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Monday, October 5, 1998
Chapter 5: Springs
Chapter 6: Linear Momentum
Conservation of Momentum
Impulse
Let’s first figure out the
force delivered by the
motor...
A 600 kg elevator starts
from rest and is pulled
upward by a motor with
a constant acceleration
of 2 m/s2 for 3 seconds.
What is the average
power output of the motor
during this time period?
Fnet = ma = (600 kg)(2 m/s2)
Fnet = 1200 N = Fmotor - Fg = Fmotor - mg
1200 N = Fmotor - (600 kg)(10 m/s2) = Fmotor - 6000 N
Fmotor = 6000 N + 1200 N = 7200 N
Now we need to
determine the work
done by the motor...
W = F Ds
A 600 kg elevator starts
from rest and is pulled
upward by a motor with
a constant acceleration
of 2 m/s2 for 3 seconds.
What is the average
power output of the motor
during this time period?
But we don’t know Ds, so….
s = s0 +v0t +0.5at2 = 0 + 0 + 0.5(2 m/s2)(3 s)2 = 9 m
W = (7200 N)(9 m) = 64800 J
W 64800 J
P

 21600 W
Dt
3s
In addition to the gravity, there are other
mechanisms to store POTENTIAL ENERGY.
One of them is...
Sir Robert Hooke unlocked the secret
of the spring...
A spring resting in its natural state, with a
length l exerts no horizontal force on anything!
However, if we compress or stretch the spring
by some amount x, then the spring is observed
to exert a Force in the opposite direction.
x
l
Hook discovered this force could be modeled
by the mathematical expression
F = - kx
Notice that this force operates along a linear line!
Force
Which means that if we looked at the plot
of Force versus compression/stretching x...
x
Slope of this line
is -k, where k is
the spring constant.
Force
If we look at the work done by an applied
force which compresses the spring through
a distance (-x1)...
F1
W  FDs  Fx1  ( F1  0) x
1
2
W  ( k )( x1 ) x1  kx
1
2
1
2
2
x
-x1
Work done BY the
external force ON
the spring.
This energy is stored in the spring...
Potential Energy of a spring is
PE spring
1 2
 kx
2
So, for spring problems, we have a new
TOTAL MECHANICAL ENERGY given by
KE PE g  PEspring
And it is THIS quantity which will be conserved
absent other, outside forces.
Momentum & Collisions
The linear momentum of an object of mass m
moving with velocity v is defined as the
product of the mass and the velocity:


p  mv


p  mv
Notice that momentum is a vector quantity,
which means that it must be specified with
both a magnitude and direction.
Also notice that the direction of the
momentum vector is necessarily parallel
to the velocity vector.


p  mv


[ p]  [m][v ]

[ p]  kg (m / s)
OR

[ p]  N s
The units suggest
a relationship between
force and momentum.


p  mv
What happens when we apply a force
to an object?
It accelerates.
Its velocity changes.
Its momentum changes.
The force imparts momentum.


p  mv
By how much will the momentum change?
That depends upon the length of time over
which the force is applied to the object.
  
v  v0  aDt



mv  mv0  ma Dt



mv  mv0  FDt



mv  mv0  FDt

 
p  p0  FDt
Change in
momentum

 
Dp  FDt  I
Impulse
The impulse of a force on an object equals the
change in momentum of that object.
Notice that impulse is a vector quantity as well!