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...
F net
= ma = (600 kg)(2 m/s 2 )
A 600 kg elevator starts from rest and is pulled upward by a motor with a constant acceleration of 2 m/s 2 for 3 seconds.
What is the average power output of the motor during this time period?
F net
= 1200 N = F motor
- F g
= F motor
- mg
1200 N = F motor
- (600 kg)(10 m/s 2 ) = F motor
- 6000 N
F motor
= 6000 N + 1200 N = 7200 N

Now we need to determine the work done by the motor...
A 600 kg elevator starts from rest and is pulled upward by a motor with a constant acceleration of 2 m/s 2 for 3 seconds.
What is the average power output of the motor during this time period?
W = F
D s But we don’t know D s
, so….
s = s
0
+v
0
t +0.5at 2 = 0 + 0 + 0.5(2 m/s 2 )(3 s) 2 = 9 m
W = (7200 N)(9 m) = 64800 J
P

W
D t

64800 J
3 s

21600 W

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.
l x
Hook discovered this force could be modeled by the mathematical expression
F = - kx
Notice that this force operates along a linear line!

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.

If we look at the work done by an applied force which compresses the spring through a distance (x
1
) ...
F
1
W
 D 
Fx

1
W
 
2
( k )(

)
1 1
 1
2 kx x
1
( F
2 1

2
0 ) x
-x
1
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 kx
2
2
So, for spring problems, we have a new
TOTAL MECHANICAL ENERGY given by
 g

spring
And it is THIS quantity which will be conserved absent other, outside forces.

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)
 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.
mv v


 v
 mv
0
0


 a t
  
D mv

 mv

0


D

Change in momentum mv

 mv

0


D
 p

 p
0


D
D
 p


F t

I

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!