Dependence of work on force. Mechanical work

Mechanical work. Units of work.

In everyday life, we understand everything by the concept of “work”.

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 body moves in the direction opposite to the direction of the applied force, for example, the sliding friction force, then this force does 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. the path is equal to the lifting height.

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 (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 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.

At the moment, it is believed that with the help of levers three thousand years ago, during the construction of the pyramids in Ancient Egypt, heavy stone slabs were moved and raised 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 are used to gain strength, that is, to increase the force acting on the body several times.

Simple mechanisms are found both in household and in all complex industrial and industrial 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 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.

A lever is a rigid body that 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.

To find the arm of the force, you need to lower the perpendicular from the fulcrum to the line of action of the 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.

Various weights are suspended from the lever (see figure) 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 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 the levers were used by the Egyptians? Or does the word “established” play an important role here?)

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 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 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. Cutting paper does not require much force, and a long blade makes it easier to cut in a straight line. Shears 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 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.

Many machines have different types of levers. 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 of a 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 of the body 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 increase 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 power 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). Paths traversed by the points of 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 use various mechanisms in order to win in strength or in travel, 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. under these ideal conditions, the 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, the total work done by a mechanism is always slightly greater than the 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.

Whatever mechanism we take, the useful work done with its help always constitutes only a part of the total 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 efficiency of the 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 of h1 = 0.08 m, while the point of application of the driving force drops to a height of 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 the “golden rule” applies 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 factories and factories, machines and machines are driven by electric motors, which consume electrical energy (hence the name).

