Lecture 4: Energy

Lecture 4: Energy

Serway

and
Vuille

Ch. 5

Outline

•
Work
-
Energy theorem

•
Gravitational potential energy

•
Spring potential energy

•
Conservation of energy

•
Power

Introduction to Energy

•
Energy

is

one

of

the

most

important

concepts

in

science,

originating

from

thermodynamics
.

•
According

to

the

ideas

of

modern

physics,

energy

is

the

substance

from

which

all

things

in

the

Universe

are

made

up
.

•
Energy

is

also

defined

as

the

ability

of

a

system

to

perform

work
.

•
The

most

important

property

of

energy

is

that

it

is

a

conserved

quantity
.

Introduction to Energy

•
Energy

is

present

in

the

universe

in

many

different

forms

–
Mechanical

–
Chemical

–
Electromagnetic

–
Nuclear

–
Thermodynamic

(Heat)

•
In

this

chapter,

we

focus

on

mechanical

energy,

which

is

the

sum

of

kinetic

energy

and

potential

energy

Work

•
In

physics,

work

involves

applying

a

force

to

an

object

while

moving

it

a

certain

distance
.

•
Therefore,

the

work

W

done

on

an

object

by

a

constant

force

F

during

a

linear

displacement

along

the

x
-
axis

is


𝑊
=



∆


•
The

SI

unit

of

work

is

Joule

(J)
.


•
Note

that

1

J

=

1

N

∙

m

=

1

kg

m
2
/s
2

Work

•
Work

is

done

only

by

the

component

of

the

force

acting

parallel

to

the

object’s

direction

of

motion
.

•
The

work

W

done

on

an

object

by

a

constant

force

F

during

a

displacement

x

is

𝑊
=
𝑭
∙
∆

=
𝑭
∆

cos
𝜃

Example 1

•
A

block

of

mass

m

=

2
.
50

kg

is

pushed

a

distance

d

=

2
.
0

m

along

a

frictionless

horizontal

table

by

a

constant

applied

force

of

magnitude

F

=

16
.
0

N

directed

at

an

angle

θ

=

25
.
0
⁰

below

the

horizontal
.

•
Determine

the

work

done

by

–
(a)

The

applied

force

–
(b)

The

normal

force

exerted

by

the

table

–
(c)

The

force

of

gravity

–
(d)

The

net

force

on

the

block

Example 1 Solution

•
The

work

done

by

the

applied

force

is


𝑊
=
𝑭
∙
∆

=


cos
25
.
0
°
∆

=
16
.
0

𝑁
2
.
0


cos
25
.
0
°
=
19
.
0




•
The

work

done

by

the

normal

force

and

the

force

of

gravity

is

𝑊
=
𝑭
∙
∆

=



∆

+




∆


=
0


•
The

work

done

by

the

net

force

on

the

block

is

from

the

applied

force

alone
.

Work
-
Energy Theorem

•
The

work
-
energy

theorem

relates

the

speed

of

an

object

to

the

net

work

done

on

it

by

external

forces
.

•
It

can

be

shown

that

the

net

work

done

on

an

object

is

related

to

its

speed

as

follows
:

𝑊
𝑛
=
1
2

𝑣
2
−
1
2

𝑣
0
2


Work
-
Energy Theorem

𝑊
𝑛
=
1
2

𝑣
2
−
1
2

𝑣
0
2

•
The

RHS

can

be

interpreted

as

the

energy

associated

with

the

object’s

motion

and

is

defined

as

the

kinetic

energy
.

•
Therefore,

the

net

work

done

on

an

object

is

equal

to

the

change

in

an

object’s

kinetic

energy
.



Work
-
Energy Theorem

•
Based

on

the

work
-
energy

theorem,

we

can

define

work

as

the

amount

of

energy

transferred

to

an

object

when

a

force

acts

upon

it
.

•
When

W

>

0

(W

<

0
),

then

the

mechanical

energy

of

the

object

increases

(decreases
)
.

•
Based

on

the

work
-
energy

theorem,

we

can

define

kinetic

energy

as

the

work

a

moving

object

can

do

in

coming

to

rest
.


