WorkWork

Readings: Chapter 111

Readings: Chapter 11

Newton’s Second Law: Net Force is zero – acceleration is zero –velocity is constant – kinetic energy is constant

210 0 const const2netF a v K mv= ⇒ = ⇒ = ⇒ = =

Newton’s Second Law: If Net Force is not zero then: What is the relation between the change of Kinetic energy and the Net Force?

v Δ

aF

vnet

vF ma mt

Δ= =

Δ1ΔnetF

212net

vF v t m v t mv v m vt

ΔΔ = Δ = Δ = Δ

Δ1 212netF s m v KΔ = Δ = Δ

2s v tΔ = ΔWhere - displacement

vA B

anetF

B As x xΔ = −

If thenF const2 21 1( )

2 2net B A B AF x x K mv mv− = Δ = −

If then netF const=

2 2This equation gives the relation between the displacement and velocity. It is the same equation as

2 2 2 ( )B A B Av v a x x− = −

for the motion with constant acceleration (constant net force)

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for the motion with constant acceleration (constant net force)

vA B

anetF

B As x xΔ = −If then we have integral

netF const≠

2 21 12 2

Bx

net B Ax

F dx K mv mv= Δ = −∫Ax

If the net force depends also on the velocity then the relation becomes more complicatedcomplicated

B

A

x

netx

F dx∫netF sΔ or is the Work done by net forceBx

4

B

A

netx

W F dx= ∫

Then Work isThen Work is

cossW F s F sθ= Δ = Δs

This combination is called the dot product (or scalar product)

cosF s F sθΔ Δ

of two vectors

cosF s F sθΔ = ⋅ Δ

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Dot Product a Dot product is a scalar:

cosa b ab θ⋅ =θ

b

θ

If we know the components of the vectors then the dot product can be calculated as

y

ya a

x x y ya b a b a b⋅ = + xxa xb

y

b

a

yb b

If vectors are orthogonal then dot product is zero a

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g p090 cos 0θ θ= ⇒ = 0a b⋅ = b

090

Dot Product: properties cosa b ab θ⋅ =

aθ

( ) ( ) cosca b c a b cab θ⋅ = ⋅ =

( )a a b a b a b+ ⋅ = ⋅ + ⋅b

1 2 1 2( )a a b a b a b+ ⋅ = ⋅ + ⋅

Dot product is positive if 090θ < cos 0θ >

Dot product is negative if 090θ > cos 0θ <

The magnitude of is 5, the magnitude of is 2, the angle is

a bθ 060angle is

What is the dot product of and

θ 060a b

70 15 2cos 60 5 2 5

2a b⋅ = ⋅ = ⋅ ⋅ =

Work produced by a force, acting on an object, is the dot product of the force and displacement

W F s= ⋅ Δ

finalpoint

initial

W F s= ⋅ Δ∫Work has the same units as the energy:

2

1 1 1 1kg m kg mJ N m m⋅ ⋅= ⋅ = =initial

point

Example: What is the work produced by the tension force?

2 21 1 1 1J N m ms s

= ⋅ = =

Example: What is the work produced by the tension force?

10T N=

100Δ 100s mΔ =

045θ =

Δ0

cos10 100cos 45

W T s T s θ= ⋅ Δ = Δ =

= ⋅ =

8

sΔ 0 00cos 5707 707N m J= ⋅ =

0W F FΔ Δ >

final initialK K K W F sΔ = − = = ⋅ Δ

0W F s F s= ⋅ Δ = Δ >

final initialK K>f iv v>

( 0)a >f f i

0W F s= ⋅ Δ >

K K( 0)a >

final initialK K> f iv v>

0W F s= ⋅ Δ =

final initialK K=f iv v=

( 0)a =

f i

0W F s= ⋅ Δ <

9final initialK K< f iv v<( 0)a <

Work produced by a NET FORCE is equal to a change of KINETIC ENERGY (it follows from the second Newton’s law)ENERGY (it follows from the second Newton s law)

net net final initialW F s K K K= ⋅ Δ = Δ = −

The NET FORCE is equal to (vector) sum of all forces acting on the bj tobject

then1 2 ...netF F F= + +

then

1 2 1 2 1 2( ...) ... ...netW F F s F s F s W W K= + + ⋅ Δ = ⋅ Δ + ⋅ Δ + = + + = Δ

so the sum of the works produced by all forces acting on the object is equal to the change of kinetic energy,

101 2 ...W W K+ + = Δ

Example: What is the change of kinetic energy? Or what is the final velocity of the block if the initial velocity is 5 m/s? The mass of the block is 5 kgof the block is 5 kg.

Forces:

n normal force

gravitational force

tension

nwT

sΔ10T N= 045θ =

w

tension T

Work produced by the net force is equal to the change of kinetic energy.100s mΔ =

w

Since then2 21 12 2net net f iW F s K mv mv= ⋅ Δ = Δ = − netF n w T= + +

W W W W T T θΔ Δ Δ Δ

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cosnet n w TW W W W n s w s T s T s θ= + + = ⋅ Δ + ⋅ Δ + ⋅ Δ = Δ

cosTW T s T s θ= ⋅ Δ = Δ0wW w s= ⋅ Δ =0nW n s= ⋅ Δ =

Example: What is the change of kinetic energy? Or what is the final velocity of the block if the initial velocity is 5 m/s? The mass of the block is 5 kgof the block is 5 kg.