A compressed spring (Fig.), when straightened, does work, raises a load to a height, or makes a cart move. A stationary load raised above the ground does not do work, but if this load falls, it can do work (for example, it can drive a pile into the ground). Every moving body has the ability to do work. Thus, a steel ball A (fig) rolling down from an inclined plane, hitting a wooden block B, moves it a certain distance. At the same time, work is done. If a body or several interacting bodies (a system of bodies) can do work, they are said to have energy. Energy - a physical quantity showing how much work a body (or several bodies) can do. Energy is expressed in the SI system in the same units as work, i.e. in joules. The more work a body can do, the more energy it has. When work is done, the energy of bodies changes. Work done is equal to the change in energy. Potential and kinetic energy. Potential (from lat. potency - possibility) energy is the energy that is determined by the relative position of interacting bodies and parts of the same body. Potential energy, for example, is possessed by a body raised relative to the surface of the Earth, because the energy depends on the relative position of it and the Earth. and their mutual attraction. If we consider the potential energy of a body lying on the Earth to be zero, then the potential energy of a body raised to a certain height will be determined by the work done by gravity when the body falls to the Earth. Let us denote the potential energy of the body E n, because E = A, and work, as we know, is equal to the product of force and path, then A = Fh, Where F- gravity. This means that the potential energy En is equal to: E = Fh, or E = gmh, Where g- acceleration of gravity, m- body mass, h- the height to which the body is raised. Water in rivers held by dams has enormous potential energy. Falling down, the water does work, driving powerful turbines of power plants. The potential energy of a copra hammer (Fig.) is used in construction to carry out the work of driving piles. When opening a door with a spring, work is done to stretch (or compress) the spring. Due to the acquired energy, the spring, contracting (or straightening), does work, closing the door. The energy of compressed and untwisted springs is used, for example, in watches, various wind-up toys, etc. Any elastic deformed body has potential energy. The potential energy of compressed gas is used in the operation of heat engines, in jackhammers, which are widely used in the mining industry, in road construction, excavation of hard soil, etc. The energy that a body possesses as a result of its movement is called kinetic (from the Greek. kinema - movement) energy. The kinetic energy of a body is denoted by the letter E To. Moving water, driving the turbines of hydroelectric power plants, expends its kinetic energy and does work. Moving air, the wind, also has kinetic energy. What does kinetic energy depend on? Let's turn to experience (see figure). If you roll ball A from different heights, you will notice that the greater the height the ball rolls from, the greater its speed and the further it moves the block, i.e., it does more work. This means that the kinetic energy of a body depends on its speed. Due to its speed, a flying bullet has high kinetic energy. The kinetic energy of a body also depends on its mass. Let's do our experiment again, but we'll roll another ball of greater mass from the inclined plane. Bar B will move further, i.e. more work will be done. This means that the kinetic energy of the second ball is greater than the first. The greater the mass of a body and the speed at which it moves, the greater its kinetic energy. In order to determine the kinetic energy of a body, the formula is used: Ek = mv^2 /2, Where m- body mass, v- speed of body movement. The kinetic energy of bodies is used in technology. The water retained by the dam has, as already mentioned, great potential energy. When water falls from a dam, it moves and has the same high kinetic energy. It drives a turbine connected to an electric current generator. Due to the kinetic energy of water, electrical energy is generated. The energy of moving water is of great importance in the national economy. This energy is used using powerful hydroelectric power plants. The energy of falling water is an environmentally friendly source of energy, unlike fuel energy. All bodies in nature, relative to the conventional zero value, have either potential or kinetic energy, and sometimes both together. For example, a flying airplane has both kinetic and potential energy relative to the Earth. We became acquainted with two types of mechanical energy. Other types of energy (electrical, internal, etc.) will be discussed in other sections of the physics course. Conversion of one type of mechanical energy into another. The phenomenon of transformation of one type of mechanical energy into another is very convenient to observe on the device shown in the figure. By winding the thread onto the axis, the device disk is lifted. A disk raised upward has some potential energy. If you let go of it, it will spin and begin to fall. As it falls, the potential energy of the disk decreases, but at the same time its kinetic energy increases. At the end of the fall, the disk has such a reserve of kinetic energy that it can rise again to almost its previous height. (Part of the energy is spent working against the frictional force, so the disk does not reach its original height.) Having risen up, the disk falls again and then rises again. In this experiment, when the disk moves downward, its potential energy turns into kinetic energy, and when it moves up, the kinetic energy turns into potential energy. The transformation of energy from one type to another also occurs when two elastic bodies collide, for example, a rubber ball on the floor or a steel ball on a steel plate. If you lift a steel ball (rice) above a steel plate and release it from your hands, it will fall. As the ball falls, its potential energy decreases, and its kinetic energy increases, as the speed of the ball increases. When the ball hits the plate, both the ball and the plate will be compressed. The kinetic energy that the ball had will turn into potential energy of the compressed plate and the compressed ball. Then, thanks to the action of elastic forces, the plate and the ball will take their original shape. The ball will bounce off the slab, and their potential energy will again turn into the kinetic energy of the ball: the ball will bounce up at a speed almost equal to the speed it had at the moment it hit the slab. As the ball rises upward, the speed of the ball, and therefore its kinetic energy, decreases, while the potential energy increases. Having bounced off the plate, the ball rises to almost the same height from which it began to fall. At the top point of the rise, all its kinetic energy will again turn into potential. Natural phenomena are usually accompanied by the transformation of one type of energy into another. Energy can be transferred from one body to another. For example, when archery, the potential energy of a drawn bowstring is converted into the kinetic energy of a flying arrow.

Note that work and energy have the same units of measurement. This means that work can be converted into energy. For example, in order to raise a body to a certain height, then it will have potential energy, a force is needed that will do this work. The work done by the lifting force will turn into potential energy.

The rule for determining work according to the dependence graph F(r): the work is numerically equal to the area of ​​the figure under the graph of force versus displacement.

Angle between force vector and displacement

1) Correctly determine the direction of the force that does the work; 2) We depict the displacement vector; 3) We transfer the vectors to one point and obtain the desired angle.

In the figure, the body is acted upon by the force of gravity (mg), the reaction of the support (N), the force of friction (Ftr) and the tension force of the rope F, under the influence of which the body moves r.

Work of gravity

Ground reaction work

Work of friction force

Work done by rope tension

Work done by resultant force

The work done by the resultant force can be found in two ways: 1st method - as the sum of the work (taking into account the “+” or “-” signs) of all forces acting on the body, in our example
Method 2 - first of all, find the resultant force, then directly its work, see figure

Work of elastic force

To find the work done by the elastic force, it is necessary to take into account that this force changes because it depends on the elongation of the spring. From Hooke's law it follows that as the absolute elongation increases, the force increases.

