Transcript 356 Linear Kinetics

Force
• force - a push or pull acting on a body
• forces are vector quantities
– magnitude (size)
– direction (line of action)
– point of application
BOX
• measured in Newtons (N)
– 1 N = (1 kg) (1 m/s2)
pt of application
F
• English units (lb)
– 1 lb = (1 slug) (1 ft/s2)
line of action
Free-Body Diagrams
A free body diagram illustrates all of the
external forces acting on an object.
If whole body is considered to be “the system”
Examples of external forces
Weight (gravity)
Ground Reaction Force (GRF)
Friction
Fluid Resistance
Examples of internal forces
Joint Reaction Force
Muscle Force
air resistance
mg
FR (GRF)
Contact Forces
NEWTON'S LAWS OF
MOTION
Philosophiae Naturalis Principia
Mathematica (1686)
Newton’s 1st Law of Motion
• Law of Inertia
– a body at rest will remain at rest and a body in
motion will remain in motion and move at a
constant velocity until a non-zero resultant
external force is applied to it.
Inertia is the resistance of an object to motion - the amount of
resistance to linear motion varies directly with the mass of the object.
When an object is in motion its resistance to change in motion is
determined by its velocity as well as its mass. Momentum is the
mass of an object multiplied by the velocity (p = mv).
Newton’s 1st Law of Motion
V
W
W
Fr
Fp
Ry
if Ry = W
then resultant force = 0
if v = 0 and SF = 0
STATIC EQUILIBRIUM
Ry
Fr = resistive force
Fp = propulsive force
if v = 0 and SF = 0
DYNAMIC EQUILIBRIUM
Newton’s 2nd Law of Motion
• Law of acceleration
– when a non-zero resultant external force is
applied to a body, the body will accelerate in
the direction of this force. The acceleration is
proportional to the force and inversely
proportional to the body’s mass
F2
SF
m
m
F1
a  SF
a
1
m
a  k1
SF
m
where k1 = constant of proportionality
SF  k ma
`
where k2 
1
k1
a
SF  k2 ma
If m is measured in kg
and a is measured in m/s2
the SI unit for force is “newton” (N)
1 N = 1 kg x 1 m/s2
where k2=1
SF = ma
Alternate Expression of
LAW OF ACCELERATION
"The rate of change of momentum of a body is
proportional to the applied force and takes place in
the direction in which the force acts."
mv f - mv i
• SF =
t
• SFt = mDv (impulse/momentum relation)
Dv
• SF = m
= ma
t
If an elephant and a feather fall
from the same height in the
absence of air resistance then the
resultant or net force acting on
each object is simply their
weight. Since W = mg then the
acceleration they experience is
SF = ma but SF = W = mg
mg = ma
or
a=g
Now consider when air resistance is
present. This force would larger on the
elephant simply because is a bigger object.
But the weight of the elephant is also
significantly larger than that of the feather.
In fact, relative to weight the air resistance
acting on the feather is larger than on the
elephant. This affects the resultant force
acting on each object such that the resultant
force acting on the feather is much closer to
0 N. Thus the feather will have a much
lower acceleration.
This example further demonstrates the change in resultant force due to air
resistance. Notice that initially air resistance due to the body falling
through the air reduces the magnitude of the acceleration but it remains a
downward acceleration. Eventually you reach a point where the air
resistance equals your body weight. This is known as terminal speed and
would be well over 100 mph for a human body.
To allow you to land without hurting yourself you deploy your parachute.
This greatly changes the resultant force such that the net force actually
points upward such that the acceleration is upward. This successfully
decreases your speed to a more manageable level for landing.
Newton’s 3rd Law
• “Law of action and reaction”
• when one body exerts a force on another
body, the second exerts an equal but
opposite force back on the 1st body
• “for every action there is an equal and
opposite reaction”
Rresistance
Aresistance
Rpropulsion
Apropulsion
Note: these forces (action and reaction) are never applied
on the same body -- it takes two bodies for a pair of
action/reaction forces to exist
LAW OF GRAVITATION
"All bodies attract one another with a force
proportional to the product of their masses and
inversely proportional to the square of the distance
between them."
