Latin letter A. A more strict definition of work sounds like this.

The work of force is called physical quantity, equal to the product of the magnitude of the force and the distance traveled by the body in the direction of the force.

Figure 2. Work is a physical quantity

The formula is valid when a constant force acts on the body.

In the international system of SI units, work is measured in joules.

This means that if under the influence of a force of 1 newton a body moves 1 meter, then 1 joule of work is done by this force.

The unit of work is named after the English scientist James Prescott Joule.

Fig 3. James Prescott Joule (1818 - 1889)

From the formula for calculating work it follows that there are three possible cases when work is equal to zero.

The first case is when a force acts on a body, but the body does not move. For example, a house is subject to a huge force of gravity. But she does not do any work because the house is motionless.

The second case is when the body moves by inertia, that is, no forces act on it. For example, spaceship moves in intergalactic space.

The third case is when a force acts on the body perpendicular to the direction of movement of the body. In this case, although the body moves and a force acts on it, there is no movement of the body in the direction of the force.

Figure 4. Three cases when work is zero

It should also be said that the work done by a force can be negative. This will happen if the body moves against the direction of the force. For example, when crane with the help of a cable, lifts a load above the ground, the work of gravity is negative (and the work of the elastic force of the cable directed upward, on the contrary, is positive).

Suppose, when executing construction work the pit must be filled with sand. It would take a few minutes for an excavator to do this, but a worker with a shovel would have to work for several hours. But both the excavator and the worker would have completed the same job.

Fig 5. The same work can be completed in different time

To characterize the speed of work done in physics, a quantity called power is used.

Power is a physical quantity equal to the ratio of work to the time it is performed.

Power is indicated by a Latin letter N.

The SI unit of power is the watt.

One watt is the power at which one joule of work is done in one second.

The power unit is named after the English scientist and inventor steam engine James Watt.

Fig 6. James Watt (1736 - 1819)

Let's combine the formula for calculating work with the formula for calculating power.

Let us now remember that the ratio of the path traveled by the body is S, by the time of movement t represents the speed of movement of the body v.

Thus, power is equal to the product of the numerical value of the force and the speed of the body in the direction of the force.

This formula is convenient to use when solving problems in which a force acts on a body moving with a known speed.

Bibliography

1. Lukashik V.I., Ivanova E.V. Collection of physics problems for grades 7-9 educational institutions. - 17th ed. - M.: Education, 2004.
2. Peryshkin A.V. Physics. 7th grade - 14th ed., stereotype. - M.: Bustard, 2010.
3. Peryshkin A.V. Collection of problems in physics, grades 7-9: 5th ed., stereotype. - M: Publishing House “Exam”, 2010.

1. Internet portal Physics.ru ().
2. Internet portal Festival.1september.ru ().
3. Internet portal Fizportal.ru ().
4. Internet portal Elkin52.narod.ru ().

Homework

1. In what cases is work equal to zero?
2. How is the work done along the path traveled in the direction of the force? In the opposite direction?
3. How much work is done by the frictional force acting on the brick when it moves 0.4 m? The friction force is 5 N.

Mechanical work. Units of work.

IN everyday life By the concept of “work” we mean everything.

In physics, the concept Job somewhat different. It is a definite physical quantity, which means it can be measured. In physics it is studied primarily mechanical work .

Let's look at examples of mechanical work.

The train moves under the traction force of an electric locomotive, and mechanical work is performed. When a gun is fired, the pressure force of the powder gases does work - it moves the bullet along the barrel, and the speed of the bullet increases.

From these examples it is clear that mechanical work is performed when a body moves under the influence of force. Mechanical work is also performed in the case when a force acting on a body (for example, friction force) reduces the speed of its movement.

Wanting to move the cabinet, we press hard on it, but if it does not move, then we do not perform mechanical work. One can imagine a case when a body moves without the participation of forces (by inertia); in this case, mechanical work is also not performed.

So, mechanical work is done only when a force acts on a body and it moves .

It is not difficult to understand that the greater the force acts on the body and the longer the path that the body travels under the influence of this force, the greater the work done.

Mechanical work is directly proportional to the force applied and directly proportional to the distance traveled .

Therefore, we agreed to measure mechanical work by the product of force and the path traveled along this direction of this force:

work = force × path

Where A- Job, F- strength and s- distance traveled.

A unit of work is taken to be the work done by a force of 1N over a path of 1 m.

Unit of work - joule (J ) named after the English scientist Joule. Thus,

1 J = 1N m.

Also used kilojoules (kJ) .

1 kJ = 1000 J.

Formula A = Fs applicable when the force F constant and coincides with the direction of movement of the body.

If the direction of the force coincides with the direction of motion of the body, then this force does positive work.

If the movement of the body occurs in the direction opposite to the direction of the applied force, for example, the force of sliding friction, then this force makes negative work.

If the direction of the force acting on the body is perpendicular to the direction of movement, then this force does no work, the work is zero:

In the future, speaking about mechanical work, we will briefly call it in one word - work.

Example. Calculate the work done when lifting a granite slab with a volume of 0.5 m3 to a height of 20 m. The density of granite is 2500 kg/m3.

