1

⇒ ⇔ ×→ ≈ ^

10.2 Kinetic and gravitational potential energy p. 269-274 10.3 Closer look at gravitational potential energy p. 274-278

10.4 Restoring forces and Hooke’s law p. 278-280

10.5 Elastic potential energy p. 280-284

10.6 Elastic collisions p. 284-288

Chapter 10: Energy

10.7 Energy Diagrams p. 288-293

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10.2 Kinetic and gravitational potential energy p. 269-274

Chapter 10: EnergyPh

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Chapter 10: Energy

Let us start with an object in free-fall

The sum of kinetic energy K

and Gravitational potential energy U

is not changed by free fall.

V = V - 2g(y - y )

Which can be re-written as

V + 2gy = V + 2gy

f

f

i

i if

f i2 2

2 2

Multiply left hand side and right hand side by m/2

m V + m g y = m V + m g y

12

12

2 2

f if i

From Kinematics we know that

Final total energy Initial total energy=

g (K + U ) = (K + U ) g gfinal initial

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Chapter 10: Energy

K = m v

U = m g y

Kinetic energy

Gravitational potential energy

212

g

FACTS:

Kinetic energy is is always positive regardlessif the particle is moving up or down or left or right (it involves velocity of particle squared v )

Gravitational potential energy can be negative (it involves position of the particle y)

Unit of energy is Joule with 1 Joule = 1 kg m / s = 1 J2 2

2

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Rank in order, fromlargest to smallest, thegravitational potentialenergies of balls 1 to 4.

1. (Ug)1 > (Ug)2 > (Ug)3 > (Ug)4 2. (Ug)4 > (Ug)3 > (Ug)2 > (Ug)1

3. (Ug)1 > (Ug)2 = (Ug)4 > (Ug)3

4. (Ug)3 > (Ug)2 = (Ug)4 > (Ug)1

5. (Ug)4 = (Ug)2 > (Ug)3 > (Ug)1

Chapter 10: Energy

Cliction 10.1

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Chapter 10: Energy

Energy bar charts

There are no numerical scaleson a bar chart, but YOU shoulddraw the bar heights proportionalto the amount of each type of energy.

When the pebble reaches its highest point (v=0 which implies K=0; chart below)

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Chapter 10: Energy

Since U = mgy The value of U depends on where you choose to PUTThe ZERO of your coordinate system!g

g

U = (1kg)(9.81)(1m) = 9.8 J

U = (1kg)(9.81)(0m) = 0!

gg

HOWEVER, they measure the samechange in U since:

Betty U - U = mg(y - y ) = mg(0-1) = - 9.8 J Bill U - U = mg(y - y ) = mg(-1-0) = - 9.8 J

g

gf

gf

gi

gi

f

f

i

i

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Chapter 10: Energy

When they apply energy conservationthey both AGREE and find that the rock hits the ground with speed:

V = -2(U - U )/m = 4.43 m/sgf gif

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10.3 Closer look at gravitational potential energy p. 274-278

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Chapter 10: Energy

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The change in gravitational potential energy dependsonly on the height difference, not on the shape of thesurface.

This means we can use energy conservation to find thespeed of a sled, for instance (Ex. 10.3 in textbook)

Show that V = 10.1 m/sfinal

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A small child slides down the four frictionless slides A–D.Each has the same height. Rank in order, from largest tosmallest, her speeds vA to vD at the bottom.

1. vA = vB = vC = vD

2. vD > vA = vB > vC

3. vD > vA > vB > vC

4. vC > vA = vB > vD

5. vC > vB > vA > vD

Chapter 10: EnergyCliction 10.2

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Chapter 10: Energy

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Conservation of Mechanical Energy is also valid if theparticle moves on a frictionless surface (e.g., a roller-coaster) regardless of the shape.

For ENERGY CALCULATIONS, the y-axis isspecifically a vertical axis. Gravitationalpotential energy depends on the HEIGHTabove the earth’s surface. A tilted coordinatesystem, such as we often used in dynamicproblems, does NOT work for problems withgravitational potential energy!

