I have a mechanics question that you have to solve with differential equations:
A particle of mass $0.4kg$ is moving at a speed of $10ms^{-1}$ when it enters an viscous liquid at a point, B. Inside the liquid the resistance force is proportional to the velocity, and initially equal to $1.2N$. Apart from the resistance an the weight, no other forces are acting on the particle.
Part A) Show that the velocity of the particle satisfies the differential equation $\frac{dv}{dt}=-0.3v$.
So I got the answer wrong, of course.
My working out is:
$R\:=kv$
$1.2\:=10k$
$0.12\:=k$
By $F=ma,$
$0.4g-0.12v\:=\:0.4\frac{dv}{dt}$
So $g-0.3v = \frac{dv}{dt}$
I don't understand how they managed to eliminate g? Or is my whole process wrong? What am I missing or fail to infer?
To have a compatible answer with the proposed question we have to consider the movement as horizontal.
After entering horizontally, the horizontal dynamics are described as
$$ m \ddot x + k\dot x = 0 $$
or
$$ m\dot v_x + k v_x = 0 $$
solving this we obtain
$$ v_x = v_x^0 e^{-\frac km t} $$
but according to the initial conditions $v_x^0 = 10$ and also
$$ -F_0 = m \dot v_x(0) = -m \frac km v_x^0\Rightarrow 1.2 = k v_x^0 = k \times 10 $$
so we have
$$ k = 0.12\Rightarrow \frac km = 0.3 $$
hence the horizontal dynamics are dictated by
$$ \dot v_x = -0.3 v_x $$