8.5

7.9 8.1 8.2 8.3 8.4

7.6 7.7 7.8

• •

•

• •

• • •

Gravitational Potential Energy Elastic Potential Energy

Isolated Systems

– – –

Potential Energy of a System Conservative and Non-Conservative Forces Relationship Between Conservative Forces and Potential Energy Energy Diagrams and Equilibrium Energy in Non-Isolated Systems Energy for Isolated Systems Kinetic Friction Changes in Mechanical Energy for Non-conservative forces Power

Finding the Force (Gradient) Turning Points Equilibrium Points

Power Reading Energy Diagrams

–

Conservation of Mechanical Energy Work Done by Kinetic Friction (Non-Conservative Forces) Conservation of Energy (General)

– –

Potential Energy Conservative Forces Determining Potential Energy Values

Copyright R. Janow – Spring 2012

Potential Energy & Energy Conservation SJ 8th Ed.: Chap 7.6 – 7.8, 8.1 – 8.5

Physics 111 Lecture 07

1 2

r r

Wsp =

• Total Energy is constant if system is isolated

Etot ≡ Emech + Eint

Copyright R. Janow – Spring 2012

∆Etot = Wext ( = 0 sometimes)

• Total Energy = Mechanical + Thermal and Internal

∆Emech = Wnc ( = 0 sometimes)

r Emech ≡ K + U(r )

• Mechanical Energy is constant when the work due to non-conservative forces is zero

• Mechanical Energy = Kinetic + Potential

• Conservative Forces 0.

A

F

Path 1

B

F

Conservative and Non-Conservative Forces

FORCE DUE TO REST OF SYSTEM

mg ∆ y ( −+ mg ∆y

falling

rising

r Fext

Copyright R. Janow – Spring 2012

r r r ∆Wext = Fext ( r ) o ∆y = ∆Ug = Ugyf ) − Ug ( yi ) = mg(yf − yi )

mg, so book moves up (speed remains ~ zero)

r F • ext is an external force, slightly greater than

Action of external, non-conservative force:

• The zero point of PE here is arbitrary – choose a reference level

∆ Ug =

POTENTIAL ENERGY CHANGE

r ∆y = ± ∆yˆj

• PE increases when Fg does negative work • PE decreases when Fg does positive work

r Fg ( r ) = − mgˆj constant force r r r r ∆Ug (r ) = − ∆Wg = − Fg ( r ) o ∆y

• Ther system is the apple plus the Earth

WORK DONE ON PARTICLE BY CONSERVATIVE FORCE

r r r r dWg = Fg ( r ) o d r ≡ − dUg (r )

Example: Gravitational Potential Energy

∆W = ∫

r r F o dr

in general

r r ∆W = F o ∆ r constant force

v rf

∆U = − ∆Wc

Work done by conservative force on ANY path

r r dU = − Fc o d r

dU Fcx = − dx

Copyright R. Janow – Spring 2012

similar for y & z

A conservative force is the gradient of a potential energy function

Path may be chosen to make the integration simple

ri

r r r r r ∆Uf ,i = − ∫ Fc (r ) o d r = U(rf ) − U(ri ) v

Change in potential energy of system

For a conservative force there can be a potential energy function: ∆Wc depends only on the endpoints, not the path details

path r F can be the net force, a dissipative force, or a conservative force r r ∆W is the work done by F (whatever F might be)

r r dW = F o d r

Recall: The work done by non-conservative forces can depend on the path:

Finding the Potential Energy

v

Fs

Lower: • Spring force still leftward • Block acquires v leftward • Positive work done on block • Potential energy of system decreases • PE grows again as block passes x = 0

Copyright R. Janow – Spring 2012

• Potential Energy is stored reversibly in deformed shape of the spring • Usually choose x = 0 as zero of U(x)

Fs

v

Upper: • Block starts at x = 0 with v rightward • Spring compresses, Fs is leftward • Negative work done on block • Potential energy of system increases, KE decreases • Block stops at some point

2

CHANGE IN SPRING POTENTIAL ENERGY

r 1 Uel (x) = kx 2

FORCE SPRING EXERTS ON BLOCK

r r r F(x) = −kx

WORK DONE ON BLOCK BY SPRING

r r r r dWs = Fs (x) o dx ≡ − dUs (x)

Block is attached to a spring

Example: Elastic Potential Energy due to a Spring

∆Us (f, i) =

1 kx f2 2

−

i

x 1 2 f kx 2 x

1 kxi2 2

∆Us (f, i) = + ∫ k x dx = + xi

xf

1 kx 2 2 indefinite integral

Us (x) =

potential function

xi

∆Us (f, i) ≡ Us (x f ) − Us (xi ) = − ∫ Fs (x)dx

Fs (x) = −kx

Us (x = 0) ≡ U0 = 0

Usually choose:

