Why is work defined as $W=Fd$?

Question

I am trying to understand what work really means in physics. I seem to be missing the conceptual link. Every resource says that $W=Fd$ but that does not make sense to me.

If, say, an elastic object suspended in space where there is no drag or resisting force of any kind is pushed by a force of a certain magnitude, then it will accelerate. The amount of 'useful' energy spent would completely go into accelerating this body of a particular mass for as long as the force is applied.

First of all, why isn't work $W = mat$ (which is the equation of momentum) for some time $t$.

Why is work $W = mas$ (for a force acting in the same direction of the motion) for some displacement $s$.

Since momentum and energy are both conserved, could it have been that it was a matter of convention how these two quantities were defined? (I.e why wasn't work defined as $W=mat$)?

Answer 1

The first thing you need to understand: you are applying the creation of physics definitions backwards. You are asking, "why isn't work given by this equation?", but this question doesn't make sense if you think about it. It is not the case in physics where we think, "Hmm... I want to define something called "work". What should it's equation be?" This doesn't make sense, as the only use an equation has in physics is how useful it is in describing the world around us. So it is fine to ask "how is the concept of work that is defined in this way useful?", but a question of "why isn't work defined to be this instead?" is not a valid question.

So why is this definition of work useful? There are many reasons, but the simplest case, and the one where it is usually first introduced to students, is that the net work done on an object is equal to the change in the object's kinetic energy, which is another useful concept that has been defined in physics. In an equation, this is $$W_\text{net}=\Delta K$$ where $K$ is the kinetic energy $K=\frac12mv^2$ for an object with mass $m$ moving with a speed of $v$. A more general derivation can be made for variable acceleration in multiple dimensions, but as an introduction you can use your constant acceleration equations to arrive at this for when $W=Fd$ is valid.$^*$ You will probably encounter more instances where the concepts of "energy" and "work" become useful, but essentially whenever we are talking about changes in some type of energy work must be involved. The definition of work then is essential when thinking about energy.

What about your other expression? As described above, there is no reason to call this "work" over what we already have, but it can still be useful. $mat$ is, in fact, a change in momentum. In general for constant mass systems, $$\mathbf F=ma=m\frac{\text d\mathbf v}{\text dt}=\frac{\text d\mathbf p}{\text dt}$$ i.e. forces cause changes in momentum over time. For constant acceleration motion, we arrive at your expression: $F\Delta t=ma\Delta t=\Delta p$. This has a special name: impulse. So just like how work deals with a force being applied over a distance, impulse deals with a force being applied over some time period. They are both useful concepts, but to ask why one is called "work" over the other isn't a question worth asking here.

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$^*$Just use your kinematic equation $v^2-v_0^2=2ad$ and Newton's second law $F=ma$ to easily show that $W=\Delta K$ for motion under constant acceleration in one direction.

It is worth noting that the equation $W=Fd$ is only valid when the force is constant in magnitude and along the direction of displacement during the entire path of interest. The more general definition of work takes this idea and breaks up a general path into small pieces where $W=Fd$ applies. Then all of the "little works" are "added" together in an integral $$W=\int_C\mathbf F\cdot\text d\mathbf x$$ where $C$ is the path we are adding up the work on, $\mathbf F$ is the force we are interested in, and $\text d\mathbf x$ is the very small displacement for one of the "little works" we are adding up.

Answer 2

The concepts of work and the associated conservation of energy were found to be useful (and hence formalized) for solving certain types of problems in mechanics. Over time it was discovered that they also gave consistent results in a wide variety of situations. So much so that many of us have come to think of energy as something which exists in the physical world. The definition of work is consistent with your experience of work. If your car breaks down and you have to push it to a service station, the bigger it is, and the farther you have to go, the more work you will do. While you are pushing the car, the road exerts a normal force on the car, but you don't think of the road as doing work. Similarly, the concept of momentum was found to be useful in collisions which involved equal and opposite forces acting for the same length of time.

Answer 3

In physics, generally, first we make observation of a phenomena then we try to define different physical quantities that are involved there and with these quantities we try to qualitatively as well as quantitatively explain the phenomenon.

But for quantitative explanation of the phenomenon we first need to relate the different quantities that we have defined so as to make proper use of them. And so is the case here.

We first define two quantities force and energy. Force acts as method to transfer energy from one system to another. (Like transport vehicles transfer commodities across different places). (As for energy, it is a number (with units) that we define for a system which we find useful in solving the dynamics of the problem)

So the question arises how do we relate energy and forces?

The answers is by defining a quantity called work

Work is the amount of energy transferred to or from a system by a force.

Now from this definition it is clear that if a system initially has energy $E_i$ and reaches a state with energy $E_f$ then the work done($W$) is given by:

$$W = E_f- E_i = \Delta E$$

Now how do we relate work to force (and in due process to energy)?

The answer is that we observed that when a constant force is $\mathbf F$ acts on an object for a displacement $\mathbf d$ then the work done is:

$$W = \mathbf F \cdot \mathbf d$$

Which when generalized for a variable force looks like:

$$ W = \int \mathbf F(x) \cdot \mathbf dx$$

It is important to note that work is not dependent on time the way you might think. I have shown this in my answer here.

Now via this definition of work we are able to relate two important quantities energy and force and this is the reason we define it that way.

Answer 4

I am trying to understand what work really means in physics. I seem to be missing the conceptual link.

The short answer is work in physics is a means of energy transfer between objects that is the result of the action of a net force on an object through a distance. A differential amount of work is $dW=F.ds$. (The other means of energy transfer is heat, which results from a temperature difference between objects). The units of work (and heat) are energy units (N.m; J). Your equation $W=mat$ is for momentum, with units (N.s; kg.m/s$^2$).

An important distinction between momentum and work is that momentum is a property of a system that is conserved, whereas work is not a property of a system. The energy that the work transfers is a property of the systems undergoing the transfer, and is overall conserved.

Hope this helps.