PHY 113 C General Physics I 11 AM – 12:15 P M MWF Olin 101 Plan for Lecture 7: Chapter 7 -- The notion of work and energy Definition of work

PHY 113 C Fall 2013 -- Lecture 7 19/17/2013

PHY 113 C General Physics I11 AM – 12:15 PM MWF Olin 101

Plan for Lecture 7:

Chapter 7 -- The notion of work and energy1. Definition of work2. Examples of work3. Kinetic energy; Work-kinetic energy

theorem4. Potential energy and work; conservative

forces


7.3,7.15,7.31,7.34


Webassign questions for Assignment 6 -- #1

52. Consider a large truck carrying a heavy load, such as steel beams. … Assume that a 10,000-kg load sits on the flatbed of a 20,000-kg truck initially moving at vi=12 m/s. Assume that the load on the truck bed has a coefficient of static friction of mS=0.5. When the truck is braked at constant force, it comes to rest in a distance d. What is the minimum stopping distance d such that the load remains stationary relative to the truck bed throughout the breaking?

via

f

m

mgnfmaf SS mm if


Webassign questions for Assignment 6 -- #1 -- continued

mgnfamf SS mm if

via m

f

gamgam SS mm or

iclicker exercise --Do we have enough information to calculate a?

A. YesB. No


Webassign questions for Assignment 6 -- #1 -- continued

22

2

2 :algebra someAfter 21

:on acceleraticonstant For

ii

ii

i

vtvxtxa

attvxtx

atvtva

via m

f


A block of mass 3 kg is pushed up against a wall by a force P that makes an angle of q=50o

with the horizontal. ms=0.25. Determine the possible values for the magnitude of P that allow the block to remain stationary.

0cos :forces Horizontal0sin :forces Vertical s

NPmgPN

qqmf

mg

N

f



0cos :forces Horizontal0sin :forces Vertical s

NPmgPN

qqmf

mg

N

f

N 48.57N 72.31

50cos25.050sinN 8.93

cossin :for Solving

0sincoscos

s

s

ooP

mgPP

mgPPPN

qmq

qqmq



F

ra

aF

ˆ :case In this2

rv

m


Preparation for the introduction of work:

Digression on the definition of vector “dot” product

q

0 then ,90 if that Note

cos

BABA

o

AB

q

q


Digression: definition of vector “dot” product -- continued

q

(scalar) 5.37120cos)15)(5(

120 ,15 ,5 :Example

cos

o

o

BA

BAq

q

BA

AB


Digression: definition of vector “dot” product – component form

q

yyxx

yxyx

BABA

BBAA

BA

jiBjiA ˆˆ and ˆˆ Suppose

Note that the result of a vector dot product is a scalar.

103412 ˆ31̂ and ˆ4ˆ2 :Example

BAjiBjiA


rFr

rdW

f

i

fi

Definition of work: F

dr

ri rj

cal 0.239 J 1

JoulesmetersNewtons Work : workof Units


Units of work:

work = force · displacement = (N · m) = (joule)

• Only the component of force in the direction of the

displacement contributes to work.• Work is a scalar quantity.• If the force is not constant, the integral form must be used.• Work can be defined for a specific force or for a combination of

forces

rFr

rdW

f

i

11 rFr

rdW

f

i

22 212121 )( WWdWf

i

rFFr

r


iclicker question: A ball with a weight of 5 N follows the trajectory shown. What is the work done by gravity from the initial ri to final displacement rf?

(A) 0 J (B) 7.5 J (C) 12.5 J (D) 50 J

1m1m

2.5m

ri

rf

10 m


mg

ri

rf

W=mg(rfri)<0

mg

ri

rf

W=mg(rfri)>0

Gravity does negative work:

Gravity does positive work:

0r 0r


Work done by a variable force: rFr

rdW

f

i

fi

f

i

f

i

x

xxfi dxFdW rF

r

r

:case In this


Example:

JmNmNdxFdWf

i

f

i

x

xxfi 2525)4)(5( 2

1 rFr

r


Example – spring force: Fx = - kx


x

F

Positive work

Negative work


Detail:

22212

21 if

x

x

x

x

x

xxfi

xxkkx

dxkxdxFdW

f

i

f

i

f

i

f

i

rFr

r


More examples:

Suppose a rope lifts a weight of 1000N by 0.5m at a constant upward velocity of 4.9m/s. How much work is done by the rope?

(A) 500 J (B) 750 J (C) 4900 J (D) None of these

Suppose a rope lifts a weight of 1000N by 0.5m at a constant upward acceleration of 4.9m/s2. How much work is done by the rope?

