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ENERGY, WORK, AND
POWER
Ozkaya and Nordin
p. 255-271
p. 295-311
• principle of conservation of energy
• potential energy
• kinetic energy
• energy generation and concentric contraction of
muscle
• energy absorption and eccentric contraction of
muscle
• joint power
What is energy?
Conservation of energy
• Energy is an abstract thing that reflects “capacity to
do work.” We have formulas for calculating energy,
but little understanding of the mechanisms or
reasons for the various formulas.
• There are several kinds of energy: gravitational
potential energy, elastic potential energy, kinetic
energy, heat energy, electrical energy, chemical
energy, radiant energy, nuclear energy, and mass
energy.
• The units for energy are Joules.
•
(1 [J] = 1 [N][m] = 1 [kg][m2][s-2])
• Energy is a scalar quantity.
The principle of conservation of energy states
that energy can not be created nor destroyed.
ETOTAL
! EKINETIC + $
#
&
= # EPOTENTIAL +& = cons tant
#
&
" EHEAT + ... %
In musculoskeletal biomechanics, we are
concerned primarily with kinetic energy and
potential energy.
We will use the symbols T (or KE ) to designate
kinetic energy, and U (or PE) to designate
potential energy.
Potential energy (PE)
Example #1: Conservation of energy
Gravitational potential energy: the energy that an
object has because of its position in space,
relative to the earth:
A block weighing 100 N rests on a frictionless inclined
plane that is a 3-4-5 triangle. A rope attached to the block
runs over a pulley, and attaches to another block W. How
heavy must W be to balance the 100 N weight on the
plane?
gravitational PE = weight ! height = mgh
no friction
W
This is an example of the general principle that a
change in energy is equal to force times the
distance the force acts through.
3
5
4
10
0N
We could solve this problem using !F = 0. We will
show that it can also be solved using conservation of
energy.
Example #1 (cont)
Example #2: Conservation of energy
The balanced system can move up and
down and maintain static equilibrium.
(T1 + U1 ) = (T2 + U2 )
T1 = T2 = 0
W
U1 = ! mi ghi
3
Weights of 60 lb and 100 lb are placed on a massless bar as
shown below. What weight W at the end of the bar is
required to keep the bar balanced, disregarding the weight
of the bar?
Y
i
4
= W " 3 + 100 " 0 = 3W
U2 = ! mi ghi
10
0N
X
10
0N
i
= W " ( #2) + 100 " 3 = 300 - 2W
3W = 300 - 2W
300
W=
= 60 N
5
5
5
3
Y
W
4
X
We could solve this problem using !M = 0. We will
show that it can also be solved using conservation of
energy.
Example #2 (cont)
Kinetic energy
Imagine the weight drops an
arbitrary distance of 4
inches. The 100 lb block
will then rise 1 inch, and the
60 lb block will rise 2
inches. If the rod is in a state
of balance, the
corresponding change in
gravitational potential
energy must be zero.
Kinetic energy (KE) is the energy associated with
movement of a mass. There are two types of kinetic
energy: translational and rotational.
translational
1
kinetic energy = mv 2
2
of a particle
m
v
Energy transfer during walking
energy (J)
"
(B)
(A)
Example #3: Conservation of energy
total
KEFORWARD
A high jumper achieves a vertical take-off velocity of
2.53 m/s. How high will her whole-body centre of
gravity rise above the ground?
PE
(A) midstance: vertical
height of COG (and PE) is
path of
maximal, horizontal
COG
velocity (and KE) is
minimal
(B) heel strike: vertical
height of COG (and PE) is
minimal, horizontal
velocity (and KE) is
maximal
time (s)
v
v
Last lecture, we solved
this problem using the
constant acceleration
equations. We will
now show that it can
be solved more simply
using conservation of
energy.
