Relocation work. Mechanical work and power of force

The horse pulls the cart with some force, let's denote it F traction. Grandpa, who is sitting on the cart, presses on her with some force. Let's denote it F pressure The cart moves in the direction of the horse's pulling force (to the right), but in the direction of the grandfather's pressure force (down), the cart does not move. Therefore, in physics they say that F traction does work on the cart, and F the pressure does not do work on the cart.

So, work done by a force on a body mechanical work- a physical quantity, the modulus of which is equal to the product of the force and the path traveled by the body along the direction of action of this force s:

In honor of the English scientist D. Joule, the unit of mechanical work was named 1 joule(according to the formula, 1 J = 1 N m).

If a certain force acts on the considered body, then a certain body acts on it. That's why the work of a force on a body and the work of a body on a body are complete synonyms. However, the work of the first body on the second and the work of the second body on the first are partial synonyms, since the modules of these works are always equal, and their signs are always opposite. That is why the “±” sign is present in the formula. Let's discuss signs of work in more detail.

Numerical values ​​of force and path are always non-negative values. In contrast, mechanical work can have both positive and negative signs. If the direction of the force coincides with the direction of motion of the body, then the work done by the force is considered positive. If the direction of the force is opposite to the direction of motion of the body, the work done by the force is considered negative.(we take "-" from the "±" formula). If the direction of motion of the body is perpendicular to the direction of the force, then such a force does no work, that is, A = 0.

Consider three illustrations on three aspects of mechanical work.

Doing work by force may look different from the point of view of different observers. Consider an example: a girl rides in an elevator up. Does it do mechanical work? A girl can do work only on those bodies on which she acts by force. There is only one such body - the elevator car, as the girl presses on her floor with her weight. Now we need to find out if the cabin goes some way. Consider two options: with a stationary and moving observer.

Let the observer boy sit on the ground first. In relation to it, the elevator car moves up and goes some way. The weight of the girl is directed in the opposite direction - down, therefore, the girl performs negative mechanical work on the cabin: A virgins< 0. Вообразим, что мальчик-наблюдатель пересел внутрь кабины движущегося лифта. Как и ранее, вес девочки действует на пол кабины. Но теперь по отношению к такому наблюдателю кабина лифта не движется. Поэтому с точки зрения наблюдателя в кабине лифта девочка не совершает механическую работу: A dev = 0.

Mechanical work. Units of work.

AT everyday life By "work" we mean everything.

In physics, the concept Work somewhat different. This is a certain physical quantity, which means that it can be measured. In physics, the study is primarily mechanical work .

Consider examples of mechanical work.

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

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

Wanting to move the cabinet, we press on it with force, but if it does not move at the same time, then we do not perform mechanical work. One can imagine the case when the 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 the body and it moves .

It is easy to understand that the greater the force acting on the body and the longer the path that the body passes under the action of this force, the greater the work done.

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

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

work = force × path

where BUT- Work, F- strength and s- distance traveled.

A unit of work is the work done by a force of 1 N on a path of 1 m.

Unit of work - joule (J ) is named after the English scientist Joule. In this way,

1 J = 1N m.

Also used kilojoules (kJ) .

1 kJ = 1000 J.

Formula A = Fs applicable when the power F is constant and coincides with the direction of motion 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 motion 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 does negative work.

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

In the future, speaking of 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 / m 3.

Given:

ρ \u003d 2500 kg / m 3

Solution:

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

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

F = 9.8 N/kg 1250 kg ≈ 12250 N.

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

Answer: A = 245 kJ.

Levers.Power.Energy

It takes different engines to do the same work. different time. For example, a crane at a construction site lifts hundreds of bricks to the top floor of a building in a few minutes. If a worker were to move these bricks, 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-share - a lot of shares), this work will be done for 40-50 minutes.

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

Power is equal to the ratio of work to the time for which it was completed.

To calculate the power, it is necessary to divide the work by the time during which this work is done. power = work / time.

where N- power, A- Work, t- time of work done.

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

N cf = A/t . The unit of power was taken as the power at which work in J is done in 1 s.

This unit is called the watt ( Tue) 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 engineering - 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 flow of water 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).

The force of gravity acting on water:

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

Work done 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 (motor of an electric razor, sewing machine) 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 contains some data about the engine, including its power.

Human power under normal working conditions is on average 70-80 watts. Making jumps, running up the stairs, a person can develop power up to 730 watts, and in some cases even more.

From the formula N = A/t it follows that

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

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 condition of the problem and solve it.

Given:

Solution:

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

Answer A= 21 kJ.

simple mechanisms.

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

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

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

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

Devices used to transform power are called mechanisms .

Simple mechanisms include: levers and its varieties - block, gate; inclined plane and its varieties - wedge, screw. In most cases, simple mechanisms are used in order to obtain a gain in strength, i.e., to increase the force acting on the body by several times.

