Chapter 10. EnergyThis pole vaulter can lift

herself nearly 6 m (20 ft) off

the ground by transforming

the kinetic energy of her

run into gravitational

potential energy.potential energy.

Chapter Goal: To introduce

the ideas of kinetic and

potential energy and to

learn a new problem-

solving strategy based on

conservation of energy.

Topics:

• A “Natural Money” Called Energy

• Kinetic Energy and Gravitational

Potential Energy

• A Closer Look at Gravitational Potential

Chapter 10. Energy

• A Closer Look at Gravitational Potential

Energy

• Restoring Forces and Hooke’s Law

• Elastic Potential Energy

• Elastic Collisions

• Energy Diagrams

Money-Energy AnalogyFrom the Parable of the Lost Penny

Money—Energy AnalogyFrom the law of conservation of energy

Kinetic and Potential EnergyThere are two basic forms of energy. Kinetic energy is an energy of motion

Gravitational potential energy is an energy of positionposition

The sum K + Ug is not changed when an object is in freefall. Its initial and final values are equal

Kinetic and Potential Energy

This is the conservation law for free fall

motion: the quantity

Free-Fall motion

2 2 2 ( )fy iy i f

v v g y y− = −− = −− = −− = −

2 21 1

2 2fy f iy i

v gy v gy+ = ++ = ++ = ++ = +

7

motion: the quantity

has the same value before and after the

motion.

21

2y

v gy++++

Free-Fall Motion

Then

ivrrrr

2 2 2 ( )fy iy i f

v v g y y− = −− = −− = −− = −

fx ixv v====

2 2 2 21 1 1 1

2 2 2 2fx fy f ix iy i

v v gy v v gy+ + = + ++ + = + ++ + = + ++ + = + +

8

fvrrrr

2 2 2 2

2 21 1

2 2f f i i

v gy v gy+ = ++ = ++ = ++ = +

Conservation law: the quantity

has the same value before and after the

motion.

21

2v gy++++

Frictionless surface: acceleration y

iy

fy

αααα

sina g αααα====arrrr

s 2 2 2f i

v v as− =− =− =− =

Motion with constant acceleration:

sin

i fy y

sαααα

−−−−====

y y−−−−

9

2 2 2 sin 2 2sin

i f

f i i

y yv v g gy gyαααα

αααα

−−−−− = = −− = = −− = = −− = = −

2 21 1

2 2f f i i

v gy v gy+ = ++ = ++ = ++ = +Conservation law: the quantity

has the same value before and after the

motion.

21

2v gy++++

y

iy

fy

Conservation law: the quantity

has the same value at the initial and the

final points

21

2v gy++++

10

In all cases the velocity at the final point is the same (FRICTIONLESS MOTION)

21

2v gy++++

21

2mv mgy++++Conservation law: or

21

2K mv==== Kinetic Energy – energy of motion

gU mgy==== Gravitational Potential Energy – energy of position

11The units of energy is Joule:

2

2

mJ kg

s====

mech gE K U= += += += + Mechanical Energy

Conservation law of mechanical energy (without friction):

constantmech

E K U= + == + == + == + =

12

13

Zero of potential energy

y y

0

,1iy

,2iy

y

, ,1 ,1 , ,1i g i f g fK U K U+ = ++ = ++ = ++ = +

, ,2 ,2 , ,2i g i f g fK U K U+ = ++ = ++ = ++ = +

,1 , ,1 , ,1( )

f i g i g fK K U U= + −= + −= + −= + −

14

0

,2fy,1f

y

,2 , ,2 , ,2( )

f i g i g fK K U U= + −= + −= + −= + −

,1 , ,2 , ,2 ,2 ,2

,1 ,1 , ,1 , ,1 ,2

( ) ( )

