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ATMO 551
Fall 2010
Impulse and Work
Normally Newton’s second law is written as
F=ma
which means the force on an object is equal to the object’s mass times its acceleration, where
bold indicates a vector. Consider gravity. The acceleration of gravity is independent of the
object being pulled. The force of gravity is proportional to the mass of the object being pulled.
So two objects with very different masses will accelerate identically but the forces on the two
objects will be quite different.
The F=ma equation can be written in terms of momentum, p. p is a vector defined as
p=mv
where m is mass and v is the velocity (vector). Therefore,
F = m a = m dv/dt = dp/dt
So force equals the time rate of change of momentum of an object.
Impulse
For understanding the pressure exerted by a gas, we need to think in terms of impulses.
An impulse, I, is defined as a finite change in momentum. So
I = p
But
t2
t2
t1
t1
 dp  
p  p2  p1 
dp
dt 
dt
t2
 Fdt
t1
so
I

t2
 Fdt
t1
Now notice that the time average of a quantity, A, is given as

A
1
t 2  t1
t2
 Adt
t1
from which it follows that

t2
t2
t1
t1
At 2  t1   A  dt 
 Adt
So

1
Kursinski 08/30/10
ATMO 551
Fall 2010
I
t2
 Fdt  F t
2
 t1   Ft  p
t1
So an impulse is equal to the average force over a time interval, t, times the length of that time
interval. We will use this relation later on understanding the pressure exerted by a gas at the
microscopic level.

Work
Kinetic energy, EK, is
1
E K  mv 2
2
Now consider the following quantity, the dot product of the force and infinitesimal change in
position:
F  dx  F dx  F dy  F dz
x
y
z
Now consider just one component or consider the direction, x, as the direction along which the
force is pushing. So the direction, x, is aligned with F.

dv
dv dx
dv
d v 2 
m
F dx  ma dx  m
dx  m
dt  m v dt  m  dt  dv 2 
dt
dt dt
dt
dt  2 
2
So F  dx is equal to the change in kinetic energy, dEK. This quantity is defined as the change in
work, dW.
 Now integrate F dx along the path, x, where x is in the direction of the acceleration.

E K  W 
x2
 F d 
x1
v 2  v 2 x 2  v 2 x1 
m
d
  2  m 2  2  E K x 2   E K x1


v x1 
v x 2 
So the force on an object times the distance over which it pushes on the object equals the change
in the kinetic energy of an object and this process is called Work.

2
Kursinski 08/30/10