Download Physics 231 Topic 4: Energy and Work Wade Fisher September 17-21 2012

Physics 231
Topic 4: Energy and Work
Wade Fisher
September 17-21 2012
MSU Physics 231 Fall 2012
1
Quiz for extra credit
A block is sliding down a slope with constant velocity. Assuming
that the only forces in play are the normal force, gravity and
friction, which is NOT true:
a) The component of the gravitational force acting on the block
parallel to the slope equals the frictional force on the block
b) There is no net force acting on the block
c) The acceleration equals zero
d) The normal force equals the total gravitational force on the
block
n = -FgL
FgP = mg cos

Fg=mg
Fg//=mgsin

MSU Physics 231 Fall 2012
2
Recall from Last Week
Force and Mass
Newton’s Laws of Motion
 Inertia, F=ma, Equal/Opposite Reactions
Application of Newton’s Laws
 Force diagrams
 Component Forces
Friction and Drag
 Kinetic, static, rolling frictions
Uniform circular motion
MSU Physics 231 Fall 2012
3
Car Driving in a Circle
A car is driving along a circular, flat road with radius of 240m. The
coefficient of static friction between the car’s tires and the road is
0.84. Find the maximum speed the car can travel without sliding.
∑F = Fstatic = ma
= sn = smg
Acceleration required for
uniform circular motion:
ac = v2/R
Fstatic = mac
smg = mv2/R
v = (sRg)
v = ( 0.84(240 m)(9.81 m/s2) ) = 44 m/s
MSU Physics 231 Fall 2012
4
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
MSU Physics 231 Fall 2012
5
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
Units: N·m = Joule
1 calorie = 4.184 J
F

Fcos
1 Calorie = 4184 J
= 1 kcal
x
MSU Physics 231 Fall 2012
6
Work and Energy
When work is done,
energy can be stored.
Example: Lifting a ball
a distance y creates
“Potential Energy”.
Where does the energy
go in this case?
Answer: the energy
goes into friction (the
ground/sled heat up!)
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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
MSU Physics 231 Fall 2012
8
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
MSU Physics 231 Fall 2012
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Clicker Question: Friction and Work
A box is being pulled
a) friction does no work at all
across a rough floor at a
b) friction does negative work
constant speed. What can
c) friction does positive work
you say about the work
done by friction?
Friction acts in the opposite direction to
N Displacement
the displacement, so the work is
Pull
f
negative.
Or using the definition of work (W = F
(Δx)cos ), because =180º, then W < 0.
MSU Physics 231 Fall 2012
mg
10
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.
Or using the definition of work (W = F
T
(Δs)cos ), because  = 90º, then W = 0.
v
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Power: The rate of energy transfer
Work (the amount of energy transfer) is independent of time.
W=(Fcos)x … no time in here!
To measure how fast we transfer the energy we define:
Power = P = 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 m ) / ( steplength )
= (36 J)  ( L m) / (1.5 m) = (24L) J
Power = W/t = 24L/t = 24vaverage = 70 Watts
vaverage=2.9 m/s
B) F=P/v so F=24 N
MSU Physics 231 Fall 2012
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Quiz 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?
Wnet = ∑Fx
a)
b)
c)
d)
e)
don’t know
100 J
200 J
300 J
500 J
Wnet=Fx=(50-30)10=200 J
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Example
A toy-rocket of 5.0 kg, after the initial acceleration
stage, 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 = mrocketg 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 W (=3.3 horsepower)
or
P=(Fcos)v = mg(h/t) =5.09.81100/2=2453 W
MSU Physics 231 Fall 2012
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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 work on the
d) four forces
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
f
f does negative work
mgsin