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

Lever arm. The balance of forces on the lever.

Consider the simplest and most common mechanism - the lever.

The lever is a rigid body that can rotate around a fixed support.

The figures show how a worker uses a crowbar to lift a load as a lever. In the first case, a worker with a 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. For this, he rotates the crowbar around an axis passing through the only motionless breaking point - its fulcrum O. Strength F, with which the worker acts on the lever, less force P, so the worker gets gain in strength. With the help of 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 O(fulcrum) is located between the points of application of forces BUT and AT. The other figure shows a diagram of this lever. Both forces F 1 and F 2 acting on the lever are directed in the same direction.

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

The length of this perpendicular will be the shoulder of this force. The figure shows that OA- shoulder strength F 1; OV- shoulder strength F 2. The forces acting on the lever can rotate it around the axis in two directions: clockwise or counterclockwise. Yes, power 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 action of forces applied to it can be established experimentally. At the same time, 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.

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

On the basis of such experiments, the condition (rule) of the balance of the lever was established.

The lever is in equilibrium when the forces acting on it are inversely proportional to the shoulders 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 Fig.).

The rule for the balance of the lever was established by Archimedes around 287-212. BC e. (But didn't the last paragraph say that the levers were used by the Egyptians? Or is the word "established" important here?)

It follows from this rule that a smaller force can be balanced with a leverage of a larger force. Let one arm of the lever be 3 times larger than the other (see Fig.). Then, applying a force of, for example, 400 N at point B, it is possible to lift a stone weighing 1200 N. In order to lift an even heavier load, it is necessary 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 arm of the lever, which is 2.4 m, if the smaller arm is 0.6 m?

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

Given:

Solution:

According to the lever equilibrium rule, 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 by applying a force of 600 N to the lever. But at the same time, the arm on which the worker acts is 4 times longer than that 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 the smaller force must be longer than the shoulder of the greater force.

Moment of power.

You already know the lever balance rule:

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

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

F 1l 1 = F 2 l 2 .

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

The product of the modulus of the force rotating the body and its arm is called moment of force; it is denoted by the letter M. So,

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

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

M1 = M2

Indeed, in the experiment we have considered, (§ 56) the acting forces were equal to 2 N and 4 N, their shoulders, respectively, were 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. A moment of force of 1 N is taken as a unit of moment of force, the shoulder of which is exactly 1 m.

This unit is called newton meter (N m).

The moment of force characterizes the action of the force, and shows that it depends simultaneously on the modulus of the force and on its shoulder. Indeed, we already know, for example, that the effect of a force on a door depends both on the modulus of the force and on where the force is applied. The door is easier to turn, the farther from the axis of rotation the force acting on it is applied. It is better to unscrew the nut with a long wrench than with a short one. 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 lever rule (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 on the road is required.

We have a gain in strength when working with scissors. Scissors - it's a lever(rice), 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 squeezing the scissors. Opposing force F 2 - the resistance force of such a material that is cut with scissors. Depending on the purpose of the scissors, their device is different. Office scissors, designed for cutting paper, have long blades and handles that are almost the same length. It does not require much force to cut paper, and it is more convenient to cut in a straight line with a long blade. Scissors for cutting sheet metal (Fig.) have handles much longer than the blades, since the resistance force of the metal is large and to balance it, the shoulder operating force have to increase significantly. Even more difference between the length of the handles and the distance of the cutting part and the axis of rotation in wire cutters(Fig.), Designed for wire cutting.

Levers different kind many cars have. A sewing machine handle, bicycle pedals or hand brakes, car and tractor pedals, piano keys 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 of a drilling machine, etc.

The action of lever balances is also based on the principle of the lever (Fig.). The training scale shown in figure 48 (p. 42) acts as equal-arm lever . AT decimal scales the arm to which the cup with weights is suspended is 10 times longer than the arm carrying the load. This greatly simplifies the weighing of large loads. When weighing a load on a decimal scale, multiply the weight of the weights by 10.

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

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

Application of the law of balance of the lever to the block.

Block is a wheel with a groove, reinforced in the holder. A rope, cable or chain is passed along the gutter of the block.

Fixed block such a block is called, the axis of which is fixed, and when lifting loads it does not rise and does not fall (Fig.

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

F = P/2 .

In this way, the movable block gives a gain in strength by 2 times .

This can also be proved using the concept of moment of force. When the block is in equilibrium, the moments of forces F and R are equal to each other. But the shoulder of strength F 2 times the shoulder strength R, which means that the force itself F 2 times less power R.

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

Equality of work when using simple mechanisms. The "golden rule" of mechanics.

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

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

Having balanced on the lever two forces of different modulus F 1 and F 2 (fig.), set the lever in motion. It turns out that for the same time, the point of application of a smaller force F 2 goes a long way s 2, and the point of application of greater force F 1 - smaller 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 win in strength, but at the same time we lose the same amount on 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 are equal to each other:

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

So, when using the leverage, the win in the work will not work.

