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
9/18/2003
PHY 113 -- Lecture 7
1
9/18/2003
PHY 113 -- Lecture 7
2
Surface friction force models
¾Static friction
f = −Fapplied if | f | < µs N
¾Kinetic friction
| f | = µs N
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PHY 113 -- Lecture 7
3
N
f
mg
F sin θ
N−mg−F sin θ = 0
f = µk N
F cos θ − f = ma
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PHY 113 -- Lecture 7
4
without friction : mg sin θ = ma
D = 12 at 2
with friction : mg sin θ − µ k mg cos θ = ma f
t f = 2t
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⇒ 4a f = a
D = 12 a f t 2f
⇒ 4( g sin θ − µ k g cos θ ) = g sin θ
PHY 113 -- Lecture 7
5
Energy Î work, kinetic energy
Force Î effects acceleration
A related quantity is Work
rf
Wi → f = ∫ F ⋅ dr
ri
A·B = AB cosθ
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
rf
ri
W=−mg(rf−ri)<0
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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
9/18/2003
PHY 113 -- Lecture 7
11
Why is work a useful concept?
Consider Newton’s second law:
Ftotal = m a Î Ftotal · dr= m a · dr
rf
rf
rf
rf
rf
dv
d v 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 Î Ftotal · dr= m a · dr
rf
rf
rf
tf
tf
dv
d v 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
2
dt
⎠ 2
ti
i ⎝2
vi
ÎWtotal = 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
14
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PHY 113 -- Lecture 7
15
Examples of the work—kinetic energy relation:
Suppose Ftotal = constant= F0
Wtotal = Kf – Ki
Î 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
Îvf2 =vi2 +2a0 (xf-xi)
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PHY 113 -- Lecture 7
16
More examples of work—kinetic energy relation without friction:
gsinθ
Assume block of mass m is sliding down a
g
h
frictionless surface at an angle of θ
L
θ
kinematic analysis: vf2 =vi2 +2a0 (xf-xi) = 0 +2(g sinθ)L = 2gh
energy analysis: Wtotal = mgh = ½ mvf2
9/18/2003
PHY 113 -- Lecture 7
17
More examples of work—kinetic energy relation with friction:
gsinθ
g
Assume block of mass m is sliding down a
fk
surface with a kinetic friction coefficient of
µk at an angle of θ
h
L
θ
kinematic analysis: vf2 =vi2 +2a0 (xf-xi) = 2(g sinθ−µkg cosθ)L
energy analysis: Wtotal = mgh −µkmg cosθL = ½ mvf2
9/18/2003
PHY 113 -- Lecture 7
18
-Ftrack-mg=-mvt2/R
If Ftrack=0, then mg(h-2R)= ½ mvt2
Îh=5/2R
h
v
2R
9/18/2003
PHY 113 -- Lecture 7
19
Suppose that the
coefficient of
friction between the
skis and the snow is
µk=0.2. What is the
stopping distance
d?
L
h
mgh- µkmgcosθL- µkmgd=0
d=h/µk -Lcosθ
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PHY 113 -- Lecture 7
20
θ
L
1
2
A ball attached to a rope is initially at an angle θ. After being
released from rest, what is its velocity at the lowest point 2?
2 gL(1 − cosθ )
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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