Your comments
I feel like this could work brilliantly in a nice sized stadium seating lecture hall. Point to a
student in the back/upper seats, say they look like they have a lot of potential and ask them
to come down front for a demonstration. When they get there say, "No, I've misjudged. You
don't have much potential at all." Send them back. When they're half way back up, stop them
and say "Well, wait, now I think I see some potential in you again!". Repeat until they catch
on or give up.
This prelecture was definitely the hardest so far. not awful, but kind of confusing
It was fairly simple stuff - it's just that there are SO MANY EQUATIONS.
I feel i invest too much time in studying this lecture videos, however dont comprehend more
then 60 % of the material?
There was so much discussed in this lecture, I am very overwhelmed. I would like all of it to
be gone over in class.
What speed does it take for something hurtling towards the earth to become a fireball?
If mass is irrelevant for a roller coaster negotiating a loop, why are there different height
restrictions for rides?
cool story bro
I think the class should be given potential energy in the form of donuts because they have
the potential to be delicious.
The relationships between potential and kinetic energy. Also, more AC/DC or the likes. That
woke me up instantly.
Gradebook & Midterm & Office Hours
“Prelectures lack examples.. all this conceptual crap is NOT on
the exam so... correlate accordingly so my mom doesn't ask why
I failed all the midterms and finals because the lectures don't
help and the prelectures are just as helpless -__-...”
Thanks for reminding me to tell you all about the exam:
Around half the problems purely conceptual:
Understanding concepts is in fact the key to solving all of the problems.
Have a look at the practice exams.
The first exam is in 2 weeks (Wednesday Feb 22 at 7pm – see planner)
If you have a conflict you can sign up for the conflict exam in your grade-book.
Don’t forget – we have extra office hours during exam week.
Mechanics Lecture 8, Slide 2
Physics 211
Lecture 8
Today's Concepts:
a) Potential Energy
b) Mechanical Energy
Mechanics Lecture 8, Slide 3
Work done by a Spring
“Can you spend more time
explaining the spring
problems.”
If you’re worried about the sign:
Use formula to get the magnitude
Use picture to get the sign
Clicker Question
A box attached at rest to a spring at its equilibrium length. You
now push the box with your hand so that the spring is compressed
a distance D, and you hold the box at rest in this new location.
D
During this motion, the spring does:
A) Positive Work
B) Negative Work
C) Zero work
Clicker Question
A box attached at rest to a spring at its equilibrium length. You
now push the box with your hand so that the spring is compressed
a distance D, and you hold the box at rest in this new location.
D
During this motion, your hand does:
A) Positive Work
B) Negative Work
C) Zero work
Clicker Question
A box attached at rest to a spring at its equilibrium length. You
now push the box with your hand so that the spring is compressed
a distance D, and you hold the box at rest in this new location.
D
During this motion, the total work done on the box is:
A) Positive
B) Negative
C) Zero
Homework Problem
Wspring = ½ k x2
= ½ (4111) (0.456)2
= 427 J
Mechanics Lecture 8, Slide 8
Homework Problem
Wtot = DK
Wspring = 427J = ½ mv2
v2 = 2 * 427 / 12
v = 8.44 m/s
Mechanics Lecture 8, Slide 9
Homework Problem
Wfriction = -mmgd
Wfriction = - (0.45)(12)(9.81)(2.1)
Wfriction = - 111.2 J
Mechanics Lecture 8, Slide 10
Homework Problem
Wtot = DK
Wspring + Wfriction = 427 J - 111.2 J = ½ mv2
v2 = 2 * (427-111.2) / 12
v = 7.25 m/s
Mechanics Lecture 8, Slide 11
Summary
DK  Wtotal
DU  W
E  K U
DE  WNC
Lecture 7
Work – Kinetic Energy theorem
Lecture 8
For springs & gravity
(conservative forces)
Total Mechanical Energy
E = Kinetic + Potential
Work done by any force other than
gravity and springs will change E
Mechanics Lecture 8, Slide 12
Relax. There is nothing new here
It’s just re-writing the work-KE theorem:
In P211 everything
except gravity
and springs is NC
DK  Wtot
 Wgravity  Wsprings  WNC
DK  DU gravity  DU springs
Define Mechanical
Energy: E  K  U
 DU gravity  DU springs  WNC
 WNC
DK  DU  WNC
DE  WNC
DE  0
If other forces aren't doing work
“So what exactly is the point of potential energy then? Also, why is
'U' the character used to represent it?”
Mechanics Lecture 8, Slide 13
Homework Problem
DE = WNC
DUspring + ½ mv2 = Wfriction
-427 J + ½ mv2 = - 111.2
v2 = 2 * (427-111.2) / 12
v = 7.25 m/s
Mechanics Lecture 8, Slide 14
Finding the potential energy change:
Use formulas to find the magnitude
Check the sign by understanding the
problem…
Mechanics Lecture 8, Slide 15
CheckPoint
Three balls of equal mass are fired simultaneously with equal speeds
from the same height h above the ground. Ball 1 is fired straight up,
ball 2 is fired straight down, and ball 3 is fired horizontally. Rank in
order from largest to smallest their speeds v1, v2, and v3 just before
each ball hits the ground.
2
A) v1 > v2 > v3
B) v3 > v2 > v1
C) v2 > v3 > v1
D) v1 = v2 = v3
1
3
h
“They have the same U and K to start with and delta_U is the same”
“I'm so confused- didn't we cover this in when we went over kinematics?”
CheckPoint
A box sliding on a horizontal frictionless surface runs into a fixed spring,
compressing it a distance x1 from its relaxed position while momentarily
coming to rest.
If the initial speed of the box were doubled, how far x2 would the spring
compress?
A) x2
 2 x1
B)
x2  2 x1
C)
x2  4 x1
x
Mechanics Lecture 8, Slide 17
CheckPoint
x
1 2
K  mv
2
1 2
U  kx
2
A) x2
 2 x1
B)
x2  2 x1
C)
x2  4 x1
A) Because w is proportional to x^2, I thought the
doubled speed would result in square root of x.
B) Both velocity and the distance compressed are
squared so they will changes with a 1 to 1 ratio
C) because the v is squared, therefore doubling it,
quadruples the result
Mechanics Lecture 8, Slide 18
DE  DK  DU  0
1 2
DK   mv
2
1 2
DU  kx
2
1 2 1 2
mv  kx
2
2
m
xv
k
Mechanics Lecture 8, Slide 19
Vertical Springs
M
kx2
x
DU = ½ kd2 - mgd
= ½ ky’2
Equilibrium without mass
d
Equilibrium with mass
They both look the same !!
Mechanics Lecture 8, Slide 20
From Tuesday’s lecture
“Why did you lie about the apple?”
I didn’t - we can explain this now:
DE = WNC
DK + DU
0
Whand
DU = Whand
The change in potential energy of the apple is due to
just the work done on it by your hand as you lift it.
From Tuesday’s lecture
“Why did you lie about the apple?”
The Work – Kinetic Energy
theorem says the same thing:
Wtot = DK
Whand + Wgravity =
Whand
0
=
- Wgravity
Whand =
DUgravity
Same result: The change in potential energy of the
apple is due to just the work done on it by your hand
as you lift it.
CheckPoint
In Case 1 we release an object from a height above the surface of the earth equal to 1
earth radius, and we measure its kinetic energy just before it hits the earth to be K1.
In Case 2 we release an object from a height above the surface of the earth equal to 2
earth radii, and we measure its kinetic energy just before it hits the earth to be K2.
Compare K1 and K2.
A)
B)
C)
D)
K2 = 2K1
K2 = 4K1
K2 = 4K1/3
K2 = 3K1/2
wrong
Mechanics Lecture 8, Slide 23
Lets look at U…
For gravity:
RE
2RE
GM e m
U (r )  
+U 0
r
3RE
“Why would you drop a ball from the height 2x the Earth's radius?
Seems impractical.”
Mechanics Lecture 8, Slide 24
Clicker Question
GM e m
U (r )  
r
What is the potential energy of an object
of on the earths surface:
RE
A) Usurface =
2RE 3RE
GMem
0
GMem
B) Usurface =
RE
GMem
C) Usurface =
2RE
RE
Mechanics Lecture 8, Slide 25
Clicker Question
GM e m
U (r )  
r
What is the potential energy of a object starting at the
height of Case 1?
A)
B)
C)
GM e m
U1  
RE
GM e m
U1  
2 RE
GM e m
U1  
3RE
RE
RE
2RE 3RE
Mechanics Lecture 8, Slide 26
Clicker Question
GM e m
U (r )  
r
What is the potential energy of a object starting at the
height of Case 2?
A)
B)
C)
GM e m
U2  
RE
GM e m
U2  
2 RE
GM e m
U2  
3RE
RE
RE
2RE 3RE
Mechanics Lecture 8, Slide 27
GM e m
GM e m
GM e m
U surface  
U2  
U1  
What is the change
inECase 1?
RE in potential2 R
3RE
What is the change in potential in Case 1?
 1
1  1 GM e m
A) DU case1  GM e m 
 