Given:

ρ = 2500 kg/m 3

Solution:

where F is the force that must be applied to uniformly lift the slab up. This force is equal in modulus to the force Fstrand acting on the slab, i.e. F = Fstrand. And the force of gravity can be determined by the mass of the slab: Fweight = gm. Let's calculate the mass of the slab, knowing its volume and the density of granite: m = ρV; s = h, i.e. path equal to height rise.

So, m = 2500 kg/m3 · 0.5 m3 = 1250 kg.

F = 9.8 N/kg · 1250 kg ≈ 12,250 N.

A = 12,250 N · 20 m = 245,000 J = 245 kJ.

Answer: A =245 kJ.

Levers.Power.Energy

Different engines require different times to complete the same work. For example, a crane at a construction site lifts hundreds of bricks to the top floor of a building in a few minutes. If these bricks were moved by a worker, it would take him several hours to do this. Another example. A horse can plow a hectare of land in 10-12 hours, while a tractor with a multi-share plow ( ploughshare- part of the plow that cuts the layer of earth from below and transfers it to the dump; multi-ploughshare - many ploughshares), this work will be completed in 40-50 minutes.

It is clear that a crane does the same work faster than a worker, and a tractor does the same work faster than a horse. The speed of work is characterized by a special quantity called power.

Power is equal to the ratio of work to the time during which it was performed.

To calculate power, you need to divide the work by the time during which this work was done. power = work/time.

Where N- power, A- Job, t- time of work completed.

Power is a constant quantity when the same work is done every second; in other cases the ratio A/t determines the average power:

N avg = A/t . The unit of power is taken to be the power at which J of work is done in 1 s.

This unit is called the watt ( W) in honor of another English scientist, Watt.

1 watt = 1 joule/1 second, or 1 W = 1 J/s.

Watt (joule per second) - W (1 J/s).

Larger units of power are widely used in technology - kilowatt (kW), megawatt (MW) .

1 MW = 1,000,000 W

1 kW = 1000 W

1 mW = 0.001 W

1 W = 0.000001 MW

1 W = 0.001 kW

1 W = 1000 mW

Example. Find the power of the water flow flowing through the dam if the height of the water fall is 25 m and its flow rate is 120 m3 per minute.

Given:

ρ = 1000 kg/m3

Solution:

Mass of falling water: m = ρV,

m = 1000 kg/m3 120 m3 = 120,000 kg (12 104 kg).

Gravity acting on water:

F = 9.8 m/s2 120,000 kg ≈ 1,200,000 N (12 105 N)

Work done by flow per minute:

A - 1,200,000 N · 25 m = 30,000,000 J (3 · 107 J).

Flow power: N = A/t,

N = 30,000,000 J / 60 s = 500,000 W = 0.5 MW.

Answer: N = 0.5 MW.

Various engines have powers ranging from hundredths and tenths of a kilowatt (electric razor engine, sewing machine) up to hundreds of thousands of kilowatts (water and steam turbines).

Table 5.

Power of some engines, kW.

Each engine has a plate (engine passport), which indicates some information about the engine, including its power.

Human power under normal operating conditions is on average 70-80 W. When jumping or running up stairs, a person can develop power up to 730 W, and in some cases even more.

From the formula N = A/t it follows that

To calculate the work, it is necessary to multiply the power by the time during which this work was performed.

Example. The room fan motor has a power of 35 watts. How much work does he do in 10 minutes?

Let's write down the conditions of the problem and solve it.

Given:

Solution:

A = 35 W * 600s = 21,000 W * s = 21,000 J = 21 kJ.

Answer A= 21 kJ.

Simple mechanisms.

Since time immemorial, man has used various devices to perform mechanical work.

Everyone knows that a heavy object (a stone, a cabinet, a machine tool), which cannot be moved by hand, can be moved with the help of a sufficiently long stick - a lever.

On this moment it is believed that with the help of levers three thousand years ago during the construction of the pyramids in Ancient Egypt moved and raised heavy stone slabs to great heights.

In many cases, instead of lifting a heavy load to a certain height, it can be rolled or pulled to the same height along an inclined plane or lifted using blocks.

Devices used to convert force are called mechanisms .

Simple mechanisms include: levers and its varieties - block, gate; inclined plane and its varieties - wedge, screw. In most cases simple mechanisms used to gain strength, that is, to increase the force acting on the body several times.

Simple mechanisms are found in both household and all complex industrial and factory machines that cut, twist and stamp large sheets steel or draw the finest threads from which fabrics are then made. The same mechanisms can be found in modern complex automatic machines, printing and counting machines.

Lever arm. Balance of forces on the lever.

Let's consider the simplest and most common mechanism - the lever.

The lever is solid, which can rotate around a fixed support.

The pictures show how a worker uses a crowbar as a lever to lift a load. In the first case, the worker with force F presses the end of the crowbar B, in the second - raises the end B.

The worker needs to overcome the weight of the load P- force directed vertically downwards. To do this, he turns the crowbar around an axis passing through the only motionless the breaking point is the point of its support ABOUT. Force F with which the worker acts on the lever is less force P, thus the worker receives gain in strength. Using a lever, you can lift such a heavy load that you cannot lift it on your own.

The figure shows a lever whose axis of rotation is ABOUT(fulcrum) is located between the points of application of forces A And IN. Another picture shows a diagram of this lever. Both forces F 1 and F 2 acting on the lever are directed in one direction.

