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Unit 4: Conservation of EnergyUnit 4: Conservation of Energy

• This is the halfway point! Chapters 1- 6 covered This is the halfway point! Chapters 1- 6 covered kinematics and dynamics of both linear and kinematics and dynamics of both linear and rotational motion. rotational motion.

• Chapters 7-12 focus on the conservation laws of Chapters 7-12 focus on the conservation laws of energy, momentum, and angular momentum and energy, momentum, and angular momentum and their applications.their applications.

• Although we can’t get into details, the conservation Although we can’t get into details, the conservation laws are deeply associated with symmetries of laws are deeply associated with symmetries of nature.nature.– Symmetry wrt to time leads to energy conservationSymmetry wrt to time leads to energy conservation– Symmetry wrt to position leads to momentum conservation.Symmetry wrt to position leads to momentum conservation.

• We’ll see that the application of the conservation We’ll see that the application of the conservation laws provides a new way to understand and solve laws provides a new way to understand and solve problems of motion.problems of motion.

• To start, we need to define work and then we’ll look To start, we need to define work and then we’ll look into the definition of energy.into the definition of energy.

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The Definition of WorkThe Definition of Work

• Merriam-Webster’s: Merriam-Webster’s: – 11:: activity in which one exerts strength or faculties to do  activity in which one exerts strength or faculties to do

or perform something or perform something – Followed by 10 more definitions!Followed by 10 more definitions!

• In physics, work has an exact definition In physics, work has an exact definition associated with the a force as it acts on an object associated with the a force as it acts on an object over a distance.over a distance.

• To be precise To be precise work is defined to be the product of work is defined to be the product of the magnitude of the displacement times the the magnitude of the displacement times the force parallel to the displacementforce parallel to the displacement..

• When you think about it the colloquial definition When you think about it the colloquial definition and the quantitative definition are not totally and the quantitative definition are not totally inconsistent. If you push that couch a distance inconsistent. If you push that couch a distance along the floor you’ve applied a force the entire along the floor you’ve applied a force the entire distance, and it sure feels like work!distance, and it sure feels like work!

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• F is the forceF is the force• d is the d is the

displacementdisplacement• is the angle is the angle

between the force between the force and the and the displacementdisplacement

• Note work is a scalar Note work is a scalar quantityquantity

• Unit is N-m = Joule(J)Unit is N-m = Joule(J)

cosFdW

dFW

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Limiting CasesLimiting Cases

• Motion and force in Motion and force in the same directionthe same direction

• = 0 and cos= 0 and cos=1=1• W=FdW=Fd• Example pushing Example pushing

the couch with the couch with 500N for 2m. The 500N for 2m. The work would be work would be 1000 Joules1000 Joules

• Holding a 20kg Holding a 20kg mass object: mass object: – Standing Still: Standing Still:

• The force you apply The force you apply equals mg=196Nequals mg=196N

• But d=0But d=0• W=FdcosW=Fdcos = F(0)=0 = F(0)=0

– Walking forward at Walking forward at constant velocity:constant velocity:• Force = 196NForce = 196N• d is nonzerod is nonzero• But cosBut cos= cos90 = = cos90 =

00• W=FdcosW=Fdcos = =

Fd(0)=0Fd(0)=0

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Example 1: Work on a CrateExample 1: Work on a Crate

• A 50-kg crate is pulled 40m by a 100N-force A 50-kg crate is pulled 40m by a 100N-force acting at a 37acting at a 37oo angle. The force of friction is 50N angle. The force of friction is 50N

• Determine the work done by the pulling, Determine the work done by the pulling, frictional, and net forces.frictional, and net forces.

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• Pick x along the displacement Pick x along the displacement vector.vector.

• The free body diagram shows 4 The free body diagram shows 4 forces:forces:

– FFGG=mg=50kg*9.8m/s=mg=50kg*9.8m/s22=490N=490N

– FFNN=490N=490N

– FFpp=100N (given)=100N (given)

– FFfrfr=50N (given)=50N (given)

• And the work:And the work:

– WWGG=(F=(FGG)(40m)cos(90)=0)(40m)cos(90)=0

– WWNN=(F=(FNN)(40m)cos(90)=0)(40m)cos(90)=0

– WWPP=(100N)(40m)cos(37)=3200J=(100N)(40m)cos(37)=3200J

– WWFF=(50N)(40m)cos(180)=-2000N=(50N)(40m)cos(180)=-2000N

• Note that the force pulling the mass Note that the force pulling the mass does positive work and the force of does positive work and the force of friction does negative work.friction does negative work.

• The net work is he sum or +1200JThe net work is he sum or +1200J

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Example 2: Work on a BackpackExample 2: Work on a Backpack

• A backpacker carries a 15.0 -kg pack up an A backpacker carries a 15.0 -kg pack up an inclined hill of 10.0m height.inclined hill of 10.0m height.

• What is the work done by gravity and the net What is the work done by gravity and the net force on the backpack? force on the backpack?

• Assume the hiker moves at a constant velocity Assume the hiker moves at a constant velocity up the hill.up the hill.

