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Урок 92. Комбинированные задачи динамики (ч.2)

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Урок 90. Движение по окружности (ч.2)

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Урок 131. Задачи на работу, мощность, КПД (ч.2)

Урок 131. Задачи на работу, мощность, КПД (ч.2)

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Урок 114. Работа. Теорема о кинетической энергии

Урок 114. Работа. Теорема о кинетической энергии

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Урок 130. Задачи на работу, мощность, КПД (ч.1)

Урок 130. Задачи на работу, мощность, КПД (ч.1)

Channel: Павел ВИКТОР

Физика

Механика

Работа

Кинетическая энергия

Теорема о кинетической энерги

Решение задач

Ришельевский лицей

00:00:11

on Tuesday you and I will have to

00:00:13

do independent work, it is

00:00:16

quite difficult because it raises

00:00:18

many different topics at once, so

00:00:22

today we will do version 0 of this

00:00:25

independent work at home, you will also

00:00:27

watch the broadcast and I hope that by

00:00:30

next Tuesday you will be able to

00:00:32

complete it without difficulty, so the topic of the

00:00:35

lesson is

00:00:39

preparation

00:00:45

for for independent work,

00:00:48

the topic of independent work conventionally

00:00:52

sounds like this: work and the theorem is

00:00:57

kinetic energy, although we

00:00:59

will need

00:01:00

kinematics and Newton’s second law in

00:01:03

general, now you will see now we will see

00:01:06

what the task is, let’s

00:01:20

do this first, the left half of the screen

00:01:23

will be a board, the

00:01:24

right half of the screen will be tasks and

00:01:26

we will fill out the table, but now

00:01:33

let's read the task itself,

00:01:38

under the influence of some constant force

00:01:41

f, a body weighing 2 kilograms moved

00:01:45

from point 1 to point 2,

00:01:47

while the body's velocity vector changed

00:01:49

from v1 to v2 as shown in the figure

00:01:52

and then draw the vector in the figure for task 1

00:01:57

movement of the body and the force

00:01:59

applied to the body, taking

00:02:01

into account the proposed scale. and

00:02:03

applying force, select point 2 and 2, a

00:02:06

large table that we will now

00:02:08

analyze, well, let’s first

00:02:10

see that yes, but even so, under the influence of a

00:02:14

constant force of a body weighing 2 kilograms, the

00:02:16

mass we know has moved from

00:02:19

point 1 here. 1 to point 2 then the

00:02:24

velocity vector changed from v1 to v2

00:02:27

as shown in the figure

00:02:29

here vector v 1 is shown here is vector v 2

00:02:33

and the scale you see is 4 cells and here it

00:02:37

says 4 meters per second,

00:02:38

that is, one cell corresponds to

00:02:40

one meter per second

00:02:42

in the figure you need to depict the vector of

00:02:46

displacement of the body and the vector of the force

00:02:48

acting on the body,

00:02:49

well, it’s clear since the speed of the body

00:02:51

has changed, which means some force is acting on it, you

00:02:53

will need to find this force

00:02:55

and show the vector of this force, taking into

00:02:58

account the

00:02:59

scale, we also have the drawings here are

00:03:01

4 cells 1 newton

00:03:04

for you we have to do similar

00:03:07

work; it will differ only in

00:03:10

appearance, but the content of

00:03:13

the task itself will be exactly the same.

