Confused (1 Viewer)

[azureus88]

A particle is moving vertically downward in a medium which exerts a resistance to the motion whithc is proportional to the speed of the particle. The particle is released from rest at O and at time t its position is at a distance x below O and its speed is v.

isnt the equation of motion v(dv/dx)=g-(kv/m)

but the only way the question works out is when the equation is v(dv/dx)=g-kv

Can someone please explain?

 

[Drongoski]

is 'm' the mass?

when something 'a' is proportional to 'v' say, it simply means:

where 'k' is simply the 'constant of proportionality'.

in the case of yr example: acceleration = v(dv/dx) = g - kv where positive is taken to be downwards. equally valid you could have used: v(dv/dx) = g + kv in which case this k would simply have an opp sign to the earlier version. In the conventional approach, u use 'g - kv' where k is deemed positive and the '-' reflect that the resistance is going against the direction of 'g'.

anyway you should not have had the 'm' in yr formula.

 

[azureus88]

The question says "resistance is proportional to velocity". This implies R=kv. If we choose downwards as positive then F=mg-kv=ma (F being total force and "a" being acceleration)

rearranging the equation, v(dv/dx)=a=g-(kv/m)

Isnt that right?

yeh, "m" is mass by the way.

 

[Drongoski]

yours is not quite wrong; it simply means 'my' k is now yr (k/m) and this is just (albeit a composite one) another constant of proportionality. You should still be able to solve the problem. The way we usually formulate the soln here is like this:

a = g - kv

Therefore: ma = mg - mkv

rather than your way, although it is not incorrect.

and u don't need to use the force version to solve yr problem; i.e. simply use a = v(dv/dx) = g - kv

Got it ?

 

[jet]

Normally, what i do is start with yours, so v(dv/dt) = g - (kv/m)

Then, to make it easier, i say let k/m = K, but there should be an m in there im quite sure. Yours should be right azureus. Where was the question from?

 

[azureus88]

yeh ok i get it. Not sure where the question is from, it was just from a worksheet i happen to have.