msu physics 231 fall 2015 1 physics 231 topic 5: energy and work alex brown october 2, 2015

MSU Physics 231 Fall 2015 1

Physics 231Topic 5: Energy and Work

Alex BrownOctober 2, 2015


What’s up? (Friday Sept 26)

1) The correction exam is now open. The exam grades will be sent out after

that on Wednesday Oct 7.

2) Homework 04 is due Tuesday Oct 13th and covers Chapters 5 and 6. It is a

little longer that usual so you may want to start early.


Key Concepts: Work and Energy

Work and Energy

Work and it’s association with forces Constant forces (driving up a hill)

Variable forces (stretching a spring)

Kinetic Energy Work-Energy Theorem

Relationship to velocity

Potential Energy

Conservation of Mechanical Energy

Power

Covers chapter 5 in Rex & Wolfson


What is Energy?

Motion – kinetic energy

Ability to produce motion – potential energy

How to we transfer energy?


Work and Energy

Work: ‘Transfer of energy’

Quantitatively: The work W done by a constant force on an object is the product of the force along the direction of displacement and the magnitude of displacement.

W = (Fcos) x = Fx x

Units: N·m = Joule

1 calorie = 4.184 J

1 Calorie = 4184 J= 1

kcal

F

Fcos

x


Non-constant force/angleW=(Fcos)x: what if Fcos is not constant while covering x?

Example: what if or F changes while pulling the block?

Area=A=(Fcos)x

x

x

Fcos

x

The work done is the area under the graph of Fcos vs x

W=(A)=total area

Fcos

x

F


A person drags a block over a floorwith a force parallel to the floor.

After 4 meters, the floor turnsrough and instead of a force of 2N a force of 4N must be applied.The force-distance diagram shows the situation.

How much work did the person do over 8 meter?a) 0 J b) 16 J c) 20 J d) 24 J

e) 32 J

4 8 m

2N

4N

0

Force

distance

Work: area under F-x diagram: 4x2 + 4x4 = 24 J

Example

4 m


Work and Energy

Positive work is done when the angle is less than 90 degrees, energy goes into the object

1) energy can be stored (potential energy increases)

2) motion can be created (kinetic energy increases)


Work and Energy

Negative work – when the angle between the force and the displacement is more the 90 degrees, energy is removed:

1) stored energy can be decreased (potential energy decreases)

2) motion can be reduced (kinetic energy decreases)

Special case for friction force, the angle is 180 degrees; potential or kinetic energy is removed and heat is created

No work – when the angle between the force and the displacement is equal to 90 degrees


Work and EnergySled is pulled across a surface at constant speed.Where does the energy go in this case?

Answer: two forces are acting; the energy goes into friction (the ground/sled heat up!)


v

T

No work is done because the force acts in a

perpendicular direction to the displacement.

Using the definition of work W = F (Δs) cos

because = 90º, then W = 0.

Clicker Question: Tension and Work

a) tension does no work at

all

b) tension does negative

work

c) tension does positive

work

A ball tied to a string is being

whirled around in a circle.

What can you say about the

work done by tension?


Power: The rate of energy transfer

Work (the amount of energy transfer) is independent of time.

W=(Fcos) x … total over all time

To measure how fast we transfer the energy we define:

Power = P = (Work/time) = (W/t) (J/s=Watt)

P = (Fcos)·x/t = (Fcos)·vaverage

1 Watt = 0.00134 horsepower


While running, a person dissipates about 0.60 J of mechanical energy per step per kg of body mass.

A) If a 60 kg person develops a power of 70 Watt during a race (distance = L), how fast is she running (1 step=1.5 m)?

B) What is the force the person exerts on the road?

W=Fx P=W/t=Fv

A) Work per step:(0.60 J/kg)(60 kg) = 36 J

Work during race: (36 J) [( L ) / ( step-length )]= (36 J) ( L) / (1.5 m) = (24L) J/m

Power = W/t = 24 L/t = 24 v = 70 Wattsv =2.9 m/s

B) F=P/v so F=24 N

A Runner


Net Work

When many forces are acting the net work done by all of them is the sum of each term

Wnet = (Fx1 + Fx2 …..) x =

= Fx,net x = (Fnet cos) x


Question

A worker pushes a wheelbarrow with a force of 50 N over a distance of 10 m. A frictional force acts on the wheelbarrow in the opposite direction, with a magnitude of 30 N.What net work is done on the wheelbarrow?

a) 0b) 100 Jc) 200 Jd) 300 Je) 500 J

Wnet = Fx,net x

= (50-30) (10) = 200 J


ExampleA toy-rocket of 5.0 kg, after the initial acceleration stage, the speed is constant and travels 100 m in 2 seconds.

