# Where is the element of time in the definition of work?

In $\text{Work} = m \times a \times \text{distance}$.

From this equation, How can we know the duration of the force that have been applied in order to move the object over that distance?

• If I buy 3 pizzas for 8 dollars each, I pay 24 dollars. More general, the price you pay for n pizzas is n_pizza * 8_dollars = total_prize. Where is the element of time here? E.g. how long do I need to work to have the money for one pizza? It does not appear at all! Time does not have to be a part of every equation ;) Some relations answer just different questions, independent of time. And, as here, leave it free to other factors (e.g. you can earn a lot/month or a few/month, in both cases you can buy the pizzas... it is independent/another question) – Mayou36 Oct 22 '17 at 9:06
• @Mayou36 That is not an equivalent analogy, because there really isn't a time factor there, but acceleration does have it. – physicsnewbie Oct 22 '17 at 9:15
• Acceleration does not have a time factor either. It just has time in it's unit. But I understand your confusion. Think of the pizza example again and add a "per month" to every unit. Like: 3 pizzas per month, 24 dollars per months, how much do I need to work per month... and all the relations stay the same. Or basically, I have multiplied both sides of the equation with the same magnitude (or rather, divided). But, the "per month" has nothing to do with how long I have to work, although it's time as well. Same goes for the $t^{-2}$ in the acceleration. – Mayou36 Oct 22 '17 at 9:52
• Whether an Acceleration is constant or variable, it still describes a change over time, and that's what i am missing here, because force has acceleration as a factor. – physicsnewbie Oct 22 '17 at 10:06
• So following your argumentation, energy also describes a change over time, right? Because energy has the $t^{-2}$ in it as well. See, it's getting tricky here. The simple answer is the one given above. Read it through again, you will see that just because time occurs in the unit does not mean there is a direct interpretation of how much time is needed for... The other thing is a more general question about units. And this requires some more in-depth understanding with Noethers theorem or, the simple observation, that energy just is conserved. And has time in it's unit. – Mayou36 Oct 22 '17 at 10:13

In the simplest case of a straight line movement with a constant force starting from a zero speed, from energy conservation

$$W=\dfrac{mv^2}{2}=\dfrac{m(2v_a)^2}{2}=2m\left(\dfrac{d}{t}\right)^2$$

Where $v_a$ is the average speed. The solution for time follows

$$t=\sqrt{\dfrac{2m}{W}}\cdot d$$

• Why did you write that W equals the Kinetic Energy equation ? – physicsnewbie Oct 21 '17 at 23:44
• Yes, according to the energy conservation law, all work in this case goes to increase the kinetic energy. – safesphere Oct 21 '17 at 23:47

First of all: in general you cannot substitute $F_r=ma$ into $W=Fd$, the reason being that the work equation can be applied for any force, while Newton's 2nd law refers to the resultant force $F_r$. Also, if the force $F$ isn't constant, and we are in 3D, then work is calculated as $W=\int \vec{F}\cdot d\vec{x}$.

As for where's time in $W=Fd$, there's the time it takes for the distance $d$ to be covered, but, yes, time isn't really playing a role in this equation and, in general:

• you cannot know the duration of the interaction from the work of the force alone.

Time does appear in the related concept of power (time rate of work: $P_\text{avg}=W/\Delta t$). So you can calculate $\Delta t$, given total $W$ and $P_\text{avg}$.

• I see, but how can work be calculated if we don't know the duration of the force? isn't that a false value ? – physicsnewbie Oct 21 '17 at 23:38
• No, it's not false, it's consistent with the rest of the theory, e.g., with the Energy-work relation. Once you've gone trough at least the very basic of mechanics, things should become clearer. – stafusa Oct 22 '17 at 1:41

Work is defined as $W=\int \vec F\cdot d\vec \ell$. It may be that the force is time-dependent so let's write $\vec F=\vec F(t)$. You can then introduce time explicitly by writing $$d\vec \ell =\frac{d\vec \ell}{dt}dt = \vec v(t) dt\, .$$ In this case you simply end up with $$W=\int \vec F(t)\cdot \vec v(t) dt\, .\tag{1}$$ This last expression can be useful in some cases to show that a force does no work when it is perpendicular to $\vec v(t)$. Thus the magnetic force $\vec F_{mag}= q\vec v\times \vec B$ is automatically perpendicular to $\vec v$ by the properties of the cross product, and can be seen to do no work as $$\left(q\vec v\times \vec B\right)\cdot \vec v \equiv 0\, .$$

• I am a little confused, isn't a force always time dependent ? – physicsnewbie Oct 22 '17 at 0:08
• @physicsnewbie how about the force of gravity near the Earth: constant and equal in magnitude to $g$. – ZeroTheHero Oct 22 '17 at 0:09
• But when we are not moving or falling, wouldn't that mean there is no work done on our bodies from the force of gravity? – physicsnewbie Oct 22 '17 at 0:12
• @physicsnewbie sorry first that would be $mg$, not $g$. If you are not moving $\vec v=0$ then there is no displacement so no work is done. – ZeroTheHero Oct 22 '17 at 0:16
• @Noumenon $d\vec \ell$ is a directed line element, tangent at one point along the path to the path connecting the initial and final position vectors which are the bounds in the integral defining the work. It is sometimes denoted $d\vec x$ or (as in this wiki page: en.m.wikipedia.org/wiki/Line_element) $d\vec s$. – ZeroTheHero Oct 22 '17 at 19:32