Mass falling in Earth's Gravity help

[kraigandrews]

Homework Statement

A particle falls in Earth's gravity. The force of air resistance is F = −αv v where v is the velocity, v is the speed, and α is a constant. (The direction of the force is opposite to the velocity, and the magnitude of the force is αv^2.) If the terminal speed is 20.2 m/s and the mass is 2.6 kg, determine the constant α.

Homework Equations

F=mdv/dt

The Attempt at a Solution

m(dv/dt)=mg-aplhav
thus v(t)=(mg/aplha)(1-e^(-(alpha/m)t))
so aplha if I am correct should be (mg/Vterminal)=1.263, however I am not sure of the units for this. My guess would be kg/s however my homework programs says that is incorrect, and also I feel I may have neglected the [/B]v[/B] and the magnitude of the opposing force causing a wrong answer.
If someone could shed some light on this it is greatly appreciated.

 

[rock.freak667]

From your formula it seems that you need to use

F=-αv² so you'd have

ma = mg - αv²

 

[kraigandrews]

If that's the case then what does the F=-αv mean? To me, the question is pretty confusing as far as what goes where.

 

[gneill]

At terminal velocity the velocity is constant. So the acceleration must be nil. What can you say about the forces acting in that case?

 

[kraigandrews]

it would be dv/dt=a=0 thus the net forces are balanced, I guess my confusion is what is the question asking for as the resistive force F=-αv or F=-αv^2?

 

[gneill]

it would be dv/dt=a=0 thus the net forces are balanced, I guess my confusion is what is the question asking for as the resistive force F=-αv or F=-αv^2?

Well, I'd go with the magnitude of the resistive force and make the appropriate assumption about its direction for a body falling straight down

I do find the notation vv a bit odd. I mean, why not v with a unit vector for direction? Maybe it's just me...

 

[Dick]

The magnitude of F = −αv v, where the second v is a vector is -av^2. And it points in a direction opposite to the vector v if a is a positive constant.

 

[Dick]

Well, I'd go with the magnitude of the resistive force and make the appropriate assumption about its direction for a body falling straight down

I do find the notation vv a bit odd. I mean, why not v with a unit vector for direction? Maybe it's just me...

I think the problem statement says the magnitude is av^2. So I don't think they mean the vector v to be a unit. As you said.

 

[gneill]

I think the problem statement says the magnitude is av^2. So I don't think they mean the vector v to be a unit.

Hi Dick, That was my point; it wasn't a (magnitude)*(unit-vector) construct, but rather a magnitude*vector (vv as opposed to v²u).

On reflection, I can see the utility of doing it this way when the magnitude you want is the square of the vector. vv rather than v²*(v/v).

 

[Dick]

Hi Dick, That was my point; it wasn't a (magnitude)*(unit-vector) construct, but rather a magnitude*vector (vv as opposed to v²u).

On reflection, I can see the utility of doing it this way when the magnitude you want is the square of the vector. vv rather than v²*(v/v).

Hi gneill, yeah, I realized what you meant after I posted. Nothing wrong with the unit vector notation either. At least it makes the velocity dependence clearer than it was to kraigandrews.