Why is work defined with respect to distance rather than time? [duplicate]

Question

The common way of finding the work done on some object is by applying the equation: force*displacement.

However, suppose we apply a force of F newtons on an object of mass M for a duration of T seconds. We can then express the work done in terms of T and F, by expressing the change in velocity in terms F and T and then by applying the work-energy theorem.

If we do some further substitutions (Express displacement in terms of F, initial velocity and T), we can show that the above 2 methods produce the same result for a given force (according to what I calculated, do correct me if I am wrong).

Why is the equation force*displacement used instead of the latter case? To me, calculating the work done with respect to time is more intuitive because the increase in momentum of an object (under a constant force) is the same per unit of time. On the other hand, the increase in momentum per unit of displacement of an object gets progressively smaller due to the acceleration of the object.

Answer 1

Suppose you know that a constant force $F$ is applied for a distance $d$ then you know the work done is equal to $Fd$ - no further information is needed.

But suppose I tell you that a force $F$ is applied for a time $t$, how are you going to calculate the work done? The problem is that the work done depends on the velocity of the object because it is given by:

$$ W = \int_0^t F~v(t')~dt' $$

So you need to know the velocity $v(t)$ as a function of time. So for example for a free mass you need to know the initial velocity at time zero and you need the mass of the object so you can calculate its acceleration and hence how the velocity changes with time. If the object is not free, e.g. if it's sliding on a rough surface, you now also need all the details of the frictional and drag forces.

Of course this calculation can be done, but it turns a very simple calculation into a potentially very complicated one. That doesn't mean we would never consider the motion as a function of time. For example the force could be time dependent or the mass could be time dependent (this happens in rocket flight as fuel mass decreases). But if there's a simple way to do a calculation you should always choose the simple way, and learning how to spot the simplest way is an important skill for physicists.

Answer 2

The ${\bf F} \cdot {\bf l }$ definition can be recast in the form of ${\bf F} \cdot {\bf v } dt$ just parametrizing the work differential form with time. But this rewriting is a mathematically equivalent way of expressing the same thing.

The key point to realize is that, whatever form one is choosing, the path used for evaluating the work and the concept of work have no direct relation with the actual trajectory corresponding to the real motion under the force F alone. This is particularly evident with the first form because there in the ${\bf F} \cdot {\bf l }$ form there is no room for the initial velocities.

The difference between evaluating the work along the real trajectory or along an arbitrary path traveled with arbitrary speed can be appreciated as soon more than one force are applied to the same body. In such a case, the power of the usual definition of work clearly emerges. On the one side, one can evaluate the individual contribution to the work of each force, even though trajectory and velocity are not the same which would be present if only one of the forces would be acting. On the other side, in the case of conservative forces, one can use additional forces to constraint the body to move over the most convenient path for evaluating work (and then potential energy).