The drag force on a sphere of radius r moving at speed 0.15nm/s through a fluid

Question

The drag force on a sphere of radius r moving at speed 0.15nm/s through a fluid with viscosity n is D-->= (6pienrv, direction opposite motion).We can model a paramecium as a sphere of diameter 50um , with a mass of 6.5*10^-11kg. Water has a viscosity of 0.0010 N*s/m^2.

A paramecium swimming at a constant speed of 0.15nm/s ceases propelling itself and slows to a stop. At the instant it stops swimming, what is the magnitude of its acceleration?


If the acceleration of the paramecium in Part A were to stay constant as it comes to rest, approximately how far would it travel before stopping?

Explanation / Answer

Force at the instant the paramecium stops swimming is:

(D=-6pieta rv=-6pi*0.0010*50*10^{-6}*0.15*10^{-9}=-1.4137*10^{-16}N)

Now, acceleration will be:

(a=D/m = -rac{-1.4137*10^{-16}}{6.5*10^{-11}}=-2.175*10^{-6}m/s^2)

Now,

if the acceleration was constant, the the distance travelled is

(d=-v^2/2a=rac{(0.15*10^{-9})^2}{2*2.175*10^{-6}}=5.17*10{-15}m)

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