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Physics 231
Topic 5: Energy and Work
Alex Brown
October 2, 2015
MSU Physics 231 Fall 2015
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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.
MSU Physics 231 Fall 2015
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MSU Physics 231 Fall 2015
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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
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What is Energy?
Motion – kinetic energy
Ability to produce motion – potential energy
How to we transfer energy?
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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
F
Units: N·m = Joule
1 calorie = 4.184 J
1 Calorie = 4184 J
= 1 kcal

Fcos
x
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Non-constant force/angle
W=(Fcos)x: what if Fcos is not constant while covering x?
Example: what if  or F changes while
pulling the block?
F

Fcos
Fcos
x
Area=A=(Fcos)x
x
W=(A)=total
area
x
x
The work done is the area under the graph of Fcos vs x
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Example
4m
A person drags a block over a floor
with a force parallel to the floor.
Force
4N
2N
After 4 meters, the floor turns
rough and instead of a force of
2N a force of 4N must be applied.
0
4
The force-distance diagram shows the situation.
8m
distance
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
Work: area under F-x diagram: 4x2 + 4x4 = 24 J
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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)
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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
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Work and Energy
Sled 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!)
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Clicker Question: Tension and Work
A ball tied to a string is being
whirled around in a circle.
What can you say about the
work done by tension?
a) tension does no work at all
b) tension does negative work
c) tension does positive work
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.
T
v
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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
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A Runner
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=Fx
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 Watts
v =2.9 m/s
B) F=P/v so F=24 N
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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
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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)
b)
c)
d)
e)
0
100 J
200 J
300 J
500 J
Wnet = Fx,net x
= (50-30) (10) = 200 J
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Example
A 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?
h=100m
A)
W = (Fcos) h = mrocket g h
= (5.0 kg)(9.81 m/s2)(100 m) = 4905 J
Force by engine must balance gravity!
B)
P=W/t = 4905/2=2453 Watt (=3.3 horsepower)
or
P=(Fcos)v = mg(h/t) =5.09.81100/2=2453 Watt
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Clicker Question: Force and Work
a) one force
A box is being pulled up a rough incline
b) two forces
by a rope connected to a pulley. How
c) three forces
many forces are doing non-zero work
d) four forces
on the box?
W = (Fcos)·x
e) no forces are doing work
Any force not perpendicular
to the motion will do work:
N
N does no work
T
T does positive work
Mg sin
fk does negative work

mg sin does negative work
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mg
fk
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Potential Energy
Potential energy (PE): energy associated with the position
of an object within some system.
Gravitational potential energy: Consider the work done by
the gravity in case of a falling object: W = (Fcos)·x
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 a
convenient way.
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(A)
v f  v0  at
(B)
x  vo t  at
(C)
x  v f t  at
(D)
x  12 (v0  v f )t  v t
(E)
x 
1
2
1
2
1
2a
2
2
(v f  v0 )
2
2
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Kinetic energy:
Consider an object that changes speed only
x=100m
t=0
VO
t=2s
Vf
W = Fcos x = (ma) x
(E)
… used Newton’s second law
x = (vf2-v02)/2a
W=½m(vf2-v02) = ½mvf2 - ½mv02
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
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Conservation of energy
Mechanical energy = Potential Energy + Kinetic energy
ME = PE + KE
Mechanical energy is conserved if:
• the system is closed (no energy can enter or leave)
• the forces are ‘conservative’ (see soon)
We’re not talking about this!
Heat, chemical energy
(e.g battery or fuel in an engine)
Are sources or sinks of internal
energy (not mechanical).
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Example of closed system
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?
t=0 s
h=35 m
v=0 m/s
t>0 s
v at h=0?
At launch:
ME = mgh + ½mv2
0.29.8135 + 0 = 68.67 J
At ground:
ME = mgh + ½mv2
= 0 + ½0.2v2 = 0.1 v2 J
Conservation of ME:
68.67 J = 0.1 v2 J
v = 26.2 m/s
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A
energy
Clicker Quiz!
B
In the absence of friction,
which energy-time diagram
is correct?
total energy
time
C
energy
potential energy
energy
time
kinetic energy
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time
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Where is the kinetic energy…
1) highest?
2) lowest ?
Parabolic
Motion

t=0
A
t=1
B
t=2
C
t=3
D
MSU Physics 231 Fall 2015
t=5
E
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Where is the potential energy…
1) highest?
2) lowest ?
Parabolic
Motion