Y
h
g
Example #4: Conservation of energy
Example #3 (cont)
1
Y
g
y1 = h = ?
v =0 m
y1
h
0
y0 = 0
vy0 = 2.53 m s
(T + U ) 0 = ( T + U )1
1
mv y0 2 + 0 = 0 + mgh
2
v y0 2 2.532
h=
=
= 0.33 m
2g 2(9.81)
Example #4 (cont)
1
T1 = mv1 2 = 0
2
U1 = mgh1
h1 = 2 ! 5 9 (1.50 ) = 1.67 m
)
U2 = mgh2 = 0
2
Recall that when a spring of stiffness k [N/m] is stretched a
distance x from its equilibrium (i.e., zero-force or resting)
length, the force it develops is:
h1
T2 = T1 + U1 " U2 = 850.2 J
1
2T2
T2 = mv 2 2 # v2 =
2
m
2 ! 850.2
v2 =
= 5.71 m s
52
v2
A second form of potential energy (in addition to
that due to gravity) is elastic potential energy.
1
U1 = 52 ! 9.81!1.67 = 850.2 J
1
Elastic potential energy: the work
done in deforming a spring
(T1 + U1 ) = (T2 + U2 )
(
s
A gymnast (of height 150 cm
and body mass 52 kg) on the
high bar is modeled as a
simple pendulum. She
descends from position 1,
where her COG is directly
above the bar and her velocity
is negligible, and releases the
bar at position 2, where her
COG is directly below the bar.
What is her linear velocity v2
at release?
y
v2
2
F = !kx
The corresponding elastic potential energy stored in the
spring is:
2
2
1
1 ( F2 # F1 )
2
2
"1F ! ds = "1#kx dx = # 2 k( x2 # x1 ) = # 2 k
2
2
Example: Bouncing ball
A : U gravitational = mgh = Etotal
A
1
2
B : T = mv B = E total
2
m
k
C : U elastic
h
1
2
D : T = mv D = E total
2
!
!
!
1
1 Fmax 2
2
= kx max =
= E total
2
2 k
B
m
vB
D
C
m
m
xmax
k
k
vD
Fmax
(a) Derive an equation for vB in terms of g and h.
Application: Predicting impact
forces during falls.
m
x
k
(b) Derive an equation for Fmax in terms of m, k, and vB.
A
h
B C D
Applied Load: Conservation of energy approach
The dynamics of impact during
a fall on the hip are
adequately represented by
a mass spring model.
“Pelvis release experiments”
provide safe measures of
the body’s effective mass
(m) and stiffness (k)
during a simulated fall on
the hip.
m = ~30 kg (or approx.
one-half body mass)
k = ~30000 N/m
conservation of energy to predict (a) the impact
velocity, and (b) the peak force.
m
m
k
x
k
m
k vimp
force plate
Fall
Initiation
Initial
Contact
1
2
m
k
Fmax
Peak
Force
3
Rotational kinetic energy (cont)
Rotational kinetic energy
Now vi = ri!˙. Therefore, the total
vi=ri#
mi
#
ri
kinetic energy can be written as
A rigid body can be
considered as a system
of i point masses, or
particles. If this system
rotates about a fixed axis
0, its total kinetic energy is
1
T = ! mi vi 2 .
i 2
Example. For a fall from a height of 1 m, use
vi=ri#
mi
#
ri
0
1
1
T = " mi vi 2 = " ( mi ri 2 )# 2
2 i
i 2
Define the moment of inertia ( I )
with respect to 0 as
I0 = " mi ri 2
0
i
Therefore, the total kinetic
energy of the rotating body is
1
T = I 0# 2
2
rotational
1
kinetic energy = I 0! 2
of a rigid body 2
Example #5: Conservation of
energy.
Example #5 (cont)
We previously determined from
1
1
(T1 + U1 ) = (T2 + U2 ) that
Consider the same problem
as in example #4. What is
the gymnast’s angular
velocity (#2) when she
releases the bar at instant 2?
T2 = 850.2 Joules. Now
1
1
T2 = mv 2 2 = I 0! 2 2
2
2
and
#2=?
