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Frank L. H. Wolfs Department of Physics and Astronomy, University of Rochester, Lecture 09,

Physics 141.Lecture 9.

Frank L. H. Wolfs Department of Physics and Astronomy, University of Rochester, Lecture 09, Page 2

Outline.

• Course information:• Homework set #4 and lab report # 2.

• Continue our discussion of Chapter 6:• Energy• Potential energy


Course information.Homework and labs.

• Homework # 4 is due on Friday October 7 at 12 pm (noon).

• Lab report # 2 is due on Wednesday October 5 at 12 pm(noon).

• Homework set # 5 is due on Friday October 14 at 12 pm(noon).

End of Course Information - back to Physics!

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The Personal Response System (PRS).Quiz Lecture 9 – All answers are correct.


The energy principle.

• The energy principle states that the change in energy of asystem (∆Esystem) is equal to the work done by thesurroundings (Wsurr) and the energy flow (Q) between thesystem and surroundings due to a difference in temperature:

∆Esystem = Wsurr + Q

• The work W done by the force F is defined as

where q is the angle between the force F and thedisplacement d.

• If Q > 0 J, energy flows into the system (e.g. Tsys < Tsurr). IfQ < 0 J, energy flows out of the system (e.g. Tsys > Tsurr).

W =F id = Fd cosθ



• We have already seen that there is a connection between the work doneby a force and the change in the speed of the object:

• If W > 0 J: speed increases

• If W = 0 J: speed remains constant

• If W < 0 J: speed decreases

Assume v1 = 0 m/s

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• Consider the bus starting from rest (v1 = 0 m/s) and having anacceleration a = F/m (assume v << c). The velocity at a later time t willbe equal to

• This relation can be used to determine the time t at which the busreaches a certain velocity v2: t = v2/a.

v t( ) = v0 + at = at

Assume v1 = 0 m/s



• The displacement at this time t is equal to

• The work done by the force F during this period is equal to

KineticEnergy K

d = x t = v2

a⎛

⎝⎜

⎞

⎠⎟ = x0 + v0t +

12 at2 = 1

2 a v2

a⎛

⎝⎜

⎞

⎠⎟

2

= 12

v22

a

W =

Fid = ma 1

2v2

2

a⎛

⎝⎜

⎞

⎠⎟ = 1

2 mv22

Assume v1 = 0 m/s



• Assuming that there is no heat flow in or out of the bus, we concludethat for low velocities (non-relativistic velocities):

The net work done on an object is equal to the change in its kinetic energy.

• In the case of the bus: Fnetd = ∆E = (1/2)mv22 - (1/2)mv1

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Assume v1 = 0 m/s

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• The relation between the change in the energy E and thework W (in the absence of heat flow) can be taken as thedefinition of energy:

• Consider that the particle, subjected to a force F, moves adistance dx along the x axis. The change in energy as aresult of this motion will be

ΔE =W

ΔE =

F idr = FxΔx =

Δpx

Δt⎛

⎝⎜⎞

⎠⎟Δx


The energy principle.• The previous relation can be rewritten as

• We know how p is defined, and we now have to determinewhat expression for E will satisfy the energy principle.

• Since E is differentiated with respect to x we want to rewritedv/dt in the following manner:

ΔEΔx

=Δpx

Δt or, if Δt → 0,

dEdx

=dpx

dt

dpx

dt= d

dtmv

1− v2 / c2

⎛

⎝⎜

⎞

⎠⎟ =

m

1− v2 / c2( )3/2

dvdt

dvdt

=dvdx

dxdt

= vdvdx

⇒ dEdx

=dpx

dt=

mv

1− v2 / c2( )3/ 2

dvdx



• By differentiating the following expression for E withrespect x we can show that this expression satisfies thework-energy theorem:

• The is the energy of a particle of rest mass m, moving withvelocity v.

• We note that even if the particle is not moving, it will have anon-zero energy, equal to mc2. This energy is called the restenergy of the particle.

E = mc2

1− v2 / c2

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• When the particle is moving, its energy is larger than its restenergy, and this excess of energy is called the kineticenergy K of the particle:

• In the non-relativistic limit, v << c, we find for K

K = E − mc2 = mc2 1

1− v2 / c2−1

⎛

⎝⎜

⎞

⎠⎟

K mc2 1+ 1

2v2

c2 −1⎛

⎝⎜⎞

⎠⎟= 1

2mv2


Relativistic energy and momentum.A few remarks.

• Using our definitions of energy and momentum we concludethat

• These two relations can be combined in the following way

E2 =

mc2( )2

1− v2 / c2 p2 =mv( )2

1− v2 / c2

E2 − p2c2 =

mc2( )2

1− v2 / c2 1− v2 / c2{ } = mc2( )2


Relativistic energy and momentum.A few remarks.

• Consider the relation between E and p:

• The right-hand side depends only on the rest mass of the particle,which is the same in each reference frame.

• Since the velocity of the particle is a relative parameter, bothenergy and momentum will change when we move to differentreferences frames.

• However, the previous discussion shows that the differencebetween E2 and (pc)2 is an invariant quantity (independent of thereference frame being used).

• If the particle has no mass (e.g. a photon) then the right-hand sideis zero, we we conclude that E = pc. We thus see that we can stillassign a linear momentum to a massless particle: p = E/c.

