Midterm 2 will be held on November 8. Covers units 4-9
Classical Mechanics
Lecture 7
Today’s Concepts:
Work & Kinetic Energy
JFB Rotunda
David Matthew Stephens
Lei Shan (Mike)
Mechanics Lecture 7, Slide 1
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:
W  K

r2
 
W   F dl

r1
1 2
K  mv
2
Mechanics Lecture 7, Slide 2
The Dot Product
Mechanics Lecture 7, Slide 3
Work –Kinetic Energy Theorem: Work on sliding box
Forces that act perpendicular to
displacement perform no work!!!
Mechanics Lecture 7, Slide 4
Work –Kinetic Energy Theorem in 2-d
Mechanics Lecture 7, Slide 5
Work-Kinetic Energy Theorem
If there are several forces acting then W is the work
done by the net (total) force:
WNET  K
 W1  W2  ...
You can just add up the
work done by each force
WNET  WTOT
Mechanics Lecture 7, Slide 6
Work done by Weight
Mechanics Lecture 7, Slide 7

r2
 
F

d
l


K

Derivation – not so important
Concept – very important

r1
r2
F

dl


K

r1
A force pushing over some distance
will change the kinetic energy.
q

F
Mechanics Lecture 7, Slide 8
Work done by gravity near the Earth’s surface
mg
Mechanics Lecture 7, Slide 9
Work done by gravity near the Earth’s surface
WTOT  W1  W2  ... WN
 
 
 
 mg  dl1  mg  dl2  ... mg  dlN
dlN
dl1
mg
dl2
dy1
dl1
dx1
mg
Mechanics Lecture 7, Slide 10
Work done by gravity near the Earth’s surface
WTOT  W1  W2  ... WN
 
 
 
 mg  dl1  mg  dl2  ... mg  dlN
 mgdy1  mgdy2... mgdyN
 m g  y
dlN
y
dl1
mg
dl2
Wg  m g  y
Mechanics Lecture 7, Slide 11
Different paths…
A.
B.
C.
D.
Three objects having the same mass begin
at the same height, and all move down the
same vertical distance H. One falls straight
down, one slides down a frictionless
inclined plane, and one swings on the end
of a string.
In which case
does the object
have the biggest
net work done
on it by all forces
during its
motion?
H
Free Fall
A) Free Fall
Frictionless incline
B) Incline
C) String
String
D) All the same
0%
0%
0%
0%
Mechanics Lecture 7, Slide 12
A.
Clicker Question
B.
C.
D.
Three objects having the same mass begin at the same
height, and all move down the same vertical distance H.
One falls straight down, one slides down a frictionless
inclined plane, and one swings on the end of a string.
What is the relationship between their velocities when they
reach the bottom?
H
Free Fall
Frictionless incline
A) vf > vi > vp
B) vf > vp > vi
String
C) vf  vp  vi
0%
0%
0%
0%
Mechanics Lecture 7, Slide 13
CheckPoint / Clicker Question
H
Free Fall
A) vf > vi > vp
Frictionless incline
B) vf > vp > vi
Only gravity will do work: Wg   K
String
C) vf  vp  vi
Wg  mgH
 K  1/2 mv22
v  2 gH
v f  vi  v p  2 gH
Mechanics Lecture 7, Slide 14
Work –Kinetic Energy Theorem Applied
Mechanics Lecture 7, Slide 15
Work done by a Spring
iˆ


  Fspring  dl
Wspring


Fspring  kx  kxiˆ


dW  Fspring  dl

dl  dxiˆ
dW   kxdx
x2
W12   dW  k  xdx
x1
W1 2
1
  k ( x22  x12 )
2
Physics 211 Lecture 7, Slide 16
Work done by a Spring


Fspring   kx   kxiˆ


dW  Fspring  dl

dl  dxiˆ
dW   kxdx
x2
W12   dW   k  xdx
x1
1
W12   k ( x22  x12 )
2
Physics 211 Lecture 7, Slide 17
Work done by a Spring: Negative Work?!!!
Physics 211 Lecture 7, Slide 18
Work done by a Spring: Conservative Force!
Conservative Force
Net Work over closed path = 0
Physics 211 Lecture 7, Slide 19
Work done by Gravitational Force
Wgravity


  Fgravity  dl

GM E m
Fgravity  
rˆ
2
r
Mechanics Lecture 7, Slide 20
Work done by Gravitational Force
Wgravity


