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Why does the acceleration decrease whereas velocity increase for a freely falling spherical body inside a viscous liquid

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    $\begingroup$ The spherical body will eventually reach a terminal velocity. $\endgroup$ – MaxW Oct 10 '19 at 15:16
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When a spherical body is falling through a viscous liquid, there are essentially two forces acting on it:

  1. Apparent weight (gravity $-$ buoyant force) -downwards

  2. Viscous Drag (or Stoke's Force) -upwards

The first force is constant in both magnitude and direction through the course of the motion.

The viscous drag on the other hand depends on the velocity and acts vertically upwards.

$$F_s = 6rv\pi\eta $$

$\eta $ - viscosity of the liquid

$v $ - velocity of the body

$r $ - radius of the body

Initially the viscous force is zero as the velocity is zero. The only force on the body is the apparent weight which causes the velocity to increase over time. Now that the velocity is increasing, the viscous drag increases causing the net force to decrease over time (as the apparent weight and viscous force are in opposite directions).

Therefore the velocity increases but the acceleration keeps on decreasing.

There comes a time when the acceleration is zero i.e. the Apparent weight is equal to the Viscous drag. The velocity remains constant after this point of time (also known as Terminal Velocity).

Note - here "Apparent Weight" is the combined effect of Gravity and bouyancy.

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