Transcript Physics 211 - University of Utah

Midterm this Friday March 13
Classical Mechanics
Midterm 2 Review
Force and Energy







Newton’s Laws of Motion
Forces and Free Body Diagrams
Friction
Work and Kinetic Energy
Conservative Forces and Potential Energy
Work and Potential Energy
Knowledge of Units 1-3 will be useful
“Playing with Blocks”
Mechanics Lecture 9, Slide 1
Midterm Exam


Multiple choice…but show your work and justification.
Calculations
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Forces and Free-Body Diagrams
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Weight and Pulleys

Springs

Friction

Gravitational

Normal
Work Calculations
Conservation of Energy
Conceptual questions…like checkpoint problems.
Bring calculators and up to five sheets of notes
Mechanics Lecture 8, Slide 2
Midterm Exam 2
Sample exam
Folder under
Files menu on
PHYS1500
Canvas Website
Phys 1500 Exams https://utah.instructure.com/courses/320947/files
- Spring 2013: http://www.physics.utah.edu/~springer/phys1500/exams/MidtermExam2.pdf
- Solutions: http://www.physics.utah.edu/~springer/phys1500/exams/MidtermExam2Soln.pdf
- Long Sample: https://utah.instructure.com/courses/320947/files/45779670/download?wrap=1
- Long Solutions:https://utah.instructure.com/courses/320947/files/45802143/download?wrap=1
Phys 2210 Exams
- Practice : http://www.physics.utah.edu/~woolf/2210_Jui/rev2.pdf
- Spring 2015: http://www.physics.utah.edu/~woolf/2210_Jui/ex2.pdf
Mechanics Lecture 8, Slide 3
Force and Energy Summary


Fnet  ma
Emechanical  Wnonconservative
Mechanics Lecture 8, Slide 4
Determining Motion
Force

Unbalanced Forces  acceleration
(otherwise objects velocity is constant)


Fnet  ma

Energy
Total Energy  Motion, Location


 Fnet
a
m
Determine Net Force acting on
object
Emechanical  K  U
Emechanical  Wnonconservative

rf

Work


Fnet   Fi
i

F12  F21

Use kinematic equations to
determine resulting motion
a  a0
v(t )  a0t  v0
x(t ) 
1 2
a0t  v0t  x0
2
Wnet


  Fnet  dl  K

r0
U  Wnet  K
 Emechanical  U  K  0

Motion from Energy conservation
Emechanical, final  K f  U f
Emechanical, final  K i  U i  Emechanical
K f  K i  U i  Emechanical  U f
vf 
2
Ki  U i  Emechanical  U f 
m
Mechanics Lecture 8, Slide 5
Unit 4: Newton’s Laws
Mechanics Lecture 4, Slide 6
Unit 5: Forces and Free-Body Diagrams

FFloor ,box

abox
Box

FMan,box

FEarth ,box
Mechanics Lecture 5, Slide 7
Inventory of Forces
 Weight
 Normal Force
 Tension
 Gravitational
 Springs
 …Friction
Mechanics Lecture 5, Slide 8
Mechanics Lecture 5, Slide 9
Force Summary


Fnet  ma


Fnet   Fi
i
Mechanics Lecture 5, Slide 10
Unit 6: Friction
Mechanics Lecture 6, Slide 11
Friction
Mechanics Lecture 6, Slide 12
Unit 7: Work and Kinetic Energy
Mechanics Lecture 7, Slide 13
Work-Kinetic Energy Theorem
The work done by force F as it acts on an object that
moves between positions r1 and r2 is equal to the
change in the object’s kinetic energy:
But again…!!!

r2
W  K
 
W   F dl

r1
1 2
K  mv
2
Mechanics Lecture 7, Slide 14
The Dot Product
Mechanics Lecture 7, Slide 15
Vectors!!!
Mechanics Lecture 8, Slide 16
Unit 8: Conservative Forces & Potential Energy
Mechanics Lecture 7, Slide 17
Unit 8: Conservative Forces & Potential Energy
Mechanics Lecture 8, Slide 18
Unit 9:Work and Potential Energy
Mechanics Lecture 8, Slide 19
Energy Conservation Problems in general
For systems with only conservative forces acting
Emechanical  0
Emechanical is a constant
Emechanical  Ki  Ui  K f  U f  K (t )  U (t )
Mechanics Lecture 8, Slide 20
Gravitational Potential Problems
 
r  rE

rE
 
r  rM
 conservation of mechanical energy
can be used to “easily” solve
problems.
Emechanical  K  U 

r

rM
 Add potential energy from each
source.
GM E m
U Earth (rE )  
rE
U Moon (rM )  
1
mv (h) 2  U (h) gravity
2
 Define coordinates: where is
U=0?
U (r )  
GM E m
 0 as r  
r
GM M m
rM