Instead a y-axis and s-axis coordinate systemMight be used.

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A box slides along the frictionless surface shown in thefigure. It is released from rest at the position shown. Is thehighest point the box reaches on the other side at level a, atlevel b, or level c?

1. At level a2. At level b3. At level c

Chapter 10: Energy

Cliction 10.3

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Helpful mental picture:

Chapter 10: Energy

SYSTEM

Motional Energy = “Kinetic Energy” = KStored (internal) energy = “Potential Energy” = U

Mechanical Energy = K + U

ENVIRONMENT

CashSavings account

Work (w) orEnergy Transfer

Getting paycheck or Paying bills

The basic energy model

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10.4 Restoring forces and Hooke’s law p. 278-280

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Chapter 10: Energy

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Systems that exhibit restoring forces are called ELASTIC.

Simplest example: spring force

A spring has an equilibri-um length L0 when it isrelaxed

You have to apply a force in order to stretch or squeezethe spring. The more you stretch/squeeze the spring, themore force you have to apply

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Chapter 10: Energy

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The force Fsp that a stretched/compressed spring exertson an object are well described by Hooke's law:

(Fsp)s = - k (s -se)

where s is the coordinate alongthe axis of the spring, s is posi-tive when the spring is stretched.

se is the equilibrium position k is the spring constant

k depends on the details of thespring and has units N/m

→

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The graph shows forceversus displacement forthree springs. Rank inorder, from largest tosmallest, the springconstants k1, k2, and k3.

1. k3 > k2 > k1

2. k1 = k3 > k2

3. k2 > k1 = k3

4. k1 > k2 > k3

5. k1 > k3 > k2

Chapter 10: Energy

Cliction 10.4

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Chapter 10: Energy

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Important features of (Fsp)s = - k (s -se):

(Fsp)s< 0 when the spring is stretched (Fsp)s> 0 when the spring is compressed |(Fsp)s| grows linearly with |s-se|

This means the spring force triesto restore equilibrium positionregardless of whether it is stretchedor compressed

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Chapter 10: Energy

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Application: stick-slip motion.

(Ex. 10.5) A box is subject to spring force and static friction.When does it start to slip?

Calculations on blackboard ..

This is what happens in earth-quakes: the tectonic plates sticktogether at their boundaries, but theymove relative to each other

When the spring force gets too largethe slip is triggered (= earthquake)

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Chapter 10: Energy

MOST COMMON ERROR!

Assuming the spring is along the x-axis, writing

F = - kx

Is misleading and it is a common source of errors.It is true ONLY if the coordinate system in the problemis chosen such that the origin is at the equilibriumposition, x , of the free end of the spring.

In general, it should always be

F = - k(x- x )

sp

sp e

e

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10.5 Elastic potential energy p. 280-284

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Chapter 10: Energy

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Is the spring force energy conserving?

Intuitive answer: yes. If a ball issitting on a compressed spring,it will gain kinetic energy whenthe spring is released

Let's try to find a quantitativeanswer. Hooke's law

(Fsp)s = - k (s -se)

depends only on the position s of the spring's end.

The elastic potential energy is given by

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Chapter 10: Energy

⇒ ⇔ ×→ ≈ ^

We therefore have energy conservation: If an object ismoving on a spring the mechanical energy K + Us isconserved, (the elastic energy is stored in the spring)

Kf + Usf = Ki + Usi

Example 10.6: spring gun

Initially the ball (& spring)is at rest at x1 = -0.1 m (springis relaxed if x=0) ⇒

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Chapter 10: Energy

⇒ ⇔ ×→ ≈ ^

At x2 =0 the spring is relaxed and the ball loses contact toit. Energy at this moment:

Energy conservation for the ball:

K + U = K + U 2 1 s1s2

=0 =0

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A spring-loaded gun shoots a plastic ball with a speedof 4 m/s. If the spring is compressed twice as far, theball’s speed will be

1. 16 m/s.2. 8 m/s.3. 4 m/s.4. 2 m/s.5. 1 m/s.

