When the work done by the force is 0. Lever

Mechanical work. Units of work.

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

In physics, the concept Job somewhat different. This 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 occurs also in the case when a force acting on a body (for example, the force of friction) 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 way which the body passes under the influence of this force, the more work is 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 given power does positive work.

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

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

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

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

Given:

ρ = 2500 kg/m 3

Solution:

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

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

F = 9.8 N/kg · 1250 kg ≈ 12,250 N.

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

Answer: A =245 kJ.

Levers.Power.Energy

To perform the same work, different engines require 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 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 at normal conditions work on average is 70-80 W. When jumping or running up stairs, a person can develop power up to 730 W, and in in some cases and even greater.

From the formula N = A/t it follows that

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

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

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

Given:

Solution:

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

Answer A= 21 kJ.

Simple mechanisms.

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

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

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

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

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

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

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

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

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

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

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

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 levers were used by the Egyptians? Or here important role plays on the word "installed"?)

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

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

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

Given:

Solution:

According to the 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, 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 strength should be longer than the shoulder 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. No paper cutting required great strength, and with a long blade it is more convenient 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. More 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 cutting wire.

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

Examples of the use of levers are the handles of vices and workbenches, the lever 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 bodies of animals and humans. These are, for example, arms, legs, jaws. Many levers can be found in the body of insects (by reading a book about insects and the structure of their bodies), birds, and in the structure of plants.

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

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

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

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

F = P/2 .

Thus, the movable block gives a 2-fold 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 strength or path, don’t simple mechanisms give a gain in work? The answer to this question can be obtained from experience.

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

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

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

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

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

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

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

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

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

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

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

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

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

Centuries-old practice has shown that None of the mechanisms gives a gain in performance. They 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. in these ideal conditions work done by the applied force (we will call this work full), is equal to useful work on lifting loads or overcoming any resistance.

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

Part of the work is done against the friction force in the mechanism and by moving it 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 full time job 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 Greek letterη, it is read as “this”:

η = Ap / Az · 100%.

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

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

Given :

Solution :

η = Ap / Az · 100%.

Total (expended) work Az = Fh2.

Useful work Ap = Рh1

P = 9.8 100 kg ≈ 1000 N.

Ap = 1000 N · 0.08 = 80 J.

Az = 250 N · 0.4 m = 100 J.

η = 80 J/100 J 100% = 80%.

Answer : η = 80%.

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

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

Energy.

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

1. From the 7th grade physics course, you know that if a force acts on a body and it moves in the direction of the force, then the force does mechanical work A, equal to the product of the force modulus and the displacement modulus:

A=Fs.

Unit of work in SI - joule (1 J).

[A] = [F][s] = 1 H 1 m = 1 N m = 1 J.

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

It follows from the formula that mechanical work is not performed if the force is zero (the body is at rest or moving uniformly and linearly) or the displacement is zero.

Let us assume that the force vector acting on the body makes a certain angle a with the displacement vector (Fig. 65). Since the body does not move in the vertical direction, the projection of force Fy per axis Y does not do work, but the projection of force Fx per axis X does work that is equal to A = F x s x.

Because the Fx = F cos a, a s x= s, That

Thus,

Job constant force is equal to the product of the magnitudes of the force and displacement vectors and the cosine of the angle between these vectors.

2. Let us analyze the resulting work formula.

If angle a = 0°, then cos 0° = 1 and A = Fs. The work done is positive and its value is maximum if the direction of the force coincides with the direction of displacement.

If angle a = 90°, then cos 90° = 0 and A= 0. The force does not do work if it is perpendicular to the direction of movement of the body. Thus, the work done by gravity is zero when a body moves along horizontal plane. The work done by the force imparted to the body is equal to zero centripetal acceleration with him uniform motion along a circle, since this force at any point of the trajectory is perpendicular to the direction of movement of the body.

If angle a = 180°, then cos 180° = –1 and A = –Fs. This case occurs when the force and displacement are directed in opposite sides. Respectively perfect work is negative and its value is maximum. Negative work is performed, for example, by the sliding friction force, since it is directed to the side opposite direction body movements.

If the angle a between the force and displacement vectors is acute, then the work is positive; if angle a is obtuse, then the work is negative.

3. Let us obtain a formula for calculating the work of gravity. Let the body have mass m freely falls to the ground from a point A, located at a height h relative to the surface of the Earth, and after some time it ends up at a point B(Fig. 66, A). The work done by gravity is equal to

A = Fs = mgh.

IN in this case the direction of movement of the body coincides with the direction of the force acting on it, therefore the work of gravity at free fall positive.

If a body moves vertically upward from a point B exactly A(Fig. 66, b), then its movement is directed in the direction opposite to gravity, and the work of gravity is negative:

A= –mgh

4. The work done by a force can be calculated using a graph of force versus displacement.

Suppose a body moves under the influence of constant gravity. Gravity modulus graph F cord from the body movement module s is a straight line parallel to the abscissa axis (Fig. 67). Let's find the area selected rectangle. It is equal to the product of its two sides: S = F cord h = mgh. On the other hand, the work of gravity is equal to the same value A = mgh.

Thus, the work is numerically equal to the area of ​​the rectangle bounded by the graph, coordinate axes and perpendicular to the x-axis at the point h.