mgsin does negative work
MSU Physics 231 Fall 2012
mg
16
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 the rocket: W = (Fcos)·x
Wgravity = Fg cos(180o)h = -mgh
= -( mghf – mghi ) = mghi - mghf
= 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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Kinetic energy:
Consider an object that changes speed only
x=100m
t=0
VO=0
t=2s
Vf > 0
a) W = Fcosx = (ma)x … used Newton’s second law
b) v(t) = v0+at so t=(vf-v0)/a
c) x(t) = x0+v0t+½at2 so x-x0 = x = v0t+½at2
Kinetic energy: KE=½mv2
Combine b) & c)
d) ax = (v2-v02)/2
Combine a) & d)
W=½m(v2-v02)
When work is done on an
object and the only change
is its speed: The work done
is equal to the change in KE:
MSU Physics 231 Fall 2012
W = KEfinal-KEinitial
18
Conservation of energy
Mechanical energy = Potential Energy + Kinetic energy
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.
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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 a closed system!
(Fuel burning)
* There is an additional condition, see next lecture
MSU Physics 231 Fall 2012
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Clicker Question: Work and Energy
Two blocks of mass m1 and m2 (m1 > m2) slide on
a) m1
a frictionless floor and have the same kinetic
energy when they hit a long rough stretch ( > 0),
which slows them down to a stop. Which one
goes farther?
b) m2
c) they will go the
same distance
W = (Fcos)·x
W = F x
m1
With the same KE, both blocks must
have the same work done to them by
friction. The friction force is less for m2
m2
so stopping distance must be greater.
MSU Physics 231 Fall 2012
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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 + ½0.202 = 68.67 J
At ground:
ME = mgh + ½mv2
=0.29.810 + ½0.2v2 = 0.1v2 J
Conservation of ME:
68.67 J = 0.1v2 J
v = 26.2 m/s
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Where is the kinetic energy…
1) highest?
2) lowest ?
Assume height of catapult is
negligible to the maximum height
of the stone.
And what about potential energy?
Parabolic
Motion