(A) 500 J (B) 750 J (C) 4900 J (D) None of these


FPq

mg

n

fk

xi xf

Assume FP sinq <<mg

Work of gravity? 0

Work of FP? FP cos q (xf-xi)

Work of fk?mkn (xf-xi)=mk(mg- FP sin q (xf-xi)

Another example


iclicker exercise:Why should we define work?

A. Because professor like to torture students.

B. Because it is always good to do workC. Because it will help us understand

motion. D. Because it will help us solve the energy

crisis.

Work-Kinetic energy theorem.


Back to work:F

dr

ri rj

rFr

rdW

f

i

fi


Why is work a useful concept?

Consider Newton’s second law:

Ftotal = m a Ftotal · dr= m a · dr

dtdtdmdt

dtd

dtdmd

dtdmdmd

f

i

f

i

f

i

f

i

f

i

total vvrvrvrarFr

r

r

r

r

r

r

r

r

r

Wtotal = ½ m vf2 - ½ m vi

2

Kinetic energy (joules)


Introduction of the notion of Kinetic energySome more details:

Consider Newton’s second law:

Ftotal = m a Ftotal · dr= m a · dr

dtdtdmdt

dtd

dtdmd

dtdmdmd

f

i

f

i

f

i

f

i

f

i

t

t

t

ttotal vvrvrvrarF

r

r

r

r

r

r

Wtotal = ½ m vf2 - ½ m vi

2

Kinetic energy (joules)

22

21

21

21

if

f

i

t

tmvmvmddmdt

dtdm

f

i

f

i

vvvvvv

v

v


Kinetic energy: K = ½ m v2

units: (kg) (m/s)2 = (kg m/s2) m

N m = joules

Work – kinetic energy relation:

Wtotal = Kf – Ki


Kinetic Energy-Work theorem

22

21

21

if

f

itotalfi mvmvdW rF

iclicker exercise:Does this remind you of something you’ve seen recently?A. YesB. No



22

21

21

if

f

itotalfi mvmvdW rF

222

:constant is if :Note

ifif

ififtotal

f

itotalfi

total

vv

mdW

rra

rrarrFrF

F



22

21

21

if

f

itotalfi mvmvdW rF

Example: A ball of mass 10 kg, initially at rest falls a height of 5m. What is its final velocity?

i

f

h

??fv

0iv22

21

21

iffi mvmvmghW

0

smmghv f /899.9)5)(8.9)(2(2


ExampleA block, initially at rest at a height h, slides down a frictionless incline. What is its final velocity?

hh=0.5m

22

21

21

iffi mvmvmghW

0

smmghv f /13.3)5.0)(8.9)(2(2


ExampleA block of mass m slides on a horizontal surface with initial velocity vi, coming to rest in a distance d.

vi vf=0

d

1. Determine the work done during this process.2. Analyze the work in terms of the kinetic friction

force.


Example -- continuedvi vf=0

d

222

21

21

21

iiffi mvmvmvW

f

gdvmgdmv

mgdfdfdxdW

ikki

k

x

xfi

f

i

f

i

2or

21

22

mm

mrFr

r


Example A mass m initially at rest and attached to a spring compressed a distance x=-|xi|, slides on a frictionless surface. What is the velocity of the mass when x=0 ?

k

smv

mxmNkkgm

xmkv

mvmvkxW

f

i

if

ififi

/63.02.05.0

5

2.0 /5 5.0For

21

21

21 222

0


Special case of “conservative” forces conservative non-dissipative

if

f

ifi UUdW rrrF

F

write topossible isit , forces edissipativ-nonFor

iiff

ifif

f

ifi

mgyUmgyU

mgymgyyymgdW

)( and )(

:Earth of surfacenear gravity of Example

rr

rF


2

212

21

2212

21

)( and )(

:force spring of Example

iiff

if

x

x

f

ifi

kxUkxU

kxkxdxxkdWf

i

rr

rF

k


iclicker exercise:Why would you want to write the work as the difference between two “potential” energies?

A. Normal people wouldn’t.B. It shows a lack of imagination.C. It shows that the work depends only on the

initial and final displacements, not on the details of the path.

dxdUF

dU

x

ref

that Note

:functionenergy potential Define

rFrr

r


(constant) 21

21

:Or21

21

22

22

EUmvUmv

mvmvUUW

iiff

ififfi

rr

rr

Work-Kinetic Energy Theorem for conservative forces:


Energy diagrams 22

21

21 kxmvE

2max2

1221

2max2

1

max

,0(0)

:0 when :Note ,0

: when :Note

kxmvU

x kxEv

xx


Example: Model potential energy function U(x) representing the attraction of two atoms

PHY 113 C General Physics I 11 AM – 12:15 P M MWF Olin 101 Plan for Lecture 7: Chapter 7 -- The notion of work and energy Definition of work

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