t=0
A
t=1
B
t=2
C
t=3
D
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t=5
E
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A swing
If released from rest, what is
the velocity of the ball at the
lowest point?
30o
h
L=5m
(PE+KE) = constant
PErelease=mgh (h=5-5cos(30o))
=6.57m J
KErelease=0
PEbottom=0
KEbottom=½mv2
½mv2=6.57m so v=3.6 m/s
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Ball on a track
A
h
end
B
h
end
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
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Conservation of mechanical energy
Mechanical energy = potential energy + kinetic energy
In a closed system, mechanical energy is conserved*
V=100 m/s
ME = PE+KE = mgh + ½mv2 = constant
What about the accelerating rocket?
h=100m
M=5 kg
VO=0
At launch:
ME = 5*9.81*0 + ½5*02 = 0 J
At 100 m height:
ME = 5*9.81*100 + ½5*1002 = 29905 J
We did not consider Fuel burning
-Another source of energy
that is not mechanical energy
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MSU Physics 231 Fall 2015
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Roller coaster
KE
PE TME NC
KE
With friction
PE TME NC
KE PE TME NC
MSU Physics 231 Fall 2015
KE
PE TME NC
KE
PE TME NC
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Conservative Forces
A force is conservative if the work done by the force when
Moving an object from A to B does not depend on the path
taken from A to B.
Example: work done by gravitational force
Using the stairs:
Wg = mghf-mghi = mg(hf-hi)
h=10m
Using the elevator:
Wg = mghf-mghi = mg(hf-hi)
The path taken (longer or
shorter) does not matter: only
the displacement does!
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Non-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.
object on rough surface
top view
Example: Friction
You have to perform more work
against friction if you take the
long path, compared to the short
path. The friction force changes
kinetic energy into heat.
(In this example difference forces
are applied so that the KE is the
same at the end.)
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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 conservative
Example: 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
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Overview
Equations of kinematics
x(t)=Xo + Vo t + ½at2
v(t)=Vo + at
Newton’s second Law
F=ma
Conservation of mechanical
energy MEi = MEf
Enc=0
Closed system
Work-energy Theorem
Enc=MEi - MEf
Work
W=(Fcos)x
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A
energy
Clicker Quiz!
B
When there is friction,
which mechanical energy-time
diagram is correct?
total energy
time
C
energy
potential energy
energy
time
kinetic energy
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time
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Question
Old faithful geyser in Yellowstone park shoots water hourly
to a height of 40 m. With what velocity does the water
leave the ground?
KE + PE = KE + PE
i
a)
b)
c)
d)
e)
7.0 m/s
14 m/s
20 m/s
28 m/s
don’t know
Step 1:
Step 2:
Step 3:
i
f
f
mv2
KE = ½
PE = mgh
At ground level:
ME = ½mv2 + mgh
= ½mv2 + 0 = ½mv2
At highest point:
ME = ½mv2 + mgh
= 0 + m*9.81*40 = 392m
Conservation of energy:
Step 4:
½mv2 = 392m  v2 = 2x392
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Conservation of mechanical energy
What is the speed of m1 and m2 when
they pass each other?
ME = (PE1+PE2+KE1+KE2)=constant
5 kg
M1
3 kg
4.0 m
M2
196.2 = 156.8 + 4.0v2
v = 3.13 m/s
At time of release:
PE1i = m1gh1i = 5.00*9.81*4.00
PE2i = m2gh2i = 3.00*9.81*0.00
KE1i= ½m1vi2 = 0.5*5.00*(0.)2
KE2i = ½m1vi2 =0.5*3.00*(0.)2
= 196.2 J
= 0.00 J
= 0.00 J
= 0.00 J
Total
= 196.2 J
At time of passing:
PE1f = m1 gh1f = 5.00*9.81*2.00
PE2f = m2 gh2f = 3.00*9.81*2.00
KE1f = ½m1v2 = 0.5*5.00*(v)2
KE2f = ½m2v2 = 0.5*3.00*(v)2
Total
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= 98.0 J
= 58.8 J
= 2.5v2 J
= 1.5v2 J
=156.8+4.0v2 J
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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?
MEi = MEf + Enc
Enc = ffrictionx = 5.00*2.00 = 10.0 J
5 kg
M1
196.2 = 156.8 + 10.0 + 4.0v 2
3 kg
4.0 m
v=2.7 m/s
M2
Without Friction:
v = 3.13 m/s
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Question!
5 kg
M1
3 kg
4.0 m
In the absence of friction, when
m1 starts to move down:
TRUE 1) potential energy is
transferred from m1 to m2
2) potential energy is
TRUE
transformed into kinetic
energy
FALSE 3) m1 and m2 have the same
kinetic energy
M2
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Question
A 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
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Quiz
A 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?
h1
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)
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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)
b)
c)
d)
e)
0.
8.2 J
9.8 J
18 J
27.8 J
MEi
– MEf
= Enc
(KEi+PEi) - (KEf+PEf) = Enc
kinetic energy:
potential energy:
½mv2
mgh g=9.81 m/s2
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
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Question
An 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)
b)
c)
d)
e)
0J
30 J
90 J
120 J
don’t know
Two components of velocity at start:
vox = vocos(30o) = 34.6 m/s
voy= vosin(30o) = 20 m/s
At highest point: only horizontal velocity
vx = vox = 34.6 m/s
vy = 0 m/s
kinetic energy: ½mv2 = ½(0.15)(34.6) 2 = 90 J
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Question
An 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 h
h = 22.4 m
ie, the
ball travels 20.4 m higher than the
player’s height
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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 + PE
Conserved if only conservative forces are present MEi = MEf
Not conserved in the presence of non-conservative forces
MEi = MEf + Enc
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Question: Free Fall 1
Two stones, one twice the mass of
the other, are dropped from a cliff.
2Just before hitting the ground, what
is the VELOCITY of the heavy stone
compared to the light one?
a) quarter as much
b) half as much
c) the same
d) twice as much
e) four times as much
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.
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Quiz: Free Fall 2
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?
a) quarter as much
b) half as much
c) the same
d) twice as much
e) four times as much
KEi + PEi = KEf + PEf
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.
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