A) What is the work done by the engine?

B) What is the power of the engine?A)W = (Fcos) h = mrocket g h

= (5.0 kg)(9.81 m/s2)(100 m) = 4905 J Force by engine must balance gravity!

h=

10

0m

B) P=W/t = 4905/2=2453 Watt (=3.3 horsepower)

or

P=(Fcos)v = mg(h/t) =5.09.81100/2=2453 Watt


Clicker Question: Force and Work

N T

mg

displacementAny force not perpendicularto the motion will do work:

N does no work

T does positive work

fk does negative work

mg sin does negative work

a) one force

b) two forces

c) three forces

d) four forces

e) no forces are doing

work

A box is being pulled up a rough incline

by a rope connected to a pulley. How

many forces are doing non-zero work

on the box?

Mg sin

W = (Fcos)·x

fk


Potential EnergyPotential energy (PE): energy associated with the positionof an object within some system.

Gravitational potential energy: Consider the work done bythe gravity in case of a falling object:

Wgravity = Fg cos(0o) h = mg h = mg hi – mg hf

= PEi - PEf

The ‘system’ is the gravitational field of the earth. PE = mgh

Since we are usually interested in the change in gravitational potential energy, we can choose the ground level (h=0) in aconvenient way.

W = (Fcos)·x


atvv f 0

221 attvx o

221 attvx f

tvtvvx f )( 021

)( 20

2

21 vvx fa

(A)

(B)

(C)

(D)

(E)


Kinetic energy:Consider an object that changes speed only

W = Fcos x = (ma) x … used Newton’s second law

x=100mt=0VO t=2s

Vf

(E) x = (vf2-v0

2)/2a

Kinetic energy: KE=½mv2

When work is done on an object and the only change is its speed: The work done is equal to the change in KE: W = KEfinal - KEinitial = KEf - KEi

W=½m(vf2-v0

2) = ½mvf2 -

½mv02


Conservation of energyMechanical energy = Potential Energy + Kinetic energyME = PE + KE

Mechanical energy is conserved if:• the system is closed (no energy can enter or leave)• the forces are ‘conservative’ (see soon)

Heat, chemical energy (e.g battery or fuel in an engine)Are sources or sinks of internal energy (not mechanical).

We’re not talking about this!


Example of closed system

At launch:ME = mgh + ½mv2

0.29.8135 + 0 = 68.67 J

At ground:ME = mgh + ½mv2

= 0 + ½0.2v2 = 0.1 v2 J

t=0 sh=35 mv=0 m/s

t>0 sv at h=0?

An object (0.2 kg) is dropped from a height of 35 m. Assuming no friction, what is the velocity when it reaches the ground?

Conservation of ME:68.67 J = 0.1 v2 J v = 26.2 m/s


Clicker Quiz!

time

en

erg

y

In the absence of friction,which energy-time diagramis correct?

potential energy

total energy

kinetic energy

A

time

en

erg

y

time

B

C

en

erg

y


t=0 t=3t=2t=1 t=5

Where is the kinetic energy…1) highest?2) lowest ?

A EDCB

ParabolicMotion


t=0 t=3t=2t=1 t=5

Where is the potential energy…1) highest?2) lowest ?

A EDCB

ParabolicMotion


A swing

h

30o

L=5m

If released from rest, what isthe velocity of the ball at the lowest point?

(PE+KE) = constant

PErelease=mgh (h=5-5cos(30o)) =6.57m JKErelease=0

PEbottom=0KEbottom=½mv2

½mv2=6.57m so v=3.6 m/s


Ball on a trackA

B

In which case has the ball the highest velocity at the end?A) Case A B) Case B C) Same speed

In which case does it take the longest time to get to the end?A) Case A B) Case B C) Same time

h

h

end

end


Conservation of mechanical energyMechanical energy = potential energy + kinetic energy

In a closed system, mechanical energy is conserved

ME = PE+KE = mgh + ½mv2 = constant

h=

10

0m

What about the accelerating rocket?