Chapter 10: Energy

Cliction 10.5

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Chapter 10: Energy

⇒ ⇔ ×→ ≈ ^

Energy conservation also applies if several conservativeforces are applied. We simply have to add all theirpotential energies

Example (10.7): a box that is launched vertically by aspring

Forces applying to the box:weight w and spring force Fsp

Energy conservation:

→ →

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Chapter 10: Energy

⇒ ⇔ ×→ ≈ ^

Initial mechanical energy, taking the equilibrium positionof the spring to be ye = 2m:

Mechanical energy at highest point (v2 =0)

Us2 = 0 because the box is separated from the spring.Al-ways use your commonsense when calculating.

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Chapter 10: Energy

⇒ ⇔ ×→ ≈ ^

Energy conservation allows us to find y2:

It is very important that energy conservation andmomentum conservation are independent

Momentum conservation is valid if a system is closed(no net external force acts on it) Energy conservation holds if only conservative forcesact on the system we will come back to this in ch.11

Often it is possible to employ both concepts to onesystem

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Chapter 10: Energy

⇒ ⇔ ×→ ≈ ^

Example 10.8: Pushingtwo boxes apart on africtionless table.

Initially the two boxes areboth at rest. Initial total momentum:

Initial kinetic energy of both boxes:

Initial elastic energy stored in the spring:

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Chapter 10: Energy

⇒ ⇔ ×→ ≈ ^

Final total momentum:

Final kinetic energy:

Final elastic energy Usf = 0 because spring is relaxed

We want to find the final velocities (vfx)1 and (vfx)2

Momentum conservation ⇒

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Chapter 10: Energy

⇒ ⇔ ×→ ≈ ^

Energy conservation ⇒

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End of Week 11

Chapter 10: Energy

34

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10.6 Elastic collisions p. 284-288

Chapter 10: EnergyPh

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Chapter 10: Energy

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In a perfectly elastic collision both momentum and energyare conserved. This is very similar to the two boxes with aspring discussed previously.

On a molecular level, an elastic collision corresponds tocompression and relaxation of molecular springs

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Chapter 10: Energy

⇒ ⇔ ×→ ≈ ^

Perfectly elastic collisions also allow us to understand thebehavior of real gases (If you have seen the van-der-Waals equation before: the volume and additional pressurefactor are determined by collisions between the molecules)

Understanding perfectly elastic collisions is a matter ofstraightforwardly applying energy and momentumconservation

Let's consider the case whentwo particles 1, 2 collide andparticle 2 is at rest before thecollision (if it is not at rest wecan switch to a referenceframe where it is at rest)

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Chapter 10: Energy

⇒ ⇔ ×→ ≈ ^

Momentum conservation ⇒

Energy conservation: before and after the collision thereare no forces acting on the particles ⇒

(2)

(1)

We are looking for the finalvelocities. Eq. (1) ⇒

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Chapter 10: Energy

⇒ ⇔ ×→ ≈ ^

Insert this into Eq. (2) ⇒

The (vix)12 terms cancel, leaving us with

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Chapter 10: Energy

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The solution (vfx)2 =0 simply means no change: theparticles miss each other

Relevant solution:

Interesting special cases:

m1 = m2 ⇒ (vfx)1 =0 and (vfx)2 = (vix)1 . For equal mas-ses momentum is completely transferred to particle 2

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Chapter 10: Energy

⇒ ⇔ ×→ ≈ ^

m1 >> m2 ⇒ (vfx)1 ≈ (vix)1 and (vfx)2 ≈ 2 (vix)1 . Particle 1 is hardly stopped by particle 2

m1 << m2 ⇒ (vfx)1 ≈ - (vix)1 and (vfx)2 ≈ 0. Particle 1 is re- flected (particle 2 like a wall)

This is quite relevant for research in many areas. Forinstance, if you mix a gas of cold (= slow) light atomswith one of hot (= fast) heavy atoms, the light atomic gasis quickly heated. But cooling it with cold heavy atomsdoesn't work.