m1* m2
• FG = G
d2
, G = 6.67X10-11 N-m2/kg2
Mass and Weight
• g depends on the distance from the earth’s
center
r
M
r
M
Earth is not a true sphere
but rather an ellipsoid:
it is fatter at the equator
m = your mass
M = earth’s mass
r = radius (distance to
(center of earth)
G = gravitational constant
Newton’s Law of Gravitation
GMm
1
Fg  2
 Fg  2 and Fg  m
r
r
Fg = gravitational force
If m = your mass then Fg = your weight (W)
r
M
Fg 
g
GMm
r2
GM
r2
W  mg
Latitude effect = g is smaller at equator
larger at poles
Altitude effect = g is smaller at high altitude
larger at low altitude
Weight
• W = mg
– weight is a force
– weight = mass x acceleration due to gravity
• Units
– N = kg x m/s2
– weight in Newtons = mass in kg x 9.81 m/s2
– a 1 kg mass weighs 9.81 N
The Relevance of Newton’s Laws
These 4 laws allow all motion in the
universe to be described and predicted
as long as the relative speeds of the
objects are small compared to the
speed of light.
Ground Reaction Forces
FV
FML
FAP
Force Platform
Contact Forces
Impulse - Momentum
Relationship
F
 ma 
F
m
F
 ma
v f  vi
t
F
 t  m(v f  vi )
F
 t  mv f  mvi
J
 p f  pi
J
 Dp
Contact Forces
Impulse-Momentum
J Dp
“Impulse = change in momentum”
Units
F t  N s
(kg m/ s2)s
 kg m/ s
mv kgm/ s
Same units!
Contact Forces
Impulse/Momentum
(SFt = mDv)
• The impulse/momentum relationship describes the
effects of a force over a period of time.
• An impulse (SF*t) causes a mass to change its
velocity
• A 10 N force is applied to a 2 kg mass for 3
seconds. What is the change in velocity?
Contact Forces
Impulse/Momentum
• How long would it take to stop a 7 kg mass that has a
velocity of 10 m/s if you are capable of producing a
maximum instantaneous force of 35 N?
Contact Forces
Impulse/Momentum
• It has been determined that a force over 200 N can
injure the hand. What is the shortest period of
time it will take to stop a 2 kg object traveling at
75 m/s if the hand is to be protected?
Contact Forces
Impulse/Momentum
• The braking impulse of a subject running
across a force platform is -10 N-s. The
propelling impulse during the same time
period is 2 N-s. What is the change in
velocity of the subject if she has a mass of
55 kg?
F
y
0
0
time
.8
Contact Forces
F
o
r
c
e
(N)
5
4
3
2
impulse = area
1
0
0
1
2
3
4
5
Time (s)
• If a 10 kg object is exposed to the above impulse
what will be the change in velocity?
Contact Forces
F
Impulse = area under F v. t curve
Jp=positive impulse
t
Jn=negative impulse
JNET = NET IMPULSE
JNET = Jp + Jn
Contact Forces
Recall that the net force is
SF = GRF - W
and by Newton’s 2nd Law
SF = ma
so GRF - W = ma
W
If GRF > W then a >0
If GRF < W then a < 0
GRF
When the jumper initiates the countermovement
he/she speeds up in the negative direction
this means that the body experiences a
negative acceleration
thus GRF < W
W
GRF
body weight
GRF
0
As the jumper nears the bottom of the
countermovement he/she slows down in the
negative direction
this means that the body experiences a
positive acceleration
thus GRF > W
W
GRF
lowest point
body weight
GRF
0
When the jumper begins the upward portion of
the countermovement he/she speeds up in the
positive direction
this means that the body experiences a
positive acceleration
W
thus GRF > W
GRF
lowest point
body weight
GRF
0
takeoff
Total Net Impulse =
Negative Net Impulse +
Positive Net Impulse
SJ = mDv = m(vf - vi)
this can be solved to find the
takeoff velocity (final velocity
of countermovement phase)
vtakeoff = SJ/m since vi = 0
Knowing vtakeoff allows you to compute jump height
Knowing vtakeoff allows you to
compute jump height
vtop = 0 m/s
vf2 = vi2 + 2ad
where
vf = vtop
vi = vtakeoff
vtakeoff
SO - jump height is increased by
increasing the total net impulse
… not just the net force
jump height
Conservation
of Momentum
Momentum represents the total quantity of motion
possessed by a body or system. The momentum of a
system cannot be altered without an external force.
Momentum = mass x velocity
Total momentumbefore = total momentumafter
for example - if there are 2 objects in the system
m1v1 + m2v2 = m1u1 + m2u2
v = velocity before
u = velocity after
subscripts represent object number
Example:
A 100 kg astronaut is moving at a speed of 9 m/s
and runs into a stationary astronaut (mass = 150
kg).