Us (x f ) = kx 2 + U0

1 2

potential well

positive parabola

reference level

The potential energy function itself is:

Copyright R. Janow – Spring 2012

∆Us is positive if |xf| > |xi| whether spring is stretched or compressed

• Substitute:

xf

• 1 dimensional motion along x

• Hooke’s Law restoring force

Potential Energy Function for the Block-Spring System

U(x) = mgy = mg f(x)

PE function can be written in terms of x:

y = f (x)

In example, roller coaster car forced to stay on the track described by

ri

r r ∆Ug (f, i) = + mg ∫ j o d r = mg(yf - yi ) v

v rf

mgyf

x

Copyright R. Janow – Spring 2012

But between endpoints rf, ri the potential difference is as above due to path independence

U’ = 0

mgyi

E

Gravitation is conservative so ∆U depends only on endpoints, not path Replace path by staircase-like segments: • horizontal steps that are workless – ∆U = 0 • vertical risers for which ∆Ui = mg∆ ∆yi Ug(x) • can choose reference level – only U=0 differences of U are meaningful

WORK DONE BY GRAVITY ON PARTICLE

r Over small regions near the Earth’s surface g is constant v r r rf r r r r r Fg ≡ −mg j ∆Ug (f, i) = − ∫ Fg (r ) o d r = Ug (rf ) − Ug (ri ) r r v ri g ≡ −g j

Application to Gravitation

C, D, E, A, B E, D, C, B, A A, B, C, D, E E, A, C, D, B B, D, C, A, E A

B

C

D

A. B. C. D. E.

C, D, A, B D, C, B, A B, C, A, D A, C, D, B B, D, C, A

A

D

E

Copyright R. Janow – Spring 2012

B C

7-2: In the figure the block is shown at its equilibrium point. The elastic potential energy of the block varies as it stretches or compresses the spring. Arrange points A, B, C, D, E in order of increasing potential energy.

A. B. C. D. E.

7-1: In the figure, the gravitational potential energy of a particle moving along the track varies. Arrange points A, B, C, D, E in order of decreasing potential energy.

Where is the Potential Energy Largest?

INCLUDES THERMAL & CHEMICAL ENERGY

Mass Electrical EM Transfer Transfer Radiation

W + Q + Tmw + Tmt + Tet + Ter

Mech Waves

Copyright R. Janow – Spring 2012

Mechanical Work: Include only Work due to Non-conservative forces

∑ (EnergyTransfers) =

Heat

Types of energy transfers into or out of a system

• The total energy of the Universe is constant – energy can not be created or destroyed. Energy transfers must balance the accounts of systems

∆Etot = 0

∆Etot = ∑ (EnergyTransfers)

Isolated systems: • Energy does not cross the boundary of the system • Total energy of the system is constant

Non- isolated systems: • Energy can cross the system boundary • Total energy of the system changes

Etot can change only if energy transfers into or out of the system

Etot = K tot + Utot + Eint

• Kinetic Energy: Associated with movement of members of a system • Potential Energy: Determined by the configuration of the system • Internal Energy: Related to the temperature of the system

Total Energy Etot of a system - Definition

Change in total energy

∆Etot ≡ Wnc

+

forces

∑U

Potential energy is exchanged reversibly with kinetic energy. if no other forces are acting ∆K tot = − ∆Utot Copyright R. Janow – Spring 2012

The potential energy includes only conservative forces (PE’s exist) acting on system Typically all conservative forces are included in the mechanical energy e.g., Ug Us Uelec Umag …

particles

∑K

For a system:

Emech ≡

Emech ≡ K + U

For a particle:

Work done by nonconservative forces

System may not be isolated Wnc includes work done by contact forces and friction Treat these as external to the system, neglect other transfers Often neglect changes in internal energy

Mechanical Energy - Definition

• • • •

Non-conservative forces may be doing mechanical work on system

Energy Conservation (for mechanics)

Emech = constant

ADVANTAGE OF ENERGY APPROACH:

or

tot,i

∆K tot = − ∆Utot

+ Utot,i = K tot,f + Utot,f

Copyright R. Janow – Spring 2012

• Need to look only at final and initial states • Details of forces may often be ignored