The shortest distance between the fulcrum and the straight line along which the force acts on the lever is called the arm of force.

The length of this perpendicular will be the arm of this force. The figure shows that OA- shoulder strength F 1; OB- shoulder strength F 2. The forces acting on the lever can rotate it around its axis in two directions: clockwise or counterclockwise. Yes, strength F 1 rotates the lever clockwise, and the force F 2 rotates it counterclockwise.

The condition under which the lever is in equilibrium under the influence of forces applied to it can be established experimentally. It must be remembered that the result of the action of a force depends not only on its numerical value (modulus), but also on the point at which it is applied to the body, or how it is directed.

Suspended from the lever (see figure) on both sides of the fulcrum various loads so that the lever remained in balance every time. The forces acting on the lever are equal to the weights of these loads. For each case, the force modules and their shoulders are measured. From the experience shown in Figure 154, it is clear that force 2 N balances the force 4 N. In this case, as can be seen from the figure, the shoulder of lesser strength is 2 times larger than the shoulder of greater strength.

Based on such experiments, the condition (rule) of lever equilibrium was established.

A lever is in equilibrium when the forces acting on it are inversely proportional to the arms of these forces.

This rule can be written as a formula:

F 1/F 2 = l 2/ l 1 ,

Where F 1And F 2 - forces acting on the lever, l 1And l 2 , - the shoulders of these forces (see figure).

The rule of lever equilibrium was established by Archimedes around 287 - 212. BC e. (but in the last paragraph it was said that levers were used by the Egyptians? Or here important role plays on the word "installed"?)

From this rule it follows that a smaller force can be used to balance a larger force using a lever. Let one arm of the lever be 3 times larger than the other (see figure). Then, by applying a force of, for example, 400 N at point B, you can lift a stone weighing 1200 N. To lift an even heavier load, you need to increase the length of the lever arm on which the worker acts.

Example. Using a lever, a worker lifts a slab weighing 240 kg (see Fig. 149). What force does he apply to the larger lever arm of 2.4 m if the smaller arm is 0.6 m?

Let's write down the conditions of the problem and solve it.

Given:

Solution:

According to the rule of lever equilibrium, F1/F2 = l2/l1, whence F1 = F2 l2/l1, where F2 = P is the weight of the stone. Stone weight asd = gm, F = 9.8 N 240 kg ≈ 2400 N

Then, F1 = 2400 N · 0.6/2.4 = 600 N.

Answer: F1 = 600 N.

In our example, the worker overcomes a force of 2400 N, applying a force of 600 N to the lever. But in this case, the arm on which the worker acts is 4 times longer than the one on which the weight of the stone acts ( l 1 : l 2 = 2.4 m: 0.6 m = 4).

By applying the rule of leverage, a smaller force can balance a larger force. In this case, the shoulder of lesser force should be longer than the shoulder of greater strength.

Moment of power.

You already know the rule of lever equilibrium:

F 1 / F 2 = l 2 / l 1 ,

Using the property of proportion (the product of its extreme members is equal to the product of its middle members), we write it in this form:

F 1l 1 = F 2 l 2 .

On the left side of the equality is the product of force F 1 on her shoulder l 1, and on the right - the product of force F 2 on her shoulder l 2 .

The product of the modulus of the force rotating the body and its shoulder is called moment of force; it is designated by the letter M. This means

A lever is in equilibrium under the action of two forces if the moment of the force rotating it clockwise is equal to the moment of the force rotating it counterclockwise.

This rule is called rule of moments , can be written as a formula:

M1 = M2

Indeed, in the experiment we considered (§ 56), the acting forces were equal to 2 N and 4 N, their shoulders respectively amounted to 4 and 2 lever pressures, i.e. the moments of these forces are the same when the lever is in equilibrium.

The moment of force, like any physical quantity, can be measured. The unit of moment of force is taken to be a moment of force of 1 N, the arm of which is exactly 1 m.

This unit is called newton meter (N m).

The moment of force characterizes the action of a force, and shows that it depends simultaneously on both the modulus of the force and its leverage. Indeed, we already know, for example, that the action of a force on a door depends both on the magnitude of the force and on where the force is applied. The easier it is to turn the door, the farther from the axis of rotation the force acting on it is applied. It’s better to unscrew the nut long wrench than short. The easier it is to lift a bucket from the well, the longer the handle of the gate, etc.

Levers in technology, everyday life and nature.

The rule of leverage (or the rule of moments) underlies the action of various kinds of tools and devices used in technology and everyday life where a gain in strength or travel is required.

We have a gain in strength when working with scissors. Scissors - this is a lever(fig), the axis of rotation of which occurs through a screw connecting both halves of the scissors. Acting force F 1 is the muscular strength of the hand of the person gripping the scissors. Counterforce F 2 is the resistance force of the material being cut with scissors. Depending on the purpose of the scissors, their design varies. Office scissors, designed for cutting paper, have long blades and handles that are almost the same length. No paper cutting required great strength, and with a long blade it is more convenient to cut in a straight line. Cutting scissors sheet metal(Fig.) have handles much longer than the blades, since the resistance force of the metal is great and to balance it, the arm of the acting force has to be significantly increased. The difference between the length of the handles and the distance of the cutting part and the axis of rotation is even greater wire cutters(Fig.), designed for cutting wire.

