Is work a vector quantity in physics?

[Amik]

Homework Statement

In the solution

Relevant Equations

W=FD

I am so confused.If F and d are both vector quantity.How come W is a scalar quantity?

 

[Orodruin]

Because the inner product of two vectors is a scalar.

 

[Amik]

But how come this is negative

 

[Orodruin]

Because the force acts in the opposite direction of the displacement. The inner product between two vectors that are pointing in opposite directions is negative.

 

[Amik]

I might be stupid here.I think scalar quantity can only be positive?and work is scalar quantity?

 

[Orodruin]

I think scalar quantity can only be positive?

No, this is incorrect.

Why do you have this impression?

 

[Amik]

The picture I sent you in the first picture said this

 

[Orodruin]

The picture I sent you in the first picture said this

It is a very bad quality picture. Please take the habit of writing out relevant quotes.

 

[Amik]

I am sorry.The last line of the first picture said work is a scalar quantity.

 

[Orodruin]

I am sorry.The last line of the first picture said work is a scalar quantity.

Sorry, but this is not writing out the relevant quote. Writing out the quote means citing it word by word. What you wrote is your interpretation.

In terms of the issue, your interpretation is not what the author intended. There is nothing preventing a general scalar from taking a negative value.

 

[Amik]

"note that work, unlike momentum. is a scalar quantity----It has a magnitude but not a direction".
I think scalar quzntity can only be positive.Because for example, speed is a scalar quantity and it has no direction.And speed can only be positive.

 

[Orodruin]

I think scalar quzntity can only be positive.

You thinking so does not make it so. You are simply wrong about this. Correcting this misconception is fundamental for your future understanding.

Because for example, speed is a scalar quantity and it has no direction.And speed can only be positive.

So what? Particular scalar quantities may fall into some restricted range because of their definitions. This does not mean that all scalars do. For example, work is a scalar quantity that may be negative. Electric potential is a scalar quantity that may be negative. There are many other examples.

 

[Amik]

I understand it now.Thanks!

 

[Amik]

It is like temperature it is scalar quantity but it can either positive or negative.

 

[Orodruin]

It is like temperature it is scalar quantity but it can either positive or negative.

If measured in °C or °F, yes. If measured in K (as is more typical in physics), it is positive.

 

[ehild]

I think scalar quantity can only be positive?and work is scalar quantity?

A scalar can be both positive and negative. The numbers are scalars, and "2" is positive, "-5" is negative.
The magnitude or absolute value is positive or zero.
Work, the scalar product of force and displacement that is, the magnitude of force multiplied by the magnitude of displacement and multiplied by the cosine of the angle enclosed by the force and displacement, can be positive, negative or zero. according to the angle.
Speed is the magnitude of the velocity vector, it can not be negative.

 

[haruspex]

A scalar can be both positive and negative. The numbers are scalars, and "2" is positive, "-5" is negative.
The magnitude or absolute value is positive or zero.

You are overlooking this part of the quoted text

"... It has a magnitude but not a direction"

This is what has misled Amik. It should say that scalars have a sign and a magnitude but not a direction.
@Amik, what textbook is this?

 

[lightlightsup]

Let's suppose that a 10kg ball is launched straight upwards and it has traveled straight up for 10 meters.
What is the work done by gravity on this ball, so far?