Example 2

•
The

driver

of

a

1
.
00

x

10
3

kg

car

traveling

on

the

interstate

at

35
.
0

m/s

slams

on

his

breaks

to

avoid

hitting

a

second

vehicle

in

front

of

him
.


•
After

the

brakes

are

applied,

a

constant

kinetic

friction

force

of

magnitude

8
.
00

x

10
3

N

acts

on

the

car
.

•
(a)

At

what

minimum

distance

should

the

brakes

be

applied

to

avoid

a

collision

with

the

other

vehicle?

•
(b)

If

the

distance

between

the

vehicles

is

initially

only

30
.
0

m,

at

what

speed

would

the

collision

occur?

Example
2

Solution

•
To

determine

the

minimum

stopping

distance,

we

apply

the

work
-
energy

theorem

to

the

car

𝑊
𝑛
=
1
2

𝑣
2
−
1
2

𝑣
0
2

−

𝑘
∆

=
−
1
2

𝑣
0
2

∆

=

𝑣
0
2
2

𝑘
=
1
.
00



10
3
𝑘

35
.
0


/

2
2
8
.
00



10
3
𝑁
=
76
.
6



Example 2 Solution

•
At

the

given

distance

of

30
.
0

m,

the

car

is

too

close

to

truck

and

thus,

will

collide

into

the

truck
.

•
To

determine

the

speed

at

impact,

we

use

the

work

energy

theorem

𝑊
𝑛
=
1
2

𝑣
2
−
1
2

𝑣
0
2

−

𝑘
∆

=
1
2

𝑣
2
−
1
2

𝑣
0
2

𝑣
=
𝑣
0
2
−
2


𝑘
∆

=
35
.
0


/

2
−
2
8
.
00



10
3
𝑁
30
.
0


1
.
00



10
3
𝑘

=
27
.
3


/


Example 3

•
A

man

pushing

a

crate

of

mass

m

=

92
.
0

kg

at

a

speed

v

=

0
.
850

m/s

encounters

a

rough

horizontal

surface

of

length

l

=

0
.
65

m
.


•
If

the

coefficient

of

kinetic

friction

between

the

crate

and

rough

surface

is

0
.
358

and

he

exerts

a

constant

horizontal

force

of

275

N

on

the

crate,

find

•
(a)

The

magnitude

and

direction

of

the

net

force

on

the

crate

while

it

is

on

the

rough

surface

•
(b)

The

net

work

done

on

the

crate

while

it

is

on

the

rough

surface

•
(c)

The

speed

of

the

crate

when

it

reaches

the

end

of

the

rough

surface
.

Example 3 Solution

•
To

determine

the

net

force,

we

first

should

make

a

free

body

diagram
.

•
Using

Newton’s

2
nd

law

gives






=


−


=
0


𝑎

=




=
−

𝑘
+


𝑎𝑝𝑝

•
The

net

force

will

be

in

the

x
-
direction

given

as

𝑭
𝑛
=




=
−
𝜇
𝑘


+


𝑎𝑝𝑝
=
0
.
358
92
.
0

𝑘

9
.
80


2
+
275

𝑁
=
−
47
.
8

𝑁

•
The

net

work

will

be

in

the

x
-
direction,

given

as

𝑾
𝑛
=
𝑭
𝑛
∙
∆


?(
𝑛
∆

=
−
47
.
8

𝑁
0
.
65


=
−
31
.
05



n

m
g

f
k

F
app

Example 3 Solution

•
To

determine

the

final

velocity,

we

apply

the

work
-
energy

theorem
.

𝑊
𝑛
=
1
2

𝑣
2
−
1
2

𝑣
0
2

𝑣
=
𝑣
0
2
+
2
𝑊
𝑛

=
0
.
850


2
+
2
−
31
.
05


92
.
0

𝑘

=
0
.
22


/


n

m
g

f
k

F
app

Gravitational Potential Energy

•
Gravitational

potential

energy

is

the

energy

associated

with

an

object’s

position

in

the

gravitational

field
.

•
Consider

a

book

of

mass

m

that

falls

Δ
y
.


