PHY2053: Lecture 13
Elastic Forces - Hooke’s Law
Work done by elastic forces
Elastic Potential Energy
Power
Elastic Forces
PHY2053, Lecture 13, Elastic Forces, Work and Potential Energy; Power
2
Objectives
1) model the force produced by a
compressed / extended spring
2) how does one model a system
with multiple springs?
3) is the elastic force conservative?
4) if so, define potential energy of the spring
- include potential energy of the spring into energy
conservation calculations
5) define concept of power
PHY2053, Lecture 13, Elastic Forces, Work and Potential Energy; Power
3
Hooke’s Law
• Discovered by Robert Hooke
• c/a 1676 “Ut tensio, sic vis”
“As the extension, so the force”
of
many
• Excellent
W = Kdescription
K
=
K
f
i
elastic systems
Fspring
Fx =
k
x
• Force is proportional to, and in
∆x
the opposite direction of, the
spring extension (compression)
PHY2053, Lecture 13, Elastic Forces, Work and Potential Energy; Power
mg
4
Example #1: Real Spring
100
2.10
2.01
200
2.90
3.02
k = 96.5 ± 3.9 N/m
FG [N]
mass [g]
fitted ∆x
∆x [cm]
value [cm]
x0 = 1.0 ± 0.2 cm
5
4.5
4
3.5
300
4.20
4.03
3
2.5
400
5.00
5.04
2
1.5
500
6.10
6.05
K=
97
N/m
x0 =
1
cm
1
0.02
0.03
PHY2053, Lecture 13, Elastic Forces, Work and Potential Energy; Power
0.04
0.05
0.06
Δx [cm]
5
Work done by an
⇥ 2,1
F
F
∆r
⇥ 1,2
F
Recall the definition of work:
W =F
W = Kf
r cos( )
Ki = K
X
Force opposes deformation → work is always
⇥ i = 0negative
F
F
∆x
Fx =
x
x
ik changes
Important:
Force
with
X(variable force)
extension
W
Fi r cos( 1i ) = 0
2
W0!x = i Fi xi =
kx
F
2
i X
[Convention: x = 0 for relaxed spring]
r
Fi cos(1 i ) = 20
PHY2053, Lecture 13, Elastic Forces, Work and Potential Energy; Power
U
= W
= k x6
•
i
i
Elastic potential energy
• Elastic forces are conservative forces
• It makes sense to define “elastic” potential energy
• Same convention as for gravity:
• In context of energy conservation:
PHY2053, Lecture 13, Elastic Forces, Work and Potential Energy; Power
7
Newt
0
1000
500
0
•
•
•
•
s
the
ative
ities
d on
ng
ate
ly
re 1.
Example #3: Ballista
Anterior Tibialis
Posterior Tibialis
Peroneus Longus
measure
of material
properties
is gainedto
by store elastic potential
AA better
ballista
uses
torsion
springs
normalizing the overall tendon strength by crossenergy,
reportedly
had
range
>2500 m
sectional area.
Thus, the results
shown
in Figure
were divided
cross-sectional
and the
1/2
talentbycaliber
[1 area
talent
~ results
32 kg]
presented as MegaPascals (MPa) or N/mm2. The
What
wasin Figure
the 3spring
constant
results shown
indicate the
similarity k if ballista had a 500m
between posterior and anterior tibialis tendons as
range?
[∆x
~
2.0
m]
well as the higher ultimate stress value for the
peroneus was
longus. the
Note tension
that both thein
posterior
What
the retracting
tibialis and peroneus longus exhibit strength that is
cable
atcomparable
full extension?
[made
of tendons]
sufficiently
to anterior tibialis
tendons.
Figure 3. Tendon strength in Newtons per cross-sectional area
e
rry
of
150
Mpa
100
ACL,
50
0
Anterior Tibialis
Posterior Tibialis
Peroneus Longus
PHY2053,
Lecture
Elastic
Forces,by
Work
and Potential
A similar
study13,
was
performed
Pearsall,
et al. 17 in Energy; Power
8
Fx =
k
x
Power
1
Welastic,0!x = k
1 2 2
F x =
kx
Uelastic,0!x =
W0!x =
(17
x
2
(18
• Recall the unit “furlong” 1 2 1 1 2
- distance
UUelastic,x1!x2
= k x2 = k kxx1
• Related to farming
= W
2
22
i
i
i
elastic,0!x
2
elastic,0!x
2
(19
that a team of oxen could plough
1 2 1 2
without resting
Uelastic,x1!x2 = k x2
k x1
(20
2
2
4 oxen would plough a furlong
Ui,grav
Ui,elastic
[ + WN
Ef = Kf + Uf,grav + Uf
C ]a=
in half+the
time needed
for
pair
how+ quickly
getting
done?
Ei =Power:
Ki + Ui,grav
Ui,elastic [is+work
WN C ] =
Ef = K
f + Uf,grav + Uf,elastic (21
6
Unit: 1 Watt = 1 Joule / second [ 1 kWh = 3.6×10 J ]
•
•
•
=
E
W
F r cos
r
=E
=
cos = F vav co
W
F r cos = F r
Ptav =
=
=
=F
cos
= F vav cos
(22
t
t
t
t
t
t
t
P = F v cos
PHY2053, Lecture 13, Elastic Forces, Work and Potential Energy; Power
9
Example #4: TGV Speed Record
The TGV high speed train
holds a speed record of 550
km/h. The traction system can
generate 19.6 MW of power.
The limiting factor for the
train’s top speed is air drag.
Compute the force due to air drag at top speed.
Compare that to the weight of a killer whale (6 tons)
If the air drag force were maintained, what distance
would be needed for the train to stop? The mass of the
train is 380 tons. [in reality it took 70 km to stop]
PHY2053, Lecture 13, Elastic Forces, Work and Potential Energy; Power
10
Summary
- Elastic forces (springs) are modeled by Hooke’s Law:
- Combining multiple springs [from Hooke’s Law]:
- Elastic forces are conservative, potential energy
- Power - rate of work done over time:
PHY2053, Lecture 13, Elastic Forces, Work and Potential Energy; Power
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