Levers various types available on many cars. The handle of a sewing machine, the pedals or handbrake of a bicycle, the pedals of a car and tractor, and the keys of a piano are all examples of levers used in these machines and tools.

Examples of the use of levers are the handles of vices and workbenches, the lever drilling machine etc.

The action of lever scales is based on the principle of the lever (Fig.). The training scales shown in Figure 48 (p. 42) act as equal-arm lever . IN decimal scales The shoulder from which the cup with weights is suspended is 10 times longer than the shoulder carrying the load. This makes weighing large loads much easier. When weighing a load on a decimal scale, you should multiply the mass of the weights by 10.

The device of scales for weighing freight cars of cars is also based on the rule of leverage.

Levers are also found in different parts bodies of animals and humans. These are, for example, arms, legs, jaws. Many levers can be found in the body of insects (by reading a book about insects and the structure of their bodies), birds, and in the structure of plants.

Application of the law of equilibrium of a lever to a block.

Block It is a wheel with a groove, mounted in a holder. A rope, cable or chain is passed through the block groove.

Fixed block This is called a block whose axis is fixed and does not rise or fall when lifting loads (Fig.).

A fixed block can be considered as an equal-armed lever, in which the arms of forces are equal to the radius of the wheel (Fig): OA = OB = r. Such a block does not provide a gain in strength. ( F 1 = F 2), but allows you to change the direction of the force. Movable block - this is a block. the axis of which rises and falls along with the load (Fig.). The figure shows the corresponding lever: ABOUT- fulcrum point of the lever, OA- shoulder strength R And OB- shoulder strength F. Since the shoulder OB 2 times the shoulder OA, then the strength F 2 times less force R:

F = P/2 .

Thus, the movable block gives a 2-fold gain in strength .

This can be proven using the concept of moment of force. When the block is in equilibrium, the moments of forces F And R equal to each other. But the shoulder of strength F 2 times the leverage R, and, therefore, the power itself F 2 times less force R.

Usually in practice a combination of a fixed block and a movable one is used (Fig.). The fixed block is used for convenience only. It does not give a gain in force, but it changes the direction of the force. For example, it allows you to lift a load while standing on the ground. This comes in handy for many people or workers. However, it gives a gain in strength 2 times greater than usual!

Equality of work when using simple mechanisms. "Golden rule" of mechanics.

The simple mechanisms we have considered are used when performing work in cases where it is necessary to balance another force through the action of one force.

Naturally, the question arises: while giving a gain in strength or path, don’t simple mechanisms give a gain in work? The answer to this question can be obtained from experience.

By balancing two different magnitude forces on a lever F 1 and F 2 (fig.), set the lever in motion. It turns out that at the same time the point of application of the smaller force F 2 goes further s 2, and the point of application of the greater force F 1 - shorter path s 1. Having measured these paths and force modules, we find that the paths traversed by the points of application of forces on the lever are inversely proportional to the forces:

s 1 / s 2 = F 2 / F 1.

Thus, acting on the long arm of the lever, we gain in strength, but at the same time we lose by the same amount along the way.

Product of force F on the way s there is work. Our experiments show that the work done by the forces applied to the lever is equal to each other:

F 1 s 1 = F 2 s 2, i.e. A 1 = A 2.

So, When using leverage, you won’t be able to win at work.

By using leverage, we can gain either power or distance. By applying force to the short arm of the lever, we gain in distance, but lose by the same amount in strength.

There is a legend that Archimedes, delighted with the discovery of the rule of leverage, exclaimed: “Give me a fulcrum and I will turn the Earth over!”

Of course, Archimedes could not cope with such a task even if he had been given a fulcrum (which should have been outside the Earth) and a lever of the required length.

To raise the earth just 1 cm, the long arm of the lever would have to describe an arc of enormous length. It would take millions of years to move the long end of the lever along this path, for example, at a speed of 1 m/s!

A stationary block does not give any gain in work, which is easy to verify experimentally (see figure). Ways, passable points application of forces F And F, are the same, the forces are the same, which means the work is the same.

You can measure and compare the work done with the help of a moving block. In order to lift a load to a height h using a movable block, it is necessary to move the end of the rope to which the dynamometer is attached, as experience shows (Fig.), to a height of 2h.

Thus, getting a 2-fold gain in strength, they lose 2-fold on the way, therefore, the movable block does not give a gain in work.

Centuries-old practice has shown that None of the mechanisms gives a gain in performance. They apply the same various mechanisms in order to win in strength or on the way, depending on the working conditions.

Already ancient scientists knew a rule applicable to all mechanisms: no matter how many times we win in strength, the same number of times we lose in distance. This rule has been called the "golden rule" of mechanics.

Efficiency of the mechanism.

When considering the design and action of the lever, we did not take into account friction, as well as the weight of the lever. in these ideal conditions work done by the applied force (we will call this work full), is equal to useful work on lifting loads or overcoming any resistance.

In practice, perfected by the mechanism full time job always a little more useful work.

Part of the work is done against the frictional force in the mechanism and by moving its individual parts. So, when using a movable block, you have to additionally do work to lift the block itself, the rope and determine the friction force in the axis of the block.