•
The

net

work

done

by

gravity

is

𝑊

=
𝑭
𝒈
∙
∆


=
−


∆


=
−
∆
𝑃


•
Using

the

work
-
energy

theorem,

we

have

𝑊
𝑛
=
−
∆
𝑃

=
∆



Spring Force and Hooke’s Law

•
The

figure

shows

a

spring

in

its

equilibrium

position,

where

the

spring

is

neither

compressed

nor

stretched
.

•
Experimentation

has

shown

that

the

spring

force

follows

Hooke’s

law





=
−
𝑘
.

•
The

spring

constant

k

is

a

measure

of

the

stiffness

of

the

spring

and

is

determined

by

how

the

spring

was

formed,

its

material

composition,

and

the

thickness

of

the

wire
.

•
This

force

is

called

a

restoring

force

because

the

spring

always

exerts

a

force

in

a

direction

opposite

the

displacement,

restoring

the

spring

to

its

original

position
.

Elastic Potential Energy

•
The

elastic

potential

energy

is

stored

energy

arising

from

the

work

done

to

compress

or

stretch

the

spring
.

•
Consider

a

horizontal

spring

and

mass

at

the

equilibrium

position

shown

here
.

•
The

work

done

by

the

spring

when

compressed

by

an

applied

force

from

equilibrium

to

a

displacement

x
.

𝑊


=
𝑭


∙
∆

=




=
−
1
2
𝑘

2
=
−
∆
𝑃



•
Using

the

work
-
energy

theorem,

we

have

𝑊
𝑛
=
−
∆
𝑃


=
∆



Mechanical Energy Conservation

•
When

only

conservative

forces

(such

as

the

gravitational

force

and

spring

force)

are

present,

then

the

work
-
energy

theorem

can

be

written

as

𝑊
𝑛
=
𝑊

+
𝑊


=
∆



−
∆
𝑃


−

∆
𝑃



=
∆



∆


+
𝑃

+
𝑃


=
0

•
This

is

called

the

conservation

of

mechanical

energy

and

it

is

one

of

the

most

important

concepts

in

classical

mechanics
.

•
Conservation

of

mechanical

energy

says

that

the

mechanical

energy

(the

sum

of

the

kinetic

energy

and

potential

energy)

remains

constant

at

all

times
.

Example 4

•
A

0
.
500
-
kg

block

rests

on

a

horizontal,

frictionless

surface
.

The

block

is

pressed

back

against

a

spring

having

a

constant

of

k

=

625

N/m,

compressing

the

spring

by

10
.
0

cm

to

point

A
.

Then

the

block

is

released
.

•
(a)

Find

the

maximum

distance

d

the

block

travels

up

the

frictionless

incline

if

θ

=

30
.
0
⁰
.

•
(b)

How

fast

is

the

block

going

when

halfway

to

its

maximum

height?

Example 4 Solution

•
Conservation

of

mechanical

energy

gives



𝑖
+
𝑃

,
𝑖
+
𝑃


,
𝑖
=



+
𝑃

,

+
𝑃


,


1
2

𝑣
𝑖
2
+




𝑖
+
1
2
𝑘

𝑖
2
=
1
2

𝑣

2
+





+
1
2
𝑘


2

1
2
𝑘

𝑖
2
=

ℎ
=

𝑑
sin
30
.
0
°

𝑑
=
𝑘

𝑖
2
2


sin
30
.
0
°
=
625
𝑁

0
.
1


2
2
0
.
500

𝑘

9
.
8


/

2
sin
30
.
0
°
=
1
.
28



Example 4 Solution

•
To

determine

the

velocity

of

the

block

when

it’s

halfway

to

its

maximum

height,

we

use

conservation

of

mechanical

energy



𝑖
+
𝑃

,
𝑖
+
𝑃


,
𝑖
=



+
𝑃

,

+
𝑃


,



1
2
𝑘

𝑖
2
=
1
2

𝑣

2
+


𝑑
sin
30
.
0
°
2


𝑣

=
𝑘

𝑖
2
−

𝑑
sin
30
.
0
°

=
625
𝑁

0
.
1


2
−
0
.
500

𝑘

9
.
80


/

2
1
.
28


sin
30
.
0
°
0
.
500

𝑘

=
2
.
50


/


Example 5

•
A

ball

of

mass

m

=

1
.
80

kg

is

released

from

rest

at

a

height

h

=

65
.
0

cm

above

a

light

vertical

spring

of

force

constant

k
.


