Lecture 6A
"Don't try to solve serious matters in the middle of the night.”
--Philip K. Dick

Quiz 3 will include all of Chapter 5.
My suggestion is to do the homework, class examples, and book examples.
Don’t forget your quiz code number or a
Scantron for Friday or you will be penalized.

Example
The masses shown to the right are connected by a massless string over a frictionless, massless pulley and are released from rest. Find (a) the velocity of the 7.0kg mass just before it hits the floor, (b) the maximum height above the ground reached by the 4.0kg mass.
Answer
First, you must define a coordinate system.
Let’s choose the up as the positive y-direction and let the ground be y   =   0.

Answer
Are there any non-conservative forces at work in this problem?
No, the only force performing work on the system is the gravitational force. So:
Δ E mec
=   Δ KE   +   Δ PE   =   0
Δ KE   =   – Δ PE
The masses start at rest and have the same velocity just before the 7.0kg mass hits the ground such that:
Δ KE   =   KE f,7kg
+   KE f,4kg
Δ KE   =   (1/2)m
7kg
(v f
) 2   +   (1/2)m
4kg
(v f
) 2
•
Δ KE   =   (1/2)(m
7kg
+   m
4kg
)(v f
) 2

•
Answer
The potential energy of the 7kg mass will decrease while the potential energy of the 4kg mass will increase.
Δ PE   =   Δ PE
7kg
+   Δ PE
4kg
Δ PE   =   –(m
7kg
)gh i,7kg
+   (m
4kg
)gh f,4kg
Δ PE   =   –(7kg)(9.8N/kg)(5m)   +   (4kg)(9.8N/kg)(5m)
•
Δ PE   =   –343J   +   196J = –147J
•
Going back to the conservation of energy equation:
•
•
Δ KE   =   – Δ PE =   +147J
Δ KE   =   (1/2)(11kg)(v f
) 2   =   147J v f
=
( )
11kg
= 5.2ms
part (a)

Answer
After the 7kg mass hits the ground, energy is no longer conserved as the ground imparts a nonconservative force on the system.
But, since the string will just go slack the 4kg mass will still have the same velocity upward which will now be affected by gravity.
!
KE
4 kg
= "!
PE
4 kg
KE
4 kg
=
KE f
KE i
=
0
1
2 m
4 kg v i
2
!
PE
4 kg
=
PE f
"
PE i
= m
4 kg gh f
" m
4 kg gh i
!
PE
4 kg
= m
4 kg
( f
" h
) i

Answer
Going back to conservation of energy equation:
!
1
2 m
4 kg v i
2 = !
m
4 kg
( f
!
h
) i v i
2
= h f
!
h i h f
= h i
2 g
Putting all of the numbers gives us: h f
=
5.0m
+
(
5.2m/s
) 2
2 9.8N/kg
)
+ v i
2
2 g h f
=
5.0m
+
1.4m
=
6.4m

Work and energy are not explicitly time dependent.
It could take you 10 seconds to do 1 Joule of work or 10 hours. Work will never tell you how long it would take.
But you can turn to average power, P avg
, which is the rate at which work is done.

The SI unit for power is: [1 Joule/sec]   = [1
Watt].
Another unit that is commonly used is horsepower,   hp, where: 1hp   =   746W.
If we have a constant force we can write:

Example
An elevator raises 100 passengers (each averaging 660N in weight) to a height of 150m in 60.0s at constant speed. What average power is required of the force performing the lifting?
Answer
First, you must define a coordinate system.
Let’s choose the upward direction as positive and make the lowest point to be y   =   0.

Answer
The lift opposes the force of gravity such that there is a constant speed up.
This means that there is zero acceleration and the forces are balanced. Thus,
F lift
=   F g
=   (100 pass.)(660N/pass.)   =   66,000N
This force of lift is creating the power to move the passengers.
P   =   F y v y
=   F lift
( Δ y/t)   =   (66,000N)(150m/60.0s)
P   =   1.65x10
5 W or 165kW

•
Three rocks of equal mass are thrown with identical speeds from the top of the same building. Rock X is thrown vertically downward, rock Y is thrown vertically upward, and rock Z is thrown horizontally. Which ball has the greatest speed just before it hits the ground?
Assume air resistance is negligible.
Y
A) Rock X.
Z
B) Rock Y.
X
C) Rock Z.
•
D) Both Rock X and Rock Y.
•
E) All rocks hit the ground with the same speed.

All three rocks have the same initial energy.
There is nothing but conservative forces involved.
All three rocks thus have the same final energy.
With an appropriate definition of PE, all three rocks have nothing but KE in the end.
Therefore, all three rocks have the same speed.

Quiz 3 will include all of Chapter 5.
My suggestion is to do the homework, class examples, and book examples.
Don’t forget your quiz code number or a
Scantron for Friday or you will be penalized.

When applying a force to an object to change its motion, the time in which the force is applied is important in determining the end results.
Impulse, I, is a vector that measures how much force has changed motion:
Impulse points in the direction of the average force.
Impulse is measured in units of kg(m/s) or Ns.
A large impulse leads to a large change in velocity.

So when a bat strikes a ball it will have a large impulse because it has a large average force even though it has a small time period over which it is interacting.
Another useful quantity in measuring your motion is the linear momentum vector, p:
Momentum points in the direction of velocity.
Momentum is measured in units of kg(m/s).

We can relate impulse and change in momentum by examining Newton’s 2nd Law:
This is the Impulse-momentum theorem, that impulse is equivalent to the change in momentum.