We didn’t take what mechanism we did, useful work The work accomplished with its help always constitutes only a part of the complete work. This means, denoting useful work by the letter Ap, total (expended) work by the letter Az, we can write:

Up< Аз или Ап / Аз < 1.

The ratio of useful work to total work is called the coefficient useful action mechanism.

The efficiency factor is abbreviated as efficiency.

Efficiency = Ap / Az.

Efficiency is usually expressed as a percentage and is denoted by the Greek letter η, read as “eta”:

η = Ap / Az · 100%.

Example: A load weighing 100 kg is suspended on the short arm of a lever. To lift it, a force of 250 N is applied to the long arm. The load is raised to a height h1 = 0.08 m, and the point of application driving force dropped to a height h2 = 0.4 m. Find the efficiency of the lever.

Let's write down the conditions of the problem and solve it.

Given :

Solution :

η = Ap / Az · 100%.

Total (expended) work Az = Fh2.

Useful work Ap = Рh1

P = 9.8 100 kg ≈ 1000 N.

Ap = 1000 N · 0.08 = 80 J.

Az = 250 N · 0.4 m = 100 J.

η = 80 J/100 J 100% = 80%.

Answer : η = 80%.

But " Golden Rule"is carried out in this case as well. Part of the useful work - 20% of it - is spent on overcoming friction in the axis of the lever and air resistance, as well as on the movement of the lever itself.

The efficiency of any mechanism is always less than 100%. When designing mechanisms, people strive to increase their efficiency. To achieve this, friction in the axes of the mechanisms and their weight are reduced.

Energy.

In plants and factories, machines and machines are driven by electric motors, which consume electrical energy(hence the name).

What does it mean?

In physics, “mechanical work” is the work of some force (gravity, elasticity, friction, etc.) on a body, as a result of which the body moves.

Often the word “mechanical” is simply not written.
Sometimes you can come across the expression “the body has done work,” which in principle means “the force acting on the body has done work.”

I think - I'm working.

I'm going - I'm working too.

Where is the mechanical work here?

If a body moves under the influence of a force, then mechanical work is performed.

They say that the body does work.
Or more precisely, it will be like this: the work is done by the force acting on the body.

Work characterizes the result of a force.

The forces acting on a person perform mechanical work on him, and as a result of the action of these forces, the person moves.

Work is a physical quantity equal to the product of the force acting on a body and the path made by the body under the influence of a force in the direction of this force.

A - mechanical work,
F - strength,
S - distance traveled.

Work is done, if 2 conditions are met simultaneously: a force acts on the body and it
moves in the direction of the force.

No work is done(i.e. equal to 0), if:
1. The force acts, but the body does not move.

For example: we exert force on a stone, but cannot move it.

2. The body moves, and the force is zero, or all forces are compensated (i.e., the resultant of these forces is 0).
For example: when moving by inertia, no work is done.
3. The direction of the force and the direction of movement of the body are mutually perpendicular.

For example: when a train moves horizontally, gravity does no work.

Work can be positive and negative

1. If the direction of the force and the direction of motion of the body coincide, positive work is done.

For example: the force of gravity, acting on a drop of water falling down, does positive work.

2. If the direction of force and movement of the body is opposite, negative work is done.

For example: the force of gravity acting on a rising balloon, does negative work.

If several forces act on a body, then the total work done by all forces is equal to the work done by the resulting force.

Units of work

In honor of the English scientist D. Joule, the unit of work was named 1 Joule.

In the International System of Units (SI):
[A] = J = N m
1J = 1N 1m

Mechanical work is equal to 1 J if, under the influence of a force of 1 N, a body moves 1 m in the direction of this force.

When flying from a person's thumb to his index finger
the mosquito does work - 0.000 000 000 000 000 000 000 000 001 J.

The human heart performs approximately 1 J of work per contraction, which corresponds to the work done when lifting a load weighing 10 kg to a height of 1 cm.

GET TO WORK, FRIENDS!

1. Mechanical work ​\(A \) ​ is a physical quantity equal to the product of the force vector acting on the body and the vector of its displacement:​\(A=\vec(F)\vec(S) \) ​. Work is a scalar quantity, characterized by numerical value and unit.

A unit of work is taken to be 1 joule (1 J). This is the work done by a force of 1 N along a path of 1 m.

\[ [\,A\,]=[\,F\,][\,S\,]; [\,A\,]=1Н\cdot1m=1J\]

2. If the force acting on the body makes a certain angle ​\(\alpha \) ​ with the displacement, then the projection of the force ​\(F \) ​ onto the X axis is equal to ​\(F_x \) ​ (Fig. 42).

Since ​\(F_x=F\cdot\cos\alpha \) ​, then \(A=FS\cos\alpha \) .

Thus, the work of a constant force is equal to the product of the magnitudes of the force and displacement vectors and the cosine of the angle between these vectors.

3. If force ​\(F\) ​ = 0 or displacement ​\(S \) ​ = 0, then mechanical work is zero ​\(A \) ​ = 0. Work is zero if the force vector is perpendicular to the displacement vector, t .e. ​\(\cos90^\circ \) ​ = 0. Thus, the work done by the force that imparts centripetal acceleration to the body during its uniform motion in a circle is equal to zero, since this force is perpendicular to the direction of motion of the body at any point of the trajectory.