By using the lever, we can win either in strength or in distance. Acting by force on the short arm of the lever, we gain in distance, but lose in strength by the same amount.

There is a legend that Archimedes, delighted with the discovery of the rule of the lever, exclaimed: "Give me a fulcrum, and I will turn the Earth!".

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

To raise the earth by only 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!

Does not give a gain in work and a fixed block, which is easy to verify by experience (see Fig.). Ways, passable points application of forces F and F, are the same, the same are the forces, which means that the work is the same.

It is possible to measure and compare with each other the work done with the help of a movable block. In order to lift the load to a height h with the help of 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.

In this way, getting a gain in strength by 2 times, they lose 2 times on the way, therefore, the movable block does not give a gain in work.

Centuries of practice has shown that none of the mechanisms gives a gain in work. Various mechanisms are used in order to win in strength or on the way, depending on the working conditions.

Already ancient scientists knew the rule applicable to all mechanisms: how many times we win in strength, how many times we lose in distance. This rule has been called the "golden rule" of mechanics.

The efficiency of the mechanism.

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

In practice, the total work done by the mechanism is always somewhat greater than the useful work.

Part of the work is done against the friction force in the mechanism and by moving its individual parts. So, using a movable block, you have to additionally perform work on lifting the block itself, the rope and determining the friction force in the axis of the block.

Whatever mechanism we choose, the useful work accomplished with its help is always only a part of the total work. So, denoting the useful work by the letter Ap, the full (spent) work by the letter Az, we can write:

Up< Аз или Ап / Аз < 1.

The ratio of useful work to full work is called the efficiency of the mechanism.

Efficiency is abbreviated as efficiency.

Efficiency = Ap / Az.

Efficiency is usually expressed as a percentage and denoted by the Greek letter η, it is read as "this":

η \u003d Ap / Az 100%.

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

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

Given :

Solution :

η \u003d Ap / Az 100%.

Full (spent) work Az = Fh2.

Useful work Ап = Рh1

P \u003d 9.8 100 kg ≈ 1000 N.

Ap \u003d 1000 N 0.08 \u003d 80 J.

Az \u003d 250 N 0.4 m \u003d 100 J.

η = 80 J/100 J 100% = 80%.

Answer : η = 80%.

But " Golden Rule" is performed in this case too. 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%. By designing mechanisms, people tend to increase their efficiency. To do this, friction in the axes of the mechanisms and their weight are reduced.

Energy.

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

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

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

This definition seen in Figure 1. eighteen . one .

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 . eighteen . one . Work 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 displacements Δ s i 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), after which it goes into the integral.

The graphic representation of the work is determined from the area curvilinear figure, located under the graph F s (x) of Figure 1. eighteen . 2.

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

An example of a coordinate-dependent force is the elastic force of a spring, which obeys Hooke's law. To stretch the 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. eighteen . 3 .

Picture 1 . eighteen . 3 . Stretched spring. The direction of the external force F → coincides with the direction of displacement s → . F s = k x , where k is the stiffness of the spring.

F → y p p = - F →

The dependence of the module of the external force on the coordinates x can be shown on the graph using a straight line.

Picture 1 . eighteen . four . Dependence of the module of the external force on the coordinate when the spring is stretched.

From the above figure, it is possible to find work 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 a spring is compressed. Both cases show that the elastic force F → y p p is equal to the work of the external force F → , but with the opposite sign.

Definition 2

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

Picture 1 . eighteen . 5 . model of mechanical work.

Determination of power

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

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

Definition 4

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

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In physics, the concept of "work" has a different definition than that used in Everyday life. In particular, the term "work" is used when physical strength causes the object to move. In general, if a powerful force causes an object to move very far, then a lot of work is done. And if the force is small or the object doesn't move very far, then only a little work. 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's important to first determine in which direction the object is moving, as well as where the force is being applied from. 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're taller than the cart) to move it forward. In this section, however, we will deal with situations in which the force (effort) and displacement of an object have the same direction. For information on how to find work when these items not have the same direction, read below.

   • To make this process easy to understand, let's follow an example task. Let's say a toy car 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 done by the entity.

2. Find the offset of the object. The first variable D or offset we need for the work formula is usually easy to find. Displacement is simply the distance that a force has caused an object to move from its original position. AT learning tasks this information, as a rule, is either given (known) or it can be deduced (found) from other information in the problem. AT real life all you have to do to find the offset is measure the distance the objects move.

   • Note that 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 down the track, then we can use 2 meters for our "D" value in the formula.

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

   • Note that force units must be in Newtons to calculate the work formula.
   • In our example, let's assume 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 the 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 × Distance. Once you know the amount of force acting on your object and the distance it has been moved, the rest is easy. Just multiply these two values ​​by each other to get the work value.

