Question

3. A marble of mass 10 g falling through a tank of liquid experiences a resistive force R. proportional to its velocity. Specifically, R =-bv, where b = 3.5 × 10-2 kg s-1. The marble is dropped from rest and begins to accelerate downward (a) What is the marble's terminal velocity in this liquid? (b) How mich time elapses before the marble has reached 95% of its terminal velocity? (c) How far does it travel during this time?

Please make work legible and clear I want to understand how to do it. Please also only use Newtonian mechanics

Answer 1

Applying Newton's second law

$F_{N}=m\cdot a$

but the net force acting on the marble is

$F_{N}=W-b\cdot v=m\cdot g-b\cdot v$

where:
$R=-b\cdot v$   is the resistive force

$W=m\cdot g$   is the weight of the marble

combining equations

$m\cdot a=m\cdot g-b\cdot v$

but remember that $a=\frac{dv}{dt}$

replacing in equation

$m\cdot \frac{dv}{dt}=m\cdot g-b\cdot v$   dividing by m $\frac{dv}{dt}= g-\frac{b}{m}\cdot v$

part A.

speed is terminal when the acceleration is zero, ie. $a= \frac{dv}{dt}=0$

Expression of terminal velocity $0= g-\frac{b}{m}\cdot v_{T}\Rightarrow v_{T}=\frac{g\cdot m}{b}$

where:

the mass is $m=10g\cdot \frac{1kg}{10^{3}g }=10\cdot 10^{-3}kg$

the constant $b=3.5\cdot 10^{-2}kg\cdot s^{-1}$

acceleration of gravity $g=9.81\frac{m}{s^{2}}$

evaluate numerically

$v_{T}=\frac{g\cdot m}{b}=\frac{9.81\frac{m}{s^{2}}\cdot 10\cdot 10^{-3}kg}{ 3.5\cdot 10^{-2}kg\cdot s^{-1}}=2.80\frac{m}{s}$

The terminal velocity is

$v_{T}=2.80\frac{m}{s}$

Part B

The expression that satisfies the equation $\frac{dv}{dt}= g-\frac{b}{m}\cdot v$   

considering
initial velocity $v_{i}=0\frac{m}{s}$
initial time $t_{i}=0s$

$V=V_{T}\left ( 1-e^{-\frac{t }{\tau }} \right )$

where:

terminal velocity

$v_{T}=2.80\frac{m}{s}$

time constant

$\tau =\frac{ m}{b }=\frac{ 10\cdot 10^{-3}kg}{ 3.5\cdot 10^{-2}kg\cdot s^{-1}}=0.29s$

if you want to know when

$V=0.95V_{T}$

substituting in the equation

$0.95\cdot V_{T}=V_{T}\left ( 1-e^{-\frac{t }{\tau }} \right )$

simplifying

$0.95= 1-e^{-\frac{t }{\tau }}$

expression for t

$0.95= 1-e^{-\frac{t }{\tau }} \Rightarrow e^{-\frac{t }{\tau }}=1-0.95\Rightarrow e^{-\frac{t }{\tau }}=0.05$

applying logarithmic properties

$ln\left (e^{-\frac{t }{\tau }} \right )=ln\left (0.05 \right )\Rightarrow -\frac{t }{\tau }=ln\left (0.05 \right )\Rightarrow t=-\tau \cdot ln\left ( 0.05 \right )$

evaluate numerically $t=-\tau \cdot ln\left ( 0.05 \right )=-0.29s\cdot ln\left ( 0.05 \right )=0.87s$

finally $t=0.87s$

Part C

the position of the object is equal to $v=\frac{dx}{dt}$

you can write the equation $V=V_{T}\left ( 1-e^{-\frac{t }{\tau }} \right )$ how $\frac{dx}{dt}=V_{T}\left ( 1-e^{-\frac{t }{\tau }} \right )$

organizing

$dx=V_{T}\left ( 1-e^{-\frac{t }{\tau }} \right )dt$

The position is equal to

$\int_{x_{0}}^{x}dx=V_{T}\int_{t_{0}}^{t}\left ( 1-e^{-\frac{t }{\tau }} \right )dt$

organizing integrals

$\int_{x_{0}}^{x}dx=V_{T}\int_{t_{0}}^{t} 1dt-V_{T}\int_{t_{0}}^{t}e^{-\frac{t }{\tau }}dt$

solving each integral

$\left [x \right ]_{x_{0}}^{x}=\left [V_{T}\cdot t+V_{T}\cdot \tau \cdot e^{-\frac{t }{\tau }} \right ]_{t_{0}}^{t}$   considering $x_{0}=0$   and $t_{0}=0$

position equation

$x=V_{T}\cdot t+V_{T}\cdot \tau \cdot e^{-\frac{t }{\tau }}$

evaluating in $t=0.87s$

$x=V_{T}\cdot t+V_{T}\cdot \tau \cdot e^{-\frac{t }{\tau }}=2.80\frac{m}{s}\cdot 0.87s+2.80\frac{m}{s}\cdot 0.29s\cdot e^{-\frac{0.87 }{ 0.29s}}=2.44m\dotplus 16.31m=18.75m$

$x=18.75m$