00:03:15

Now let’s see what they are asking you to

00:03:17

find in this,

00:03:19

turn the picture over yourself; you need to find the

00:03:22

module of the body’s displacement; the

00:03:25

module of the applied force; the projection of the force on

00:03:29

the direction of movement; the work of the force; the

00:03:32

module of the initial impulse of the body; the module

00:03:35

of the final It’s easy to look for the body’s momentum in the direction

00:03:38

because it coincides with the

00:03:40

direction of the velocity, then the

00:03:43

impulse module of the forces applied to the body in

00:03:46

section 12 the

00:03:47

initial kinetic energy the final

00:03:50

kinetic energy the acceleration module the time of

00:03:52

movement the acceleration projection on the x axis the

00:03:56

angle between the acceleration vector of the body and the

00:03:59

y axis the angle between the force vector and

00:04:02

vector of body movement, that is, 14

00:04:05

questions, which side to take on this,

00:04:09

let's read one small

00:04:12

note in small handwriting, what is

00:04:15

usually written in small print is

00:04:17

the most important, especially on advertising

00:04:19

posters, a

00:04:21

hint, the proposed task does not

00:04:24

necessarily have to be completed in the order in

00:04:27

which they are listed, so let's look at

00:04:31

drawing, what is the easiest thing to do first of all,

00:04:35

of course, it is easiest to first

00:04:38

construct a displacement vector, what is

00:04:41

called a displacement vector, a vector

00:04:43

drawn from the start to the end

00:04:46

point, we take it and draw it, we take a pen or

00:04:51

pencil and draw it, we connect the

00:04:53

initial position of the body with the final

00:04:55

position of the body, this will be the

00:04:58

displacement vector like this Now we need to draw a

00:05:04

displacement vector,

00:05:06

I draw it and then we will work with this

00:05:10

in more detail, so with a vector means the

00:05:15

initial coordinates of the body, by the way,

00:05:19

since Roma has already asked this question, let’s

00:05:22

look at the drawing and

00:05:23

immediately find the projections of the

00:05:26

displacement vector onto the coordinate axes, this

00:05:28

may be useful to us later sx

00:05:33

projection displacement vector on

00:05:36

coordinate axes

00:05:37

what is the projection of a vector this is the

00:05:39

end coordinate minus the

00:05:42

origin coordinator to the end of the

00:05:44

displacement vector look at the axes the

00:05:48

scale is in meters see in meters it means the

00:05:52

end coordinate is 35 meters the

00:05:55

start coordinate is 15 so we can write os x

00:06:00

equals 35 meters minus 15 meters

00:06:05

equals how many 20 meters similarly

00:06:10

let's find the coordinates let's find the

00:06:14

projection of the displacement vector onto the y axis with

00:06:18

y equals

00:06:20

the y coordinate of the end of the vector

00:06:24

the coordinate of the end of the vector 1515 this is

00:06:29

the end of the vector

00:06:30

its y coordinate 15 division price here 5

00:06:34

means 15 meters minus the coordinates of the beginning

00:06:42

55 minus 55 meters

00:06:46

it turns out what is the projection minus 40

00:06:52

meters, so now let's see if there is

00:06:55

any benefit from what we have learned,

00:06:57

read and here the very first question sounds

00:06:59

like this: find the modulus of displacement of the body pore,

00:07:03

which means we can write

00:07:04

s equals the square root of s x

00:07:10

square plus c y square and now

00:07:13

we substitute the numbers s equals the

00:07:17

square root of s x square 20 squared

00:07:22

plus 40 I won’t write the minus sign

00:07:25

since it is raised to the square of

00:07:27

meters approximately equals now

00:07:30

let’s see how many 20 roots sleep this

00:07:34

sounds great but physicists prefer

00:07:36

decimal fractions and so the root of 20

00:07:41

squared plus 40 squared 20 roots with

00:07:47

5 thank you press the sd button and

00:07:50

get 44 with a fraction we will round to

00:07:54

three significant figures,

00:07:56

which means we got approximately forty-

00:07:59

four and seven forty-four and seven meters

00:08:04

now we turn the table over and see the

00:08:10

body displacement module symbol with the

00:08:13

value

00:08:14

forty-four and seven further units

00:08:19

of measurement what

00:08:21

meters we got the results into the world

00:08:24

so let's move on what other questions are asked to us the

00:08:27

modulus of the applied force we probably wo

00:08:31

n’t be able to answer this question right

00:08:33

away beyond the projection of forces and the direction of

00:08:37

movement will probably also be a

00:08:38

bit difficult to work with the force modulus of the initial

00:08:41

impulse just a minute yes yes two kilogram

00:08:50

modulus of momentum

00:08:52

and modulus of momentum we are just asked

00:08:55

to find this fifth sixth point

00:08:57

our modulus of the initial impulse modulus of the

00:08:59

final everything means let's do this

00:09:01

remember that impulse n

00:09:04

is a physical quantity

00:09:06

equal to the product of the mass of a body by its

00:09:09

speed the

00:09:10

mass of the body is known two kilograms

00:09:12

now we need to know

00:09:15

We look at the speed value at the figure again and, taking into account

00:09:19

the scales, we determine the speed value.

