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Honors Physics Chapter 11

Energy and Its Conservation

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Chapter 11

Turn in Chapter 10 homework, worksheet, and Lab Report

Take Quiz 11 Lecture Q&A

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NGSS

HS-PS3-1.

Create a computational model to calculate the change in the energy of one component in a system when the change in energy of the other component(s) and energy flows in and out of the system are known.

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Forms of Energy

Kinetic energy: energy due to motion– Linear kinetic energy: translational motion– Rotational kinetic energy: rotation

Potential energy: energy stored in a system due to state, position, or configuration– Gravitational potential energy: due to height of object– Spring (elastic) potential energy: due to compression or stretch

of spring– Chemical, nuclear, …

Work: energy transfer by means of force

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Kinetic Energy, KE

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2KE mv

m: mass v: velocity

Unit:

KE 2m v

2m

kgs

2

2

mkg

s

2

mkg m

s

N m Joule J

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Work-Kinetic Energy Theorem

Wnet: net work done on an object

KE: kinetic energy of that object

Work and energy are related by

netW KE f iKE KE

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11 1537.85 10 , 25.0 10 , ,4.2 10 ?nuclear

mm kg v KE

sE J

KE

202.45 10 J

2

2 11 3 201 17.85 10 25.0 10 2.45 10

2 2

mmv kg J

s

Example: Pg287pp3A comet with a mass of 7.85 1011 kg strikes Earth at a speed of 25.0 km/s. Find the kinetic energy of the comet in joules, and compare the work that is done by Earth in stopping the comet to the 4.2 1015 J of energy that was released by the largest nuclear weapon ever built.

415

5.8 104.2 10

nuclearnuclear

EE

J

8

34.20 4.20 10 , 965m

m g kg vs

) ?a KE

) ?b W

) 0.75 , ?c d m F

3) 1.5 0.015 , 0, 1.96 10 , ?f id d cm m KE KE J F

Practice: A rifle can shoot a 4.20-g bullet at a speed of 965 m/s.a) Find the kinetic energy of the bullet.b) What work is done on the bullet if it starts from rest?c) If the work is done over a distance of 0.75 m, what is the average force on the bullet?d) If the bullet comes to rest by pushing 1.5 cm into metal, what is the magnitude and direction of the average force it exerts?

KE 2

2 3 31 14.20 10 965 1.96 10

2 2

mmv kg J

s

W KE f iKE KE 31.96 10fKE J

W F Fd3

31.96 102.6 10

0.75

W JN

d m

W F KE Fd f iKE KE iKE3

51.96 101.3 10

0.015iKE J

Nd m

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Gravitational Potential Energy(PE, U, or Ug)

Gravitational Potential Energy:

Unit:

m: mass g = 9.8 m/s2

h = height

PE mgh

PE m g h 2

m

kg ms

N m Joule J

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Frame of Reference

In the formula PE = mgh, upward has been defined as the positive direction for h.

We are still free to choose where h = 0. Reference level, height, or point: h = 0

PE has no physical significance, only ΔPE has physical significance, and ΔPE does not depend on the choice of Reference level.

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Example:

Is it possible for a system to have negative potential energy?

a) No, because the kinetic energy of a system must equal its potential energy.

b) Yes, since the choice of the zero of potential energy is arbitrary.

c) No, because this would have no physical meaning.

d) Yes, as long as the total energy is positive.

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Practice:

An acorn falls from a tree. Compare its kinetic energy K, to its potential energy U.

a) K decreases and U decreases.

b) K increases and U decreases.

c) K increases and U increases.

d) K decreases and U increases.

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h = 0

h1 = 45 m

h2 = ?

1

2

90 , 45 ,m kg h m

h

Let h = 0 at initial position.

1PE

2PE

ExampleA 90-kg rock climber first climbs 45 m upward to the top edge of a quarry, then, from the top, descends 85 m to the bottom. Find the potential energy of the climber at the edge and at the bottom, using the initial height as the reference level.

0

2

1

45 85 40 ,m m m

1 2? ?PE PE

2 41 90 9.8 / 45 4.0 10mgh kg m s m J

2 42 90 9.8 / 40 3.5 10mgh kg m s m J

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Practice Pg291pp7

If a 1.8-kg brick falls to the ground from a chimney that is 6.7 m high, what is the change in its potential energy?