Problem:
What is the velocity of the astronauts after
the collision?
System = both astronauts
External Forces = none
Therefore the momentum must be conserved.
Total momentum (p):
pbefore =
pafter =
v=?
A 75 kg rugby player is moving at 2 m/s when he
runs into a 100 kg player running at –1.5 m/s. Which
direction will the resulting collision travel (-, 0, +)?
If they collide in mid-air (0).
If they collide on the ground the players will be able
to exert external forces and the larger player will
probably be able to exert larger external forces (+).
pbig fish-before + plittle fish-before = pbig fish-after + plittle fish-after
pbig fish-before + plittle fish-before = pbig fish-after + plittle fish-after
Joint Reaction Force
The net force acting across a joint
e.g. when standing, the thigh exerts a downward force
on the shank, conversely, the shank exerts an
upward force on the thigh of equal magnitude
NOTE: this JRF does not include
the forces exerted on the joint
by the muscle crossing the
joint. The force that includes
all of the forces crossing the joint
is known as the bone-on-bone
force
Contact Forces
Block
Pulling Force
Friction Force
Normal Force
(weight of the block)
• There is an interaction between the surface
of the block and the table.
• The friction force always opposes motion.
• The pulling force must be greater than the
frictional force to move the block.
Contact Forces
Normal Force = force perpendicular to surface
When surface is horizontal the
perpendicular direction is vertical so
the normal force is simply the weight
of the object
Weight
Normal Force
FN
q
When surface is inclined the perpendicular
direction is NOT vertical so the normal
force is only a component of the object’s
weight.
Friction Force
maximal static friction force
Static
impending motion
Dynamic
Applied Force
• At some point the pulling force will be great enough so
that the friction force cannot prevent movement.
Contact Forces
Friction
• The coefficient of static friction is expressed as:
Fmax
ms 
Fnormal
where ms= coefficient of static friction
Fnormal = normal force
Fmax = maximal static friction force
•The coefficient of friction is a dimensionless number. It is
unaffected by the mass of the object or the contact area.
•The greater the magnitude of ms the greater the force
necessary to move the object.
Friction
• As the block moves along the table, there still
is a frictional force that resists motion.
• Sliding and rolling friction are types of
dynamic friction.
md 
F friction
FNormal
where md= coefficient of dynamic friction
N = normal force
Ffriction = force resisting motion
Contact Forces
Friction
•
It has been found experimentally that md < ms.
•
md depends of the relative speed of the
surfaces.
• At speeds from 1 cm/s to several m/s, md is
approximately constant.
Contact Forces
Why is it easier to pull a desk than push it?
When you push you usually have a downward
component of force -- so the normal
force is increased and therefore the
frictional force is increased.
P
h
R = W+v
v
W
W
Contact Forces
When you pull you usually have an upward
component of force -- so the normal
force is reduced and therefore the
frictional force is reduced.
P
v
R = W-v
h
W
W
Contact Forces
Fluid Resistance
• The transmission of energy from an object
passing through a fluid to the fluid is known
as fluid resistance.
• The resistance of an object passing through
a fluid increases as the speed of the object
increases and as the viscosity of the fluid
increases.
Contact Forces
Surface and Form Drag
• Surface drag is a result of the friction
between the surface and the fluid.
• The fluid closest to the object (boundary
layer) rubs against the object creating
friction.
• Kyle (1989) reported that wearing loose
clothing can increase surface drag from 2%
to 8%.
Contact Forces
Surface Drag
Van Ingen Schenau
(1982) reported a
10% reduction in
surface drag when a
speed skater wears a
smooth body suit.
Contact Forces
Form Drag
Form drag occurs when air is driven past an
object and is diverted outward creating a low
pressure region behind the object.
low pressure
high pressure
Contact Forces
Form Drag
Low form drag
The orientation of the
object will affect the
frontal area and will
play an important role in
the amount of form
drag.
High form drag
Contact Forces
.5m2 (upright)
.42m2 (touring)
.34m2 (racing)
The second cyclist can ride within the low
pressure zone of the first cyclist and thus
lower the pressure difference and the drag.
This is called drafting.
frontal area
Contact Forces
Flow Type
laminar
At low velocities laminar
flow occurs. The
separated
boundary layer remains
attached to the surface.
During separated
flow the boundary layer
fully turbulent
separates toward the
back of the object and a
low pressure region
is formed. During fully turbulent flow the
boundary layer becomes turbulent and the
size of the pocket is decreased.