00:09:22

The scales are shown on the right, here they are, which

00:09:25

means look at 4 cells 4 meters per

00:09:29

second, that is, in fact, one cell

00:09:31

1 meter per second, we count one time

00:09:35

two three 4 5 6 7 8 9 10 meters per second

00:09:40

two times two three 4 5 we’ve already found

00:09:44

the speeds, let’s write them down in one

00:09:49

equals 10 meters per second in 2

00:09:55

equals 5 meters per second

00:09:58

is and now if we multiply these

00:10:01

values by the mass of the body, then we find out

00:10:04

the modulus of the impulse n 1 equals v1

00:10:11

multiplied by mass that we get

00:10:14

20 about what units

00:10:18

kilogram multiplied by meters divided by

00:10:20

second

00:10:21

p2 not vector by scalar equals 10

00:10:29

kilograms per meter per second squared

00:10:32

we can enter this data yes per meter per

00:10:36

second habit kilogram per meter per

00:10:38

second this is Newton, now we put it in

00:10:42

the table and so the modulus of the initial impulse

00:10:46

p 120 kilograms we will write 20 kilograms per meter

00:10:55

per second the final modulus 10

00:11:00

units and kilograms per meter per second

00:11:05

correctly, Zhenya Dovganyuk immediately suggested that

00:11:08

since we know the mass and speed, we can

00:11:10

find the kinetic energy,

00:11:12

well so let's find the kinetic

00:11:14

energy, remember that the kinetic

00:11:16

energy is calculated using this formula

00:11:18

m squared in half, so we can find the

00:11:24

kinetic energy and the first kinetic

00:11:29

1 is equal to our body has a mass of 2

00:11:33

kilograms, this is written in clear text,

00:11:35

here are two kilograms, which means we

00:11:40

get 1 2 by 2 kilograms

00:11:45

multiplied by b square 1 2 I already wrote

00:11:50

if your square in the first case, our

00:11:52

speed was from 10, then it will be 100

00:11:55

square meters per second squared and

00:11:58

this turns out to be 100 joules in the second

00:12:04

case and the kinetic second is equal to 1

00:12:08

2 by 2 kilogram

00:12:11

multiplied by 5 squared by 25 by 25

00:12:17

meters squared per second squared

00:12:19

this will be 25 joules

00:12:24

great we can put these results

00:12:27

in the table

00:12:28

initial kinetic energy 100

00:12:31

joules

00:12:32

we write here 100 joules final

00:12:40

kinetic energy 25 joules so

00:12:48

now we can find a job why

00:12:54

where did you decide from Roma that now we can

00:12:57

find work on the kinetic

00:13:01

energy theorem, the total work of all forces

00:13:04

acting on a body is equal to the change in its

00:13:06

kinetic energy, there is only one

00:13:08

force here, so the work of this

00:13:11

one and only force is equal to the

00:13:13

change in the kinetic energy of the body,

00:13:16

I will have to erase something here I’ll

00:13:19

erase this, I’ll erase

00:13:25

the rest for now I won’t touch it, and so

00:13:28

the work equals the change,

00:13:33

that is, the final value minus the

00:13:35

initial I genetic 2 minus e

00:13:40

kinetic

00:13:41

first we write take a equals the

00:13:47

final value 25 joules -100

00:13:54

complains it turns out minus 75 joules

00:13:59

the work is negative, this is already tells us

00:14:04

that the angle between the direction of the

00:14:07

force vector and the displacement vector is so

00:14:11

obtuse, we need to be mentally prepared for this, let’s

00:14:14

enter this data well, so

00:14:18

the work of force is

00:14:21

75

00:14:23

joules, so the next thing we can do is

00:14:28

let’s see what we haven’t answered yet,

00:14:30

what questions we haven’t answered yet

00:14:36

modulus of the impulse of force so where is it

00:14:42

here here and here is the modulus of the impulse of force

00:14:47