•
The

ball

strikes

the

top

of

the

spring

and

compresses

it

a

distance

d

=

9
.
00

cm
.


•
Neglecting

any

energy

losses

during

the

collision,

find


•
(a)

t
he

speed

of

the

ball

just

as

it

touches

the

spring

•
(b)

t
he

force

constant

of

the

spring

Example 5 Solution

•
Because

there

are

no

energy

losses

during

the

collision,

only

conservative

forces

are

present

and

thus,

mechanical

energy

conservation

can

be

applied
.

•
It

may

be

helpful

to

break

this

problem

into

two

phases

–
Stage

1
:

Free

fall

under

gravity

–
Stage

2
:

Compression

of

the

spring

Example 5 Solution

•
For

stage

1
,

the

gravitational

force

is

the

only

force

present

and

thus,

we

can

use

mechanical

energy

conservation
.



𝑖
+
𝑃

,
𝑖
=



+
𝑃

,


1
2

𝑣
𝑖
2
+




𝑖
=
1
2

𝑣

2
+






𝑣

=
2




𝑖
=
2
9
.
8


/

2
0
.
65


=
3
.
56


/


Example 5 Solution

•
For

stage

2
,

both

the

gravitational

force

and

spring

force

are

present
.

•
Using

mechanical

energy

conservation,

we

have





𝑖
+
𝑃

,
𝑖
+
𝑃


,
𝑖
=



+
𝑃

,

+
𝑃


,



1
2

𝑣
𝑖
2
=
−


𝑑
+
1
2
𝑘
𝑑
2


𝑘
=

𝑣
𝑖
2
+
2

𝑑
𝑑
2
=
1
.
80

𝑘

3
.
56


2
+
2
1
.
80

𝑘

9
.
80


/

2
0
.
09


0
.
09


2
=
3
.
22



10
3

𝑁
/


Work and
Nonconservative

Forces

•
When

nonconservative

forces

(such

as

kinetic

friction)

are

present,

then

the

work
-
energy

theorem

can

be

written

as

𝑊
𝑛
=
𝑊
𝑛

+
𝑊

+
𝑊


=
∆



𝑊
𝑛
−
∆
𝑃


−

∆
𝑃



=
∆



𝑊
𝑛
=
∆
𝑃

+
∆
𝑃


+
∆



•
If

mechanical

energy

is

not

conserved

in

a

system,

then

it

must

leave

the

system

or

be

transformed

into

a

form

of

non
-
mechanical

energy

(such

as

internal

energy)
.

•
When

positive

(negative)

work

is

done

on

the

system,

energy

is

transferred

from

the

environment

(system)

to

the

system

(environment
)
.

Example
6

•
A

skier

starts

from

rest

at

the

top

of

a

frictionless

incline

of

height

20
.
0

m
.


•
At

the

bottom

of

the

incline,

the

skier

encounters

a

horizontal

surface

where

the

coefficient

of

kinetic

friction

between

skis

and

snow

is

0
.
210
.

•
(a)

Find

the

skier’s

speed

at

the

bottom
.

•
(b)

How

far

does

the

skier

travel

on

the

horizontal

surface

before

coming

to

rest?

Example
6

Solution

•
We

can

split

this

problem

into

two

parts
:

–
Trip

1
:

Going

from

points

A

to

B

under

the

influence

of

gravity

–
Trip

2
:

Going

from

points

B

to

C

under

the

influence

of

kinetic

friction

Example
6

Solution

•
For

trip

1
,

we

can

find

the

speed

at

point

B

by

using

the

energy

conservation

equation



𝑖
+
𝑃
𝑖
=



+
𝑃


1
2

𝑣
𝑖
2
+




𝑖
=
1
2

𝑣

2
+






𝑣

=
2




𝑖
=
2
9
.
8


/

2
20


=
1
9
.
8


/


Example
6

Solution

•
For

trip

2
,

we

can

use

the

work
-
energy

theorem

to

determine

the

distance

traveled

from

B

to

C
.

