Your comments
So what do we do when one of the r's is 0 in the equation GmM(1/r-1/r)?
Do we need to derive all of these potential energy formulas?
I don't understand springs
This was the first lecture I actually learned something that I had never seen before, regarding
the new definition of potential energy. It's confusing, but cool
Whoever got so mad about the apple problem (e.g. the guy who's text got put up during the
lecture) needs to shape up and learn some humility. Whoever has the nerve to tell a
professor "you should have known better" on a CHECKPOINT QUESTION needs to
reconsider their priorities. We're all here because we're smart, you don't need to go
around being an arrogant … (it got better but I can’t put it here)
Could we go over mechanical energy and conservative forces?
Overall, everything made sense, I would just like it if you quickly explained again how
potential energy is mixed in with kinetic energy and work.
I found everything to be difficult because it is the first time I am learning these concepts.
Still waiting to find out where to purchase "It is what it has to be to do what it does" tshirts/mugs/assorted merchandise.
Gradebook & Midterm & Office Hours
A little commentary on the exam coming up on Oct.
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 Oct. 3 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 situation and knowledge
of definition of work 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
D K  W total
D U  W
E  K U
D E  W NC
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:
D K  W tot
In P211 everything
except gravity
and springs is NC
 W gravity  W springs  W N C
  D U gravity  D U springs  W N C
D K  D U gravity  D U springs  W N C
Define Mechanical
Energy: E  K  U
D K  D U  W NC
D E  W NC
D E  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
Ef - Ei = Wfriction
(Uf + Kf) - (Ui + Ki) = Wfriction
½ mv2 - 427 J = - 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”
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) x 2 
2 x1
B) x 2  2 x1
C) x 2  4 x1
x
Mechanics Lecture 8, Slide 17
CheckPoint
K 
1
x
mv
2
2
U 
1
kx
2
2
A) x 2 
2 x1
B) x 2  2 x1
C) x 2  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
DK  
1
mv
2
DU 
2
1
kx
2
2
1
mv 
2
2
1
kx
2
2
xv
m
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 in a
different way now and still get the
same answer:
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?”
Last time we used the Work –
Kinetic Energy theorem to explain
this and it 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.
Fun with balls rolling along tracks
track A
track B
A) Ball on track A wins the race
B) Ball on track B wins the race
C) It’s a tie—balls arrive at the end at the same time
I will ask for volunteers to explain the reasoning leading to your selection.
Fun with balls rolling along tracks
Explanation for why ball B wins race
Let’s now consider a slightly different scenario
It’s getting funner and funner
A) Ball on track A wins the race
B) Ball on track B wins the race
C) It’s a tie—balls arrive at the end at the same time
I will ask for volunteers to explain the reasoning leading to your selection.
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 26
Lets look at U…
For gravity:
U (r )  
GM em
r
RE
2RE
+U 0
3RE
Mechanics Lecture 8, Slide 27
Clicker Question
U (r )  
GM em
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 28
Clicker Question
U (r )  
GM em
r
What is the potential energy of a object starting at the
height of Case 1?
A)
B)
U1  
GM em
U1  
GM em
U1  
GM em
C)
RE
RE
2RE 3RE
RE
2 RE
3RE
Mechanics Lecture 8, Slide 29
Clicker Question
U (r )  
GM em
r
What is the potential energy of a object starting at the
height of Case 2?
A)
B)
C)
U2  
GM em
U2  
GM em
U2  
GM em
RE
RE
2RE 3RE
RE
2 RE
3RE
Mechanics Lecture 8, Slide 30
GM e m
GM em
GM em
U surface  
U2  
U1  
What is the change
in potential2 R
inECase 1?
RE
3RE
What is the change in potential in Case 1?
A) D U case1
 1
1  1 GM em
 GM em 


 2 Re Re  2 Re
B) D U case1
 1
1 
1 GM em
 G M em 


2 Re
 Re 2 Re 
RE
Mechanics Lecture 8, Slide 31
GM e m
GM em
GM em
U surface  
U2  
U1  
What is the change
in potential2 R
inECase 2?
RE
3RE
What is the change in potential in Case 2?
A) D U case 2
 1
1 
1 GM em
 G M em 


3 Re
 Re 3 Re 
B) D U case 2
 1
1 
2 GM em
 GM em 


3 Re
 Re 3 Re 
RE
Mechanics Lecture 8, Slide 32
D U case1  
GM em
2 Re
What is the ratio
D U case 2  
DK 2
DK1

2G M e m
3 Re
DU 2
A) 2
DU 1
B) 4
2
4
3


1
3
2
C) 4/3
D) 3/2
Mechanics Lecture 8, Slide 33
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 34