Contact Forces
Factors Affecting Flow Type
•
•
•
•
•
size
shape
surface roughness
viscosity of the fluid
flow velocity
Contact Forces
Airfoil
The particles following the path from D1 to D2
will be more spread out than particles following
the path from C1 to C2 because of the greater
distance. This creates a low pressure region
above the airfoil.
Bernoulli’s Principle, 1738
Contact Forces
Lift
Fair resistance
Flift
Flift = 1/2(ClArv2)
Fdrag = 1/2(CdArv2)
Fdrag
direction of
movement
Lift always acts perpendicular to drag.
Contact Forces
• The lift-to-drag ratio is critical (i.e. the larger the
ratio, the more effective the airfoil is in flight).
• L/D ratio is dependent on the angle that the airfoil
makes with the incoming air (this is called the
ANGLE OF ATTACK).
• Increasing the angle of attack increases the L/D
ratio to a point; beyond that point the angle
becomes too steep and the airfoil stalls
• typical angles of attack:
airfoil - below 15o
javelin - 10o
Contact Forces
Angle of
Attack
(degrees)
0
10
20
25
27
28
29
30
35
40
45
50
60
70
80
90
Lift
(N)
0.00
4.33
10.64
12.83
13.80
13.80
11.01
11.21
10.12
8.50
8.90
8.62
6.88
4.77
2.55
0.00
Drag
(N)
1.17
1.50
4.13
5.79
6.88
7.41
7.94
8.18
8.74
9.55
11.13
12.22
14.98
16.43
16.88
17.73
Lift
Drag
0.00
2.89
2.58
2.22
2.01
1.86
1.39
1.37
1.16
0.89
0.80
0.71
0.46
0.29
0.15
0.00
Lift to drag ratios
for the discus.
Adapted from Aerodynamic
Factors Which Influence
Discus Flight, Ganslen.
Contact Forces
Rotating Objects
Direction of air flow
low pressure
zone
Rotating objects can also
create a pressure difference.
high pressure
zone
Magnus Effect,Contact
1852
Forces
intended direction
of flight
actual direction
of flight
low pressure
zone
high pressure
zone
Contact Forces
The golf club imparts backspin on the golf ball and
increases the length of the drive.
Contact Forces
Dimples on a golf ball increase the velocity of the boundary
layer and can dramatically influence the length of a drive.
Depth of Dimple
(mm)
0.05
0.10
0.15
0.20
0.25
0.30
Carry
(m)
107
171
194
204
218
206
Length of Drive
(m)
134
194
212
218
239
219
From The Mechanics of Sport, E. Bade.
Contact Forces
Terminal Speed
An object falling through a fluid reaches its
terminal speed when the drag force is equal to its
weight. This results in a net force of zero and thus
no further acceleration takes place.
weight
drag force
Contact Forces
Estimated Terminal Speeds of
Selected Spheres
Weight Diameter
K
Terminal V
Ball
(N)
(cm) (Drag Factor)
(m/s)
16-lb shot
71.27
1.86
0.00014
145.28
Baseball
1.43
1.14
0.0016
42.47
Golf ball
0.45
0.66
0.0018
40.23
Basketball
5.84
3.73
0.007
20.12
Ping-Pong ball
0.03
0.58
0.04
8.94
Adapted from Sport
Science by Peter J.
Brancazio.
C DrD 2 g
K
8W
VT 
g
K
CD:
r:
D:
W:
VT:
coefficient of drag
fluid density
sphere diameter
weight of sphere
terminal speed
Contact Forces
Centripetal vs. Centrifugal Force
• Centripetal force (center
seeking force) = mass x
centripetal acceleration
Fc  mac
2
vt
m
r
 m 2r
• Centrifugal force (center fleeing force)
-- reaction to the centripetal force;
applied to the other body
Consider Newton’s second law of motion:
SF = ma
Now substitute centripetal acceleration. In centripetal motion the
centripetal acceleration is linked to a centripetal force. You can
think of this force as being responsible for holding the object in a
circular path.
Example
you make a right turn in your car
you feel the driver door push on you to the right
(toward the center of the curvature of your
curved path)
the door applies a centripetal force to you,
you apply a centrifugal force which is
equal and opposite to the centripetal force
door
you
Fcf
Fcp
Forces occurring
along a curved path
Pressure
• localized effect of a force being applied to
an area of a certain surface
Pressure 
W
xxxxx
x x x x x P F
s
xxxxx
A
Rn
if W = 100 lb, A = 500 in2
Force
Area
Bird’s eye view
x = pt of application for a force
P
100 lb
2  0.2 psi

0
.