To calculate the work of the elastic force during the transition of a spring (body) from an undeformed state to a deformed state, use the formula

Power

A scalar quantity that characterizes the speed of work (an analogy can be drawn with acceleration, which characterizes the rate of change in speed). Determined by the formula

Efficiency

Efficiency is the ratio of the useful work done by a machine to all the work expended (energy supplied) during the same time

The efficiency is expressed as a percentage. The closer this number is to 100%, the higher the machine's performance. There cannot be an efficiency greater than 100, since it is impossible to do more work using less energy.

The efficiency of an inclined plane is the ratio of the work done by gravity to the work expended in moving along the inclined plane.

The main thing to remember

1) Formulas and units of measurement;
2) The work is performed by force;
3) Be able to determine the angle between the force and displacement vectors

If the work done by a force when moving a body along a closed path is zero, then such forces are called conservative or potential. The work done by the friction force when moving a body along a closed path is never equal to zero. The friction force, unlike the force of gravity or elastic force, is non-conservative or non-potential.

There are conditions under which the formula cannot be used
If the force is variable, if the trajectory of movement is a curved line. In this case, the path is divided into small sections for which these conditions are met, and the elementary work on each of these sections is calculated. The total work in this case is equal to the algebraic sum of the elementary works:

The value of the work done by a certain force depends on the choice of reference system.

You are already familiar with mechanical work (work of force) from the basic school physics course. Let us recall the definition of mechanical work given there for the following cases.

If the force is directed in the same direction as the movement of the body, then the work done by the force

In this case, the work done by the force is positive.

If the force is directed opposite to the movement of the body, then the work done by the force

In this case, the work done by the force is negative.

If the force f_vec is directed perpendicular to the displacement s_vec of the body, then the work done by the force is zero:

Work is a scalar quantity. The unit of work is called the joule (symbol: J) in honor of the English scientist James Joule, who played an important role in the discovery of the law of conservation of energy. From formula (1) it follows:

1 J = 1 N * m.

1. A block weighing 0.5 kg was moved along the table 2 m, applying an elastic force of 4 N to it (Fig. 28.1). The coefficient of friction between the block and the table is 0.2. What is the work acting on the block?
a) gravity m?
b) normal reaction forces?
c) elastic forces?
d) sliding friction forces tr?

The total work done by several forces acting on a body can be found in two ways:
1. Find the work of each force and add up these works, taking into account the signs.
2. Find the resultant of all forces applied to the body and calculate the work of the resultant.

Both methods lead to the same result. To make sure of this, go back to the previous task and answer the questions in task 2.

2. What is it equal to:
a) the sum of the work done by all forces acting on the block?
b) the resultant of all forces acting on the block?
c) work resultant? In the general case (when the force f_vec is directed at an arbitrary angle to the displacement s_vec) the definition of the work of the force is as follows.

The work A of a constant force is equal to the product of the force modulus F by the displacement modulus s and the cosine of the angle α between the direction of the force and the direction of displacement:

A = Fs cos α (4)

3. Show that the general definition of work leads to the conclusions shown in the following diagram. Formulate them verbally and write them down in your notebook.

4. A force is applied to a block on the table, the modulus of which is 10 N. What is the angle between this force and the movement of the block if, when moving the block 60 cm along the table, this force does the work: a) 3 J; b) –3 J; c) –3 J; d) –6 J? Make explanatory drawings.

2. Work of gravity

Let a body of mass m move vertically from the initial height h n to the final height h k.

If the body moves downwards (h n > h k, Fig. 28.2, a), the direction of movement coincides with the direction of gravity, therefore the work of gravity is positive. If the body moves upward (h n< h к, рис. 28.2, б), то работа силы тяжести отрицательна.

In both cases, the work done by gravity

A = mg(h n – h k). (5)

Let us now find the work done by gravity when moving at an angle to the vertical.

