Announcements
1. Physics colloquium (Professor Brian Matthews) cancelled.
2. Scholarship opportunities for undergraduate students
3. Schedule -- First exam – Tuesday, Sept. 30th
Review Chapters 1-8 – Thursday, Sept. 25th
Extra practice problems on line
Extra problem solving sessions???
4. Some comments on friction and HW 6
5. The notion of energy and work
6/23/2016
PHY 113 -- Lecture 7
1
6/23/2016
PHY 113 -- Lecture 7
2
Surface friction force models
Static friction
f = -Fapplied if | f | < ms N
Kinetic friction
| f | = ms N
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PHY 113 -- Lecture 7
3
N
f
mg
F sin 
N-mg-F sin  = 0
f = mk N
F cos  - f = ma
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PHY 113 -- Lecture 7
4
without friction : mg sin   ma
D  12 at 2
with friction : mg sin  - m k mg cos   ma f
t f  2t
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 4a f  a
D  12 a f t 2f
 4 g sin  - m k g cos    g sin 
PHY 113 -- Lecture 7
5
Energy  work, kinetic energy
Force  effects acceleration
A related quantity is Work
rf
Wi  f   F  dr
ri
A·B = AB cos
F
dr
ri
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rj
PHY 113 -- Lecture 7
6
Units of work:
work = force · displacement = (N · m) = (joule)
•Only the component of force in the direction of the
displacement contributes to work.
•Work is a scalar quantity.
•If the force is not constant, the integral form must be used.
•Work can be defined for a specific force or for a combination
of forces
rf
W1   F1  dr
ri
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rf
rf
W2   F2  dr
W1 2   (F1  F2 )  dr  W1  W2
ri
PHY 113 -- Lecture 7
ri
7
Peer instruction question
A ball with a weight of 5 N follows the trajectory shown. What
is the work done by gravity from the initial ri to final
displacement rf?
2.5m
rf
ri
1m
1m
(a) 0 J
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10 m
(b) 7.5 J
(c) 12.5 J
PHY 113 -- Lecture 7
(d) 50 J
8
Gravity does
negative work:
rf
Gravity does
positive work:
ri
mg
mg
ri
W=-mg(rf-ri)<0
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rf
W=-mg(rf-ri)>0
PHY 113 -- Lecture 7
9
F
Positive work
x
Negative work
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PHY 113 -- Lecture 7
10
More examples:
Suppose a rope lifts a weight of 1000N by 0.5m at a constant
upward velocity of 2m/s. How much work is done by the
rope?
W=500 J
Suppose a rope lifts a weight of 1000N by 0.5m at a constant
upward acceleration of 2m/s2. How much work is done by
the rope?
W=602 J
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PHY 113 -- Lecture 7
11
Why is work a useful concept?
Consider Newton’s second law:
 Ftotal · dr= m a · dr
Ftotal = m a
rf
rf
rf
rf
rf
dv
dv dr
dv
 Ftotal  dr   ma  dr   m  dr   m  dt   m  vdt
dt
dt dt
dt
ri
ri
ri
ri
ri
Wtotal = ½ m vf2 - ½ m vi2
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Kinetic energy (joules)
PHY 113 -- Lecture 7
12
Introduction of the notion of Kinetic energy
Some more details:
Consider Newton’s second law:
Ftotal = m a
rf
 Ftotal · dr= m a · dr
rf
rf
tf
tf
dv
dv dr
dv
F

d
r

m
a

d
r

m

d
r

m

dt

m
 vdt
 total




dt
dt dt
dt
ri
ri
ri
ti
ti
tf
vf
f
dv
1
1 2 1 2


 m  vdt   mdv  v   d  mv  v   mv f - mvi
dt
2
 2
ti
vi
i 2
Wtotal = ½ m vf2 - ½ m vi2
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Kinetic energy (joules)
PHY 113 -- Lecture 7
13
Kinetic energy: K = ½ m v2
units: (kg) (m/s)2 = (kg m/s2) m
N
m
= joules
Work – kinetic energy relation:
Wtotal = Kf – Ki
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PHY 113 -- Lecture 7
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PHY 113 -- Lecture 7
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Examples of the work—kinetic energy relation:
Suppose Ftotal = constant= F0
Wtotal = Kf – Ki
 F0(xf-xi) = ½ mvf2 - ½mvi2
In this case, we also know that F0 = ma0 so that,
F0(xf-xi) = ma0 (xf-xi) = ½ mvf2 - ½mvi2
vf2 =vi2 +2a0 (xf-xi)
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PHY 113 -- Lecture 7
16
More examples of work—kinetic energy relation without friction:
gsin
Assume block of mass m is sliding down a
g
h
frictionless surface at an angle of 
L

kinematic analysis: vf2 =vi2 +2a0 (xf-xi) = 0 +2(g sin)L = 2gh
energy analysis: Wtotal = mgh = ½ mvf2
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PHY 113 -- Lecture 7
17
More examples of work—kinetic energy relation with friction:
gsin
g
Assume block of mass m is sliding down a
fk
surface with a kinetic friction coefficient of
mk at an angle of 
h
L

kinematic analysis: vf2 =vi2 +2a0 (xf-xi) = 2(g sin-mkg cos)L
energy analysis: Wtotal = mgh -mkmg cosL = ½ mvf2
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PHY 113 -- Lecture 7
18
-Ftrack-mg=-mvt2/R
If Ftrack=0, then mg(h-2R)= ½ mvt2
h=5/2R
h
v
2R
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PHY 113 -- Lecture 7
19
Suppose that the
coefficient of
friction between the
skis and the snow is
mk=0.2. What is the
stopping distance
d?
L
h
mgh- mkmgcosL- mkmgd=0
d=h/mk -Lcos
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PHY 113 -- Lecture 7
20

L
1
2
A ball attached to a rope is initially at an angle . After being
released from rest, what is its velocity at the lowest point 2?
2 gL(1 - cos )
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PHY 113 -- Lecture 7
21
Hooke’s “law” (model force)
F = -kx
xf
Wi  f   - kx dx  - 12k x 2f - xi2 
xi
xi
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PHY 113 -- Lecture 7
xf
22
Example problem:
A mass m with initial velocity v compresses a spring
with Hooke’s law constant k by a distance xf. What is xf
when the mass momentarily comes to rest?
K i  12 mv 2
Kf  0
Wi  f  - 12 kx f
Wi  f  K f - K i
xf 
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2
 12 kx f  12 mv 2
2
m
v
k
PHY 113 -- Lecture 7
23