4. The work done by a force can be either positive or negative. The work is positive ​\(A \) ​ > 0, if the angle is 90° > ​\(\alpha \) ​ ≥ 0°; if the angle 180° > ​\(\alpha \) ​ ≥ 90°, then the work is negative ​\(A \) ​< 0.

If the angle ​\(\alpha \) ​ = 0°, then ​\(\cos\alpha \) ​ = 1, ​\(A=FS \) ​. If the angle ​\(\alpha \) ​ = 180°, then ​\(\cos\alpha \) ​ = -1, ​\(A=-FS \) ​.

5. At free fall from a height ​\(h\) ​ a body of mass ​\(m\) ​ moves from position 1 to position 2 (Fig. 43). In this case, the force of gravity does work equal to:

\[ A=F_тh=mg(h_1-h_2)=mgh \]

​When a body moves vertically downward, the force and displacement are directed in one direction, and the force of gravity does positive work.

If a body rises upward, then the force of gravity is directed downward, and if it moves upward, then the force of gravity does negative work, i.e.

\[ A=-F_тh=-mg(h_1-h_2)=-mgh \]

6. The work can be presented graphically. The figure shows a graph of the dependence of gravity on the height of a body relative to the surface of the Earth (Fig. 44). Graphically, the work of gravity is equal to the area of ​​the figure (rectangle) bounded by the graph, coordinate axes and perpendicular to the abscissa axis
at point ​\(h\) ​.

The graph of the elastic force versus the elongation of the spring is a straight line passing through the origin of coordinates (Fig. 45). By analogy with the work of gravity, the work of the elastic force is equal to the area of ​​the triangle bounded by the graph, the coordinate axes and the perpendicular to the abscissa at the point ​\(x\) ​.
​\(A=Fx/2=kx\cdot x/2 \) ​.

7. The work done by gravity does not depend on the shape of the trajectory along which the body moves; it depends on the initial and final positions of the body. Let the body first move from point A to point B along the trajectory AB (Fig. 46). The work of gravity in this case

\[ A_(AB)=mgh \]

Let now the body move from point A to point B, first along the inclined plane AC, then along the base of the inclined plane BC. The work done by gravity when moving along the aircraft is zero. The work of gravity when moving along the AC is equal to the product of the projection of gravity onto the inclined plane ​\(mg\sin\alpha \) ​ and the length of the inclined plane, i.e. ​ \(A_(AC)=mg\sin\alpha\cdot l \)​. Product ​\(l\cdot\sin\alpha=h \) ​. Then \(A_(AC)=mgh \) . The work of gravity when moving a body along two different trajectories does not depend on the shape of the trajectory, but depends on the initial and final positions of the body.

The work of the elastic force also does not depend on the shape of the trajectory.

Suppose that a body moves from point A to point B along the trajectory ACB, and then from point B to point A along the trajectory BA. When moving along the ACB trajectory, gravity does positive work; when moving along the BA trajectory, the work of gravity is negative, equal in magnitude to the work when moving along the ACB trajectory. Therefore, the work done by gravity along a closed path is zero. The same applies to the work of the elastic force.

Forces whose work does not depend on the shape of the trajectory and is equal to zero along a closed trajectory are called conservative. Conservative forces include gravity and elasticity.

8. Forces whose work depends on the shape of the path are called non-conservative. The friction force is non-conservative. If a body moves from point A to point B (Fig. 47) first along a straight line and then along a broken line ACB, then in the first case the work of the friction force ​\(A_(AB)=-Fl_(AB) \) ​, and in the second ​\(A_(ABC)=A_(AC)+A_(CB) ​, \(A_(ABC)=-Fl_(AC)-Fl_(CB) \) .

Therefore, work ​\(A_(AB) \) ​ is not equal to work ​\(A_(ABC) \) ​.

9. Power is a physical quantity equal to the ratio of work to the period of time during which it is performed. Power characterizes the speed at which work is done.

Power is denoted by the letter ​\(N\) ​.

Unit of power: ​\([N]=[A]/[t] \) ​. ​\([N] \) ​ = 1 J/1 s = 1 J/s. This unit is called the watt (W). One watt is the power at which 1 J of work is performed in 1 s.

10. The power developed by the engine is equal to: ​\(N = A/t \) ​, ​\(A=F\cdot S \) ​, whence ​\(N=FS/t \) ​. The ratio of movement to time is the speed of movement: ​\(S/t = v\) ​. Where ​\(N = Fv \) ​.

From the resulting formula it is clear that with a constant resistance force, the speed of movement is directly proportional to the engine power.

In various machines and mechanisms, mechanical energy is converted. Due to energy, work is done during its transformation. In this case, only part of the energy is spent on performing useful work. Some of the energy is spent doing work against friction forces. Thus, any machine is characterized by a value indicating what part of the energy transferred to it is used usefully. This quantity is called efficiency factor (efficiency).

The efficiency coefficient is a value equal to the ratio of useful work ​\((A_p) \) ​ to all completed work \((A_s) \) : ​\(\eta=A_p/A_s \) ​. Efficiency is expressed as a percentage.