so how to find the modulus of the impulse of force by

00:14:51

Pythagorean losses this is a mathematical

00:14:54

technique and what do we

00:14:56

rely on from users by Fedora we

00:15:01

rely on Newton’s second law in

00:15:03

impulse form change in momentum body

00:15:07

is equal to the momentum vector of the forces acting

00:15:12

on the body, it was not in vain that I said the words vector

00:15:15

this is a vector relationship, so don’t

00:15:18

fall for the bait that I just

00:15:22

emphasized these are

00:15:25

impulse modules, but we have a completely

00:15:27

different relationship: the vector relationship

00:15:30

f by t vector equals the final impulse

00:15:37

p2 vector minus the initial impulse p1 is

00:15:41

a vector, so if we want to do everything

00:15:45

correctly, we need to subtract two vectors,

00:15:48

for this we need to know how these vectors

00:15:51

are directed, how the impulse vectors are directed,

00:15:53

just like the velocity vectors,

00:15:57

so let’s draw somewhere on the side now; it’s

00:16:00

not necessary; it’s not necessary to

00:16:03

do this exactly to scale; let’s depict it

00:16:05

vector of the initial

00:16:06

and knights to the initial and final

00:16:08

impulse the final impulse p2 is

00:16:12

directed to the right p2 the

00:16:21

initial impulse is directed downward let's

00:16:25

do this the initial impulse is directed

00:16:29

downward because in one is directed downward

00:16:32

b2 vector b one vector we need to find

00:16:40

the impulse of the force

00:16:42

this is the difference of these two vectors how do you

00:16:45

clean I built two vectors

00:16:47

specially from one point

00:16:49

to connect the end of these

00:16:52

vectors with each other and where the vector of the force impulse will be directed is where they

00:16:58

take it from p2

00:17:00

here this is the vector f by t the module of

00:17:06

this vector as Dima suggested, we

00:17:09

can find it using the Pythagorean theorem,

00:17:11

which means we can write down that

00:17:17

ft is equal to the square root of the sum of the

00:17:25

squares of the legs of n 1 square plus d2

00:17:29

quadra

00:17:30

that is, ft is

00:17:34

equal to the square root of p 1

00:17:39

square 20 squared + p 2 square 10

00:17:47

squared

00:17:49

approximately equals now

00:17:51

let's see how much so now let's check the

00:17:56

square root of 20 squared plus

00:17:59

10 squared equals but thanks for that

00:18:03

it’s more interesting to us so twenty

00:18:05

two point four tenths so I

00:18:08

’ll write twenty two point four tenths on the board in

00:18:11

parentheses we’ll indicate the units of measurement in which

00:18:16

units the impulse of force is measured

00:18:19

you can write newton multiplied by a

00:18:22

second it’s clear why but you can write

00:18:26

in such units as

00:18:28

the impulse is measured, this will be correct,

00:18:30

kilogram per meter

00:18:32

divided by second, so we enter the result in the table:

00:18:35

22 and 422 and 4 kilograms

00:18:45

kilogram per meter per second, so what

00:18:51

to do next and how to complete the list of

00:18:58

questions, please announce the entire list

00:19:00

and so I I I’ll write and say that you also need to

00:19:04

find the force module, the

00:19:06

projection of force on the direction of movement, the acceleration

00:19:08

module, the time of movement, the

00:19:11

acceleration project on the x axis, the angle between the

00:19:13

accelerations on the y axis and the angle between the force and the

00:19:16

displacement vector, the projection of force on the

00:19:21

x axis,

00:19:22

and since then you will need to look for the

00:19:24

acceleration module, then probably it will be necessary to

00:19:27

find the projection of the force on the y axis.

