Why is work defined as $W=Fd$?

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 $W=mat$ )?

$\begingroup$ It would be nice, if you could use Mathjax and define the symbols you are using. Regarding your question, please think of a string and consider here $W=F\cdot d$ and $F\cdot t$. Alternatively, think of carrying a heavy weight or putting the wait onto a rolling wagon. Does the wagon really perform work on the weight? $\endgroup$

$\begingroup$ A definition can’t explain anything. Definitions are just definitions. But a physical relationship between two defined quantities (say, work and kinetic energy) can. $\endgroup$

$\begingroup$ @G.Smith, agreed. Novice physics students tend to want to justify new physics definitions with a math proof, rather than just memorize the definition and use it. There is a misconception that physics is math, when in reality, mathematical models are used to represent physical observations of the real world. This is subtle for a novice, but it means that physics concepts drive the math, rather than math driving the physics. $\endgroup$

$\begingroup$ @Semoi . regarding your question, does the wagon really perform work on the weight? You probably meant, does the weight of an object in a wagon perform work on the object moving horizontally? The answer is no, regardless of whether the wagon was accelerating horizontally or moving horizontally at a constant speed. $\endgroup$

4 Answers 4

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 $$W_\text=\Delta K$$ 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, $$\mathbf F=ma=m\frac=\frac$$ 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.

$^*$ 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 $\text dW=\mathbf F\cdot\text d\mathbf x$ 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.