# B Would it be correct to think of force, work and power as...?

1. Oct 29, 2016

### Ehden

Of course, work is the change in force with respect to position, and power is the change in work with respect to time, while or both velocity and acceleration change with respect to time. However, is it reasonable to think of force, work and power in a similar fashion as position, velocity and acceleration?

2. Oct 29, 2016

### Simon Bridge

The "work is the change in force with respect to position" would be written $W=\Delta F/\Delta x$ ... does this look like the equation of work to you?

$P=\Delta W/\Delta t$ ... average power but ok.
You are thinking that velocity is change in position over time, so if force is analagous to a position, then you are thinking that work is analagous to velocity and power is analagous to acceleration?
This does not work for two main reasons:
1. They do not have the kind of analogy you have asserted.
You need to show that work is to force what velocity is to position. Let's see:
Velocity is defined: $\vec v = d\vec r/dt$ where $\vec r$ is the position vector.
By comparison, $W=-\int \vec F\cdot d\vec r$ ... ie, velocity is a vector, work is not; velocity involves a time derivative, work does not.
[the closest you can get is: $W=-\int \vec F\cdot \vec v\; dt$ ]
2. They are physically quite different things. This is quite important - the similarity involved is a deepity: insofar as it is true, it is trivial; insofar as it is deep, it is not true. The reason we have these concepts is so we can understand Nature better - "they both use a time derivative" (in case you want to make the analogy for other relations) does not help with that.

Notes: the equation for work says that work is the area under the force-displacement graph
It is better to think of work as a change in energy - like when: $\vec F=-\vec\nabla U$

Last edited: Oct 29, 2016
3. Oct 29, 2016

### Ehden

I feel like an idiot...

I forgot that Work is the integral of Force with change in position, not the derivative of Force with respect to position. Hence I thought it would be analogous to think of force, work and power as position, velocity and acceleration despite the different derivatives.

I asked this before, but, I've had trouble trying to find intuitive sense for integral. Derivatives seems much more intuitive though, such as how velocity is a slope, mps, a change in position within an interval of time.

4. Oct 29, 2016

### Simon Bridge

An integral is a sum when the parts of the sum are very small.
It is used to find areas when the shape is bounded by a smooth curve.

So the acceleration is the slope of the velocity-time graph, while the displacement is the area under the graph inside the time interval of interest.
Similarly, the slope of the acceleration time graph is called the jerk, and the change in velocity is the area under the acceleration time graph in the time interval of interest.

The derivative is kinda easier to wrap the mind around - even though it is actually the slope of the curve at a single point (you normally need two points to get a slope).

Note: even if you had some quantitiy (B) that was a derivative of something else (A) with position ... say B=dA/dx, then you had C=dB/dt (this is what you had, incorrectly, with A=force, and B=work etc - but lets say there is something that these relations do work for... even if that was the case:)
You still cannot say that B is to A what v is to x ... since v=dx/dt and B is not dA/dt.
You can say that B=(dA/dx)(dx/dt)/v = (dA/dt)/v ... which should make it clear why B is probably a different thing from velocity (the time rate of change of A is, B multiplied by velocity)

Bottom line is: you cannot do physics by analogy.

5. Oct 29, 2016

### Simon Bridge

If you have a curve described by y=f(x) and you want to find the area above the x axis and below the curve, between lines x=a and x=b, with a<b, ...
Then ... you will notice that the area of the little box at position (x,y) with small width dx and small height dy has a small area dA equals base times height dA=dxdy
... all we need to do is add up the area of all these boxes in the regeon.
You can see that $\int \;dA = A$ - that was easy - those symbols just mean that the sum of all the little areas in the region is the area of the whole region: make sense?

But you want A in terms of the x and y constraints of the problem.
That means we have to add up all the dxdy things ...
At position x, the dxdy boxes are stacked from y=0 to y=f(x) ... so we add those up: $\int_0^{f(x)}\;dy\;dx = f(x)\;dx$ ... that is the area of a stack of boxes width dx and height f(x) at position x.
This area will be different for different x's - so we have to add them up from x=a to x=b like this: $A=\int_a^bf(x)\; dx$ ... look familiar?