00:19:31

The proposal is well accepted. It’s not for nothing that I left the

00:19:35

projections of the displacement of the body on the board and so

00:19:42

we know about the movement of the body, we know the

00:19:46

projections of the displacement of the body and we know the

00:19:49

initial and final speed,

00:19:51

we want to find the acceleration for this, what is the

00:19:55

formula for us a formula for the

00:19:57

projection of displacement that does not include

00:20:00

time is useful, this is the so-called 3 formula as

00:20:03

we call it among ourselves, namely with x

00:20:06

equals the fraction at x square minus at 0

00:20:14

x square divided by 2a x

00:20:17

and similarly we can write the formula

00:20:19

for c y with y equals

00:20:23

y squared minus 0 and y squared by 2a

00:20:28

y now we need to find these

00:20:33

values you x squared at 0 x squared you

00:20:36

igres at 0 y if we know them

00:20:39

we can find the projection of acceleration

00:20:41

from here we express what x equals it

00:20:45

will be we know x squared minus

00:20:48

0 and x squared 2s x asics we

00:20:53

know and y equals y squared

00:20:59

minus 0 y squared 2s y so now

00:21:05

look at the drawing and please tell me

00:21:09

what it equals x this is the final value of the

00:21:18

velocity projection on the x axis the projection on the

00:21:21

x axis of the final speed is the final

00:21:23

speed projection on the x axis how many

00:21:25

meters per second 5 meters per second 5

00:21:31

meters per second the projection of the same

00:21:34

speed on the y axis projection of the final

00:21:38

speed on the y axis look at the figure

00:21:40

vector v 2 goes parallel axis x the

00:21:43

projection onto the y axis is equal to zero,

00:21:46

we go further to 0 and x the

00:21:50

projection of the initial velocity on the x axis the

00:21:54

initial velocity v1 the projection and and is equal to

00:21:57

zero at 0 y at 0 y equals 0 this is

00:22:06

one for us, how many is wrong -10 because the vector

00:22:14

is directed against the axis y minus 10 meters per

00:22:19

second, well, however, an error in the sign

00:22:21

will not affect the answer since there are squares here, well,

00:22:24

let's count

00:22:25

and x is equal to the fraction in the denominator 2 with x

00:22:31

two by 22 by 20 meters in the numerator there is a

00:22:37

difference of squares you x square 25

00:22:43

meters squared per second squared

00:22:45

minus

00:22:46

0 and x this will be 0 equals

00:22:52

25 forties so 25 forties is well

00:22:58

divisible 25 divided by 40 will be 5 8

00:23:03

holina 620 50 625 meters per second

00:23:10

squared this is the projection onto the x axis

00:23:13

it can be immediately entered into the table and so

00:23:17

0 625 meters second squared is ready,

00:23:26

well, now we’ll be consistent,

00:23:29

we’ll find the projection onto the y axis, the projection of

00:23:32

acceleration onto the y axis is equal to the fraction

00:23:35

denominator 2 multiplied by minus 40 minus

00:23:42

40 meters in the numerator in y square

00:23:48

0 we don’t write anything in 0

00:23:53

y square is -10 punish this table 100

00:23:57

meters squared per second squared

00:24:00

equals minus 100 minus 100 correct

00:24:06

minus 100 equals how many 100 eighths

00:24:11

1.25

00:24:14

1.25 meters per second squared

00:24:18

for us this information is so

00:24:20

useful that it makes sense here at the

00:24:23

bottom of our drawing of this save it in the table

00:24:28

and so x equals 0 625

00:24:36

meters second squared I duplicate

00:24:39

and y is greater than

00:24:44

625 I understand what I say and duplicate it but

00:24:47

let them be nearby know if you want to

00:24:50

have information about the vector it is useful to

00:24:52

keep it all in one place so a y

00:25:00

1 .25 per second squared, so

00:25:04

what next can we do: the acceleration module is like

00:25:08

any vector,

00:25:09

and the acceleration vector has a module

00:25:12

that can be calculated using the formula

00:25:15

following from the Pythagorean theorem: a

00:25:18

equals the square root of a x

00:25:23

square plus a y square,

00:25:25

that is, a equals the square root from

00:25:29

0 625 squared plus 125 squared

00:25:39

meters per second squared,

00:25:43

take a calculator and calculate this is the root

00:25:48

of 0

00:25:51

625 squared plus 125 squared

00:25:56

equals, convert the decimal fraction 1

00:26:02

397 with kopecks, round up and please

00:26:06

correctly to three significant figures 140 140 is

00:26:14

simply 397 here so and so this is the

00:26:18

acceleration module is one, 40 meters per second

00:26:24

squared

00:26:26

the next thing you can do right away is

00:26:30

the force find we know the acceleration

00:26:35

we know the mass according to Newton’s second law the