( ) ( )

g g i g f i f

i f g i g f g

U U U mg y y

mg y y U U U

∆ = − = − =∆ = − = − =∆ = − = − =∆ = − = − =

= − = − = ∆= − = − = ∆= − = − = ∆= − = − = ∆

Only the change of potential energy has the physical meaning

, ,i g i f g fK U K U+ = ++ = ++ = ++ = +

2

0

1

2i

K mv====

, 0g iU mgy====

, 10

g fU mgy= == == == =

15

, 10

g fU mgy= == == == =

2

1

1

2f

K mv====

2 2

1 0 0

1 1

2 2mv mv mgy= += += += +

2

1 0 02 4 100 10.2 /v v gy m s= + = + ≈= + = + ≈= + = + ≈= + = + ≈

Restoring Force: Hooke’s Law

16

spF k s= − ∆= − ∆= − ∆= − ∆

spring constant

Hooke’s Law:

Restoring Force: Hooke’s Law

spF k s= − ∆= − ∆= − ∆= − ∆

The sign of a restoring force is always opposite to the sign of displacement

17

Restoring Force: Elastic Potential Energy

spF k s= − ∆= − ∆= − ∆= − ∆

21( )

2s

U k s= ∆= ∆= ∆= ∆

18

The elastic potential energy is always positive

Restoring Force: Elastic Potential Energy

21( )

2s

U k s= ∆= ∆= ∆= ∆

2 21 1( ) constant

2 2s

K U mv k s+ = + ∆ =+ = + ∆ =+ = + ∆ =+ = + ∆ =

19

Restoring Force: Elastic Potential Energy 21( )

2s

U k s= ∆= ∆= ∆= ∆

202 21 1( ) constant

2 2s

K U mv k s+ = + ∆ =+ = + ∆ =+ = + ∆ =+ = + ∆ =

Law of Conservation of Mechanical Energy

2 21 1( ) constant

2 2g s

K U U mv mgy k s+ + = + + ∆ =+ + = + + ∆ =+ + = + + ∆ =+ + = + + ∆ =

21

(((( ))))2 ,2 2 ,2 1 ,1 1 ,1fx ix fx ixm v m v m v m v− = − −− = − −− = − −− = − −

1 ,1 2 ,2 1 ,1 2 ,2ix ix fx fxm v m v m v m v+ = ++ = ++ = ++ = +

,1 ,2 ,1 ,2ix ix fx fxp p p p+ = ++ = ++ = ++ = +

22

,1 ,2 ,1 ,2ix ix fx fxp p p p+ = ++ = ++ = ++ = +

, ,ix total fx totalp p====

The law of conservation of momentum

Elastic Collisions

Perfectly Elastic Collisions

A A

,A ivrrrr

vrrrr

,A fvrrrr

During the collision the kinetic energy will be transformed into elastic energy and then elastic energy will transformed back into kinetic energy

Perfectly Elastic Collision: Mechanical Energy is Conserved (no internal friction)

2 2

, ,

1 1

2 2

1 1

A A i B B im v m v+ =+ =+ =+ =

24

B

B

,B ivrrrr

,B fvrrrr

2 2

, ,

1 1

2 2A A f B B f

m v m v= += += += +

Perfectly inelastic collision:

A collision in which the two objects stick together and move with a

common final velocity – NO CONSERVATION OF MECHANICAL

ENERGY

Perfectly Elastic Collisions

A

B

A

,A ivrrrr

,B ivrrrr

,A fvrrrr

vrrrr

Conservation of Momentum and Conservation of Energy

isolated system

25

B

,B fv

2 2 2 2

, , , ,

1 1 1 1

2 2 2 2A A i B B i A A f B B f

m v m v m v m v+ = ++ = ++ = ++ = +

, , , ,A A i B B i A A f B B fm v m v m v m v+ = ++ = ++ = ++ = +

r r r rr r r rr r r rr r r r

Now we have enough equations to find the final velocities.

Example:

,A iv ,

0B i

v ====

,A fv

,B fv

2 2 2

, , ,

1 1 1

2 2 2A A i A A f B B f

m v m v m v= += += += +

, , ,A A i A A f B B fm v m v m v= += += += +

, ,

A B

A f A i

A B

m mv v

m m

−−−−====

++++

26

, ,

2A

B f A i

A B

mv v

m m====

++++

is always positive, can be positive (if ) or

negative (if )

,B fv

,A fv

A Bm m>>>>

A Bm m<<<<

,0

A fv ====

, ,B f A iv v====

ifA B

m m====

Chapter 10. Summary SlidesChapter 10. Summary SlidesChapter 10. Summary SlidesChapter 10. Summary Slides

General Principles

General Principles

Important Concepts

Important Concepts

Important Concepts

Applications

Applications

Chapter 10. Clicker QuestionsChapter 10. Clicker QuestionsChapter 10. Clicker QuestionsChapter 10. Clicker Questions

Rank in order,

from largest to

smallest, the

gravitational

potential energies

of balls 1 to 4.