For a system of particles: K

For a single particle: K + U = K + U i i f f

Total mechanical energy is constant, but individual particles can exchange kinetic and potential energy

or

∆Emech = 0

Holds for isolated systems i.e., Wnc= 0 and when Eint is constant

Principle of Mechanical Energy Conservation

θ T

mg

∆s

Isolated system Conservative forces only due to gravitation No dissipation

Emech = K + Ug = constant

Copyright R. Janow – Spring 2012

Example: Conservation of Emech in a Pendulum Oscillator

Emech (init) = Emech(final) 1 mv 2f 2

vf =

SAME AS KINEMATICS FORMULAS (IN THIS CASE)

vf =

vi2 + 2g [h − y]

R. Janow Energy contains velocity squared, results do not depend onCopyright direction of v–iSpring 2012

Approach is the same, result is

Suppose ball is thrown upward instead with speed vi

Note: vi = 0

2g [h − y]

+ mgy

v f2 = vi2 + 2g [h − y]

+ mgh =

Ki + Ugi = K f + Ugf

1 mvi2 2

∆Emech = 0

Apply Mechanical Energy Conservation

A ball of mass m is dropped from rest at a height h above the ground. What is the ball’s speed when it is at height y above the ground. Neglect air resistance. The system is the ball + the Earth The system is isolated The only force acting (gravity) is conservative

Example: Ball in free Fall

Emech(top) = Emech(bot)

Same as kinematics formula

2 vbot = v 2top + 2g [ytop − ybot ]

SOLVE

USE:

1 1 2 2 mvtop + mgytop = mvbot + mgybot 2 2

K(top) + Ug (top) = K(bot) + Ug (bot)

∆Emech = 0

• Normal force N (non-conservative) does zero work • Net force (gravity) is conservative

Apply Conservation of Mechanical Energy • System (Earth + child + slide) is isolated

surface

BOTTOM

TOP

N

mg

y

θ

Copyright R. Janow – Spring 2012

vbot = 2gh

v top = 0 ybot = 0 ytop = h

The child (mass m) starts at rest at the top - a vertical distance h above the ground. The slide is frictionless. Find speed at bottom. Slope angle θ is not known, so cannot solve using Newton’s second Law alone

Example: Speed at Bottom of a Water Slide

∆K + ∆Us + ∆Ug = 0

d2 −

h = s − L − d = 2.1 m

Close! hope jumper is not too tall

m = 61 kg s = 45 m L = 25 m k = 160 N/m

s Low point Ef

Start point Ei

Note: Did not need detailed analysis of forces, acceleration, velocity Copyright R. Janow – Spring 2012

• Find low point h:

• Quadratic equation for d:

2mg 2mg d − L = 0 k k • Choose positive root: d = 17.91 m

∆Ug = − mg [L + d]

•Speed = 0 at jumper’s start and low point ∴ ∆K = 0 1 ∆Us = kd2 d = maximum Bungee stretch 2

∆Emech = Ef − Ei = 0

Mechanical Energy is conserved, meaning….

A Bungee jumper (mass m = 61 kg) steps off the cliff a distance s above the ground. He free-falls a distance L before the Bungee (exactly like a spring with constant k) begins to stretch. How close to the ground down he turn around? Does he hit? • Two conservative forces act, and no others • Gravity Ug(y) = mgy • Spring Us(x) = ½kx2 • The system (jumper + Earth + Bungee) is isolated

Example: Bungee Jumper

Wf ≡ + fk ∆x

Generalize for A) or B):

∆Eint = fk ∆x

∆Emech = Wnc = − Wf + ∑ Wother

∆Emech + ∆Eint = +F∆x

Copyright R. Janow – Spring 2012

Wother ≡ + F∆x

B) Consider system to be block + surface. Friction is internal and raises internal energy. Force F is external

∆Emech = Wnc = − fk ∆x + F∆x

A) Consider system to be block alone. Friction and contact force F are both external

vi

fs is work - less

Wother ≡ + F∆x

r r " Work" done W ≡ ∫ f o d r f k

• Block sliding rightward with friction and external force F • Normal force and weight do no work as they are perpendicular to the displacement along x-axis • No changes in gravitational potential energy • Forces are constant as block moves by ∆x

EXAMPLE

v fk ≡ kinetic friction force.