00:26:42

force the force module is equal to the product of

00:26:45

mass and acceleration or just immediately on the

00:26:49

calculator this is the number that we

00:26:59

underline that the fact that this is not 139 or

00:27:03

141, we guarantee that these are three

00:27:06

significant figures and the 3rd significant figure

00:27:08

is 0,

00:27:09

there can also be any number 12,970,

00:27:14

the same equal, so now we find

00:27:17

the force, right here the

00:27:22

acceleration module we have 1,397 even not

00:27:25

rounded so that there are no rounding errors

00:27:27

let's multiply this by mass

00:27:29

multiply by 2 equals 2,795

00:27:37

again since we are rounding to

00:27:39

three significant figures we will write

00:27:43

280 and so the force is the modulus of the applied force

00:27:47

two, 80 of course newton so well

00:27:54

what else you can find out the time the time of

00:28:00

movement how to do this the moment is not

00:28:07

the moment of force but

00:28:08

the impulse of force we really know the

00:28:11

product ft

00:28:14

here it is 20 24 and we know the modulus

00:28:18

means one of the ways to find

00:28:21

time is to divide the impulse of force by

00:28:24

time hi tack time p is equal to ft

00:28:29

divided by f like this it’s a bit of an

00:28:33

unusual formula, but that’s simply because

00:28:40

we don’t have a special letter to

00:28:42

denote the

00:28:44

impulse of force, well, let’s divide means

00:28:46

that we divide

00:28:48

ft 22 and 4 divide by 2 and 8 by 2 and 8

00:28:58

it turns out 8 seconds exactly 8 before means

00:29:04

somewhere we write here here 8 seconds

00:29:08

okay guys and some other

00:29:10

way can you suggest

00:29:15

with x equals 0 t plus and xp squared in

00:29:20

half good and simpler

00:29:22

and even simpler not through the average through

00:29:27

when what is acceleration is the

00:29:31

ratio of the change in speed to the time

00:29:33

during which it occurred look we

00:29:36

have initial velocities and final

00:29:40

velocities, and we also have acceleration, I

00:29:46

won’t do this on the board now,

00:29:48

but since we know the projection of the initial

00:29:50

and final velocities, they are already simply

00:29:52

erased and the projections of acceleration, then according to the formula in x,

00:29:58

for example, equals 0 and x plus and xt

00:30:02

we know all the values aix you 0 xv

00:30:06

from here we can find t similarly for in y

00:30:10

equals 0

00:30:12

y + a y t you can and from this formula

00:30:16

by the way you can check the author of the problems

00:30:18

whether he set the problem correctly in all

00:30:21

three cases when calculating using this formula

00:30:23

according to this formula and according to this formula, the result should be

00:30:26

the same time. When I composed this task, I

00:30:29

had to specially select a

00:30:31

condition so that all this would be performed, otherwise the

00:30:34

task would be incorrect, well, let’s see that there is

00:30:37

still time left, we know the projection of the

00:30:45

force on the direction of movement,

00:30:47

but how do we we can do this through the loss of

00:30:52

kinetic energy, we

00:30:54

have already used it, well, let's

00:30:56

really look at this question in more detail,

00:31:00

we know work, work can be calculated

00:31:06

not only by the formula f on ios by cosine

00:31:08

alpha,

00:31:09

but we also know this formula f work

00:31:13

a is equal to the projection of force on the direction

00:31:18

of movement there is no movement per module,

00:31:21

here is the displacement vector, for example, here is the

00:31:26

force vector f,

00:31:27

and here is the projection of the force onto the direction of

00:31:33

displacement,

00:31:35

we know the work, we know, we know the displacement module,

00:31:42

which means we can write that fs, hence the

00:31:45

projection of the force on the direction of displacement

00:31:48

is equal to the ratio of work

00:31:51

to the displacement module, using this formula we

00:31:55

we can find the work we know the displacement

00:31:57

we know from the tables we take and divide and so we

00:32:01

need to divide -75 joules we write minus

00:32:09

seventy-five divided by module c module

00:32:14

c we have in the first line 44 744,

00:32:20

7 we get 1 -160 68 minus

00:32:29

168 this projection of force on direction of

00:32:33

movement minus 168 what units

00:32:40

of newton is ready so now we can

00:32:46

also find the angle between the force vector and the

00:32:52

displacement vector, this angle is

00:32:55

designated by the letter b, then

00:33:02

look at beta fs divisor by f and that is,

00:33:07

cosine beta cosine tetta equals ef

00:33:13

ef divided by f well then we need to

00:33:19

find this ratio and take the arc cosine.

