Question

Find the work done in moving an object along a vector r = 3i + 2j - 5k if applied force is F = 2I - J - K

Answer 1

\[W=\int\vec F\cdot d\vec s\]so it's just the integral of the dot product

Answer 2

\[W=\int\vec F\cdot d\vec r\]

Answer 3

but there are no bounds on your integral, so perhaps they just want the dot product...?

Answer 4

yeah... the question does not have bounds it is just plane like that

Answer 5

maybe you should post this in the physics group (though I usually would help you in that dept. too)
I'm just not sure about the meaning of the question since there is no actual path, just a direction

Answer 6

ohkay i will do that now

Answer 7

I'm starting to think they just want the dot product

Answer 8

because r has a length, so it can be interpreted as a line segment
so just do\[W=\vec F\cdot\vec r\]

Answer 9

...at least that's what I think

Answer 10

ohkay... because the 1st equation you gave out is for the conservative force field

Answer 11

?

Answer 12

the first equation I gave is for the work done by a force function acting on a parametric displacement function

Answer 13

the force does not need to be conservative for that equation to hold

Answer 14

it's actually a line integral\[W=\int\limits_C\vec F\cdot d\vec s\]so it doesn't have to be conservative

Answer 15

if\[\oint\vec F\cdot d\vec s=0\]then \(\vec F\) is conservative

Answer 16

yes i agree

Answer 17

but like I say, I don't even think your question wants you to do an integral... as far as I can see

Answer 18

yep cause it does not have bounds

Answer 19

\[|\overrightarrow F|=\sqrt{(2)^2+(-1)^2+(-1)^2}=\sqrt{6}.\]\[|\overrightarrow r|=\sqrt{(3)^2+(2)^2+(-5)^2}=\sqrt{38}.\]\[W=F*r*\cos \theta\]But I don't know what is theta?

Answer 20

so where will i get the angle from

Answer 21

I think Dot product of vector will come into application.
Only i think.

Answer 22

you can just do it with the dot product method

Answer 23

\[\vec v=\langle a,b,c\rangle\]\[\vec u=\langle x,y,z\rangle\]\[\vec u\cdot\vec v=ax+by+cz\]

Answer 24

that is the definition of the euclidean inner product in \(\mathbb R^3\)

Answer 25

ohk i will try with the dot product

Answer 26

another definition*

Answer 27

@TuringTest I wanna ask a question.
Plz help:)

Answer 28

I can try, but please post separately
I will help if I know how of course :)

Answer 29

which is true?
1.) \[W=\overrightarrow F* \overrightarrow S* \cos \theta\]
2.) \[W=F*S* \cos \theta.\]

Answer 30

not sure what the second means exactly, but if F and S are magnitudes then 2) is true

Answer 31

@TuringTest must derivate the vector r ?

Answer 32

no, it is a line integral

Answer 33

...normally I mean

Answer 34

the formula is force times distance equals work basically

Answer 35

since r is a distance no derivative required

Answer 36

so i use the normak r given out ?

Answer 37

Isn't it like this "Work is dot product of force & displacement"?
\[W=\overrightarrow F . \overrightarrow S\]\[\implies W= |\overrightarrow F| \times |\overrightarrow S| \times \cos \theta= F*S* \cos \theta\]
???????????????????/

Answer 38

@TuringTest Plz:)

Answer 39

if it is a straight line, yes
but if the path curves you have to parameterize it in terms of time and that method will not work\[\vec u\cdot\vec v=\|\vec u\|\|\vec v\|\cos\theta_{uv}\]

Answer 40

w = f . r
=9 am i right?

Answer 41

that's the formal definition statement^

Answer 42

oh i see?
@kay-jay what value did u take as theta?

Answer 43

theta?

Answer 44

@kay-jay I get 9 as well

Answer 45

you do not need theta @maheshmeghwal9 there are two ways to define our dot product

Answer 46

am not using the dot product that needs an angle

Answer 47

the formula\[\vec v\cdot\vec u=\|\vec v\|\|\vec u\|\cos\theta\] is more limited than the formula I prescribed above; it only works when the vectors are constant and do not change with time
in this case we could use this formula as well, but I forget how to find the angle...
gotta look at my notes

Answer 48

@TuringTest so you think our answer might be true ?

Answer 49

Oh i see what did u both do?
That was an extraordinary shot^
Thanx for nice solution @Turingtest
:)

Answer 50

It must be true:)

Answer 51

our answer is true if I have interpreted the problem correctly
I am saying that the problem states that the given force F is acting along a vector r, and that the magnitude of vector r represents distance that it travels
if I am right about that interpretation I think that we have the right answer

Answer 52

yeah Answer depends on Interpretation.

Answer 53

PS: the only way I know to find the angle between them is to use the other formula for the dot product\[\vec u\cdot\vec v=ax+by+cz\]so it would be redundant to try to find the angle

Answer 54

am writing on monday at 14h00 so i want to represent this group with good marks. @TuringTest you are really helping... thank you and thank you also @maheshmeghwal9

Answer 55

yw:D
Best of luck for ur representation^_^
@kay-jay

Answer 56

you could say\[\vec u\cdot\vec v=\|\vec u\|\|\vec v\|\cos\theta\implies\theta=\cos^{-1}{\vec u\cdot\vec v\over\|\vec u\|\|\vec v\|}\]but see how you still have to use\[\vec u\cdot\vec v=ax+by+cz\]to find the angle
so like I say, that method is redundant at best

Answer 57

@TuringTest the one you mension now is for three dimensional

Answer 58

yes

Answer 59

Wow seems more interesting:D

Answer 60

mathematics is interesting but challenging

Answer 61

^_^

Answer 62

I think one attribute contributes to the other :)

Answer 63

yeah i think too:D