GM E m GM M m
U total (r )       
r  rE
r  rM
Mechanics Lecture 8, Slide 21
Example Problem : Block and spring
A 2.5 kg box is held released from rest 1.5 m above the ground and slides
down a frictionless ramp. It slides across a floor that is frictionless, except for
a small section 0.5 m wide that has a coefficient of kinetic friction of 0.2. At
the left end, is a spring with spring constant 250 N/m. The box compresses
the spring, and is accelerated back to the right.
What is the speed of the box at the bottom of the ramp?
What is the maximum distance the spring is compressed by the box?
Draw a free-body diagram for the box while at the top of the incline ? When the spring
is maximally compressed? When the box is sliding on the rough spot to the right?
What is the total work done by friction? Each way?
What height does the box reach up the ramp after hitting the spring once?
Where will the box come to rest?
2.5 kg
k=250 N/m
h=1.5 m
mk = 0.4
d = 0.50 m
Mechanics Review 2 , Slide 22
Example Problem: Free-Body Diagram
1) FBD
m2
N
f
m2
T
g
T
m2g
m1
m1
m1g
Mechanics Lecture 8, Slide 23
Example Problem: Free-Body Diagram
1) FBD
2) SF=ma
m2
N
T
m2
f
g
T
m2g
N = m2g
T – m m2g = m2a
m1g – T = m1a
m1
m1
m1g
add
m1g – m m2g = m1a + m2a
a=
m1g – m m2g
m 1 + m2
Mechanics Lecture 8, Slide 24
Example Problem: Free-Body Diagram
1) FBD
2) SF=ma
m2
N
f
m2
T
g
T
m1
m1
m2g
m1g
a=
m1g – m m2g
m1 + m2
m1g – T = m1a
T = m1g – m1a
T is smaller when a is bigger
Mechanics Lecture 8, Slide 25
Example Problems
Mechanics Lecture 8, Slide 26
Example Problem
Wtension
 
  T  dl  Tx
Wnet  Wtension  W friction  K
W friction  Wtension  K

1
K  m v 2f  v02
2


1
W friction  Wtension  m v 2f  v02
2

Mechanics Lecture 8, Slide 27
Block
1 2
2x
at  a  2
2
t
F
m g sin   f k
a  net 
m
m
2x 

f k  m g sin   m a  m g sin   2 
t 

x 
Mechanics Lecture 5, Slide 28
Pushing Blocks

F23net
a
Fh1
F
 h1
(m1  m2  m3  m4 ) 4m1
F23 netx  (m3  m4 )a  2m1a 
2m1 Fh1 Fh1

4m1
2
F23 net y  N  (m3  m4 ) g  2m1 g
2
F23 net  F
2
23 net x
F
2
23 net y
F 
2
  h1   2m1 g 
 2 
Mechanics Lecture 5, Slide 29
Example Problems
Emechanical  W friction
W friction   m k m gx
Emechanical, final  Emechanical,initial  W friction
m gh  m ghi  m k m gx  m ghi  m k x 
h  hi  m k x 
Mechanics Lecture 8, Slide 30
Example: Pendulum
vv 22gh
gh
hh
Conserve Energy from initial to final position.
1 2
mgh  mv
2
v  2gh
Mechanics Review 2 , Slide 31
Example Problems
Mechanics Lecture 8, Slide 32
Example: Pulley and Two Masses
A block of mass m1 = 1 kg sits atop an inclined plane of angle
θ = 20o with coefficient of kinetic friction 0.2 and is connected
to mass m2 = 3 kg through a string that goes over a massless
frictionless pulley. The system starts at rest and mass m2 falls
through a height H = 2 m.
Use energy methods to find the velocity of mass m2 just
before it hits the ground?
θ
m2
HH
=2
m
=2
2m
kg
Mechanics Lecture 19, Slide 33
Example Problems
Mechanics Lecture 8, Slide 34
Example Problems
Mechanics Lecture 8, Slide 35
Force and Energy Summary


Fnet  ma
Emechanical  Wnonconservative
Mechanics Lecture 8, Slide 36