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• There are two forces on the packThere are two forces on the pack

– FFG G = mg = 147N= mg = 147N

– The force of the hiker holding the The force of the hiker holding the pack aloft Fpack aloft FH H = mg = +147N= mg = +147N

• The work done by the hiker is:The work done by the hiker is:

– WWHH= F= FHH(d)(cos(d)(cos) = F) = FHHh= h= 147N(10.0m)= 1470J147N(10.0m)= 1470J

– Only the height matters -not the Only the height matters -not the distance traveled.distance traveled.

• The work done by gravity is:The work done by gravity is:– WWGG= = FFGG(d)(cos(180-(d)(cos(180-))= F))= FGG(d)(d)

(-cos(-cos)=-F)=-FGG(d)(cos(d)(cos)=-F)=-FGGh= -h= -147N*(10.0m)=-1470J147N*(10.0m)=-1470J

– Once again only the height Once again only the height matters -not the distance matters -not the distance traveled.traveled.

• The net work is just the sum of the The net work is just the sum of the work done by both forces or 0.work done by both forces or 0.

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A Technical Note: The Scalar A Technical Note: The Scalar ProductProduct

• Although work is a scalar it Although work is a scalar it involves two vectors: force involves two vectors: force and displacementand displacement

• There is a mathematical There is a mathematical operation called the scalar operation called the scalar product that is very useful product that is very useful for the manipulation of for the manipulation of vectors required to vectors required to calculate work.calculate work.

• The scalar or dot product is The scalar or dot product is defined asdefined as

• Work can then be Work can then be rewritten asrewritten as

cosABBA

dFFdW

cos

cosABBA

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Properties of the Scalar ProductProperties of the Scalar Product

• Since A, B, and cosSince A, B, and cos are scalars the scalar product is are scalars the scalar product is commutative:commutative:

• The scalar product is also distributive:The scalar product is also distributive:

• If we define If we define AA=A=Axxii+A+Ayyjj+A+Azzkk, and , and BB=B=Bxxii+ B+ Byyjj+B+Bzzk, k, then then

ABBAABBA

coscos

CABACBA

)(

zzyyxx BABABABA

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1111

zzyyxx

zzyzxz

zyyyxy

zxyxxx

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kkBAjkBAikAB

kjBAjjBAijBA

kiBAjiBAiiBA

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kBjAjBjAiBjA

kBiAjBiAiBiA

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100

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Example 3: Using the Dot ProductExample 3: Using the Dot Product

• A boy pulls a wagon 100m with a force of 20N A boy pulls a wagon 100m with a force of 20N at an angle of 30 degrees with respect to the at an angle of 30 degrees with respect to the ground.ground.

• How much work has been done on the wagon?How much work has been done on the wagon?

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• With the axes shownWith the axes shown– FFPP=F=Fxxii+F+Fyyjj= (F= (FPPcoscos))ii+ (F+ (FPPsinsin))jj=17N=17Nii+10N+10Njj

– dd=100m=100mii

• AccordinglyAccordingly

dFdFdF

JNmNdFW

PP

P

coscos also is first term theNote

1700)0)(0()0)(10()100)(17(

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What about Varying Forces?What about Varying Forces?

• So far we’ve only considered the work of a So far we’ve only considered the work of a constant force. constant force.

• But it’s for more common to have a force But it’s for more common to have a force varying with positionvarying with position– A traveling rocket subject to diminishing gravityA traveling rocket subject to diminishing gravity– A simple harmonic oscillatorA simple harmonic oscillator– A car with uneven accelerationA car with uneven acceleration

• We could break the motion into small We could break the motion into small enough intervals so that the work is more enough intervals so that the work is more or less constant during each interval and or less constant during each interval and then sum the work of the segments. then sum the work of the segments.

• Basically this is the idea behind an integral. Basically this is the idea behind an integral.

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• Consider an object Consider an object traveling in the xy plane traveling in the xy plane and subject to a varying and subject to a varying force.force.

• We could just break the We could just break the trajectory into small trajectory into small enough intervals such that enough intervals such that the force is more-or-less the force is more-or-less constant during each constant during each interval.interval.

• So for any particular So for any particular interval labeled “i”:interval labeled “i”:

• And the total work for And the total work for seven intervals would be:seven intervals would be:

iii lFW cos

7

1

cosi

iii lFW

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• The sum can be The sum can be graphically represented ifgraphically represented if– Each shaded rectangle Each shaded rectangle

is one element of the is one element of the sum.sum.

– The curve represents The curve represents the function Fcosthe function Fcos..

– Then, the work Then, the work approximates the area approximates the area under the curve.under the curve.

• As we make As we make llii smaller and smaller and smaller, the sum of smaller, the sum of rectangles gives a better rectangles gives a better and better estimate of the and better estimate of the area under the curve.area under the curve.

• In fact as it approaches In fact as it approaches zero we get an exact result zero we get an exact result for the area and for the for the area and for the work:work:

i

iiil

lFWi

coslim0

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1919

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2020

In closingIn closing

• The precise definition of work is given byThe precise definition of work is given by

• Next we’ll do some examplesNext we’ll do some examples• You’ll see next it can also be interpreted as You’ll see next it can also be interpreted as

the amount of energy given to an object.the amount of energy given to an object.• Which opens the door to the conservation of Which opens the door to the conservation of

energy… Test Wednesday…See you Friday!energy… Test Wednesday…See you Friday!

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