Part 1

1. The work is determined by the formula

1) ​\(A=Fv \) ​
2) \(A=N/t\) ​
3) \(A=mv\) ​
4) \(A=FS \) ​

2. The load is evenly lifted vertically upward by a rope tied to it. The work of gravity in this case

1) equal to zero
2) positive
3) negative
4) more work elastic forces

3. The box is pulled by a rope tied to it, making an angle of 60° with the horizontal, applying a force of 30 N. What is the work done by this force if the modulus of displacement is 10 m?

1) 300 J
2) 150 J
3) 3 J
4) 1.5 J

4. Artificial satellite The Earth, whose mass is equal to ​\(m\) ​, moves uniformly in a circular orbit with radius ​\(R\) ​. The work done by gravity in a time equal to the period of revolution is equal to

1) ​\(mgR \) ​
2) ​\(\pi mgR \) ​
3) \(2\pi mgR \) ​
4) ​\(0 \) ​

5. A car weighing 1.2 tons travels 800 m along a horizontal road. How much work was done by the friction force if the friction coefficient is 0.1?

1) -960 kJ
2) -96 kJ
3) 960 kJ
4) 96 kJ

6. A spring with a stiffness of 200 N/m is stretched by 5 cm. How much work will the elastic force do when the spring returns to equilibrium?

1) 0.25 J
2) 5 J
3) 250 J
4) 500 J

7. Balloons same mass slide down the slide along three different chutes, as shown in the figure. In which case will the work done by gravity be greatest?

1) 1
2) 2
3) 3
4) the work is the same in all cases

8. Work along a closed path is zero

A. Friction forces
B. Elastic forces

The correct answer is

1) both A and B
2) only A
3) only B
4) neither A nor B

9. The SI unit of power is

1) J
2) W
3) J s
4) Nm

10. What is the useful work if the work done is 1000 J and the engine efficiency is 40%?

1) 40000 J
2) 1000 J
3) 400 J
4) 25 J

11. Establish a correspondence between the work of force (in the left column of the table) and the sign of work (in the right column of the table). In your answer, write down the selected numbers under the corresponding letters.

WORK OF FORCE
A. Work of elastic force when stretching a spring
B. Work of friction force
B. Work of gravity when a body falls

WORK SIGN
1) positive
2) negative
3) equal to zero

12. From the statements below, choose two correct ones and write their numbers in the table.

1) The work of gravity does not depend on the shape of the trajectory.
2) Work is done during any movement of the body.
3) The work done by the sliding friction force is always negative.
4) The work done by the elastic force along a closed contour is not zero.
5) The work of the friction force does not depend on the shape of the trajectory.

Part 2

13. A winch uniformly lifts a load weighing 300 kg to a height of 3 m in 10 s. What is the power of the winch?

Answers

In physics, the concept of "work" has a different definition than that used in Everyday life. Specifically, the term "work" is used when a physical force causes an object to move. In general, if a strong force causes an object to move very far, then a lot of work is being done. And if the force is small or the object does not move very far, then only a small amount of work is done. The force can be calculated using the formula: Work = F × D × cosine(θ), where F = force (in Newtons), D = displacement (in meters), and θ = angle between the force vector and the direction of motion.

Steps

Part 1

1. Find the direction of the force vector and the direction of movement. To get started, it is important to first determine in which direction the object is moving, as well as where the force is being applied. Keep in mind that objects don't always move according to the force applied to them - for example, if you pull a small cart by the handle, then you apply a diagonal force (if you are taller than the cart) to move it forward. In this section, however, we will deal with situations in which the force (effort) and movement of an object have same direction. For information on how to find a job when these items Not have the same direction, read below.

   • To make this process easy to understand, let's follow an example problem. Let's say a toy carriage is pulled straight ahead by a train in front of it. In this case, the force vector and the direction of movement of the train point to the same path - forward. In the next steps, we will use this information to help find the work performed by the object.

2. Find the object's displacement. The first variable D or offset that we need for the work formula is usually easy to find. Displacement is simply the distance a force caused an object to move from its original position. IN educational tasks this information is usually either given (known) or can be inferred (found) from other information in the problem. IN real life all you have to do to find displacement is measure the distance the objects move.

   • Note that the distance units must be in meters in the formula to calculate work.
   • In our toy train example, let's say we find the work done by the train as it passes along the track. If it starts at a certain point and stops at a place about 2 meters along the track, then we can use 2 meters for our value of "D" in the formula.

3. Find the force applied to the object. Next, find the amount of force used to move the object. This is a measure of the "strength" of the force - the greater its magnitude, the more it pushes the object and the faster it accelerates. If the magnitude of the force is not provided, it can be derived from the mass and acceleration of the displacement (assuming there are no other conflicting forces acting on it) using the formula F = M × A.

   • Note that the units of force must be in Newtons to calculate the work formula.
   • In our example, let's assume that we don't know the magnitude of the force. However, let's assume that we know that the toy train has a mass of 0.5 kg and that a force causes it to accelerate at a speed of 0.7 meters/second 2 . In this case, we can find the value by multiplying M × A = 0.5 × 0.7 = 0.35 Newton.

4. Multiply Force x Distance. Once you know the amount of force acting on your object and the distance it was moved, the rest is easy. Simply multiply these two values ​​by each other to get the work value.