1.8 , 6.7 , ?m kg h m PE

PE

21.8 9.8 6.7 118

mmg h kg m J

s

f iPE PE f imgh mgh f img h h

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630 , 5.0gW F N h m )a F

PE

c) From person (food or fat).

W

PE ) ?b PE

ExampleA person weighing 630 N climbs up a ladder to a height of 5.0 m.a) What work does the person do?b) What is the increase in the gravitational potential energy of the person from the ground to this height?c) Where does the energy come from to cause this increase in the gravitational potential energy?

630 , 5.0 , ?gF N d h m W

630 5.0 3150 3.2Fd N m J kJ

mgh

mg h 630 5.0 3150 3.2N m J kJ

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Spring (Elastic) Potential Energy(PEs, PEsp, or Us)

Unit:

k: spring constant: Stiffness of spring

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2sPE kx

Hooke’s Law: Force by spring:

x: compression or stretch of spring. (x is more exact.) Displacement of spring from relaxed position

F kx F is opposite to x.stiff spring

Small k soft spring

PE 2k x

N

m2m N m Joule J

, unit: N/m

Large k

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Example:

Calculate the work required to compress an initially uncompressed spring with a spring constant of 25 N/m by 10 cm.

a) 0.25 J

b) 0.17 J

c) 0.10 J

d) 0.13 J

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Practice:

What work is required to stretch a spring of spring constant 40 N/m from x = 0.20 m to 0.25 m? (Assume the unstretched position is at x = 0.)

a) 0.80 J

b) 0.050 J

c) 0.45 J

d) 1.3 J

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(Total) Mechanical Energy, Emec or E

E KE PE

g spKE PE PE

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Conservation of Energy

When the only forces doing work are gravitational forces and/or spring force during a process, the total mechanical energy is conserved.

i fE E

i i f fKE PE KE PE

i gi si f gf sfKE PE PE KE PE PE

0E KE PE

KE PE

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85 , 8.5i

mm kg v

s

) ?ia KE

i fE E

b) Let h = 0 at initial height, then hi = 0, vf = 0, hf = ?

c) No. Mass canceled out in part b).

iKE

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2 i fmv mgh

i i f fKE PE KE PE

Let system include bike, rider, and Earth.

fh

Example: Pg297pp15A bike rider approaches a hill with a speed of 8.5 m/s. The combined mass of the bike and rider is 85 kg. Choose a suitable system.a) Find the initial kinetic energy of the system.b) The rider coasts up the hill. Assuming there is no friction, at what height will the bike come to rest?c) Does your answer depend on the mass of the bike and rider? Explain.

2

21 185 8.5 3070 3.1

2 2

mmv kg J kJ

s

2

2

2

8.53.7

22 9.8

i

mv s

mmgs

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Let h = 0 at lowest point.

85 , 4.0 , 0, 0i f im kg h m h v ) ?fa v

b) No, the mass canceled out.

i fE E

i i f fKE PE KE PE 21

2i fmgh mv

fv

PracticeTarzan, mass 85 kg, swings down from a tree limb on the end of a 20-m vine. His feet touch the ground 4.0 m below the limb.a) How fast is Tarzan moving when he reaches the ground?b) Does your answer depend on Tarzan’s mass?

i

f

l

h = 0

22 2 9.8 4.0 8.9i

m mgh m

s s

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Practice:

A lightweight object and a very heavy object are sliding with equal speeds along a level frictionless surface. They both slide up the same frictionless hill. Which rises to a greater height?

a) The lightweight object, because it weighs less.

b) The heavy object, because it has greater kinetic energy.

c) They both slide to the same height.

d) cannot be determined from the information given

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Practice:

A skier, of mass 40 kg, pushes off the top of a hill with an initial speed of 4.0 m/s. Neglecting friction, how fast will she be moving after dropping 10 m in elevation?

a) 7.3 m/s b) 49 m/s c) 15 m/s d) 196 m/s

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Collision

Momentum is conserved

Conserved

Pi = Pf

Kinetic Energy is

If the two bodies stick together after collision

Not conserved

KEi = KEf Elastic Collision

KEi KEf Inelastic Collision

v1f = v2f Completely Inelastic Collision

During completely Inelastic Collision, there is most possible loss (but usually not all) of kinetic energy.