2
lb
/
in
500 in 2
Special Force
Application
W = 110 lb *4.45 N/lb = 489.5 N
What is the avg. pressure
when standing on one foot?
In SI Units?
10 in2 = 0.00645 m2
2
 2.54 cm 
10 in 2  
  64.5 cm2  0.006 m2
 in 
489.5 N
P
 75.9 kPa
2
0.00645 m
40 in2 =
0.0258 m2
489.5 N
P
 19.0 kPa
2
0.0258 m
Special Force
Application
Pressure
distributions in
the foot
While standing, most of the
pressure is in the heel and
the forefoot
Work
• Work is a force applied over a given distance.
• W = F*d
• Units are N-m, 1 N-m = 1 Joule.
• Positive work occurs when the force is applied in
the same direction as the motion (concentric
contractions).
• Negative work occurs when the force is applied in
the opposite direction of the motion (eccentric
contractions).
Special Force
Application
direction of motion
q
• Only the component of force parallel to
the direction of motion is responsible for
work being done. If there is no movement
there is no work being done.
Special Force
Application
Internal Work vs External Work
• External work is the result of external
forces such as ground reaction forces.
• Internal work is the result of internal forces
such as muscle forces.
While running uphill there is internal work
done to rotate the segments and external
work done to raise the body to a new height.
Energy
• Energy is the capacity to do work.
• Performing positive work on an object will
increase its energy while performing
negative work will decrease its energy.
W = DE
• Energy is measured in Joules.
Special Force
Application
• Kinetic Energy (KE) is the energy that an
object possesses because of its velocity.
KE = ½mv2
• Potential Energy (PE) is the energy that an
object possesses because of its height.
PE = mgh
• Strain Energy (SE) is the energy that an
object possesses because of its deformation:
SE = ½kx2
Special Force
Application
Kinetic Energy
• Kinetic Energy (KE) - energy due to motion
KE  1 mv 2
2
unit: kg  m2 / s2  J
– e.g. A diver (mass = 70 kg) hits the water after a
dive from the 10 m tower with a velocity of 14
m/s. How much KE does she possess?
2  6860 J
KE  1 mv 2  1 (70 kg)(14 m
)
s
2
2
Special Force
Application
Potential Energy
• Potential Energy (PE) - energy due to
gravity
– PE = mgh = mass*gravity*height
– e.g. A diver (mass = 70 kg) is 10 meters above
the water. How much PE does the diver have?
PE = 70*9.8*10 = 6860 J
Special Force
Application
Strain Energy
• Strain (SE) - energy due to deformation
– SE = ½kx2, k = stiffness, x = deformation
– this type of energy arises in compressed springs,
squashed balls ready to rebound, stretched tendons
inside the body, and other deformable structures
– e.g. A muscle tendon with a stiffness of 70,000
N/m is stretched by 1 cm. How much strain
energy does it have?
SE = ½(70,000)(.01)2 = 3.5 J
Special Force
Application
Conservation of Energy
• Energy can not be created nor destroyed, it can
only change forms (von Helmholtz, 1847).
PE = mgh
Energy is often
transferred
between kinetic,
potential and
strain.
KE =.5mv2
Special Force
Application
KE = 0
PE = max
1 release
3
KE = 0
PE = max
.5m
KE = max
2 PE = 0
• If the 4 kg pendulum above is released from a
height of .5 meters what is the maximum
velocity?
PE = 4(9.81).5 = 19.62 J
KE = 19.62 J = .5(4)v2, v = 3.13 m/s
Special Force
Application
Ht (m)
v (m/s)
PE (J)
KE (J)
3.0
29.4
0
0
2.5
24.5
3.1
4.9
2.0
19.6
4.4
9.8
1.5
14.7
5.4
14.7
1.0
9.8
6.3
19.6
time
Special Force
Application
Conversion of PE to KE and vice-versa
Start with no KE b/c v=0
You can use this principle of energy conservation to solve
useful problems. For instance – how high would you need to
raise the slide in the picture below to successfully prevent the
slider from running off the end?