   • It's time to solve our example problem. Given a force value of 0.35 Newton and a displacement value of 2 meters, our answer is a matter of simple multiplication: 0.35 × 2 = 0.7 Joules.
   • You may have noticed that in the formula given in the introduction, there is an additional part to the formula: cosine (θ). As discussed above, in this example the force and direction of motion are applied in the same direction. This means that the angle between them is 0 o. Since cosine(0) = 1, we don't have to include it - we just multiply by 1.

5. Express your answer in Joules. In physics, values ​​for work (and several other quantities) are almost always given in a unit called the Joule. One joule is defined as 1 Newton of force applied per meter, or in other words, 1 Newton × meter. This makes sense - since you are multiplying distance by force, it is logical that the answer you get will have a unit of measurement equal to the unit of magnitude of your force times the distance.

   Part 2

   Calculating work using angular force

   1. Find the force and displacement as usual. Above we dealt with a problem in which an object moves in the same direction as the force that is applied to it. In reality, this is not always the case. In cases where the force and motion of an object are in two different directions, the difference between the two directions must also be factored into the equation to obtain an accurate result. First, find the magnitude of the object's force and displacement as you would normally do.

      • Let's look at another example problem. In this case, let's say we're pulling the toy train forward as in the example problem above, but this time we're actually pulling upward at a diagonal angle. We'll take this into account in the next step, but for now we'll stick to the basics: the movement of the train and the amount of force acting on it. For our purposes, let's say force has the magnitude 10 Newton and that he drove the same 2 meters forward as before.

   2. Find the angle between the force vector and the displacement. Unlike the above examples with a force that is in a different direction than the object's motion, you need to find the difference between the two directions in terms of the angle between them. If this information is not provided to you, you may need to measure the angle yourself or infer it from other information in the problem.

      • For our example problem, assume that the force being applied is approximately 60 o above the horizontal plane. If the train is still moving straight ahead (that is, horizontally), then the angle between the vector of force and the motion of the train will be 60 o.

   3. Multiply Force × Distance × Cosine(θ). Once you know the object's displacement, the amount of force acting on it, and the angle between the force vector and its motion, the solution is almost as easy as without taking angle into account. Simply take the cosine of the angle (you may need a scientific calculator for this) and multiply it by the force and displacement to find the answer to your problem in Joules.

      • Let's solve an example of our problem. Using a calculator we find that the cosine of 60 o is equal to 1/2. Including this in the formula, we can solve the problem as follows: 10 Newtons × 2 meters × 1/2 = 10 Joules.

   Part 3

   Using the Work Value

   1. Modify the formula to find distance, force, or angle. The work formula given above is not Just useful for finding work - it is also valuable for finding any variables in an equation when you already know the value of work. In these cases, simply isolate the variable you are looking for and solve the equation according to the basic rules of algebra.

      • For example, let's say we know that our train is being pulled with a force of 20 Newtons at a diagonal angle over 5 meters of track to do 86.6 Joules of work. However, we do not know the angle of the force vector. To find the angle, we simply isolate this variable and solve the equation as follows: 86.6 = 20 × 5 × Cosine(θ) 86.6/100 = Cosine(θ) Arccos(0.866) = θ = 30 o

   2. Divide by the time spent moving to find the power. In physics, the work is closely related to another type of measurement called power. Power is simply a way of defining the amount of speed at which work is carried out on a particular system over a long period of time. So to find the power, all you have to do is divide the work used to move the object by the time it takes to complete the move. Power measurements are expressed in units of W (which is equal to Joule/second).

      • For example, for the example problem in the above step, let's say it took 12 seconds to move the train 5 meters. In this case, all you have to do is divide the work done to move it 5 meters (86.6 J) by 12 seconds to find the answer to calculate the power: 86.6/12 = " 7.22 W.

   3. Use the formula TME i + W nc = TME f to find the mechanical energy in the system. Work can also be used to find the amount of energy contained in a system. In the above formula TME i = initial total mechanical energy in the TME system f = final total mechanical energy in the system and W nc = work done in communication systems due to non-conservative forces. . In this formula, if a force is applied in the direction of movement, then it is positive, and if it presses against (against) it, then it is negative. Note that both energy variables can be found using the formula (½)mv 2, where m = mass and V = volume.

      • For example, for the example problem two steps above, assume that the train initially had a total mechanical energy of 100 J. Since the force in the problem is pulling the train in a direction it was already traveling, it is positive. In this case, the final energy of the train is TME i + W nc = 100 + 86.6 = 186.6 J.
      • Note that non-conservative forces are forces whose power to affect the acceleration of an object depends on the path traveled by the object. Friction is good example- an object that is pushed along a short, straight path will feel the effects of friction for a short time, while an object that is pushed along a long, winding path to the same final location will feel more friction overall.
   • If you manage to solve the problem, then smile and be happy for yourself!
   • Practice solving as much as possible more tasks, this ensures complete understanding.
   • Keep practicing, and try again if you don't succeed the first time.
   • Study the following points regarding work:
     • The work done by a force can be either positive or negative. (In this sense, the terms "positive or negative" have their mathematical meaning, but their ordinary meaning).
     • Work done is negative when the force acts in the opposite direction to the displacement.
     • Work done is positive when the force is in the direction of displacement.