# Resistive Force

• May 24th 2011, 12:37 PM
LukeEvans
Resistive Force
A body of mass m (kg) is falling vertically under gravity g m/s^2 in a medium whose resistance to the speed of the body, v m/s, is proportional to v^n (n positive).

If the body was released from rest and has terminal velocity, v^ m/s, use Newton's 2nd law of motion to show that its acceleration, a m/s^2, may be expressed as

a = g(1-(v/v^t)^n)

(4 Marks)

I would obviously like to attempt to solve this problem, but am unsure what the first step should be. Perhaps the resistive force will need to be deducted from the velocity... many thanks to anybody who looks.
• May 24th 2011, 01:36 PM
TheEmptySet
Quote:

Originally Posted by LukeEvans
A body of mass m (kg) is falling vertically under gravity g m/s^2 in a medium whose resistance to the speed of the body, v m/s, is proportional to v^n (n positive).

If the body was released from rest and has terminal velocity, v^ m/s, use Newton's 2nd law of motion to show that its acceleration, a m/s^2, may be expressed as

a = g(1-(v/v^t)^n)

(4 Marks)

I would obviously like to attempt to solve this problem, but am unsure what the first step should be. Perhaps the resistive force will need to be deducted from the velocity... many thanks to anybody who looks.

So Newtons 2nd law says that force is equal to mass times acceleration. This gives

$ma=mg-kv^n$

I think you have a typo (I can't read you terminial velocity) so lets call the terminal velocity [tex]w[tex]

We know that the acceleration is equal to zero when you reach terminal velocity this gives

$0=mg-kw^n \iff k=\frac{mg}{w^n}$

Now if we plug this back into the first equation we get

$ma=mg-\left( \frac{mg}{w^n}\right)v^n$

$a=g-\frac{gv^n}{w^n}=g\left(1- \frac{v^n}{w^n}\right)$