Question

according to your data and the following general expression $f_{drag}=bv^{n}$. how does the force of air resistance on a coffee filter vary with the velocity of the filter? questions 1. in this lab we made a very important assumption that the average velocity is equal to the terminal velocity. in reality this is not the case. the average velocity is slightly less than the terminal velocity. why? 2. what effect does this have on your experimental value of $n$? would it make $n$ too high or too small? justify your answer.

Explanation:

Step1: Consider motion to terminal velocity

The coffee - filter accelerates from rest until it reaches terminal velocity. During the time it is accelerating, its velocity is less than the terminal velocity. The average velocity is calculated over the entire motion (including the acceleration phase), so it is the weighted - average of the lower velocities during acceleration and the terminal velocity. Since the time spent at velocities less than terminal velocity is non - zero, the average velocity is less than the terminal velocity.

Step2: Analyze effect on \(n\)

The drag force is given by \(F_{drag}=bv^{n}\). If we assume the average velocity (\(v_{avg}\)) instead of the terminal velocity (\(v_{t}\)) and we know \(v_{avg}

Answer:

1. The coffee - filter starts from rest and accelerates to terminal velocity. The average velocity is calculated over the whole motion including the acceleration phase where velocities are less than terminal velocity, so the average velocity is less than the terminal velocity.
2. The experimental value of \(n\) will be too high. Since \(F_{drag}=bv^{n}\) and using a smaller \(v\) (average velocity instead of terminal velocity) to get the same \(F_{drag}\) (equal to gravitational force at terminal velocity) requires a higher value of \(n\).