TEinitial = TEfinal
TE = PE + KE
PEinitial
+
KEinitial =
PEfinal
+
KEfinal
m(9.8 m/s2)(4.0 m)+ ½m(8 m/s)2 = m(9.8 m/s2)(h)+ ½m(0 m/s)2
h = 7.6 m
Work-Energy Relationship
• If you lift a barbell into the air you are
performing work on the barbell. You apply a
force over a distance.
• By performing work on the barbell you change
the amount of PE that the barbell has.
W = DE
PE = mgh2
PE = mgh1
Special Force
Application
• If you lift a 300 Newton barbell 1 meter you
have a change in energy of 300 J. This is equal
to the amount of work done.
W = DE
PE1 = (300)0 = 0 J
PE2 = (300)1 = 300 J
PE = mgh2
PE = mgh1
Special Force
Application
Work-Energy Relationship
W = DTE = DPE+ DKE+ DSE
When you ski down a slope you
begin with only PE that is converted
to KE as go down the hill.
Assuming friction is negligible the
TE will not change. At the bottom
of the hill you have only KE which
means you are moving FAST. In
order to slow down you have to
dissipate this kinetic energy. This
requires a change in TE. By the
work-energy relationship, work must
be performed on the skier to bring
him/her to a stop. To perform work
you need a force. In this case it is
the friction developed between the
skier and some unpacked snow at
the end of the run.
AAAHHHGGG – I’M BACK IN DRIVER’S ED!
Three people are driving the same type of car.
Driver A is traveling at 10 mph.
Driver B is traveling at 20 mph.
Driver C is traveling at 30 mph.
What can you say about the relationship of the stopping
distance between vehicles? (Is it linear or something else?)
The cars have no change in PE so only KE must be changed. Thus the
work necessary to bring the cars to rest is equal to the change in KE:
W = DKE
SF * d = ½ m(Dv)2
So the stopping distance (d) is proportional to the square of velocity
d a v2
Kinetic Energy to Strain Energy
• Kinetic energy will be used to
deform the elastic tissues. Strain
energy can then be transformed back
into kinetic energy during the
pushoff phase.
• At low speeds, kangaroos use a pentapedal form of
locomotion (four feet and a tail).
• At 6-7 km/hr the switch to hopping.
• The rate of O2 consumption increases sharply with speed
during pentapedal locomotion.
• When they begin to hop, the rate of O2 consumption
decreases with increasing speed until 18 km/hr.
Why?
Dawson and Taylor, 1973
Special Force
Application
Kinetic to Strain to Kinetic
• Kangaroos have large Achilles tendons (1.5
cm in diameter and 35 cm in length).
• It is possible that a large amount of strain
energy is stored in the tendon during the
landing and converted back into kinetic
energy as the kangaroo rebounds into the
air.
Special Force
Application
Collisions and Impacts
Elastic
Inelastic
Coefficient of Restitution
• The coefficient of restitution (e) determines the
elasticity of an impact.
• If e = 1 the impact is completely elastic. This
means that the object contains all of the energy it
had before the impact.
• If e = 0 the impact is completely plastic. This
means that the object contains none of the energy
it had before the impact.
Special Force
Application
Objects typically deform during an impact
This is referred to as the period of deformation
This is followed by a period of restitution
thus the name coefficient of restitution
Coefficient of Restitution
• e=-
• e=
relative velocity after impact
relative velocity before impact
h bounce
h drop
Power
• Power is the rate of performing work.
P=
W
Dt
P = F*s
Dt
P=F*s
Dt
P = F*v
Special Force
Application
Power
• Units are Joules/second or Watts.
• Greater power must be developed in order
to do mechanical work more quickly.
• Power can be positive or negative
depending on whether F and v point in same
general direction (+ power) or in opposite
directions (- power)
Special Force
Application
Power
• Positive power indicates that energy is being
generated and negative power indicates that
energy is being absorbed thus:
• Positive power is associated with concentric
muscular contractions, while negative power
is associated with eccentric muscular
contractions.
Special Force
Application
-1
Power (Watts·kg )
Knee Power During Running
15
10
Energy
Generation
5
0
Energy
Absorption
-5
-10
-15
-20
0
10
20
30
40
50
60
70
80
90
100
Percent of Stance
Special Force
Application
Power Example
m = 100 kg, g = 9.8 m/s2, h = 2 m
W = mgh = 100 (9.8) 2 = 1960 J
now add time
Case 1: raise the barbell slowly -- t = 5 s
P  W 1960 J 392 W
t
5s
Case 2: raise the barbell quickly -- t = 1.5 s
P  W 1960 J 1307 W
t 1.5 s
Special Force
Application