# Why Work Is Scalar Quantity ??

by Xidike
Tags: quantity, scalar, work
 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?
 HW Helper P: 6,214 The dot product of two vectors gives a scalar value. Say F= and d= then F.d = F1d1+F2d2+F3d3 which is purely a scalar. Cross-product on the other hand gives a vector.
 P: 197 Also naturaly we don't want work to be a vector as the direction of work is not important.
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
 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 ??????? Attached Thumbnails
 P: 125 The long-winded answer is that energy is the time component of the momenergy vector. You should expect one component of a 4-vector to act as a scalar. Now if you want me to connect all that to the 3+1 pre-relativity view of nature, you will have to give me time to sleep on it. http://books.google.com/books?id=PDA...heeler&f=false
Math
Emeritus
Thanks
PF Gold
P: 38,895
 Quote by Xidike I'm wondering that Why is work a scalar quantity ??? Since it is the product of the force and displacement which are both vector?
 Quote by Xidike 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 ???????
There is no such thing as the "dot product of two scalars". The dot product of two vectors is defined as $u\cdot v= |u||v|cos(\theta)$ where |u| and |v| are the lengths of the vectors and $\theta$ is the angle between them.

Since you are asking why "work" is a scalar and not a vector, what is your understanding of the definition of "work"?
 P: 72 In my mind work should be vector quantity, because it includes direction... and what's about my figured question answer ??
 P: 1,205 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.
Math
Emeritus
Thanks
PF Gold
P: 38,895
 Quote by Xidike In my mind work should be vector quantity, because it includes direction... and what's about my figured question answer ??
Again, what definition of "work" are you using?
Math
Emeritus
Thanks
PF Gold
P: 38,895
 Quote by Xidike In the following figure is the work done is Same from A to C, and B to C
Assuming a constant force, then, no it won't be. The distance from A to C is greater than the distance from B to C.

 if it is same then how it is same AND,,,,,, the Dot product of two scalar always give a scalar quantity ???????
You seem to be using words, like "work" and "dot product" without knowing their definitions.
P: 197
 Quote by Xidike In my mind work should be vector quantity, because it includes direction... and what's about my figured question answer ??
Hmm..you seem to be confused by vectors. The point is, although work might have direction, we just think it's not important and don't take it into account. Therefore, work is scalar. For every vector we can decide not to take its direction into account, only its magnitude and proclaim it scalar.
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.
P: 741
 Quote by mitch_1211 yea exactly, pushing a block to the left and performing work on it is the same as pushing it to the right
This is not really a good analogy. It implies that the sign of a scalar is never important. The sign of scalars are important. In the case of work, the work represents the change in internal energy of whatever system is doing the work. Positive work represents gaining energy and negative work represents losing internal energy.

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.
 P: 72 The formula for work done is W=F.D Cosθ Why we use cosθ ??? and why not sine ??
P: 1,205
 Quote by Darwin123 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.
You're telling me speed or length are not scalars because they cannot be negative?

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?

 Quote by Xidike The formula for work done is W=F.D Cosθ Why we use cosθ ??? and why not sine ??
Because using sine would give you a quantity that is not equal to the energy expended in the movement.
P: 2,265
 Quote by HallsofIvy There is no such thing as the "dot product of two scalars". The dot product of two vectors is defined as $u\cdot v= |u||v|cos(\theta)$ where |u| and |v| are the lengths of the vectors and $\theta$ is the angle between them.
Halls, i thought that the definition of the dot product of two vectors of the same length, $N$

$$\vec{u} = [ u_1 \ u_2 \ u_3 \ \ldots u_N ] \$$

$$\vec{v} = [ v_1 \ v_2 \ v_3 \ \ldots v_N ] \$$

is

$$\vec{u} \cdot \vec{v} \ \triangleq \ \sum_{n=1}^{N} u_n v_n$$

and from that definition you can show that (in 3 dimensions)

$$\vec{u} \cdot \vec{v} \ = \ | \vec{u}| \ | \vec{v} | \ \cos(\theta) \$$.

but it ain't all that easy in 3-dimensional 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 2-dimensional.

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.
 P: 1,205 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/(|u||v|)] !

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