E2 − p2c2 = mc2( )2

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2 Minute 32 Second Intermission

• Since paying attention for 1 hourand 15 minutes is hard when thetopic is physics, let’s take a 2minute 32 second intermission.

• You can:• Stretch out.• Talk to your neighbors.• Ask me a quick question.• Enjoy the fantastic music.• Go asleep, as long as you wake up

in 2 minutes and 32 seconds.


The energy principle (Q = 0).Multi-particle systems.

• When we are dealing with multi-particle systems we have toseparate the forces acting on the system into two groups:• Internal forces: forces that act between the particles that make up the system.

The work done by the internal forces is Wint.• External force: forces that are generated due to interactions between the

system and its surroundings. The work done by the external forces is Wext.• The change in the energy of the particles that make up the system is

equal to

• The opposite of the work done by the internal forces on the systemresults in a change in the potential energy U of the system

ΔEi

i∑ =Wext +Wint

ΔEi

i∑ + −Wint( ) = ΔEi

i∑ + ΔU =Wext


The energy principle (Q = 0).Multi-particle systems.

• The total energy of the system (Esystem) is defined as

• The work-energy theorem for the system, expressed in termsof the total energy of the system, is thus equal to

• The total energy (the energy of the system and the energy ofits surroundings) is a conserved quantity:

Esystem = Ei

i∑ +U

ΔEsystem =Wext

ΔEsystem + ΔEsurroundings = 0

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The potential energy U.

• The potential energy U is the energy associated with theinteraction between the constituents of the system.

• The change in the potential energy of a pair of particles withan interaction force F acting between them is given by

• If the configuration of the system does not change (e.g. arigid object), its potential energy will also not change.

ΔU = −

F1,2 iΔ

r1 +F2,1iΔ

r2( ) = −F2,1i Δ

r2 − Δr1( ) = −F2,1iΔ

r21


Calculating the potential energy U.One dimension.

• Per definition, the change in potential energy is related tothe work done by the force:

• The potential energy at x can thus be related to the potentialenergy at a point x0:

• When we apply conservation of energy, we are in general only concerned with changes in the potential energy, ΔU, and not the actual value of U.

• We are free to assign a value of 0 J to the potential energy when the system is in its reference configuration.

U x( ) =U x0( ) + ΔU =U x0( )− F x( )dx

x0

x

∫

ΔU = −W = − F x( )dx

x0

x

∫


Calculating the potential energy. The gravitational potential energy - Part I.

• The gravitational force close tothe surface of the Earth is -mg.

• When we move a box from thesurface (y = 0 m) to a height h (y= h) the work done by the force isequal to -(mg)h. Note: the workis negative since the force and thedisplacement are pointing inopposite directions.

• The potential energy associatedwith the position of the box isU(y) = U(0) - W = mgh.

h

U = 0 Jy = 0 m

Note: reference position is the surface. The potential energy at this position is 0 J.

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The energy principle. Example 1.

• An object with mass m is at restat time t = 0 s. It falls under theinfluence of gravity through adistance h. What is its velocity atthat point?

• Solution:• Assume Q = 0.• Work done by the gravitational

force = mgh.• Change in energy = change in

kinetic energy = (1/2)mv12.• The Energy Principle (Q = 0):

mgh = (1/2)mv12 or v1 = √(2gh)

y0 = 0 m

v0 = 0 m/s

y1 = -h

v1 = ?

y-axis

W

W


Conservation of energy.Dissipative forces.

• When dissipative forces, such as friction forces, are present,the total energy of a system will no longer be conserved andenergy will be exchanged between the system and itssurroundings.

• The amount of energy being exchanged will depend on thepath along which the system evolved, and we can not assigna potential energy with these forces (nor will a motion alongthe path in opposite direction restore the energy that wemoved from the system to its environment.

• The amount of energy dissipated by these non-conservativeforces can be calculated if we know the magnitude anddirection of these forces along the path followed by theobject.


Problems with dissipative forces.Example 2.

• A ball bearing whose mass is m isfired vertically downward from aheight h with an initial velocityv0. It buries itself in the sand at adepth d. What average upwardresistive force f does the sandexert on the ball as it comes torest?

• The initial total energy of thesystem (the bearing + the earth) isequal to

m

vo

h

df

y = 0

U = 0

Ei = mgh + 1

2mv0

2

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Problems with dissipative forces.Example 2 (continued).

• The final energy of the system isequal to just the potential energyof the bearing which is equal toEf = -mgd.

• The energy of the ball bearing isnot conserved since the frictionforce between the bearing and itsenvironment dissipates some ofthe system energy. The workdone by the friction force is equalto Wf = -fd.

• The change in the energy of thesystem is

m

vo

h

df

y = 0

U = 0

ΔE = E f − Ei = −mgd( )− mgh + 1

2mv0

2⎛⎝⎜

⎞⎠⎟= −mg h + d( )− 1

2mv0

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Problems with dissipative forces.Example 2 (continued).

• The change in the mechanicalenergy of the system is must beequal to the work done by thefriction force (assuming this isthe only dissipative force actingon the system):

• This relation can now be used tocalculate the magnitude of thefriction force f:

m

vo

h

df

y = 0

U = 0

− fd = −mg h + d( )− 1

2mv0

2

f = mg 1+ h

d⎛⎝⎜

⎞⎠⎟+

mv02

2d


That is all for today.More energy next lecture!

This object is not in a state of thermal equilibrium.

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