9/17/2013PHY 113 C Fall 2013 -- Lecture 71 PHY 113 C General Physics I 11 AM – 12:15 PM MWF Olin 101 Plan for Lecture 7: Chapter 7 -- The notion 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 if


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

mgnfamf SS if

via m

f

gamgam SS 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

2

1

:on acceleraticonstant For

ii

ii

i

vtvxtxa

attvxtx

atvtv

a

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 Horizontal

0sin :forces Vertical s

NP

mgPN

f

mg

N

f



0cos :forces Horizontal

0sin :forces Vertical s

NP

mgPN

f

mg

N

f

N 48.57

N 72.31

50cos25.050sin

N 8.93

cossin :for Solving

0sincos

cos

s

s

ooP

mgPP

mgPP

PN



F

ra

aF

ˆ :case In this2

r

v

m


Preparation for the introduction of work:

Digression on the definition of vector “dot” product

q

0 then ,90 if that Note

cos

BA

BAo

AB


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

q

(scalar) 5.37120cos)15)(5(

120 ,15 ,5 :Example

cos

o

o

BA

BA

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

BA

jiBjiA


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(rf-ri)<0

mg

ri

rf

W=-mg(rf-ri)>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

x

xfi dxFdW rFr

r

:case In this


Example:

JmNmNdxFdWf

i

f

i

x

x

xfi 2525)4)(5( 21 rF

r

r


Example – spring force: Fx = - kx


x

F

Positive work

Negative work


Detail:

22212

21 if

x

x

x

x

x

x

xfi

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


FP

q

mg

n

fk

xi xf

Assume FP sin q <<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

dtdtd

mdtdtd

dtd

mddtd

mdmdf

i

f

i

f

i

f

i

f

i

total vvrv

rv

rarFr

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 energy

Some more details:

Consider Newton’s second law:

Ftotal = m a Ftotal · dr= m a · dr

dtdtd

mdtdtd

dtd

mddtd

mdmdf

i

f

i

f

i

f

i

f

i

t

t

t

ttotal v

vrvr

vrarF

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

dtd

mf

i

f

i

vvvvv

v 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

2

1

2

1if

f

i

totalfi mvmvdW rF

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



22

2

1

2

1if

f

i

totalfi mvmvdW rF

222

:constant is if :Note

ifif

ififtotal

f

i

totalfi

total

vv

mdW

rra

rrarrFrF

F



22

2

1

2

1if

f

i

totalfi 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

0iv

22

2

1

2

1iffi 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

2

1

2

1iffi 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

2

1

2

1

2

1iiffi mvmvmvW

f

gd

vmgdmv

mgdfdfdxdW

ikki

k

x

x

fi

f

i

f

i

2or

2

1

22

rFr

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

xm

kv

mvmvkxW

f

i

if

ififi

/63.02.05.0

5

2.0 /5 5.0For

2

1

2

1

2

1 222

0


Special case of “conservative” forces conservative non-dissipative

if

f

i

fi UUdW rrrF

F

write topossible isit , forces edissipativ-nonFor

iiff

ifif

f

i

fi

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

i

fi

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.

dx

dUF

dU

x

ref

that Note

:functionenergy potential Define

rFrr

r


(constant) 2

1

2

1

:Or2

1

2

1

22

22

EUmvUmv

mvmvUUW

iiff

ififfi

rr

rr

Work-Kinetic Energy Theorem for conservative forces:


Energy diagrams 22

2

1

2

1kxmvE

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

Documents

9/17/2013PHY 113 C Fall 2013 -- Lecture 71 PHY 113 C General Physics I 11 AM – 12:15 PM MWF Olin 101 Plan for Lecture 7: Chapter 7 -- The notion of work