I0 = mr 2 = 52 " (0.835)
= 36.3 kg " m
v2 = 5.72 m/s
2
#2
2
v2
2T2
2 " 850.2
#! 2 =
=
I0
36.3
2
2
! 2 = 6.84 rad/s
Kinetic energy of a rigid body
(general)
The total kinetic energy of a rigid
body that has rotational
r
velocity ! , and linear
r
velocity vCG of its centre of
gravity, is :
1
1
T = mv CG 2 + ICG! 2
2
2
where m is the mass of the body,
and ICG is the moment of inertia
about the centre of gravity.
vCG
Work is the integral of force times
distance
#
If an object moves along a curve under the influence of a
force, the work done by the force F on the object from point 1
to point 2 is the integral of the component of force along the
curve times the differential displacement ds:
2
W1!2 = # F " ds
1
CG
=#
x2
x1
y
$
y1
$1
Fx dx + # 2 Fy dy + # 2 M d$
(2D)
Only the component of force (or moment) in the direction of
motion (equal or opposite) contributes to the work done.
The units for work are Nm or Joules, the same as energy.
The work-energy theorem states that the change between
points 1 and 2 in total mechanical energy of a system
equals the work done on the system between these points:
!E = W1"2 = (T + U )2 " (T + U )1
The work produced by a
force F is equal to the
integral of force times
displacement in the
direction (equal or
opposite) of the force
2
W1!2 = # F " ds
This is the general statement
of conservation of energy.
1
=#
x2
x1
y2
Fx dx
X-direction force FX (N)
Work performed by a force
The work-energy theorem
work = area under curve
2
= ! FX dx
1
x1
x2
X-direction displacement (m)
+ # Fy dy (2D)
y1
The work produced by a
moment M can be
determined by inserting
ds=rd" into the work
equation:
2
W1!2 = # F " ds
1
2
= # Frd$
1
2
= # Md$
1
Joint Moment M (Nm)
Work performed by a moment
work = area under curve
2
= " Md!
1
"1
"2
Joint rotation " (radians)
Example #6: work during a knee
extension exercise
The figure at left shows the
measured variation in knee
extensor moment versus
knee extension angle
during a single repetition
of exercise. Calculate the
work performed between "
= 1.57 rad and " = 3.14
rad, if the moment
averages 50 Nm over this
interval.
Eccentric and concentric contractions
Eccentric/concentric contraction (cont)
Positive and negative joint work
Joint
moment
(Nm)
CONCENTRIC
OR ECCENTRIC?
negative
work (eccentric)
A to B: eccentric
A'''
A' to B': concentric
A'' to B'': concentric
A''''
B'''
A''' to B''': concentric
flexion
flexor
extension
negative
work (eccentric)
positive
work
(concentric)
A'''' to B'''': eccentric
Joint
rotation
(radians)
L to M: concentric
L' to M': eccentric
B''''
A''
B
A'
A
B'
flexion
extension
L'
M''
L''
L'' to M'': eccentric
L''' to M''': concentric
B''
extensor
positive
work (concentric)
Positive and negative joint work (cont)
flexor
Joint
moment
(Nm)
!
extensor
The direction of joint
work depends on the
polarity of joint torque
and the direction of
joint rotation
! for example, an
extensor torque
produces positive work
if the joint is moving
into increasing
extension, and negative
work if the joint is
moving into increasing
flexion.
eccentric
contraction
of m2
concentric
contraction
of m2
Concentric contractions
tend to increase joint
angular velocity, and
increase the total
energy of the system
Eccentric contractions
tend to decrease (or
brake) joint angular
velocity, and reduce the
total energy of the
system
M'''
L'''
M'
Joint
rotation
(radians)
L
M
Power equals work done per
second
dW
dt
Units of power are Watts [N][m][s-1] or Horsepower
Power =
What are the definitions (i.e., formulas) for the
various types of kinetic and potential energy?
! What is the work-energy theorem?
! At what point in the gait cycle is kinetic energy
maximum? At what point is potential energy
maximum?
! What magnitude of work is generated by a vertical
force applied to a horizontally-moving mass?
! Identify a sample problem that could be solved with
the constant acceleration equations, or with
conservation of energy principles.
!
!W
Average power =
!t
Instantaneous power =
Review Questions
dW F ! ds
=
= F!v
dt
dt
Md"
=
= M#
dt