Answer:
a. acceleration (a) = 4.9 m/s²
b. Â [tex]V(t) = \frac{mgdsin \theta}{uA}(1-\frac{uA}{e^{md}} t)[/tex]
c. see the attached diagram below
d  V = 0.048 m/s
e. The viscosity u = 0.27 Ns/m²
Explanation:
Given that :
m = 5 kg
A = ( 0.1 m )²
d = 0.20 mm
θ = 30 °
By applying Newton's second Law ; we have
F = mgsinθ - [tex]F_f[/tex]    (since no friction ; then
ma = mg sinθ
a = gsin θ
a = 9.8 × sin 30
a = 4.9 m/s²
b) By applying Newton's second law at any instant .
ma = mgsin θ - [tex]F_f[/tex]
and [tex]F_f[/tex] = T×A = [tex]u \frac{dv}{dy}A[/tex] - [tex]u \frac{v}{d}A[/tex]
Also; ma = [tex]m \frac{dv}{dt}[/tex]
∴ ma = mg sin θ - [tex]\frac{u A}{d}V[/tex]
separating the variables ; we have
[tex]\frac{dv}{gsin \theta - \frac{uA}{md}V} = {dt}[/tex]
By integrating using limits ; we have:
[tex]\frac{m.d}{uA} \lim(1-\frac{uA}{mgdsin \theta}V)= t}[/tex]
∴ [tex]V(t) = \frac{mgdsin \theta}{uA}(1-\frac{uA}{e^{md}} t)[/tex]
c) The curve for V(t) plot is attached in the diagram below.
d)
To find the speed after 1 sec; we have:
at t = 1.0 sec;
V =[tex]\frac{5*9.81*0.0002sin (30)}{0.4*(0.1)^2}(1-e^{-\frac{0.4*0.01}{5*0.002}*0.1})[/tex]
V = 0.048 m/s
e) In order to determine the viscosity for which V(0.1) = 0.3 m/s ; we have the following.
∴[tex]v(t= 0.3) = \frac{mgd sin \theta }{uA}(1-e^{\frac{uA}{md}(t=0.1)})[/tex]
[tex]0.3 = \frac{5*9.81*sin 30*0.0002}{u(0.1)^2}(1-e^{\frac{u*0.01}{5*0.0002}(0.1)})[/tex]
The viscosity u = 0.27 Ns/m²