Let us now consider the case when the force acting on the body is directly proportional to the displacement. Such a force, as is known, is the elastic force. Its module is equal F control = k D l, where D l- lengthening of the body.

Suppose a spring, the left end of which is fixed, is compressed (Fig. 68, A). At the same time, its right end shifted to D l 1. An elastic force arose in the spring F control 1, directed to the right.

If we now leave the spring to itself, its right end will move to the right (Fig. 68, b), the elongation of the spring will be equal to D l 2, and the elastic force F exercise 2.

Let's calculate the work done by the elastic force when moving the end of the spring from the point with coordinate D l 1 to the point with coordinate D l 2. We use a dependence graph for this F control (D l) (Fig. 69). The work done by the elastic force is numerically equal to the area of ​​the trapezoid ABCD. The area of ​​a trapezoid is equal to the product of half the sum of the bases and the height, i.e. S = AD. In the trapeze ABCD grounds AB = F control 2 = k D l 2 , CD= F control 1 = k D l 1 and the height AD= D l 1 – D l 2. Let's substitute these quantities into the formula for the area of ​​a trapezoid:

S= (D l 1 – D l 2) =– .

Thus, we found that the work of the elastic force is equal to:

A =– .

5 * . Let us assume that a body of mass m moves from a point A exactly B(Fig. 70), moving first without friction along an inclined plane from a point A exactly C, and then without friction along the horizontal plane from the point C exactly B. Work of gravity on the site C.B. is zero because the force of gravity is perpendicular to the displacement. When moving along an inclined plane, the work done by gravity is:

A AC = F cord l sin a. Because l sin a = h, That A AC = Ft cord h = mgh.

Work done by gravity when a body moves along a trajectory ACB equal to A ACB = A AC + A CB = mgh + 0.

Thus, A ACB = mgh.

The result obtained shows that the work done by gravity does not depend on the shape of the trajectory. It depends only on the initial and final positions of the body.

Let us now assume that the body moves along a closed trajectory ABCA(see Fig. 70). When moving a body from a point A exactly B along the trajectory ACB the work done by gravity is A ACB = mgh. When moving a body from a point B exactly A gravity does negative work, which is equal to A BA = –mgh. Then the work of gravity on a closed trajectory A = A ACB + A BA = 0.

The work done by the elastic force on a closed trajectory is also zero. Indeed, suppose that an initially undeformed spring is stretched and its length increases by D l. The elastic force did the work A 1 = . When returning to equilibrium, the elastic force does work A 2 = . The total work done by the elastic force when the spring is stretched and returned to its undeformed state is zero.

6. The work done by gravity and elastic force on a closed trajectory is zero.

Forces whose work on any closed trajectory is zero (or does not depend on the shape of the trajectory) are called conservative.

Forces whose work depends on the shape of the trajectory are called non-conservative.

The friction force is non-conservative. For example, a body moves from a point 1 exactly 2 first in a straight line 12 (Fig. 71), and then along a broken line 132 . At each section of the trajectory the friction force is the same. In the first case, the work of the friction force

A 12 = –F tr l 1 ,

and in the second -

A 132 = A 13 + A 32, A 132 = –F tr l 2 – F tr l 3 .

From here A 12A 132.

7. From the 7th grade physics course you know that important characteristic devices that do work are power.

Power is a physical quantity equal to the ratio of work to the period of time during which it is performed:

Power characterizes the speed at which work is performed.

SI unit of power - watt (1 W).

[N] === 1 W.

A unit of power is taken to be the power at which work 1 J is completed for 1 s .

Self-test questions

1. What is work called? What is the unit of work?

2. In what case does a force do negative work? positive work?

3. What formula is used to calculate the work of gravity? elastic forces?

5. What forces are called conservative? non-conservative?

6 * . Prove that the work done by gravity and elasticity does not depend on the shape of the trajectory.

7. What is called power? What is the unit of power?

Task 18

1. A boy weighing 20 kg is carried evenly on a sled, applying a force of 20 N. The rope by which the sled is pulled makes an angle of 30° with the horizontal. What is the work done by the elastic force generated in the rope if the sled moves 100 m?

2. An athlete weighing 65 kg jumps into water from a platform located at a height of 3 m above the surface of the water. How much work is done by the force of gravity acting on the athlete as he moves to the surface of the water?

3. Under the action of an elastic force, the length of a deformed spring with a stiffness of 200 N/m decreased by 4 cm. What is the work done by the elastic force?

4 * . Prove that the work variable force numerically equal to the area of ​​the figure, limited by schedule dependence of force on coordinates and coordinate axes.

5. What is the traction force of a car engine if constant speed 108 km/h it develops a power of 55 kW?

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

Figure 1. Work in the everyday sense of the word

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

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

The work of force is a physical quantity equal to the product the magnitude of the force over the distance traveled by the body in the direction of the force.

Figure 2. Work is a physical quantity

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

IN international system SI units of work are measured in joules.

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

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

Fig 3. James Prescott Joule (1818 - 1889)

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

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

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

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

Figure 4. Three cases when work is zero

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

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

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

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

Power is a physical quantity equal to the ratio work by the time it is completed.

Power is indicated by a Latin letter N.

The SI unit of power is the watt.

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

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

Fig 6. James Watt (1736 - 1819)

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

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

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

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

Bibliography

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

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

Homework

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