MFMcGraw
• Work by a Constant Force
• Kinetic Energy
• Potential Energy
• Work by a Variable Force
• Springs and Hooke’s Law
• Conservation of Energy
• Power
Ch06 - Energy - Revised: 2/20/10 2

The three kinds of energy are: kinetic energy , potential energy , and rest energy .
Energy may be converted from one form to another or transferred between bodies.
MFMcGraw Ch06 - Energy - Revised: 2/20/10 3

MFMcGraw Ch06 - Energy - Revised: 2/20/10 4

The unit of work and energy is the joule (J).
1 J = 1 Nm = 1 kg m 2 /s 2 .
MFMcGraw Ch06 - Energy - Revised: 2/20/10 5

Only the force in the direction of the displacement that does work.

F
 r x
An FBD for the box at left: y
 r x
N
 x w F
The work done by the force F is:
W
F

F x
 r x


F cos
 
 x
Ch06 - Energy - Revised: 2/20/10 6 MFMcGraw

The work done by the normal force N is: W
N

0
The normal force is perpendicular to the displacement.
The work done by gravity ( w ) is: W g

0
The force of gravity is perpendicular to the displacement.
MFMcGraw Ch06 - Energy - Revised: 2/20/10 7

The net work done on the box is:
W net





W
F
F
F
 cos cos
W


N



 x
 x
W
 g
0

0
MFMcGraw Ch06 - Energy - Revised: 2/20/10 8

In general, the work done by a force F is defined as
W

F
 r cos
 where F is the magnitude of the force,
 r is the magnitude of the object’s displacement, and  is the angle between F and
 r (drawn tail-to-tail).
MFMcGraw Ch06 - Energy - Revised: 2/20/10 9

Example: A ball is tossed straight up. What is the work done by the force of gravity on the ball as it rises?
y
 r
FBD for rising ball: w x
W g
 w
 y cos 180

  mg
 y
MFMcGraw Ch06 - Energy - Revised: 2/20/10 10

A box of mass m is towed up a frictionless incline at constant speed .
The applied force F is parallel to the incline. y
Question: What is the net work done on the box?
F F N
An FBD for the box: x
MFMcGraw

Apply Newton’s 2 nd Law:



F x
F y

F
 w sin


N
 w cos
 w

0

0
Ch06 - Energy - Revised: 2/20/10 11

Example continued:
The magnitude of F is:
F
 mg sin

If the box travels along the ramp a distance of
 x the work by the force F is
W
F

F
 x cos 0
  mg
 x sin

The work by gravity is
W g
 w
 x cos
  
90


  mg
 x sin

MFMcGraw Ch06 - Energy - Revised: 2/20/10 12

Example continued:
The work by the normal force is:
W
N

N
 x cos 90
 
0
The net work done on the box is:
W net

W
F


W g

W
N mg
 x sin
  mg
 x sin


0

0
MFMcGraw Ch06 - Energy - Revised: 2/20/10 13

Example: What is the net work done on the box in the previous example if the box is not pulled at constant speed?

F x

F
 w sin

F
 ma
 w sin

 ma
Proceeding as before:
New Term
W net



W
F
 ma
 


W mg
 x g


W sin
F net
N
 
 x
 x
 mg
 x sin
 
0
MFMcGraw Ch06 - Energy - Revised: 2/20/10 14

MFMcGraw
K

1
2 mv
2 is an object’s translational kinetic energy.
This is the energy an object has because of its state of motion.
It can be shown that, in general
Net Work = Change in K
W net
 
K .
Ch06 - Energy - Revised: 2/20/10 15

Example: The extinction of the dinosaurs and the majority of species on
Earth in the Cretaceous Period (65 Myr ago) is thought to have been caused by an asteroid striking the Earth near the Yucatan Peninsula. The resulting ejecta caused widespread global climate change.
If the mass of the asteroid was 10 16 kg (diameter in the range of 4-
9 miles) and had a speed of 30.0 km/sec, what was the asteroid’s kinetic energy?
K

1
2 mv
2

4 .
5


10
24
J
1
2

10
16 kg

30

10
3 m/s
2

This is equivalent to ~10 9 Megatons of TNT.
MFMcGraw Ch06 - Energy - Revised: 2/20/10 16

Gravitational Potential Energy
Potential energy is an energy of position .
There are potential energies associated with different forces. Forces that have a potential energy associated with them are called conservative forces.
In general W cons
  
U
MFMcGraw
Not all forces are conservative, i.e. Friction.
Ch06 - Energy - Revised: 2/20/10 17

Gravitational Potential Energy
The change in gravitational potential energy (only near the surface of the Earth) is

U g
 mg
 y where
 y is the change in the object’s vertical position with respect to some reference point.
You are free to choose to location of this where ever it is convenient.
MFMcGraw Ch06 - Energy - Revised: 2/20/10 18

GPE Problem
The table is 1.0 m tall and the mass of the box is 1.0 kg.
Ques: What is the change in gravitational potential energy of the box if it is placed on the table?
U = 0
First: Choose the reference level at the floor. U = 0 here.