A. (Ug)1 > (Ug)2 = (Ug)4 > (Ug)3

B. (Ug)4 > (Ug)3 > (Ug)2 > (Ug)1

C. (Ug)1 > (Ug)2 > (Ug)3 > (Ug)4

D. (Ug)4 = (Ug)2 > (Ug)3 > (Ug)1

E. (Ug)3 > (Ug)2 = (Ug)4 > (Ug)1

Rank in order,

from largest to

smallest, the

gravitational

potential energies

of balls 1 to 4.

A. (Ug)1 > (Ug)2 = (Ug)4 > (Ug)3

B. (Ug)4 > (Ug)3 > (Ug)2 > (Ug)1

C. (Ug)1 > (Ug)2 > (Ug)3 > (Ug)4

D. (Ug)4 = (Ug)2 > (Ug)3 > (Ug)1

E. (Ug)3 > (Ug)2 = (Ug)4 > (Ug)1

A small child slides down the four frictionless slides A–D. Each has the

same height. Rank in order, from largest to smallest, her speeds vA to vD at

the bottom.

A. vC > vA = vB > vD

B. vC > vB > vA > vD

C. vD > vA > vB > vC

D. vA = vB = vC = vD

E. vD > vA = vB > vC

A small child slides down the four frictionless slides A–D. Each has the

same height. Rank in order, from largest to smallest, her speeds vA to vD at

the bottom.

A. vC > vA = vB > vD

B. vC > vB > vA > vD

C. vD > vA > vB > vC

D. vA = vB = vC = vD

E. vD > vA = vB > vC

A box slides along the frictionless surface shown in the figure. It is

released from rest at the position shown. Is the highest point the box

reaches on the other side at level a, at level b, or level c?reaches on the other side at level a, at level b, or level c?

A. At level a

B. At level b

C. At level c

A box slides along the frictionless surface shown in the figure. It is

released from rest at the position shown. Is the highest point the box

reaches on the other side at level a, at level b, or level c?reaches on the other side at level a, at level b, or level c?

A. At level a

B. At level b

C. At level c

The graph shows force versus

displacement for three

springs. Rank in order, from

largest to smallest, the spring

constants k1, k2, and k3.

A. k1 > k3 > k2

B. k3 > k2 > k1

C. k1 = k3 > k2

D. k2 > k1 = k3

E. k1 > k2 > k3

The graph shows force versus

displacement for three

springs. Rank in order, from

largest to smallest, the spring

constants k1, k2, and k3.

A. k1 > k3 > k2

B. k3 > k2 > k1

C. k1 = k3 > k2

D. k2 > k1 = k3

E. k1 > k2 > k3

A spring-loaded gun shoots a plastic ball with a speed of 4 m/s. If the spring is compressed twice as far, the ball’s speed will be

A. 1 m/s.

B. 2 m/s.

C. 4 m/s.

D. 8 m/s.

E.16 m/s.

A spring-loaded gun shoots a plastic ball with a speed of 4 m/s. If the spring is compressed twice as far, the ball’s speed will be

A. 1 m/s.

B. 2 m/s.

C. 4 m/s.

D. 8 m/s.

E.16 m/s.

A particle with the potential energy shown in the graph is moving to the

right. It has 1 J of kinetic energy at x = 1 m. Where is the particle’s

turning point?

A. x = 1 m

B. x = 2 mB. x = 2 m

C. x = 5 m

D. x = 6 m

E. x = 7.5 m

A particle with the potential energy shown in the graph is moving to the

right. It has 1 J of kinetic energy at x = 1 m. Where is the particle’s

turning point?

A. x = 1 m

B. x = 2 mB. x = 2 m

C. x = 5 m

D. x = 6 m

E. x = 7.5 m