VO=0

V=100 m/s

At launch:ME = 5*9.81*0 + ½5*02 = 0 J

At 100 m height:ME = 5*9.81*100 + ½5*1002 = 29905 J

M=5 kg

We did not consider Fuel burning- Another source of energy

that is not mechanical energy

*


Roller coaster

KE PE TME NC KE PE TME NC KE PE TME NC KE PE TME NC KE PE TME NCWith friction


A force is conservative if the work done by the force whenMoving an object from A to B does not depend on the pathtaken from A to B.

Example: work done by gravitational force

h=

10

m

Using the stairs:Wg = mghf-mghi = mg(hf-hi)

Using the elevator:Wg = mghf-mghi = mg(hf-hi)

The path taken (longer or shorter) does not matter: only the displacement does!

Conservative Forces


A force is non-conservative if the work done by the force when moving an object from A to B depends on the path taken from A to B.

Example: FrictionYou have to perform more work against friction if you take the long path, compared to the shortpath. The friction force changeskinetic energy into heat.(In this example difference forces are applied so that the KE is the same at the end.)

object on rough surfacetop view

Non-Conservative Forces


Conservation of mechanical energy only holds if the system is closed AND all forces are conservative

MEi-MEf=(PE+KE)i-(PE+KE)f=0 if all forces are conservativeExample: throwing a snowball from a building neglecting air resistance

MEi-MEf=(PE+KE)i-(PE+KE)f=Encif some forces are nonconservative.Enc=positive energy dissipated by non-conservative forces (in the books notation Enc = -Wf)

Example: throwing a snowball from a building taking into account air resistance


Overview

Newton’s second LawF=ma

WorkW=(Fcos)x

Equations of kinematicsx(t)=Xo + Vo t + ½at2

v(t)=Vo + at

Work-energy TheoremEnc=MEi - MEf

Conservation of mechanical energy MEi =

MEf

Enc=0 Closed system


Clicker Quiz!

time

en

erg

y

When there is friction,which mechanical energy-time diagram is correct?

potential energy

total energy

kinetic energy

A

time

en

erg

y

time

B

C

en

erg

y


QuestionOld faithful geyser in Yellowstone park shoots water hourlyto a height of 40 m. With what velocity does the water leave the ground?

a) 7.0 m/sb) 14 m/sc) 20 m/sd) 28 m/se) don’t know

At ground level: ME = ½mv2 + mgh

= ½mv2 + 0 = ½mv2

KEi + PEi = KEf + PEf

KE = ½ mv2

PE = mgh

Conservation of energy:½mv2 = 392m v2 = 2x392

so v=28 m/s

At highest point: ME = ½mv2 + mgh = 0 + m*9.81*40 = 392m

Step 1:

Step 2:

Step 3:

Step 4:


Conservation of mechanical energyWhat is the speed of m1 and m2 when

they pass each other?

At time of release:PE1i = m1gh1i = 5.00*9.81*4.00 = 196.2 JPE2i = m2gh2i = 3.00*9.81*0.00 = 0.00 JKE1i= ½m1vi

2 = 0.5*5.00*(0.)2 = 0.00 JKE2i = ½m1vi

2 =0.5*3.00*(0.)2 = 0.00 J

Total = 196.2 J

196.2 = 156.8 + 4.0v2 v = 3.13 m/s

M2

3 kg

5 kg

M1

4.0 m

At time of passing:PE1f = m1 gh1f = 5.00*9.81*2.00 = 98.0 JPE2f = m2 gh2f = 3.00*9.81*2.00 = 58.8 JKE1f = ½m1v2 = 0.5*5.00*(v)2 = 2.5v2 JKE2f = ½m2v2 = 0.5*3.00*(v)2 = 1.5v2 J

Total =156.8+4.0v2 J

ME = (PE1+PE2+KE1+KE2)=constant


Friction (non-conservative)The pulley is now not frictionless. The friction force equals 5 N. What is the speed of the objects when they pass?