U g

 mg
 y


1 .
0 kg
  mg
 y f
 y i

9 .
8 m/s
2
 
1 .
0 m

0 m

 
9 .
8 J
MFMcGraw Ch06 - Energy - Revised: 2/20/10 19

GPE Problem
Example continued:
Now take the reference level (U = 0) to be on top of the table so that y i
=

1.0 m and y f
= 0.0 m.

U g

 mg
 y

 mg
9 .
8 m/s
 y f
 y i

2
 
0 .
0 m



1 .
0 m
 
 
9 .
8 J
The results do not depend on the location of U = 0.
MFMcGraw Ch06 - Energy - Revised: 2/20/10 20

Total Mechanical Energy
Mechanical energy is E

K

U
The total mechanical energy of a system is conserved whenever nonconservative forces do no work.
That is E i
= E f or

K =

U.
Then if

K increases

U decreases and vice versa
MFMcGraw Ch06 - Energy - Revised: 2/20/10 21

Mechanical Energy Problem
A cart starts from position 4 with v = 15.0 m/s to the left. Find the speed of the cart at positions 1, 2, and 3. Ignore friction.
MFMcGraw
U
4
E
4

K
4

E
3

U
3

K
3 mgy
4

1
2 mv
4
2 v
3

 mgy
3 v
4
2 

1 mv
3
2
2
2 g
 y
4
 y
3


20 .
5 m/s
Ch06 - Energy - Revised: 2/20/10 22

Mechanical Energy Problem
U
4
E
4

K
4

E
2

U
2

K
2 mgy
4

1
2 mv
4
2 v
2

 mgy
2 v
4
2 

1 mv
2
2
2
2 g
 y
4
 y
2


18 .
0 m/s
Or use
E
3
=E
2
U
4
E
4

K
4

E
1

U
1

K
1 mgy
4

1
2 mv
4
2 v
1

 mgy
1 v
4
2 

1 mv
1
2
2
2 g
 y
4
 y
1


24 .
8 m/s
MFMcGraw Ch06 - Energy - Revised: 2/20/10
Or use
E
3
=E
1
E
2
=E
1
23

Roller Coaster Problem
A roller coaster car is about to roll down a track. Ignore friction and air resistance.
m = 988 kg
(a) At what speed does the car reach the top of the loop?
20 m y=0
U i
E i

K i

E f

U f

K f mgy i

0
 v f
 mgy
2 g f

 y i
1
2
 mv y f
2 f


19 .
8 m/s
Ch06 - Energy - Revised: 2/20/10 MFMcGraw
40 m
24

Roller Coaster Problem
Example continued:
(b) What is the force exerted on the car by the track at the top of the loop?
Apply Newton’s Second Law:
FBD for the car: y 
F y
 
N
 w
  ma r
  m v
2 r
N
 m v
2 r
N
 w
 m v
2 r
 mg

2 .
9

10
4
N
N w x
MFMcGraw Ch06 - Energy - Revised: 2/20/10 25

Roller Coaster Problem
Example continued:
(c) From what minimum height above the bottom of the track can the car be released so that it does not lose contact with the track at the top of the loop?
Using conservation of mechanical energy:
U i
E i

K i

E f

U f

K f mgy i

0
 mgy f

1
2
2 mv min
Solve for the starting height y i
 y f

2 v min
2 g
MFMcGraw Ch06 - Energy - Revised: 2/20/10 26

Roller Coaster Problem
Example continued:
What is v min
?
v = v min when N = 0. This means that
N
 w
 m v
2 r w
 mg
 m
2 v min r v min
 gr
The initial height must be y i
 y f

2 v min
2 g

2 r
 gr
2 g

2 .
5 r

25 .
0 m
MFMcGraw Ch06 - Energy - Revised: 2/20/10 27

Nonconservative Forces
What do you do when there are nonconservative forces?
For example, if friction is present

E

E f

E i

W fric
The work done by friction.
MFMcGraw Ch06 - Energy - Revised: 2/20/10 28

Gravitational Potential Energy
The general expression for gravitational potential energy is:
U
 
U
GM
1
M
2 r
 r
 


MFMcGraw Ch06 - Energy - Revised: 2/20/10 29

Gravitational Potential Energy
Example:
What is the gravitational potential energy of a body of mass m on the surface of the Earth?
U
 r