𝑊
𝑛
=
∆


=
1
2

𝑣


2
−
1
2

𝑣

2

−

𝑘
𝑑
=
1
2

𝑣


2
−
1
2

𝑣

2

−
𝜇
𝑘


𝑑
=
−
1
2

𝑣

2

𝑑
=
𝑣

2
2
𝜇
𝑘



𝑑
=
19
.
8


2
2
0
.
210
9
.
8


/

2
=
95
.
2



n

m
g

f
k

Application: The Simple Pendulum

•
There

are

only

two

forces

acting

upon

the

pendulum

bob
:

the

gravitational

and

tension

force
.


•
The

gravitational

force

does

work

on

the

pendulum,

but

it

does

not

change

the

total

gravitational

potential

energy

of

the

system
.

•
The

tension

force

does

not

do

work

on

the

pendulum

since

it

acts

perpendicular

to

the

motion

of

the

pendulum
.


•
Here,

the

total

mechanical

energy

of

the

pendulum

is

conserved
.

•
Video

demonstration

Power

•
Power

can

be

defined

as

the

rate

at

which

energy

is

transferred

into

or

out

of

a

system
.

•
Mathematically,

power

can

be

defined

𝑃
=
𝑊
∆


=

𝑣

•
The

SI

unit

of

power

is

joule

per

second,

also

called

the

watt
.

•
1

W

=

1

J/s

=

1

kg

m
2
/s
3

Example 7

•
What

average

power

would

a

1
.
00

x

10
3

kg

speedboat

need

to

go

from

rest

to

20
.
0

m/s

in

5
.
00

s,

assuming

the

water

exerts

a

constant

drag

force

of

magnitude

f

=

5
.
00

x

10
2

N

and

the

acceleration

is

constant
.

•
Find

an

expression

for

the

instantaneous

power

as

a

function

of

the

drag

force

f
d
.

Example 7 Solution

•
The

power

is

provided

by

the

engine

and

thus,

𝑃

=
𝑊
𝑛𝑖𝑛
∆



•
To

determine

the

work

done

by

the

engine,

we

use

the

work
-
energy

theorem
.

𝑊
𝑛
=
𝑊
𝑛𝑖𝑛
+
𝑊

𝑎
=
1
2

𝑣
2
−
1
2

𝑣
0
2

𝑊
𝑛𝑖𝑛
=
1
2

𝑣
2
+

𝑘
∆


Example 7 Solution

•
To

determine

the

displacement,

we

must

first

find

the

acceleration

using

kinematics

equation

1
.

𝑣
=
𝑣
0
+
𝑎


𝑎
=
𝑣
−
𝑣
0


=
20


−
0


5


=
4
.
00


/

2

•
The

displacement

can

be

calculated

using

kinematics

equation

3

𝑣
2
=
𝑣
0
2
+
2
𝑎
∆


∆

=
𝑣
2
−
𝑣
0
2
2
𝑎
=
20


2
−
0


2
2
4
.
00


/

2
=
50
.
0



Example 7 Solution

•
Calculating

the

work

done

by

the

engine

gives

𝑊
𝑛𝑖𝑛
=
1
2

𝑣
2
+

𝑘
∆

=
1
2
1
.
00



10
3

𝑘

20


2
+
5
.
00



10
2

𝑁
50
.
0


=

2
.
25



10
5


•
The

power

is

provided

by

the

engine

and

thus,

𝑃

=
𝑊
𝑛𝑖𝑛
∆


=
2
.
25



10
5


5
.
00


=
4
.
50



10
4

𝑊

•
To

relate

instantaneous

power

to

the

drag

force,

we

use

Newton’s

second

law

𝑎
=



=


𝐸
−




𝑃
=


𝐸
𝑣
=
𝑎
+



𝑣
=

𝑎
2
+
𝑎