resisted motion (1 Viewer)

[cutemouse]

Hello,

I was doing Mechanics and found some in resisted motion.

When the question gives the resistance force as kv, some books use mkv in their calculations.

Coroneos kinda explained it, if the resistance force is proportional to velocity, then you can let the resistance force be R(v)=Kv, where K=mk, so R(v) becomes R(v)=mkv... Coroneos says this makes calculations easier (undoubtably it does), but you can't change the question like that, can you?

Terry Lee on the other hand, uses R(v)=kv...

Which method is correct?

Thanks

 

[study-freak]

It's different for different textbooks (even different questions in one textbook).
In cases those textbooks make things unclear, bad luck.. Just guess or deduce from the answer... (Cambridge is the only book that clearly states what they want I think.)

But exam questions should (not guarenteed) make it clear.
I think it's usually F=mkv and a=kv.

Not so sure though.

 

[cutemouse]

Yeah, but if you could let K=kv, then that makes life easier. Would this method be acceptable in an exam?

 

[study-freak]

Well, if exam papers don't state it clearly, just go with the simplest method. If the teacher insists it is wrong, then argue about the ambiguity of the question.

 

[cutemouse]

Hmm, I wanna second opinion. Anyone?

 

[tommykins]

It is simply notation.

I can do F = ma - msv if I wanted to, but it's just convention to make it ma - mkv

the more important factor is keeping the sign convention correctly.

 

[Timothy.Siu]

if a=-kv of course u can let k=mc where k,c and m are constants

 

[cutemouse]

if a=-kv of course u can let k=mc where k,c and m are constants

Yeah but, some people may just use kv, instead of letting k=Km... and this will yield different answers.

So..?

I mean, I could just let k=0 and then solve the expression that way. Obviously that's wrong. So that's what I'm asking about here.