# Lecture 4: Energy

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27 Οκτ 2013 (πριν από 4 χρόνια και 6 μήνες)

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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

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

.

(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
𝑛𝑖𝑛



To

determine

the

work

done

by

the

engine,

we

use

the

work
-
energy

theorem
.

𝑊
𝑛
=
𝑊
𝑛𝑖𝑛
+
𝑊
 𝑎
=
1
2

𝑣
2

1
2

𝑣
0
2

𝑊
𝑛𝑖𝑛
=
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
𝑛𝑖𝑛
=
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
𝑛𝑖𝑛


=
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