t=0
A
t=1
B
t=2
C
t=3
D
MSU Physics 231 Fall 2012
E
t=5
23
MSU Physics 231 Fall 2012
24
Clicker Question: Free Fall 1
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.
MSU Physics 231 Fall 2012
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A swing
If relieved 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
MSU Physics 231 Fall 2012
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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
 Elastic PE: ½kx2 energy stored in stretched/compressed spring.
 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 KEi+PEi=KEf+PEf
Not conserved in the presence of non-conservative forces
(KEi+PEi) - (KEf+PEf) = Wnc
MSU Physics 231 Fall 2012
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Clicker Question: 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 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.
MSU Physics 231 Fall 2012
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MSU Physics 231 Fall 2012
29
Race track
KE
PE TME NC
KE
With friction
PE TME NC
KE PE TME NC
MSU Physics 231 Fall 2012
KE
PE TME NC
KE
PE TME NC
30
Practice 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
so v=28 m/s
MSU Physics 231 Fall 2012
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True or False?
5 kg
M1
3 kg
4.0 m
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 energy
M2
A. 1) and 2) and 3) are true
B. only 2) and 3) are true
C. only 1) and 2) are true
D. only 2 is true.
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Conservation of mechanical energy
What is the speed of m1 and m2 when
they pass each other?
(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:
PE1 = m1gh1 =5.00*9.81*4.00
PE2 = m2gh2 =3.00*9.81*0.00
KE1 = ½m1v2 =0.5*5.00*(0.)2
KE2 = ½m1v2 =0.5*3.00*(0.)2
=196.2 J
=0.00 J
=0.00 J
=0.00 J
Total
=196. J
At time of passing:
PE1 = m1gh1 = 5.00*9.81*2.00
PE2 = m2gh2 = 3.00*9.81*2.00
KE1 = ½m1v2 = 0.5*5.00*(v)2
KE2 = ½m1v2 = 0.5*3.00*(v)2
=98.0 J
=58.8 J
=2.5v2 J
=1.5v2 J
Total
MSU Physics 231 Fall 2012
=156.8+4.0v2 J
33
Work
How much work is done by the
gravitational force by the point when
the masses pass each other?
W = Fx
= m1g2.00 + m2g(-2.00)
= 5*9.81*2 - 3*9.81*2 = 39.2 J
5 kg
PEstart- PEpassing= PE
M1
3 kg
4.0 m
M2
= (m1gh + m2g(0)) – (m1gh/2 + m1gh/2)
= (5*9.81*4 + 0)- (5*9.81*2 + 3*9.81*2)
= (196.2) – (98.1 + 58.9)
= 39.2 J
The work done by Fg is the same as
the change in potential energy
MSU Physics 231 Fall 2012
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Clicker 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
A. 1) and 2) and 3) are true
B. only 2) and 3) are true
C. only 1) and 2) are true
D. only 2 is true.
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Springs!
Hooke’s Law:
Fs = -kx
k: spring constant (N/m)
x: distance the spring is stretched from
equilibrium
Fs (x=0) = -k(0) = 0 N
Fs (x=a) = -ka N
Ws = Fsx = (ka/2)a = ½ka2
The energy stored in a spring depends
on how far the spring has been
stretched: elastic potential energy
PEspring = ½ k x2
MSU Physics 231 Fall 2012
36
PINBALL!
The ball-launcher spring has a
constant k = 120 N/m.
A player pulls the handle 0.05 m.
The mass of the ball is 0.1 kg.
What is the launching speed?
PEspring = ½ k x2 PEgravity = mgh KE = ½ mv2 ME = PE + KE
= Constant
(PEgravity+PEspring+KEball)pull = (PEgravity+PEspring+KEball)launch
mghpull + ½kxpull2 + ½mvpull2 = mghlaunch + ½kxlaunch2 + ½mvlaunch2
½120(0.05)2 = 0.1 (9.81) ( 0.05*sin(10o) ) + ½(0.1)vlaunch2
0.15 = 8.5x10-3 + 0.05v2
v =1.7 m/s
MSU Physics 231 Fall 2012
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Clicker Question Kinetic Energy
Two cars are driving along a highway.
Car #1 has twice the mass of car #2,
but they both have the same kinetic
energy.
a)
2v1 = v2
b)
 2v1 = v2
c)
4v1 = v2
d)
v1 = v2
e)
8v1 = v2
How do their speeds compare?
Because the kinetic energy is KE=½ mv2, and the mass of car #1 is greater,
then car #2 must be moving faster. If the ratio of m1/m2 is 2, then the ratio
of v2 values must also be 2: (v2/v1)2 = 2
This means that the ratio of v2/v1 must be the square root of 2.
MSU Physics 231 Fall 2012
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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!
MSU Physics 231 Fall 2012
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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.
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40
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=Wnc if some
forces are nonconservative.
Wnc=work done by non-conservative forces.
Example: throwing a snowball from a
building taking into account air resistance
MSU Physics 231 Fall 2012
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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?
(PE+ KE)start-(PE+KE)passing = Wnc
PEstart-PEpassing-KEpassing = Wnc
Wnc = Ffrictionx = 5.00*2.00 = 10.0 J
5 kg
M1
3 kg
4.0 m
M2
Without Friction:
v = 3.13 m/s
PEstart- PEpassing
= (m1gh + m2g(0)) – (m1gh/2 + m1gh/2)
= (5*9.81*4 + 0)- (5*9.81*2 + 3*9.81*2)
= (196.2) – (98.1 + 58.9)
= 39.2 J
39.2 J - KEpassing = 10 J
KE = 29.2 J = ½(m1+m2)v2 = ½(8)v2 = 4v2
v=2.7
m/s
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Overview
Conservation of mechanical
energy
Newton’s second Law
F=ma
Work
W=(Fcos)x
Wnc=0
Closed system
Work-energy Theorem
Wnc=MEf-MEi
Equations of kinematics
X(t)=X(0)+V(0)t+½at2
V(t)=V(0)+at
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43
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
MSU Physics 231 Fall 2012
time
44
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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45
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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Extra credit 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)
MSU Physics 231 Fall 2012
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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 work is done by friction?
a)
b)
c)
d)
e)
0.
8.2 J
9.8 J
18 J
27.8 J
MEi
– MEf
= Wnc
(KEi+PEi) - (KEf+PEf) = Wnc
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
Wnc = 18-9.8 = 8.2 J
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48
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:
vx = vocos(30o) = 34.6 m/s
vy = vosin(30o) = 20 m/s
At highest point: only horizontal velocity
vx,highest = 34.6 m/s
vy,highest = 0 m/s
kinetic energy: ½mv2 = ½(0.15)(34.6) 2 = 90 J
MSU Physics 231 Fall 2012
49
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?
Two components of velocity at start:
vx = vocos(30o) = 34.6 m/s
vy = vosin(30o) = 20 m/s
Only velocity in y direction matters!
Initial: KE + PE = ½mv2 + mgh
= ½ (0.15)(20) 2 + (0.15)(9.81)(2) = 30 + 2.94
At highest point: only potential energy
KE + PE = PE = mgh = (0.15)(9.81)(h) = 1.47 h
32.94 = 1.47 h
h = 22.4 m
ie, the ball travels 20.4 m higher than the
player’s height
MSU Physics 231 Fall 2012
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For Next Week
Rex & Wolfson Chapter 6.1-6.5:
Momentum and Collisions
Homework Set 4: Covers Chapters 5
Due Wednesday 11PM
MSU Physics 231 Fall 2012
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