$W = \vec{F} \cdot \vec{d}$
$W_{done \ by \ a \ force \ on \ an \ object} = \vec{F}_{on \ the \ object} \cdot \vec{d}_{by \ the \ object \ while \ the \ force \ is \ acting \ on \ it}$
$W_{g} = \vec{F}_{g} \cdot \vec{d}$
$W_{g} = m\vec{g} \cdot \vec{d}$
$\vec{g} = (9.8)[\cos{(270^{\circ})}\hat{i} + \sin{(270^{\circ})}\hat{j}](\frac{m}{s^2})$
$m\vec{g} = (10)(9.8)[\cos{(270^{\circ})}\hat{i} + \sin{(270^{\circ})}\hat{j}](kg)(\frac{m}{s^2})$
$\vec{d} = (10)[\cos{(90^{\circ})}\hat{i} + \sin{(90^{\circ})}\hat{j}](m)$
$W_{g}= [0\hat{i} + -98\hat{j}](kg\frac{m}{s^2}) \cdot [0\hat{i} + 10\hat{j}](m)$
$W_{g} = - 980 \ kg\frac{m^2}{s^2} = -980 \ Nm = -980 \ J$
$\vec{F}_{g}$ completed $-980J$ of work relative to the displacement vector ($\vec{d}$). Herein, the negative sign mathematically symbolizes that the force ($\vec{F}_{g}$) was acting opposite to the direction of displacement. The force ($\vec{F}_{g}$) took away $980J$ of energy from the ball.

Hope this helps.
Someone, please correct me if I'm wrong.
I mentioned all the little details to show you what an elegant system this is.

 

[collinsmark]

Also, just to clear up something about the notation your book uses, and how that applies to your first equation, (10.5).

It states:
$W = Fd$.
But for this to be accurate, it is essential that you read the line below it:
"Work done by a constant force $\vec F$ in the direction of a displacement $\vec d$." Note the part about "in the direction."

It's telling you there that the equation only applies if $\vec F$ and $\vec d$ are in the same direction. If they are in the same direction, then all you need to do is multiply the magnitudes $F$ and $d$.

It should be noted that $W$, $F$, and $d$ here are all scalar quantities. And, at the risk of being redundant, that the vector quantities $\vec F$ and $\vec d$ have to be in the same direction if the $W = Fd$ equation is to apply.

Another way of looking at this is to start with the more general equation,

$W = \vec F \cdot \vec d = Fd \cos \theta,$

and since it tells you that $\vec F$ and $\vec d$ are in the same direction, then $\theta = 0$, and the equation simplifies to $W = Fd$.

---

Also, generally speaking, scalar quantities don't have a direction, but they can be positive or negative.

 

[archaic]

This is a matter of notation. Some textbooks would use a bold character to denote a vector and a normal character to denote scalars. Hence, the force vector would be $\mathbf F$ and its magnitude (which is a scalar quantity) would be $F$ or $\mathrm F$. Other textbooks, however, might denote vectors like this $\vec F$ and their magnitude as $||\vec F||$, $F$ or $\mathrm F$.
From the text before the equation you have, it is clear that your textbook uses the second notation.

 

[haruspex]

This is a matter of notation. Some textbooks would use a bold character to denote a vector and a normal character to denote scalars. Hence, the force vector would be $\mathbf F$ and its magnitude (which is a scalar quantity) would be $F$ or $\mathrm F$. Other textbooks, however, might denote vectors like this $\vec F$ and their magnitude as $||\vec F||$, $F$ or $\mathrm F$.
From the text before the equation you have, it is clear that your textbook uses the second notation.

As I noted in post #17, the root of the error seems to be in the book's statement that scalars have only magnitude. I have since found this to be almost ubiquitous on the web (even NASA) and taken it upon myself to contact the site managers where possible. So far I have had two responses; both thanked me and promised correction.

Perhaps some readers with relevant textbooks would take the time to check for this error and notify the publishers.

 

[kuruman]

I think that part of the blame for the confusion could be assigned to those (myself included) who cut corners when they know what they mean but beginners do not. Here is why. We express a vector, say a velocity, generally as $\vec v=v_x~\hat x+v_y~\hat y+v_z~\hat z$ where the components $v_i$ are scalar, algebraic quantities, i.e. they can be positive or negative. That's because the direction is denoted by the unit vectors, which are assumed strictly positive else we would run into trouble with things like $\hat x \times \hat y$. So far so good.