 2 Re Re  2 Re
1
1 
1 GM e m
B) DU case1  GM e m  

2 Re
 Re 2Re 
RE
Mechanics Lecture 8, Slide 28
GM e m
GM e m
GM e m
U surface  
U2  
U1  
What is the change
inECase 2?
RE in potential2 R
3RE
What is the change in potential in Case 2?
A) DU case 2
 1
1 
1 GM e m
 GM e m  

3 Re
 Re 3Re 
B) DU case 2
 1
1 
2 GM e m
 GM e m  

3 Re
 Re 3Re 
RE
Mechanics Lecture 8, Slide 29
GM e m
DU case1  
2 Re
What is the ratio
DU case 2
2GM e m

3Re
DK 2 DU 2

DK1 DU1
2
4
3


1
3
2
A) 2
B) 4
C) 4/3
D) 3/2
Mechanics Lecture 8, Slide 30
Checkpoint
In Case 1 we release an object from a height above the surface of the earth equal
to 1 earth radius, and we measure its kinetic energy just before it hits the earth to
be K1.
In Case 2 we release an object from a height above the surface of the earth equal
to 2 earth radii, and we measure its kinetic energy just before it hits the earth to be
K2.
Compare K1 and K2.
A) K2 = 2K1
B) K2 = 4K1
C) K2 = 4K1/3 D) K2 = 3K1/2
Mechanics Lecture 8, Slide 31