00:33:22

By the way, you remember that we are waiting for the angle

00:33:25

to be obtuse,

00:33:27

let’s try whether we can do it or not

00:33:30

using a calculator, and so we need to

00:33:33

find the angle

00:33:34

whose cosine is equal to the ratio of

00:33:37

these two quantities,

00:33:38

write shift cosine further the relation fs we have it

00:33:45

equal to minus 168 minus 1 minus

00:33:53

168 divided by the modulus of force the modulus of forces

00:33:58

we have 282 and 8 we closed the bracket we got

00:34:06

126 with the fraction 127 degrees beta

00:34:12

127 degrees we really got an

00:34:15

obtuse angle, we enter this value angle

00:34:19

beta one

00:34:20

hundred twenty-seven degrees but what All that

00:34:24

remains is to find the angle between the

00:34:27

acceleration vector of the body and the y axis,

00:34:32

how can you do this through arctangent

00:34:38

arctangent is a good thing, but it has

00:34:41

one drawback: using tangent you

00:34:43

lose information about the direction, that is,

00:34:46

you get a direction in this

00:34:49

direction and in this direction will be

00:34:51

described by one and with the same tangent,

00:34:53

but through sine or cosine this can be

00:34:55

done more correctly, let’s then

00:35:00

count both sine and cosine, but the fact is

00:35:03

that we have already figured out something,

00:35:06

namely, we found out that the acceleration vector

00:35:08

has positive projections, both of them

00:35:12

by the way and then you can also use the tangent,

00:35:14

well, in this

00:35:17

case,

00:35:18

we really know where the acceleration vector is directed,

00:35:20

it is directed, I will show in

00:35:25

Figure 1 its projections, both are positive, I

00:35:28

will start to be directed something like

00:35:30

this, then we can say that the tangent of the

00:35:33

angle between

00:35:35

in the projection between acceleration and the y axis,

00:35:40

now we will figure it out, here is the y axis and here is the

00:35:45

x axis

00:35:46

and here is the acceleration vector, which means we

00:35:54

need to find the angle between the

00:35:57

acceleration vector and the alpha axis

00:35:59

then we have this segment aix this

00:36:03

segment we have a y and then we can

00:36:07

say that tangent alpha is equal

00:36:12

to help

00:36:14

a x divide by y alex

00:36:19

divide by a y count tangent alpha

00:36:26

means we are looking for arctangent shift

00:36:29

tangent x and y I have already written them down

00:36:34

here we have them stored here so aix

00:36:38

we have 0 625 divided by 125 the

00:36:46

bracket is closed equals 26 with a fraction

00:36:49

approximately 27 degrees we write 27

00:36:55

degrees connection and we have three significant figures

00:37:02

and do those Linda agree Roma twenty-

00:37:05

six and six degrees

00:37:07

well, by the way, then you can

00:37:09

also write 00 next to the eight for

00:37:11

correctness and here you can add

00:37:15

0 but these are minor little things

00:37:18

that are evaluated they won’t, but if you

00:37:21

overdo it with indicating the number of

00:37:23

significant digits, draw everything that the

00:37:25

calculator depicted on the scale, then this will