1 1 2 2

1 2

i if

m v m vv

m m

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Elastic Collision

i fP P

Only when v2i = 0

1 21 1

1 2

12 1

1 2

2

f i

f i

m mv v

m m

mv v

m m

Consider only collisions with v2i = 0,

i fKE KE

1 2 1 2i i f fP P P P 1 1 2 2i im v m v 1 1 2 2f fm v m v

1 2 1 2i i f fKE KE KE KE 21 1 2 2

1 1

2 2i im v m v2 2 2

1 1 2 2

1 1

2 2f fm v m v

When using these formula, make sure we choose m1 and m2 such that v2i = 0.

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Example

A 2.00-g bullet, moving at 538 m/s, strikes a 0.250-kg piece of wood at rest on a frictionless table. The bullet sticks in the wood, and the combined mass moves slowly down the table.

a. Find the speed of the combination after the collision.b. Find the kinetic energy of the bullet before the

collision.c. Find the kinetic energy of the combination after the

collision.d. How much (percent) kinetic energy was lost?

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Solution

31 1 2 22.00 2.00 10 , 538 , 0.250 , 0i i

mm g kg v m kg v

s

1 2) ?f f fa v v v

1) ?ib KE

fv

i fP P

1 1 2 2 1 2i i fm v m v m m v

1iKE

3

1 13

1 2

2.00 10 5384.27

2.00 10 0.250i

mkg

m v msm m kg kg s

2

2 31 1

1 12.00 10 538 290

2 2i

mm v kg J

s

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Solution

) ?d KE

) ?fc KE

fKE 21

2 fmv 21 2

1

2 fm m v

2

312.00 10 .250 4.27 2.30

2

mkg kg J

s

2.30 290% 99.2%

290f i

i

KE KE J JKE

KE

Where did the kinetic energy go? Heat (and sound)

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31 2 1 2 24.2 4.2 10 , 104 , 835 , 0, ?i i f

mm g kg m kg v v v

s

2 fv 3

11 3

1 2

2 4.2 102835

4.2 10 104i

kgm mv

m m kg kg s

0.0674 6.74m cm

s s

PracticeAs everyone knows, bullets bounce from Superman’s chest. Suppose Superman, mass 104 kg, while not moving, is struck by a 4.2-g bullet moving with a speed of 835 m/s. If the collision is elastic, find the speed that Superman had after the collision. (Assume the bottoms of his superfeet are frictionless.)

Must choose m1 and m2 this way to use formulas.

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Practice: Pg300pp21

A 91.0-kg hockey player is skating on ice at 5.50 m/s. Another hockey player of equal mass, moving at 8.10 m/s in the same direction, hits him from behind. They slide off together.

a) What are the total energy and momentum in the system before the collision?

b) What is the velocity of the two hockey players after the collision?

c) How much energy was lost in the collision?

1 2 1 2 1 291.0 , 5.50 , 8.10 ,i i f f f

m mm m kg v v v v v

s s

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Solution: Pg300pp21

) ?, ?i ia KE P

iKE

2 2

1 191.0 5.50 91.0 8.10 4360

2 2

m mkg kg J

s s

iP

91.0 5.50 91.0 8.10 1240m m m

kg kg kgs s s

1 2i iKE KE 2 21 1 2 2

1 1

2 2i im v m v

1 2i iP P 1 1 2 2i im v m v

) ?fb v

i fP P

fv

1 2i i fP P P 1 1 2 2 1 2i i fm v m v m m v

1 2 1 2 1 291.0 , 5.50 , 8.10 ,i i f f f

m mm m kg v v v v v

s s

1 1 2 2

1 2

91.0 5.50 91.0 8.106.8

91.0 91.0i i

m mkg kgm v m v ms s

m m kg kg s

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Solution

) ?c KE

fKE

KE

21

2 fmv 2

21 2

1 191.0 91.0 6.8 4210

2 2f

mm m v kg kg J

s

4210 4360 150f iKE KE J J J