Forces:

n normal force

gravitational force

tension

nwT

sΔ10T N= 045θ =

w

tension T

2 2 21 1 1 cos2 2 2f i net imv mv W mv T s θ= + = + Δ100s mΔ =

w

cosT θ

This problem can be also solved by finding acceleration (this is the

2 2 cos2f iTv v s

mθ

= + Δ

12motion with constant acceleration). Then

2 2 2f iv v a s= + ΔcosTam

θ=

Example: Free fall motion.

Forces:

Work produced by the net force (gravitational

gravitational force wiy

p y (gforce) is equal to the change of kinetic energy.sΔ

2 21 1W F s K mv mvΔ Δ

w

y2 2net net f iW F s K mv mv= ⋅ Δ = Δ = −fy

2 21 1( )W W Δ Δ 2 2( )2 2net w i f f iW W w s mg s mg y y mv mv= = ⋅ Δ = Δ = − = −

then2 21 1mgy mv mgy mv+ = +

2 2i i f fmgy mv mgy mv+ = +

Conservation of mechanical energy

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Work produced by gravitational force can be written as the change of gravitational potential energy

Not for all forces the work can be written as the difference between the potential energy between two points. Only if the force does not depend on velocity then we can introduce potential energy as

initial finalW F s U U= ⋅ Δ = −

Potential energy depends only on the position of the object

initial final

It means also that the work done by the force does not depend on the trajectory (path) of the object:

1 2path path A BW W U U= = −

Such forces are called conservative forces

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Example 1: Gravitational Force

Conservative Forces

Example 1: Gravitational Force

U mgh=

Example 2: Elastic Force2

( )2

ff fxx x

xW Fdx kx dx k= = − = − =∫ ∫22

2

2 2

i i ix x x

fiinitial final

xxk k U U= − = −

∫ ∫

2xU k=2 2 f

2U k=

Nonconservative (dissipative) forces Example: Friction

Direction of friction force is always opposite to the direction of velocity, so the friction force depends on velocity.

Example: Friction

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The work done by friction force depends on path.

If we know force we can find the corresponding potential energy

( )final initialW F s U U U= ⋅ Δ = −Δ = − −

We assume that the potential (for example) at point A is 0, then the potential energy at

B B

B BU U F ds F ds= − ⋅ = − ⋅∫ ∫

point B is

B BA A∫ ∫

UΔIf we know potential we can find the force as s

UFs

Δ= −

ΔExample: Gravitational force: U mgy= y

mg yF mgΔ= − = −

Δ

ygy y g

yΔ

Example: Elastic force: 2

2x

U k= ( )2 / 2x

k xF kx

xΔ

= − = −Δ

16The exact relation between the force and potential:

2 xΔ( , , )

xU x y zF

x∂

= −∂

( , , )z

U x y zFz

∂= −

∂( , , )

yU x y zF

y∂

= −∂

Nonconservative (dissipative) forces

Example 1: Friction

Direction of friction force is always opposite to the direction of velocity, so the friction force depends on velocity.

The work done by friction force depends on pathThe work done by friction force depends on path.

kf1path 1 1 2 1 2( )path k k kW f s f s f s s= − − = − +

B

f

1path

2path1s

2s

3s2 3path kW f s= −

A

kf

kf

1 3

W W≠

3 1 2( )s s s≠ +

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A 1 2path pathW W≠

Example 2: External Force

The work done by the net force is equal to the change of kinetic energy.

The net force is the sum of conservative forces (gravitational force, elastic force, …) and nonconservative force (friction force ..). The work done by conservative forces can be written as the change ofwork done by conservative forces can be written as the change of potential energy. Then

W F s= ⋅ Δ =net net

conservative nonconservative

W F s

F s F s

U U F K K

= ⋅ Δ =

= ⋅ Δ + ⋅ Δ =

Δinitial final nonconservative final initialU U F s K K= − + ⋅ Δ = −

( ) ( )nonconservative final final initial initialF s U K U K⋅ Δ = + − + =

, ,

f f

mech final mech initialE E= −

Work done by nonconservative forces is equal to the change of

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Work done by nonconservative forces is equal to the change of mechanical energy of the system (sum of kinetic energy, gravitational potential energy, elastic potential energy, and so on).

Example

0v = 0initialv =

0finalv =h

Work done by friction force is equal to the change of mechanical energy:

f i ti k h fi l h i iti lW f ds E E mgh= ⋅ = − = −∫

gy

, ,friction k mech final mech initialW f ds E E mgh∫

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Nonconservative forces: friction (dissipative) forces and external forces.

Friction force decreases the mechanical energy of the system but increases the TEMPERATURE of the system – increases thermal energy of the system. Thenenergy of the system. Then

, ,friction thermal initial thermal finalF s E E⋅ Δ = −

, ,nonconservative external thermal initial thermal final

mech final mech initial

F s F s E E

E E

⋅ Δ = ⋅ Δ + − =

= −, ,mech final mech initial

, , , ,( ) ( )external mech final thermal final mech initial thermal initialF s E E E E⋅ Δ = + − +

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Work done by external forces is equal to the change of the total energy of the system (mechanical + thermal)

Work done by external forces is equal to the change of the total energy of the system (mechanical + thermal)

Thermal energy is also a mechanical energy – this is the energy ofThermal energy is also a mechanical energy this is the energy of motion of atoms or molecules inside the objects.

, , , ,( ) ( )external mech final thermal final mech initial thermal initialF s E E E E⋅ Δ = + − +

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