  Fgravity  dl

GM E m
Fgravity  
rˆ
2
r


GM E m
dW  Fgravity  dl  
rˆ  (drrˆ  rdqqˆ)
2
r
rˆ  (drrˆ  rdqqˆ)  drrˆ  rˆ  0  dr
Mechanics Lecture 7, Slide 21
Work done by Gravitational Force
dW  
GM E m
dr
2
r
r2
1
dr
2
r
r1
W12   dW  GM E m 
W12  GM E m(
1 1
 )
r2 r1
Mechanics Lecture 7, Slide 22
Work done by Gravitational Force
Mechanics Lecture 7, Slide 23
Main Points
Mechanics Lecture 7, Slide 24
Main Points
Mechanics Lecture 7, Slide 25
Main Points
Mechanics Lecture 7, Slide 26
Work-Kinetic Energy Theorem: 1-D Example
If the force is constant and the directions aren’t
changing then this is very simple to evaluate:
car
F
d

r2
   
W   F dl  F d

r1
In this case
= Fd
since cos(0)=1
This is probably what you remember from High School.
Mechanics Lecture 7, Slide 27
A.
Clicker Question
B.
C.
A lighter car and a heavier van, each initially at rest, are
pushed with the same constant force F. After both
vehicles travel a distance d, which of the following
statements is true? (Ignore friction)
F
d
F
d
car
van
A) They will have the same velocity
B) They will have the same kinetic energy
C) They will have the same momentum
0%
0%
0%
Mechanics Lecture 7, Slide 28
Work Done by Spring?
A.
B.
C.
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
0%
0%
0%
Mechanics Lecture 7, Slide 29
Work done by hand
A.
B.
C.
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
0%
0%
0%
Mechanics Lecture 7, Slide 30
Total Work on Box?
A.
B.
C.
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
0%
0%
0%
Mechanics Lecture 7, Slide 31
Clicker Question
A.
33%
33%
33%
B.
C.
In Case 1 we send an object from the surface of the earth to a height
above the earth surface equal to one earth radius.
In Case 2 we start the same object a height of one earth radius above
the surface of the earth and we send it infinitely far away.
In which case is the magnitude of the work done by the Earth’s
gravity on the object biggest?
A) Case 1
B) Case 2
C) They are the same
1 1
W  GM e m  
 r2 r1 
Mechanics Lecture 7, Slide 32
Clicker Question Solution
Case 1:
Case 2:
 1
1 
GM e m
W  GM e m
   
2R E
 2 RE RE 
1
1 
GM e m
  
W  GM e m 
2R E
  2 RE 
Same!
RE
2RE
Mechanics Lecture 7, Slide 33
Work on Two Blocks
W  (m gh)
Mechanics Lecture 7, Slide 34
Work on Two Blocks
1
1
2
W  (m2 gh)  (m1  m2 )(v22 f  v20
)  (m1  m2 )v22 f
2
2
 2m2 gh
v1 f  v2 f 
(m1  m2 )
W1 
1
1
m1 (v12f  v102 )  (m1 )v12f
2
2
W1  Tx
W1
T
x
Mechanics Lecture 7, Slide 35
Work on Two Blocks
W2 
1
1
2
m2 (v22 f  v20
)  (m2 )v22 f
2
2
Mechanics Lecture 7, Slide 36
Work on Two Blocks 2
x0  x
 
W   F  dx  Fx
x0
x0  x
 
WN   N  dx  0
x0


1
1
m v 2f  v02  m v2f
2
2
2W
vf 
m1  m2
W  Fx 
Mechanics Lecture 7, Slide 37
Work on Two Blocks 2
1
m2 v 2f
2
W
W  Fx  T x
T
W
x
W
W1net 
1
m1v 2f
2
1
m1v 2f or W1net  Fnet x  ( F  T )x
2
Mechanics Lecture 7, Slide 38
Block Sliding
x0  x
Wspring
  x0  x   1
2
  F  dx   kx  dx  k x 
2
x0
x0
1 2
m vf
2
2Wspring
Wspring 
vf 
m
Mechanics Lecture 7, Slide 39
Block Sliding
x0  x
x  x

 0

W friction   F friction  dx     k mg iˆ  dx    k mg x 
x0
x0
W friction 
1
m(v 2f  v02 )
2
1


2W friction  m v02 
2

vf  
m
1

2
2   k m gx fric   k xspring  
2
 0
vf  
m
1
 k m gx friction   k xspring 2
2
1
2
k xspring 
x friction  2
k m g
Mechanics Lecture 7, Slide 40
Block Sliding
1
2
k xspring 
x friction  2
k m g
xspring 
2 k m gx friction 
k
Mechanics Lecture 7, Slide 41
Block Sliding 2
Wspring 
1
1
mv 2  mv 2f
2
2
Wspring 
1
1
2
k x   m v2
2
2
x 
m v2f
k
Mechanics Lecture 7, Slide 42
Block Sliding 2
Wspring 

1
1
mv 2  m v 2f  v02
2
2

W friction    k m gx
x 
W friction
 k m g
Mechanics Lecture 7, Slide 43
Block Sliding 2
W friction    k m gxrough 
vspring 
x 
1
2
m(0  vspring
)
2
2  k m gxrough
m
2
m vspring
k

2  k m gxrough
k
W friction    k m gxrough
xrough  2xrough   k 
1
k
2
Mechanics Lecture 7, Slide 44
Block Sliding 2
Wfriction  k mgxrough
Mechanics Lecture 7, Slide 45