Now consider how one would normally say with an equation that "the velocity of the object is 3 m/s to the left." Formally we would substitute the values in the general expression to get $\vec v=(-3~\mathrm{m/s}) ~\hat x$. Most of us eschew the use of unit vectors in one dimension, so we take the shortcut, drop the arrow over the velocity symbol on the left and the unit vector on the right to write $v=-3~\mathrm{m/s}$ as the velocity of the object. Note that this is a vector equation because it is intended to express a velocity although there is nothing explicit to denote that fact. Suppose now someone said to you the velocity of an object is $v=-3~\mathrm{m/s}$. Question: is $-3~\mathrm{m/s}$ a scalar quantity? You would probably answer, "No, the negative sign indicates the direction, which is to the left. The scalar quantity is just 3 m/s without the direction." This is the correct answer because it would be wrong to say that a velocity is a scalar. It should be obvious then that by taking the shortcut of stripping the direction when the vector is expressed in one dimension, the quantity $(-3~\mathrm{m/s})$ that started out as a scalar is no longer a scalar.

 

[haruspex]

You would probably answer, "No, the negative sign indicates the direction, which is to the left. The scalar quantity is just 3 m/s without the direction

Not at all. The -3 is the scalar coefficient of the unit vector.

 

[PeroK]

Suppose now someone said to you the velocity of an object is $v=-3~\mathrm{m/s}$. Question: is $-3~\mathrm{m/s}$ a scalar quantity?

Mathematically, the Real (and Complex) numbers are both vectors and scalars.

If you constrain motion to one dimension, you restrict vector quantities like velocity and make them scalar-like.

 

[kuruman]

Not at all. The -3 is the scalar coefficient of the unit vector.

Yes it is. I am attempting to explain the origin of the misconception. When one writes "The velocity of the object is v = -3 m/s", the "-3 m/s" is the coefficient of a unit vector which is not explicitly shown. An expert understands how to interpret that. A beginner might think that the negative sign is actually the unit vector because after all it indicates direction "to the left". A scalar has no direction therefore to get the scalar that multiplies direction in this vector expression, one has to strip the negative sign.

If you constrain motion to one dimension, you restrict vector quantities like velocity and make them scalar-like.

I don't see it that way. There is nothing scalar-like in $\vec v=(-3~\mathrm{m/s}) ~\hat x$. It's the removal of the symbols denoting vector quantities (for convenience?) that makes the expression scalar-like.

 

[vela]

I don't see it that way. There is nothing scalar-like in $\vec v=(-3~\mathrm{m/s}) ~\hat x$. It's the removal of the symbols denoting vector quantities (for convenience?) that makes the expression scalar-like.

It is scalar-like because that vector only requires one number to specify it.

The problem seems to be the common description that scalars don't have direction, yet your example is a clear counterexample. Some scalars, like scalar components of a vector, can be interpreted as representing a direction while others, like mass and charge, can't. What scalar quantities have in common is that they can all be described by single numbers.

 

[haruspex]

Real (and Complex) numbers are both vectors and scalars

Hmm.. well, you can construct a 1D vector space from the reals (or complex numbers, or any field) in a trivial way, but it does involve a construction. It does not mean the reals are a vector space.

 

[haruspex]

The problem seems to be the common description that scalars don't have direction

I would say the problem is the common description that scalars only have magnitude, hence are nonnegative. Whether having a sign is equivalent to having direction is debatable. E.g. charge has sign, but does it have direction?

 

[PeroK]

Hmm.. well, you can construct a 1D vector space from the reals (or complex numbers, or any field) in a trivial way, but it does involve a construction. It does not mean the reals are a vector space.

$\mathbb{R}^n$ is a vector space for all $n$, including $n = 1$. No "construction" is needed. They satisfy all the axioms of a vector space, with $\mathbb{R}$ as the field of scalars.

 

[haruspex]

$\mathbb{R}^n$ is a vector space for all $n$, including $n = 1$. No "construction" is needed. They satisfy all the axioms of a vector space, with $\mathbb{R}$ as the field of scalars.