5. A small block of mass m slid along an inclined plane of length s and height h (Fig. 28.3). The inclined plane makes an angle α with the vertical.

a) What is the angle between the direction of gravity and the direction of movement of the block? Make an explanatory drawing.
b) Express the work of gravity in terms of m, g, s, α.
c) Express s in terms of h and α.
d) Express the work of gravity in terms of m, g, h.
e) What is the work done by gravity when the block moves upward along the entire same plane?

Having completed this task, you are convinced that the work of gravity is expressed by formula (5) even when the body moves at an angle to the vertical - both down and up.

But then formula (5) for the work of gravity is valid when a body moves along any trajectory, because any trajectory (Fig. 28.4, a) can be represented as a set of small “inclined planes” (Fig. 28.4, b).

Thus,
the work done by gravity when moving along any trajectory is expressed by the formula

A t = mg(h n – h k),

where h n is the initial height of the body, h k is its final height.
The work done by gravity does not depend on the shape of the trajectory.

For example, the work done by gravity when moving a body from point A to point B (Fig. 28.5) along trajectory 1, 2 or 3 is the same. From here, in particular, it follows that the force of gravity when moving along a closed trajectory (when the body returns to the starting point) is equal to zero.

6. A ball of mass m hanging on a thread of length l was deflected 90º, keeping the thread taut, and released without a push.
a) What is the work done by gravity during the time during which the ball moves to the equilibrium position (Fig. 28.6)?
b) What is the work done by the elastic force of the thread during the same time?
c) What is the work done by the resultant forces applied to the ball during the same time?

3. Work of elastic force

When the spring returns to an undeformed state, the elastic force always does positive work: its direction coincides with the direction of movement (Fig. 28.7).

Let's find the work done by the elastic force.
The modulus of this force is related to the modulus of deformation x by the relation (see § 15)

The work done by such a force can be found graphically.

Let us first note that the work done by a constant force is numerically equal to the area of ​​the rectangle under the graph of force versus displacement (Fig. 28.8).

Figure 28.9 shows a graph of F(x) for the elastic force. Let us mentally divide the entire movement of the body into such small intervals that the force at each of them can be considered constant.

Then the work on each of these intervals is numerically equal to the area of ​​the figure under the corresponding section of the graph. All work is equal to the sum of work in these areas.

Consequently, in this case, the work is numerically equal to the area of ​​the figure under the graph of the dependence F(x).

7. Using Figure 28.10, prove that

the work done by the elastic force when the spring returns to its undeformed state is expressed by the formula

A = (kx 2)/2. (7)

8. Using the graph in Figure 28.11, prove that when the spring deformation changes from x n to x k, the work of the elastic force is expressed by the formula

From formula (8) we see that the work of the elastic force depends only on the initial and final deformation of the spring. Therefore, if the body is first deformed and then returns to its initial state, then the work of the elastic force is zero. Let us recall that the work of gravity has the same property.

9. At the initial moment, the tension of a spring with a stiffness of 400 N/m is 3 cm. The spring is stretched by another 2 cm.
a) What is the final deformation of the spring?
b) What is the work done by the elastic force of the spring?

10. At the initial moment, a spring with a stiffness of 200 N/m is stretched by 2 cm, and at the final moment it is compressed by 1 cm. What is the work done by the elastic force of the spring?

4. Work of friction force

Let the body slide along a fixed support. The sliding friction force acting on the body is always directed opposite to the movement and, therefore, the work of the sliding friction force is negative in any direction of movement (Fig. 28.12).

Therefore, if you move the block to the right, and the peg the same distance to the left, then, although it will return to its initial position, the total work done by the sliding friction force will not be equal to zero. This is the most important difference between the work of sliding friction and the work of gravity and elasticity. Let us recall that the work done by these forces when moving a body along a closed trajectory is zero.

11. A block with a mass of 1 kg was moved along the table so that its trajectory turned out to be a square with a side of 50 cm.
a) Has the block returned to its starting point?
b) What is the total work done by the frictional force acting on the block? The coefficient of friction between the block and the table is 0.3.

5.Power

Often it is not only the work being done that is important, but also the speed at which the work is being done. It is characterized by power.

