According to Newton’s Second Law, the net force on an object is proportional to

Question

According to Newton’s Second Law, the net force on an object is proportional to its acceleration. That is, F = ma. Furthermore, acceleration is defined as the rate of change of velocity with time, a = dv/dt . It can be shown that air resistance acts opposite the direction of motion, and for larger velocities obeys a velocity-squared rule. That is,

|Fair| = kv^2 ,

where k is a constant. You may assume that the force due to gravity is Fg = mg where g = 9.81m/s2 is acceleration due to gravity.

a) Formulate a differential equation in velocity v describing the motion of an object falling vertically under the forces of gravity and air resistance.

Explanation / Answer

When an object is falling vertically under the forces of gravity, then by Newton’s second law:

Force = mass*acceleration, which implies that :

Fgrav = mg (where m is the mass of the object in kilograms and g is the gravitational acceleration near the earth's surface and is taken 10m/sec2 and both m and g are positive).

Now when object is vertically under the forces of gravity, the air resistance acts opposite to the direction of motion. Also we know that the force due to air resistance is proportional to the speed and is applied in the direction opposite to motion.

Let s(t) be the downward speed (in m/sec ) of the object at any time t seconds after it is dropped.
So, Fair = -ks(t) (here the positive constant k is the constant of proportionality and its units are kg/sec and t is time. k depends upon the shape of the object being dropped and the density of the atmosphere).

Now the total force acting on the falling object at any time t is given by

Ftot = Fgrav + Fair = mg - ks(t)     …….(1)

if a(t) is the acceleration on the object at time t, then a(t) = ds/dt = s'(t). We may rewrite Newton's Law as:

Ftot = ma(t)
or from (1) above, mg - ks(t) = ms'(t)

or s'(t) = (k/m)[(mg/k) - s(t)]

Now let another function p(t) = (mg/k) - s(t), such that that p'(t) = -s'(t) …..(2)

or p'(t) = (-k/m)p(t)

We know in Differential Calculus that if a function is proportional to its own derivative it should be in the form of an exponential function for any constant C.

So, p(t) = Ce(-k/m)t

or s(t) = (mg/k) - Ce(-k/m)t

or s(t) = (mg/k)(1 - e(-k/m)t)

Which is the required differential equation in speed ‘s’ describing the motion of an object falling vertically under the forces of gravity and air resistance.

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