Is Work a Scalar Quantity Despite Being Composed of Vectors?

In summary, work is considered a scalar quantity because it is defined as the dot product of force and displacement, which always results in a scalar value. While it includes direction, it is not considered important in the calculation of work. The sign of work can be important, as it represents the change in internal energy. However, in cases of pushing a block in opposite directions, the work remains the same regardless of the direction, as long as the force and displacement values are the same.
  • #1
Xidike
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Why Work Is Scalar Quantity ??

I'm wondering that Why is work a scalar quantity ? Since it is the product of the force and displacement which are both vector?
 
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  • #2


The dot product of two vectors gives a scalar value.

Say F= <F1,F2,F3> and d=<d1,d2,d3>

then F.d = F1d1+F2d2+F3d3 which is purely a scalar.

Cross-product on the other hand gives a vector.
 
  • #3


Also naturaly we don't want work to be a vector as the direction of work is not important.
 
  • #4


yea exactly, pushing a block to the left and performing work on it is the same as pushing it to the right
 
  • #5


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 ??
 

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  • #7


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

Xidike said:
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 [itex]u\cdot v= |u||v|cos(\theta)[/itex] where |u| and |v| are the lengths of the vectors and [itex]\theta[/itex] 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"?
 
  • #8


In my mind work should be vector quantity, because it includes direction...
and what's about my figured question answer ??
 
  • #9


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


Xidike said:
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?
 
  • #11


Xidike said:
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.
 
  • #12


Xidike said:
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.
 
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  • #13


mitch_1211 said:
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.
 
  • #14


The formula for work done is W=F.D Cosθ
Why we use cosθ ? and why not sine ??
 
  • #15


Darwin123 said:
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?

Xidike said:
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.
 
  • #16


HallsofIvy said:
There is no such thing as the "dot product of two scalars". The dot product of two vectors is defined as [itex]u\cdot v= |u||v|cos(\theta)[/itex] where |u| and |v| are the lengths of the vectors and [itex]\theta[/itex] is the angle between them.

Halls, i thought that the definition of the dot product of two vectors of the same length, [itex]N[/itex]

[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 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 don't know 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


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|)] !
 

1. What is a scalar quantity?

A scalar quantity is a type of measurement that has only magnitude (size or quantity) and no direction. Examples of scalar quantities include mass, time, and temperature.

2. How is work defined as a scalar quantity?

Work is defined as a scalar quantity because it only considers the magnitude of the force applied to an object and the displacement of the object in the same direction as the force. The direction of the force does not affect the amount of work done.

3. Why is work not a vector quantity?

Work is not a vector quantity because it does not take into account the direction of the force applied. In order for a quantity to be considered a vector, it must have both magnitude and direction.

4. Can work ever be a vector quantity?

No, work can never be a vector quantity because it only considers the magnitude of the force and the displacement and ignores direction. However, force, which is closely related to work, is a vector quantity because it has both magnitude and direction.

5. How is the scalar quantity of work calculated?

The scalar quantity of work is calculated by multiplying the magnitude of the applied force by the distance over which the force is applied in the direction of the force. This can be represented by the equation W = Fd, where W is work, F is force, and d is distance.

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