Power P is the ratio of the work done A to the time period t during which this work was done:

(Sometimes power in mechanics is denoted by the letter N, and in electrodynamics by the letter P. We find it more convenient to use the same designation for power.)

The unit of power is the watt (symbol: W), named after the English inventor James Watt. From formula (9) it follows that

1 W = 1 J/s.

12. What power does a person develop by uniformly lifting a bucket of water weighing 10 kg to a height of 1 m for 2 s?

It is often convenient to express power not through work and time, but through force and speed.

Let's consider the case when the force is directed along the displacement. Then the work done by the force A = Fs. Substituting this expression into formula (9) for power, we obtain:

P = (Fs)/t = F(s/t) = Fv. (10)

13. A car is traveling on a horizontal road at a speed of 72 km/h. At the same time, its engine develops a power of 20 kW. What is the force of resistance to the movement of the car?

Clue. When a car moves along a horizontal road at a constant speed, the traction force is equal in magnitude to the resistance force to the movement of the car.

14. How long will it take to uniformly lift a concrete block weighing 4 tons to a height of 30 m if the power of the crane motor is 20 kW and the efficiency of the electric motor of the crane is 75%?

Clue. The efficiency of an electric motor is equal to the ratio of the work of lifting the load to the work of the engine.

Additional questions and tasks

15. A ball weighing 200 g was thrown from a balcony with a height of 10 and an angle of 45º to the horizontal. Having reached a maximum height of 15 m in flight, the ball fell to the ground.
a) What is the work done by gravity when lifting the ball?
b) What is the work done by gravity when the ball is lowered?
c) What is the work done by gravity during the entire flight of the ball?
d) Is there any extra data in the condition?

16. A ball with a mass of 0.5 kg is suspended from a spring with a stiffness of 250 N/m and is in equilibrium. The ball is raised so that the spring becomes undeformed and released without a push.
a) To what height was the ball raised?
b) What is the work done by gravity during the time during which the ball moves to the equilibrium position?
c) What is the work done by the elastic force during the time during which the ball moves to the equilibrium position?
d) What is the work done by the resultant of all forces applied to the ball during the time during which the ball moves to the equilibrium position?

17. A sled weighing 10 kg slides down a snowy mountain with an inclination angle of α = 30º without initial speed and travels a certain distance along a horizontal surface (Fig. 28.13). The coefficient of friction between the sled and snow is 0.1. The length of the base of the mountain is l = 15 m.

a) What is the magnitude of the friction force when the sled moves on a horizontal surface?
b) What is the work done by the friction force when the sled moves along a horizontal surface over a distance of 20 m?
c) What is the magnitude of the friction force when the sled moves along the mountain?
d) What is the work done by the friction force when lowering the sled?
e) What is the work done by gravity when lowering the sled?
f) What is the work done by the resultant forces acting on the sled as it descends from the mountain?

18. A car weighing 1 ton moves at a speed of 50 km/h. The engine develops a power of 10 kW. Gasoline consumption is 8 liters per 100 km. The density of gasoline is 750 kg/m 3, and its specific heat of combustion is 45 MJ/kg. What is the efficiency of the engine? Is there any extra data in the condition?
Clue. The efficiency of a heat engine is equal to the ratio of the work performed by the engine to the amount of heat released during fuel combustion.

In physics, the concept of "work" has a different definition than the one 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

Finding the value of work in one dimension

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 problems, 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 a 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 many problems as possible to ensure 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.

Every body that makes a movement can be characterized by 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 movement.

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

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

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

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

Let's consider another example: a body weighing 2 kg, moving with an acceleration of 6 m/s2, has traveled 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, let's calculate how much force needs to be applied to impart an acceleration of 6 m/s2 to the body.

F = 2 kg * 6 m/s2 = 12 H.
Under the influence of a force of 12N, the body moved 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 and mechanisms perform the same work in different periods of time. To compare them, the concept of power is introduced.
Power is a quantity that shows the amount of work performed per unit of time.

Power is measured in Watts, in honor of 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.
Let's define the work performed by cranes. The load rises 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 by the distance traveled by the loads, that is, by 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 minute (60 seconds).

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.