Strictly speaking, $\mathbb{R}^n$ is just the set of n-tuples {(x₁,..,x_n)}. To make it a vector space you have to specify two sets, the set of vectors and the field of scalars, then define how the vector elements add and what the product of a scalar and a vector is.
That we commonly think of $\mathbb{R}^n$ as the natural vector space $\{\mathbb{R}^n, \mathbb{R}\}$ is just a convenient shorthand.
Is there only one way to do that with $\mathbb{R}^n$ as a set of vectors over the field $\mathbb{R}$?
What if I take the n-tuples $\mathbb{C}^n$ over the field $\mathbb{R}$? Doesn't that make a vector space in the obvious way?

 

[Orodruin]

Strictly speaking, $\mathbb{R}^n$ is just the set of n-tuples {(x₁,..,x_n)}. To make it a vector space you have to specify two sets, the set of vectors and the field of scalars, then define how the vector elements add and what the product of a scalar and a vector is.
That we commonly think of $\mathbb{R}^n$ as the natural vector space $\{\mathbb{R}^n, \mathbb{R}\}$ is just a convenient shorthand.
Is there only one way to do that with $\mathbb{R}^n$ as a set of vectors over the field $\mathbb{R}$?
What if I take the n-tuples $\mathbb{C}^n$ over the field $\mathbb{R}$? Doesn't that make a vector space in the obvious way?

The original discussion was regarding $\mathbb R$ itself, not $\mathbb R^n$. Any field is a vector space over itself without any additional structure as it already has addition and multiplication by scalars. I therefore do not think it makes sense to claim that $\mathbb R$ is not a vector space - it has all of the required properties. That does not mean it cannot also be made into a vector space over a different field, but in itself it has all of the required properties to be called a vector space.

 

[PeroK]

Strictly speaking, $\mathbb{R}^n$ is just the set of n-tuples {(x₁,..,x_n)}.

What is $\mathbb{R}$ for that matter? Strictly speaking it has no properties until you define it by some sort of construction. It's a set of equivalence classes of Cauchy sequences of rationals if you want to start at the beginning.

There's no reason arbitrarily to decide that $\mathbb{R}$ needs "construction" specifically to be a vector space, but no construction to be a field of scalars.

 

[haruspex]

What is $\mathbb{R}$ for that matter? Strictly speaking it has no properties until you define it by some sort of construction. It's a set of equivalence classes of Cauchy sequences of rationals if you want to start at the beginning.

There's no reason arbitrarily to decide that $\mathbb{R}$ needs "construction" specifically to be a vector space, but no construction to be a field of scalars.

$\mathbb{R}$ is defined (as I understand it) not just as a set but all the operations and axioms necessary to be a field. I accept that one could also define something else with the same name (a pun) which is a vector space constructed with the elements of the set underlying $\mathbb{R}$ as its vectors, over the field $\mathbb{R}$, with all the addition and multiplication rules defined in the obvious way.
But it is an interesting question as to whether there is a nonobvious construction that makes a different vector space.

 

[jbriggs444]

$\mathbb{R}$ is defined (as I understand it) not just as a set but all the operations and axioms necessary to be a field. I accept that one could also define something else with the same name (a pun) which is a vector space constructed with the elements of the set underlying $\mathbb{R}$ as its vectors, over the field $\mathbb{R}$, with all the addition and multiplication rules defined in the obvious way.
But it is an interesting question as to whether there is a nonobvious construction that makes a different vector space.

So the idea is that one is using R as the set of elements of the vector space and R as the set of scalars.
With ordinary real addition for the addition of two vectors and ordinary real multiplication for the product of a vector and a scalar.

Given that, I do not see that there is any freedom to define anything non-obvious. Everything is already nailed down.

 

[haruspex]

With ordinary real addition for the addition of two vectors and ordinary real multiplication for the product of a vector and a scalar.

Those are choices that nail it down.
My question is whether you could define the vector addition and scalar x vector multiplication differently and still arrive at a structure satisfying vector space axioms. Even if you can, is the resulting vector space isomorphic to the usual one?