
#1
Nov1312, 10:12 AM

P: 72

I'm wondering that Why is work a scalar quantity ??? Since it is the product of the force and displacement which are both vector?




#2
Nov1312, 11:00 AM

HW Helper
P: 6,212

The dot product of two vectors gives a scalar value.
Say F= <F_{1},F_{2},F_{3}> and d=<d_{1},d_{2},d_{3}> then F.d = F_{1}d_{1}+F_{2}d_{2}+F_{3}d_{3} which is purely a scalar. Crossproduct on the other hand gives a vector. 



#3
Nov1312, 11:41 AM

P: 197

Also naturaly we don't want work to be a vector as the direction of work is not important.




#4
Nov1312, 02:41 PM

P: 99

Why Work Is Scalar Quantity ??
yea exactly, pushing a block to the left and performing work on it is the same as pushing it to the right




#5
Nov1312, 11:55 PM

P: 72

In the following figure is the work done is Same from A to C, and B to C
if it is same then how it is same AND,,,,,, the Dot product of two scalar always give a scalar quantity ??????? 



#6
Nov1412, 12:01 AM

P: 125

The longwinded answer is that energy is the time component of the momenergy vector. You should expect one component of a 4vector to act as a scalar. Now if you want me to connect all that to the 3+1 prerelativity view of nature, you will have to give me time to sleep on it.
http://books.google.com/books?id=PDA...heeler&f=false 



#7
Nov1412, 07:26 AM

Math
Emeritus
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PF Gold
P: 38,879

Since you are asking why "work" is a scalar and not a vector, what is your understanding of the definition of "work"? 



#8
Nov1412, 09:00 AM

P: 72

In my mind work should be vector quantity, because it includes direction...
and what's about my figured question answer ?? 



#9
Nov1412, 09:07 AM

P: 1,199

Put it this way. Your mind associates work with direction because for any work to be done, there must be a directed force causing a directional displacement.
But the amount of work done itself has nothing to do with direction. If I push a block 10m east with a force of 10N eastwards, then the work done is 100 J. If I push the block the same distance south with a 10N southwards force, the work done is also 100 J. The work I have done in pushing the block in two different directions is identical, because the energy I must expend has nothing to do with my orientation in any absolute coordinates, it only matters on the angle between the force and the displacement vectors. If you're looking to associate a direction with something then you need to look for a different quantity, because work has no direction. your figured question has insufficient information, since you don't mention anything about the direction of the force. 



#10
Nov1412, 09:39 AM

Math
Emeritus
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PF Gold
P: 38,879





#11
Nov1412, 09:45 AM

Math
Emeritus
Sci Advisor
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PF Gold
P: 38,879





#12
Nov1412, 10:15 AM

P: 197

Work is defined to be a dot product (scalar product) of force and displacement. The dot product (or scalar product) of two vectors is always a scalar. In your picture, there are no forces acting on a body shown, so we cannot tell which path takes more work. You must tell us about the forces. 



#13
Nov1412, 01:01 PM

P: 741

Consider a block initially moving leftward. Suppose a spring is located to the right of the block. If the spring pushes the block, the force of the spring is leftward. The work is positive. The block speeds up, gaining kinetic energy. The spring loses tension, losing potential energy. If the spring pulls the block, the force of the spring is rightward. The work is negative. The block slows down, losing kinetic energy. The spring gets more tense, gaining potential energy. Technically, a scalar has to transform like a vector. In other words, the sign of the scalar is important. The are single numbers that aren't scalars. The speed of an object is not a true scalar. Speed is the magnitude of the velocity. Speed is never negative. So technically, it is not a scalar. Similarly, length is also a magnitude of displacement. It is never negative, either. I think that what you mean is that quantities are the sign is unimportant to quantities which vary with the magnitude of a scalar. The magnitude of a scalar is different from the scalar. One should not confuse the magnitude of a scalar for the scalar. The scalar varies with sign. The magnitude has one and only one unvarying sign. 



#14
Dec1212, 02:29 AM

P: 72

The formula for work done is W=F.D Cosθ
Why we use cosθ ??? and why not sine ?? 



#15
Dec1212, 06:14 AM

P: 1,199

The fact that a scalar is restricted to a certain part of the real line does not make it any less of a scalar. Where are you getting these (false) definitions from? 



#16
Dec1212, 10:20 AM

P: 2,265

[tex] \vec{u} = [ u_1 \ u_2 \ u_3 \ \ldots u_N ] \ [/tex] [tex] \vec{v} = [ v_1 \ v_2 \ v_3 \ \ldots v_N ] \ [/tex] is [tex] \vec{u} \cdot \vec{v} \ \triangleq \ \sum_{n=1}^{N} u_n v_n [/tex] and from that definition you can show that (in 3 dimensions) [tex] \vec{u} \cdot \vec{v} \ = \  \vec{u} \  \vec{v}  \ \cos(\theta) \ [/tex]. but it ain't all that easy in 3dimensional space. if i remember right, you need to, in 3 dimension, get a handle on the plane that is in common to to two vectors (if they are not colinear), then to get a handle on that angle. it's not too hard if it's 2dimensional. but that's what i think is the general definition of the dot product. i dunno how to answer the OPs question other than to say that the flow or motion of energy might have direction, similar to the flow of charge or matter. and there is a vector for that (Poynting or intensity) but the quantity of energy is just what it is. 



#17
Dec1412, 04:27 AM

P: 1,199

In any number of dimensions of Euclidean space, two vectors still have an angle between them, because the angle is defined as arccos[u.v/(uv)] !



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