Enc = ffrictionx = 5.00*2.00 = 10.0 J

v=2.7 m/sM2

3 kg

5 kg

M1

4.0 m

MEi = MEf + Enc

196.2 = 156.8 + 10.0 + 4.0v 2

Without Friction:v = 3.13 m/s


Question!In the absence of friction,

when m1 starts to move down:

1) potential energy is transferred from m1 to m2

2) potential energy is transformed into kinetic energy

3) m1 and m2 have the same kinetic energyM

2

3 kg

5 kg

M1

4.0 m

TRUE

TRUE

FALSE


QuestionA ball rolls down a slope as shown in the figure. The starting velocity is 0 m/s. There is some friction between the ball and the slope. Which of the following is true?

h

a) The kinetic energy of the ball at the bottom of the slope equals the potential energy at the top of the slope

b) The kinetic energy of the ball at the bottom of the slope is smaller than the potential energy at the top of the slope

c) The kinetic energy of the ball at the bottom of the slope is larger than the potential energy at the top of the slope


QuizA ball rolls down a slope and back up a more shallow slope as shown in the figure. The starting velocity is 0 m/s. There is some friction between the ball and the slope. Which is true?

h2

a) The maximum height reached on the right (h2) is the same as the height the ball started at on the left (h1)

b) The maximum height reached on the right (h2) is smaller than the height the ball started at on the left (h1)

c) The maximum height reached on the right (h2) is larger than the height the ball started at on the left (h1)

h1


Question

A ball of 1 kg rolls up a ramp, with initial velocity of 6 m/s.It reaches a maximum height of 1 m (i.e., the velocity is 0 m/s at that point). How much energy is dissipated by friction?a) 0.

b) 8.2 Jc) 9.8 Jd) 18 Je) 27.8 J

Initial: ME = ½mv2 (kinetic only) = ½x1x62 = 18 J

Final: ME = mgh (potential only) = 1x9.8x1 = 9.8 J

Enc = 18-9.8 = 8.2 J

kinetic energy: ½mv2

potential energy: mgh g=9.81 m/s2

MEi – MEf = Enc

(KEi+PEi) - (KEf+PEf) = Enc


QuestionAn outfielder who is 2M tall throws a baseball of 0.15 kg at a speed of 40 m/s and angle of 30 degrees with the field. What is the kinetic energy of the baseball at the highest point, ignoring friction?

a) 0 Jb) 30 Jc) 90 Jd) 120 Je) don’t know

Two components of velocity at start:vox = vocos(30o) = 34.6 m/svoy= vosin(30o) = 20 m/s

At highest point: only horizontal velocityvx = vox = 34.6 m/svy = 0 m/s

kinetic energy: ½mv2 = ½(0.15)(34.6) 2 = 90 J


QuestionAn outfielder who is 2m tall throws a baseball of 0.15 kg at a speed of 40 m/s and angle of 30 degrees with the field. How high does the ball go at its highest point, ignoring friction?

Initial: KE + PE = ½mv2 + mgh = ½ (0.15)(40) 2 + (0.15)(9.81)(2)

= 120 + 2.94

At highest point: KE + PE = 90 + mgh = 90 + 1.47 h

122.9 = 90 + 1.47 hh = 22.4 m ie, the ball travels 20.4 m higher than the player’s height


Work and Energy Work: W = Fcos()x Energy transfer

The work done is the same as the area under the graph of Fcos versus x

Power: P = W/t Rate of energy transfer Potential energy (PE) Energy associated with position.

Gravitational PE: mgh Energy associated with position in grav. field.

Kinetic energy KE: ½mv2 Energy associated with motion Conservative force: Work done does not depend on path

Non-conservative force: Work done does depend on path

Mechanical energy ME: ME = KE + PEConserved if only conservative forces are present MEi = MEf

Not conserved in the presence of non-conservative forces MEi = MEf + Enc


All freely falling objects fall with the same acceleration (g=9.81 m/s2).

Because the acceleration is the same for both, and the distance is the same for both, then the final velocities will be the same for both stones.

2

Question: Free Fall 1

a) quarter as much

b) half as much

c) the same

d) twice as much

e) four times as much

Two stones, one twice the mass of

the other, are dropped from a cliff.

Just before hitting the ground, what

is the VELOCITY of the heavy stone

compared to the light one?


Consider the work done by gravity to make the stone fall distance d:

KE = Wnet = F d cos

KE = mgd Thus, the stone with the greater mass has the greater KE, which is twice as big for the heavy stone.

Quiz: Free Fall 2

a) quarter as much

b) half as much

c) the same

d) twice as much

e) four times as much

Two stones, one twice the mass of

the other, are dropped from a cliff.

Just before hitting the ground, what

is the KINETIC ENERGY of the heavy

stone compared to the light one?

KEi + PEi = KEf + PEf

msu physics 231 fall 2015 1 physics 231 topic 5: energy and work alex brown october 2, 2015

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