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Chapter 10: Energy

⇒ ⇔ ×→ ≈ ^

Let's see if this works witha rebounding pendulum

This involves elastic collisionsand conversion of potential tokinetic energy

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Chapter 10: Energy

⇒ ⇔ ×→ ≈ ^

L is the length of the pendulum and θ0 the initial angle. Theinitial height of particle A is

(y0)A = L – L cos θ0

PART 1: Energy conservation ⇒

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Chapter 10: Energy

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For the collision we simply can take over our results forperfectly elastic collisions.

PART 2: Momentum conservation during the collision thenimply

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Chapter 10: Energy

⇒ ⇔ ×→ ≈ ^

PART 3: The pendulum bounces back and reaches a maximumheight (y3)A. Energy conservation implies

See last part of example 10.9 (page 289 in textbook) onhow to calculate the angle of rebound once (y ) is known.

= 0.0541 m

3 A

Angle of rebound

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Chapter 10: Energy

10.7 Energy Diagrams p. 288-293

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Chapter 10: Energy

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The particle can only go up to the point where TE = PE which is the same as Emech = Ug. Because Emech = K + Ug this means that K=0 (i.e. zero speed ⇔ maximum height). Because K = Emech- Ug = TE – PE the height differencebetween TE and PE corresponds to the kinetic energy at y

An extremely useful tool to better understand energyconservation are energy diagrams. This is simply a graphthat shows the system's total energy TE and potentialenergy PE as a function of position

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Chapter 10: Energy

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At maximum height the particle turns back: turning point

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Chapter 10: Energy

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Another example:

Elastic potential energy

Us = 0.5k(x-xe)2

Particle initially at rest, spring elongated: x is maximal

The Total Energy (TE) line is under YOUR control. YOU can move the TE lineas far up or down as you wish by changing the initial conditions, such asprojecting the particle upward with a different speed or dropping it from adifferent height. Once YOU have determined the initial conditions, YOU can usethe energy diagram to analyze the motion for that amount of total energy TE.

The TE line

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Chapter 10: Energy

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At the point where PE (=U ) has its minimum the kineticenergy is maximal. At this position the particle achieves maximum speed.

s

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⇒ ⇔ ×→ ≈ ^

Chapter 10: Energy

Combined potentials (examples)

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y=0h

y=0

E

xxeq

PEg = mgy = 0 PEg = mgh

PEg=0

PE=PEsp

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E

xxeqPEg=mgh

PE = PEg+PEsp

PEsp

51

Chapter 10: Energy

⇒ ⇔ ×→ ≈ ^

Any potential energy (PE)versus position graph looks likea mountain to a skier

If you start up high you gain alot of kinetic energy at thebottom.

MORE GENERAL (or complex) ENERGY DIAGRAMS

The particle’s kinetic energy at x is zero, hence the TE line must crossThe PE curve since: TE = K + PE = 0 + PE = PE

1

The particle’s kinetic energy at x is zero. The particle is instantaneouslyat rest, then reverses direction. 5

The particle will oscillate back and forth between x and x , following thePattern of slowing down and speeding up outlined in the diagram above.

1 5

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A particle with the potential energy shown ismoving to the right. It has 1.0 J of kinetic energyat x = 1.0 m. In the region 1.0 m < x < 2.0 m,the particle isA. speeding up.B. slowing down.C. moving at constant speed.D. There’s not enough information to say.

Chapter 10: Energy

Cliction 10.7

53

A particle with the potential energy shown ismoving to the right. It has 1.0 J of kineticenergy at x = 1.0 m. In the region 1.0 m < x <2.0 m, the particle isA. speeding up.B. slowing down.C. moving at constant speed.D. There’s not enough information to say.

Losing potential energy, thusgaining kinetic energy.

