Question

I found the v1,then what? what is the relationship between 'K' and mass.
A coffee filter, dropped from a height of 2.23 m, reaches the ground in a time of 3.23 s. When two more coffee filters are nestled within the first, the drag force remains the same, but the weight is tripled. Find the time for the combined 3 filters to reach the ground. (Neglect the brief period when the filters are accelerating up to their terminal speed.)

Answer 1

Hi! What equations do you have for air drag?

Answer 2

Also, what level class is this for?

Answer 3

I think the relationship you mention could be in this:

The speed, like \(v_1\), is the terminal velocity (we ignore the time that it takes to speed up to terminal velocity).

It is the velocity that makes the force of air drag, probably involving your \(K\), equal to the weight of the coffee filter so that there is
no net force
\(\qquad\rightarrow\) which means that there is no acceleration
\(\qquad\qquad\rightarrow\) which means that there is no change in velocity
which means the coffee filter is at terminal velocity.

So, the forces must be opposite - gravity and air drag must cancel.
Here's where the mass is important. Force of gravity on the filters: \(F_G=m\ g\).

Going back, we want consider all of the variables \(\sf at\ terminal\ velocity\), which is just where \(F_G+F_\text{drag}=0\), or maybe you prefer \(F_G=-F_\text{drag}\) (by subtracting both sides by \(F_\text{drag}\)).
So,
\(F_G=-F_\text{drag}\). That is, \(m\ g=-F_\text{drag}\).
The velocity is part of the \(F_\text{drag}\), so it will have the \(v\) you want to solve for that makes the whole equation true.

You had \(F_G=m_\text{coffee filter}\ g\) before. The force of gravity on three filters is \(m_\text{three coffee filters}\ g\). See what I did there? And \(m_\text{three coffee filters}=3m_\text{coffee filter}\).

Then \(F_{G,\ \text{on three}}=3m_\text{coffee filter}\ g\)
Now, the force of gravity is different than before. So, solve for \(v\) to get your \(v_2\) now! Or whatever you want to call it.

You can use the \(v=\dfrac{d}{t}\) equation to solve for the time of three coffee filters falling, just like you (probably) did to find \(v_1).

Good luck!

Answer 4

Thanks.
'K' is from this eqn: F=kv^2 k is constant.
It is a higher level problem.
I got the concept and context of the problem.

It was really helpful!

Answer 5

That's great! :)
So, any questions on that?

Answer 6

No, but I am struggling with another problem. The problem I have posted and bumped, and that is about hanging bricks of 203 kg. Do you have any idea? please.

Answer 7

What are the eqns and concept for solving this problem?

A load of bricks of mass M = 203 kg is attached to a crane by a cable of negligible mass and length L = 2.81 m. Initially, when the cable hangs vertically downward, the bricks are a horizontal distance D = 1.49 m from the wall where the bricks are to be placed. What is the magnitude of the horizontal force that must be applied to the load of bricks (without moving the crane) so that the bricks will rest directly above the wal