Kinetic friction forces dissipate energy by opposing the motion • Reduce the mechanical energy, converting it to heat • Work done by friction may be internal (increases Eint) or external

Friction in Energy Conservation

1

1

vf =

vi2 +

Fnet,ext

2 ∆x[ F − fk ] m

1 1 mv 2f = mvi2 − fk ∆x + F∆x 2 2

= N - mg ⇒ N = mg fk = µkN = µkmg = 0.15x 6.0x 9.8 fk = 8.2 N

∑ Fy = 0

Find fk

vi

0+

2 3.0[ 12 − 8.82 ] 6.0 Copyright R. Janow – Spring 2012

v f = 1.78 ≈ 1.8 m/s

vf =

Substitute numerical values:

∆Emech = ∆K + ∆Ug = K f − Ki = mv f2 − mvi2 2 2

∆Ug = 0

Wother = + F∆x

Gravitational potential energy is constant

Wf = − fk ∆x

∆Emech = Wnc = Wf + ∑ Wother

Apply mechanical energy conservation for a non-isolated system with friction

Find the speed of the block after it has moved a horizontal distance of 3.0 m, starting from rest. Assume: m = 6.0 kg F = 12 N µk = 0.15

Example: Block Pulled on a Rough Surface

∆Eint =

1 100x 202 + 100x9.8x(25 − 30) 2

vf = 0

1 1 mv02 + mgy0 − mv 2f − mgyf 2 2

0 =

1 mv02 + mgy0 − mgy'f 2

Values: m = 100 kg V0 = 20 m/s y0 = 25 m yf = 30 m

yf

∆Eint = 15,100 Joules

y0

v0

Copyright R. Janow – Spring 2012

y'f = y0 + v02 = 45.4 m 2g

1

How high would the car go if friction were absent?

Evaluate:

∆Eint =

∆Etot = ∆Emech + ∆Eint = Wnc = 0

• System = Earth + track + car • Friction is internal - increases Eint • Can not use Wf = fs∆x (or an integral) to evaluate work done by friction as friction force varies in an unknown way. System is isolated, so…

The roller coaster car starts with speed v0 when its height is y0 above the ground. It reaches a maximum height yf before sliding back. There is friction. Find the increase in thermal energy of the system.

Example: Roller Coaster with Friction

1 0 = − m2gh + kh2 + µkm1gh 2

h=0

or

2g h = [ m2 − µkm1 ] k

Variation: Suppose µk is known but want to find h Solve quadratic above. Roots are:

Substitute:

2

∆Eint = + W f = fkh = µkm1gh

s

m1g

m2g − kh

Copyright R. Janow – Spring 2012

What does the system do next?

µk =

∆Etot = 0 = ∆Emech + ∆Eint = ∆K + ∆Ug + ∆Us + ∆Eint 1 ∆U = kh2 ∆Ug = − m2gh ∆K = 0 1 2

m1g

Define the system to be both blocks + Earth + spring + table + cord System is isolated: total energy is constant, but friction converts potential and kinetic energy to heat (internal energy).

Cord is un-stretchable and massless. The system is initially at rest with the spring neither stretched nor compressed. The hanging mass falls a distance h and comes to rest. Find the sliding friction coefficient µk.

N

Example: Connected Blocks in Motion with Friction

P≡

∆W ∆t

dW ∆W = Lim dt ∆ t → 0 ∆ t

Pavg ≡

r r P= Fov

dW = P dt r r dW = F o d r

∆W = ∫ P dt

P not necessarily constant

Power is scalar [P] = Joules/s = Watts

Copyright R. Janow – Spring 2012

1 Watt ≡ 1 Joule/s = 0.738 ft.lb/sec 1 horsepower ≡ 1 hp = 746 Watts = 550 ft.lb/sec 1 kilowatt - hour ≡ 1000 watts x 3600 sec = 3.6 x 106 J.

r d rr dW ∴ P= = Fo dt dt Units and Conversions

Work done by force F and displacement dr

Infinitesimal work is done in time dt

Instantaneous Power

∆W = Pavg∆t

Work done in time ∆t

Power is the rate of energy transfer (using mechanical work for now). Average Power:

Power

Since v is changing, the net power is also changing

Copyright R. Janow – Spring 2012

Net power transfer is positive, so KE is increasing as is v

P1 = − 6.0 Watts - unchanged r r P2 = F2 o v = 6.0 × 3.0 × cos(60o ) = + 9.0 Watts

Now let the magnitude of F2 = 6.0 N. What changes?

Pnet = − 6.0 + 9.0 = 3.0 Watts

F2 is supplying energy

F1 is drawing energy

Net power transfer = 0, so KE is constant as is v

Since v is constant, the net power is also constant

Pnet = P1 + P2 = − 6.0 + 6.0 = 0.0 Watts

Apply instantaneous power formula

r r P= Fov r r P1 = F1 o v = 2.0 × 3.0 × cos(180o ) = − 6.0 Watts r r P2 = F2 o v = 4.0 × 3.0 × cos(60o ) = + 6.0 Watts

Two constant forces act as shown on a box sliding across a frictionless surface F1 = 2.0 N, F2 = 4.0 N, v = 3.0 m/sec Find the power transferred by each force and the net power. Is Pnet changing?

Example: Power done by forces acting on a box