Chapter 10: Energy

Cliction 10.7

54

A particle with the potential energy shown ismoving to the right. It has 1.0 J of kineticenergy at x = 1.0 m. Where is the particle’sturning point?A. 1.0 mB. 2.0 mC. 5.0 mD. 6.0 mE. It doesn’t have a turning point.

Chapter 10: Energy

Cliction 10.8

55

A particle with the potential energy shown ismoving to the right. It has 1.0 J of kineticenergy at x = 1.0 m. Where is the particle’sturning point?A. 1.0 mB. 2.0 mC. 5.0 mD. 6.0 mE. It doesn’t have a turning point.

Total energy line

Chapter 10: Energy

Cliction 10.7

56

⇒ ⇔ ×→ ≈ ^

Chapter 10: Energy

Equilibrium points

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Chapter 10: Energy

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Equilibrium points are pointswhere the force on the particlevanishes: for conservativeforces we have F = 0 wherethe potential's slope is zero

Equilibrium points can bestable or unstable

Stable equilibrium = minimum of PE The force acts like a spring force and pulls the particle backto points (x2, x4)

Unstable equilibrium = maximum of PE The force accelerates the particle away from the equilibriumpoint (x3)Ph

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A particle with this potential energy is instable equilibrium at x =A. 0.0 mB. 1.0 mC. 2.0 mD. Either A or CE. Either B or C

Chapter 10: Energy

Cliction 10.8

59

A particle with this potential energy could bein stable equilibrium at x =A. 0.0 mB. 1.0 mC. 2.0 mD. Either A or CE. Either B or C

Stable equilibrium at the minimum

Chapter 10: Energy

Cliction 10.8

60

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Chapter 10: Energy

Equilibrium positions (say x1, x2, x3 ect ..) are found by:

Say Utotal = 3 x2 - 2x then to find equilibrium positions do:

6x-2 = 0 xeq = 1/3

Derivative with respect to x

61

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Chapter 10: Energy

Equilibrium positions (say y1, y2, y3 ect ..) are found by:

Say Utotal = 2 y3 - y then to find equilibrium positions do:

6y2-1 = 0 yeq,1 = -sqrt(1/6) and yeq,2 = sqrt(1/6)

Derivative with respect to y

62

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Chapter 10: Energy

Practise on problem 10.10 in your textbook !!!

This problem is a good example of how tofind the equilibrium positions

63

Chapter 10: Energy Tactics Box 10.2 Interpreting an energy diagram

The distance from the axis to the PE curve is the particle’s potential energy.The distance from the PE curve to the TE line is its kinetic energy. These aretransformed as the position changes, causing the particle to speed up or slowdown, but the sum doesn’t change.

A point where the TE line crosses the PE curve is a turning point. The particlereverses direction.

The particle cannot be at a point where the PE curve is above the TE line.

The PE curve is determined by the properties of the system—mass, springconstant, and the like. You cannot change the PE curve. However, you can raise orlower the TE line simply by changing the initial conditions to give the particlemore or less total energy.

A minimum in the PE curve is a point of stable equilibrium. A maximum in thePE curve is a point of unstable equilibrium.Ph

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Chapter 10Reading Quiz

Chapter 10: EnergyPh

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Energy is a physical quantity with propertiessomewhat similar to

1. money.2. heat.3. a liquid.4. work.5. momentum.

Chapter 10: EnergyPh

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Hooke’s law describes the force of

1. gravity.2. tension.3. a spring.4. collisions.5. none of the above.

Chapter 10: EnergyPh

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A perfectly elastic collision is a collision

1. between two springs.2. that conserves potential energy.3. that conserves thermal energy.4. that conserves kinetic energy.5. All of 2, 3, and 4.

Chapter 10: EnergyPh

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Selected Problems

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Chapter 10: Energy

75

End of Chapter 10

IMPORTANT:

Print a copy of the SUMMARY page (p. 294)and add it here to your lecture notes.

It will save you crucial time when trying to recall:Concepts, Symbols, and Strategies

Chapter 10: EnergyPh

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