Fundamentals of Physics

2009 Physics 2111 Fundamentals of Physics Chapter 7 1

Fundamentals of Physics

Chapter 7 Kinetic Energy & Work1. Energy2. Work3. Work & Kinetic Energy4. Work Done by a Gravitational Force5. Work Done by a Spring Force6. Work Done by a General Variable Force7. Power

Review & SummaryQuestionsExercises & Problems


Energy & Work

• Energy is conserved.• A new tool to solve problems.

221 vmEnergyKinetic

2

2

11:Unitss

mkgJoule

Energy comes in many forms:• Thermal, Chemical, Atomic, Nuclear Energies

• Descriptions of Energy:• Potential Energy (Chapter 7)• Kinetic Energy - motion of a massive object


WOW!

August 10, 1972, a large meteorite skipped across the atmosphere.

JvmKE 14221 1050

bombsatomicWWIIMTKE 81.0

smv

kgm

15

104 6


“Work”

• Work (W) is energy transferred to or from an object by means of a force acting on the object.

– Energy transferred to the object is positive work.– Stepping on the gas causes positive work to be done on a car.

– Energy transferred from the object is negative work.– Stepping on the brakes causes negative work to be done on a

moving car.


Work & Kinetic Energy

Consider a force acting on an object constrained to move in the x-direction:

The force causes a change in kinetic energy; the force does work:

Units: 1 Joule = 1 kg (m / s)2 = (1 kg m / s2) m = 1 N m

dFdFWork x cos""

davv x220

2

Frictionless Wire

dFvmvm

damvmvm

davv

x

x

x

202

1221

202

1221

20

2 2


A force acting on a particle moving along a curve:

work.no doesF

s

cos

FWork

sFWork

sFWork S

090coscos 0

velocity. thechangenot doesF


“Work”

Work (W) is energy transferred to or from an object by means of a force acting on the object.

– A force does positive work when it has a vector component in the same direction as the displacement.

– A force does negative work when it has a vector component in the opposite direction as the displacement.

– A force does zero work when it is perpendicular to the displacement.

dFdFW cos


Work – Kinetic Energy Theorem

2

1

2

1

s

s S

s

s S

S

dsamdsFW

dsFW

ds

dvv

dt

ds

ds

dv

dt

dvaS

v

v

s

sdvvmds

ds

dvvmW

0

2

1

KW

KKW

vmvmW

if

202

1221

Net Work Done = Change in Kinetic Energy

chain rule for derivatives


Scalar (Dot) Product of Two Vectors

Commutative Law:

cosbaba

abba

zzyyxx babababa


Example

Tipler p 157

s FWork

NmW

mNmNW

yFxFW yx

14

5423


Example:

Cutnell p 160

sFWork net

Knet fgmF 025sin

kgm

v sm

58

6.30

NfK 70

202

12

21025sin vmvmsfgm fK

ms 57

sm

fv 19

vf = ?


Work done by the forces on the box?

F N

d m

210

20

3

0

W F d

W F d N m J

F

F

cos ( )( ) cos( ) 210 3 20 6000

W N d

W N d W

N

N g

cos( )90 00

mg

F

N


Object sliding on a frictionless floor

F F F F

F F F

F F F

net

net

net net net

x

y

x Y

1 20

30

20

30

2 2 2

50 35

50 35

sin cos

cos sin

Since v d is parallel to F W F dnet net0 0 , :

v0 0

W J153.

NF

NF

NF

0.10

00.4

00.3

3

2

1

Work = ?


Work Done by the Gravitational Force

Going up:

The work done by the gravity force reduced the kinetic energy of the object.

dgmW

dgmW

dFW

0180cos

Going down:

The work done by the gravity force increased the kinetic energy of the object.

dgmW

dgmW

dFW

00cos


Work Done Lifting & Lowering an Object

To raise an object to a new location, an applied force, F, must act to overcome the gravity force, Fg.

Consider a case where an object is at rest before and after the move.

Work- Energy Theorem:

Kf - Ki = Wa + Wg

But Kf = Ki = 0:

0 = Wa + Wg

Wa = - Wg

The work done by the applied force is the negative of the work done by the gravitational force.

Raising Lowering

dgmWa dgmWa The work done by the lifter.


Pull with constant force F.

F

F ma

F mg ma

F m g

2 0

12

Is less energy expended or less work done when using a pulley (or other such device)?

F

F

F


Pull with constant force F.

F

d2d

F ma

F mg ma

F m g

2 0

12

W F d

W m g d

( )( )2 W F d

W m g d

( )( )2

Same amount of work!

F

F


An object accelerated downwards

Work Energy Theorem:

gMT

gMaMmgT

amF

43

4

Mg

T

aa

d

00 v

dgMWT 43

dgMWg dgMWnet 41

dgv

dgMvM

WKK net

21

412

21

0

0

a = g/4drops dWork by cord ?Work by gravity?Final velocity


Work Done by a Spring Force

Compressing

Stretching

Relaxed

Hooke’s Law:

where k = spring constant (N/m)

The force varies linearly with displacement.

dkF


Work Done by a Spring Force

Compressing

Stretching

Relaxed Work done by the spring force:

W F x dx

W k x dx k x

W k x k x

S x

x

S x

x

x

x

S i f

i

f

i

f

i

f

( )

12

2

12

2 12

2

W k x xS i 12

2 0( )

Hooke’s Law: dkF

Sa WW


A block is dropped on a relaxed spring

W m g d

W k d

W W W

g

S

net g S

12

2

m g d k d mv 12

2 12

20

vk

md g d

v ms

2 2

35.

d

v

0v

ifnet KKW

m = 250 gk = 2.5 N/cmd = 12 cmWg ? Ws?v ?


Work – Kinetic Energy Theorem with a Variable Force

W F x dx m a dx

a dxdv

dtdx

dx

dtdv v dv

W m v dv mv

W KE KE

S x

x

xx

x

x

x

x

v

v

f i

i

f

i

f

i

f

i

f

( )

12

2


Problem

a x

ms

a x

ms

20

82 5

2 5

2 2.

.

W F x dx m x dx

W m x J

S x

x

x

x

i

f

i

f

( ) .

( )( ). .

2 5

10 8 8002 52

2

0

82 5

22

A 10 kg brick moves with the following variable acceleration; how much work is done on the brick in the first 8 seconds.


Power

Power = time rate at which Work is done due to a Force.

SI Units: 1 Watt = 1 Joule / s = 1 N m / s

English Units: horsepower

1 hp = 550 ft lbs / s = 746 Watts

Work = Power x Time

1 kW hr = 3.6 x 106 J

dt

dWP

For a constant force:

vFdt

dFdP

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Fundamentals of Physics