Why do we need the s-2 when describing force/energyetc?

[Femme_physics]

For force, for instance, why can't we just use m x kg without the s-2?

 

[bp_psy]

This question is very weird and my attempt of an answer will fail miserably. Force as it is defined and understood is something that is at the base of the concept of motion (the main subject of pretty much all physics). Motion is only meaningful if you have the notion of time. So there is no quantity that is equal to distance*mass because it is not really useful for describing anything about motion. If that quantity would be very useful than it would get some name.

 

[Pythagorean]

Because force and energy are associated with motion (especially if we want to do anything interesting with them) which requires a change in the system. Seconds are a unit of change.

Of course, realize that no motion is a special case of motion in this context (we don't exclude v = 0 for instance, on a graph that we'd plot motion on). And in the case of energy, there is potential energy which describes a potential for motion.

 

[Femme_physics]

Right, so there has to be a defined scale of time in order for it to work, I see. Does it mean if the same m x kg is applied at 5 seconds as opposed to 1 second, the force would be lesser.

Correct?

 

[Pythagorean]

Right, so there has to be a defined scale of time in order for it to work, I see. Does it mean if the same m x kg is applied at 5 seconds as opposed to 1 second, the force would be lesser.

Correct?

Well, it's good thinking, but there's other things to consider. Consider the momentum formulation of force:

F = dp/dt = m*(dv/dt) = m*(dx^2/dt^2)

which requires calculus to understand in full. If you haven't seen any calculus yet, we'll look at

F = m*(dv/dt)

the mass (m) multiplied by the change in velocity (dv) over the change in time (dt). To treat dv/dt like division is incorrect in general, but I'm trying to demonstrate it algebraicly.

we can change this to

F*dt = dv

So if you replace dt with 1 sec or 5 sec, you could still have the same F if you also changed dv to make it match.

But for a fixed dv, yes your statement would be true.

The way to imagine this is a chunk of clay splatting against the wall. It has some velocity, v when it hits the wall. Over the next couple milliseconds (or whatever dt is), it slows down to 0. So it's change in velocity, dv = v-0 = v. F tells you the force that the clay and the wall imparted on each other throughout the event.

 

[Pythagorean]

oh, and if you're wondering where the other s went, it's implicit in the v:

v = dx/dt (velocity is the change in position with respect to time)