   • It's time to solve our example problem. With a force value of 0.35 Newton and a displacement value of 2 meters, our answer is a simple multiplication question: 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, 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. Enter your answer in Joules. In physics, work (and a few other quantities) is 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're multiplying distance by force, it makes sense that the answer you'll get will have a unit of measure equal to the unit of your force times your 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 being applied to it. In fact, this is not always the case. In cases where the force and motion of an object are in two different directions, the difference between these two directions must also be accounted for in the equation in order 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 task. In this case, let's say we're pulling the toy train forward like in the example problem above, but this time we're actually pulling up at a diagonal angle. In the next step, we will take this into account, but for now we will stick to the basics: the movement of the train and the amount of force acting on it. For our purposes, let's say the force has the magnitude 10 Newton and that he drove the same 2 meters forward like before.

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

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

   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 the angle into account. Simply take the cosine of an angle (this may require a scientific calculator) and multiply it by force and displacement to find your answer in Joules.

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

   Part 3

   Use of work value

   1. Modify the formula to find distance, force, or angle. The work formula above is not simply useful for finding work - it's also valuable for finding any variables in an equation when you already know the value of the 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, suppose we know that our train is being pulled with a force of 20 Newtons at a diagonal angle of more than 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 extract this variable and solve the equation as follows: 86.6 = 20 × 5 × Cosine(θ) 86.6/100 = Cosine(θ) Arccos(0.866) = θ = 30o

   2. Divide by the time spent in motion to find the power. In physics, work is closely related to another type of measurement called "power". Power is simply a way of quantifying the rate at which work is being done 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 indicated in units - W (which is equal to Joule / second).

      • For example, for the example task in the step above, let's say it took 12 seconds for the train to move 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 the force is applied in the direction of movement, then it is positive, and if it presses on (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 problem example two steps above, let's assume that the train initially had a total mechanical energy of 100 joules. Since the force in the problem is pulling the train in the direction it has already passed, 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 an object's acceleration 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 generally feel more friction.
   • If you manage to solve the problem, then smile and be happy for yourself!
   • Practice solving as much as you can more tasks, it guarantees full understanding.
   • Keep practicing and try again if you don't succeed the first time.
   • Learn 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" carry their mathematical meaning, but the usual meaning).
     • The work done is negative when the force is acting in the opposite direction to the displacement.
     • The work done is positive when the force acts in the direction of travel.

Every body that moves can be described as work. In other words, it characterizes the action of forces.

Work is defined as:
The product of the modulus of force and the path traveled by the body, multiplied by the cosine of the angle between the direction of force and motion.

Work is measured in Joules:
1 [J] = = [kg* m2/s2]

For example, body A, under the influence of a force of 5 N, has passed 10 m. Determine the work done by the body.

Since the direction of movement and the action of the force are the same, the angle between the force vector and the displacement vector will be 0°. The formula is simplified because the cosine of an angle at 0° is 1.

Substituting the initial parameters into the formula, we find:
A= 15 J.

Consider another example, a body with a mass of 2 kg, moving with an acceleration of 6 m / s2, passed 10 m. Determine the work done by the body if it moved upward along an inclined plane at an angle of 60 °.

To begin with, we calculate what force must be applied to inform the body of an acceleration of 6 m / s2.

F = 2 kg * 6 m/s2 = 12 H.
Under the action of a force of 12H, the body traveled 10 m. The work can be calculated using the already known formula:

Where, a is equal to 30 °. Substituting the initial data into the formula, we get:
A= 103.2 J.

Power

Many machines of mechanisms perform the same work for a different period of time. To compare them, the concept of power is introduced.
Power is a value that shows the amount of work done per unit of time.

Power is measured in watts, after the Scottish engineer James Watt.
1 [Watt] = 1 [J/s].

For example, a large crane lifted a load weighing 10 tons to a height of 30 m in 1 minute. A small crane lifted 2 tons of bricks to the same height in 1 minute. Compare crane capacities.
Define the work performed by cranes. The load rises by 30m, while overcoming the force of gravity, so the force expended on lifting the load will be equal to the force of interaction between the Earth and the load (F = m * g). And work is the product of forces and the distance traveled by the goods, that is, the height.

For a large crane A1 = 10,000 kg * 30 m * 10 m / s2 = 3,000,000 J, and for a small crane A2 = 2,000 kg * 30 m * 10 m / s2 = 600,000 J.
Power can be calculated by dividing work by time. Both cranes lifted the load in 1 min (60 sec).

From here:
N1 = 3,000,000 J/60 s = 50,000 W = 50 kW.
N2 = 600,000 J / 60 s = 10,000 W = 10 kW.
From the above data, it is clearly seen that the first crane is 5 times more powerful than the second.