R e

 
GM
1
M
2 r
 
GM e m
R e
MFMcGraw Ch06 - Energy - Revised: 2/20/10 30

Planetary Motion
A planet of mass m has an elliptical orbit around the Sun. The elliptical nature of the orbit means that the distance between the planet and Sun varies as the planet follows its orbital path. Take the planet to move counterclockwise from its initial location.
QUES: How does the speed of a planet vary as it orbits the Sun once?
The mechanical energy of the planet-sun system is:
E

1
2 mv
2 
GmM sun r
 constant
MFMcGraw
B
Ch06 - Energy - Revised: 2/20/10 r A
31

Planetary Motion
E

1
2 mv
2 
GmM sun r
 constant
B r
At point “B” the planet is the farthest from the Sun. At point “A” the planet is at its closest approach to the sun.
A
Starting from point “B” (where the planet moves the slowest), as the planet moves in its orbit r begins to decrease. As it decreases the planet moves faster.
At point “A” the planet reaches its fastest speed. As the planet moves past point A in its orbit, r begins to increase and the planet moves slower.
MFMcGraw Ch06 - Energy - Revised: 2/20/10 32

Work can be calculated by finding the area underneath a plot of the applied force in the direction of the displacement versus the displacement.
MFMcGraw Ch06 - Energy - Revised: 2/20/10 33

Example: What is the work done by the variable force shown below?
F x
(N)
F
3
F
2
F
1
The net work is then
W
1
+W
2
+W
3.
x
1
The work done by F
1 is
The work done by F
2 is
The work done by F
3 is x
2 x
3 x (m)
W
1
W
2
W
3


F
1
 x
1

0


F
2
F
3
 x
2
 x
1
 x
3
 x
2


MFMcGraw Ch06 - Energy - Revised: 2/20/10 34

By hanging masses on a spring we find that stretch
 applied force. This is Hooke’s law.
x

F x is the magnitude of the force exerted by the free end of the spring, x is the measured stretch of the spring, and k is the spring constant (a constant of proportionality; its units are N/m).
A larger value of k implies a stiffer spring.
MFMcGraw Ch06 - Energy - Revised: 2/20/10 35

(a) A force of 5.0 N applied to the end of a spring cause the spring to stretch 3.5 cm from its relaxed length.
Ques: How far does a force of 7.0 N cause the same spring to stretch?
For springs F
 x. This allows us to write
F
1
F
2
 x x
2
1 .
Solving for x
2
: x
2

F
2 x
1
F
1

7 .
0 N
5.0
N

3 .
5 cm


4 .
9 cm.
MFMcGraw Ch06 - Energy - Revised: 2/20/10 36

Example continued:
(b) What is the spring constant of this spring?
k

F
1 x
1

5 .
0 N
3.5
cm

1 .
43 N/cm.
Or k

F
2 x
2

7 .
0 N
4.9
cm

1 .
43 N/cm.
MFMcGraw Ch06 - Energy - Revised: 2/20/10 37

An ideal spring has k = 20.0 N/m. What is the amount of work done (by an external agent) to stretch the spring 0.40 m from its relaxed length?
MFMcGraw
W

Area under curve

1
2
  
1 1

1
2 kx
1
2 
1
2

20 .
0 N/m

0 .
4 m

2 
1 .
6 J
Ch06 - Energy - Revised: 2/20/10 38

The work done in stretching or compressing a spring transfers energy to the spring.
Below is the equation of the spring potential energy.
U s

1
2 kx
2
The spring is considered the system
MFMcGraw Ch06 - Energy - Revised: 2/20/10 39

A box of mass 0.25 kg slides along a horizontal, frictionless surface with a speed of 3.0 m/s. The box encounters a spring with k = 200 N/m.
Ques: How far is the spring compressed when the box is brought to rest?
E i
U i

K i

E f

U f

K f
0

1
2 mv
2 x


1


2 kx
2 m k


 v
0

0 .
11 m
MFMcGraw Ch06 - Energy - Revised: 2/20/10 40

Power is the rate of energy transfer.
Average Power
Instantaneous Power
P av


E
 t
P

Fv cos

The unit of power is the watt. 1 watt = 1 J/s = 1 W.
MFMcGraw Ch06 - Energy - Revised: 2/20/10 41

A race car with a mass of 500.0 kg completes a quarter-mile
(402 m) race in a time of 4.2 s starting from rest. The car’s final speed is 125 m/s. (Neglect friction and air resistance.)
Ques: What is the engine’s average power output?
P av


E
 t


U
 
K
 t


K
 t

1 mv
2 f
2
 t

9 .
3

10
5
watts
MFMcGraw Ch06 - Energy - Revised: 2/20/10 42

MFMcGraw
• Conservation of Energy
• Calculation of Work Done by a Constant or
Variable Force
• Kinetic Energy
• Potential Energy (gravitational, elastic)
• Power
Ch06 - Energy - Revised: 2/20/10 43