00:37:30

be considered an error since the table is filled,

00:37:33

but we need to return to the first task;

00:37:37

we have already managed to draw the displacement module,

00:37:40

but we still need to show the vector of forces

00:37:46

acting on the body of Vitya, draw the

00:37:48

vector of displacement of the body and the force

00:37:50

applied to the body, how to do this,

00:37:53

in fact, we know the modulus of the force and we

00:37:56

know the angle that the force vector forms

00:37:59

with the displacement vector, but we don’t have non-

00:38:02

dimensional instruments, what then should we do

00:38:05

through the projections? It’s not for nothing that we wrote down the

00:38:08

acceleration projection

00:38:11

on the coordinate axes now, according to

00:38:13

Newton’s second law, if we multiply these projections of

00:38:15

acceleration by mass, then we find out

00:38:18

the projection of force; the mass is equal to two

00:38:20

kilograms, which means then fx

00:38:25

equals 2 kilograms multiplied by 0 625

00:38:31

meters per second squared, it turns out

00:38:34

how much is 1.3 newtons

00:38:38

now on the y and y axis equals 2

00:38:42

kilograms per 1.25 meters per second

00:38:47

squared will be two and a half newtons and

00:38:51

so the projection on the x axis we have one and three

00:38:55

on the y axis 2 and a half in the task it is said by the

00:39:01

point of application of the force,

00:39:03

select point two, then the force vector

00:39:06

will be displayed from here next

00:39:09

the scale will be 1 newton, this is 4

00:39:12

cells, which means one cell is a

00:39:14

quarter of a newton, one and three tenths for

00:39:19

125, I'm sorry, up to 125,

00:39:22

this is 5 cells, which means from this point we put 5

00:39:26

cells to the right one two three

00:39:29

4 5 we get exactly to the end of the vector v

00:39:32

2 and vertically we have two and a half,

00:39:35

this will be how many 10 cells one two

00:39:39

three 4 5 6 7 8 9 10

00:39:42

that means so and so and now we

00:39:47

draw a vector, I’ll even take a red

00:39:49

pen draw a force vector from point two like

00:39:56

this

00:40:00

this will be a force vector f and at the same time let's

00:40:08

show two angles

00:40:11

that asked to show the angle between the

00:40:14

acceleration vector of the body and the y axis where

00:40:18

the acceleration is directed, the force is also directed there, which

00:40:20

means the acceleration vector is directed here,

00:40:24

we show the angle with a dotted line

00:40:30

parallel to the y axis, here is the angle alpha

00:40:35

and the angle between the force vector and the displacement vector

00:40:41

guys angles are always measured between

00:40:44

positive directions, so

00:40:46

you see we have

00:40:48

a vector with directed like this, which means we

00:40:51

continue it, here is the positive direction, here is

00:40:56

the direction of the force, and here is this

00:41:00

angle, this will be the angle between the direction of the

00:41:03

force and the direction of movement, this

00:41:07

will be the angle beta,

00:41:08

the task is completely completed

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