Post on 25-Dec-2015
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Force• force - a push or pull acting on a body• forces are vector quantities
– magnitude (size)– direction (line of action)– point of application
• measured in Newtons (N)– 1 N = (1 kg) (1 m/s2)
• English units (lb)– 1 lb = (1 slug) (1 ft/s2)
BOX
Fpt of application
line of action
Free-Body Diagrams
A free body diagram illustrates all of theexternal forces acting on an object.
FR (GRF)
mg
air resistance
Contact ForcesContact Forces
If whole body is considered to be “the system”Examples of external forces
Weight (gravity)Ground Reaction Force (GRF)FrictionFluid Resistance
Examples of internal forcesJoint Reaction ForceMuscle Force
NEWTON'S LAWS OF MOTIONPhilosophiae Naturalis Principia Mathematica (1686)
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• 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.
Newton’s 1st Law of Motion
W
Ry
if Ry = Wthen resultant force = 0if v = 0 and F = 0STATIC EQUILIBRIUM
W
Ry
FpFr
Fr = resistive forceFp = propulsive forceif v = 0 and F = 0DYNAMIC EQUILIBRIUM
V
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
F1
F2
Fa
a F am
a k F
m 1
1
where k1 = constant of proportionality
F k ma where kk
` 2
1
1
m m
F k ma 2
If m is measured in kgand 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
F = 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."
• F =
• Ft = mv (impulse/momentum relation)
• F = m = mat
mv - mv if
v
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
F = ma but F = W = mgmg = maora = 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”
Rpropulsion
Apropulsion
Rresistance
Aresistance
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."
• FG = G , G = 6.67X10-11 N-m2/kg2 m1*m2
d 2
Mass and Weight
• g depends on the distance from the earth’s center
M
r
M
r
Earth is not a true spherebut rather an ellipsoid: it is fatter at the equator
m = your massM = earth’s massr = radius (distance to (center of earth)G = gravitational constant
Newton’s Law of Gravitation
FgGMm
rFg
rand Fg m
2 2
1
Fg = gravitational force
If m = your mass then Fg = your weight (W)
FgGMm
r
g GM
r
W mg
2
2
Latitude effect = g is smaller at equator larger at poles
Altitude effect = g is smaller at high altitudelarger at low altitude
M
r
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 LawsThe 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 ForcesContact Forces
Impulse - Momentum Relationship
Contact ForcesContact Forces
maF amF
t
vvmF if
)( if vvmtF
if mvmvtF
if ppJ
pJ
Impulse-MomentumJ p
“Impulse = change in momentum”Units
smkg
ssmkg
sNtF
/
)/( 2
Contact ForcesContact Forces
smkgvm /
Same units!
Impulse/Momentum(Ft = mv)• The impulse/momentum relationship describes the
effects of a force over a period of time.• An impulse (F*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 ForcesContact 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 ForcesContact Forces
• 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?
Impulse/Momentum
Contact ForcesContact 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? Fy
0
0 .8time
Contact ForcesContact Forces
• If a 10 kg object is exposed to the above impulse what will be the change in velocity?
5
4
3
2
1
0
2 54310
Time (s)
Force
(N) impulse = area
Contact ForcesContact Forces
F
t
Impulse = area under F v. t curve
Jn=negative impulse
Jp=positive impulse
JNET = NET IMPULSEJNET = Jp + Jn
Contact ForcesContact Forces
Recall that the net force is
F = GRF - W
and by Newton’s 2nd Law
F = ma
so GRF - W = ma
If GRF > W then a >0
If GRF < W then a < 0
W
GRF
W
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
body weightbody weight
GRFGRF
00
W
GRF
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
lowest point
body weightbody weight
GRFGRF
00
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
thus GRF > W
lowest point
takeoff
body weightbody weight
GRFGRF
00
W
GRF
Total Net Impulse = Total Net Impulse = Negative Net ImpulseNegative Net Impulse + +Positive Net ImpulsePositive Net Impulse
J = mJ = mv = m(vv = m(vff - v - vii))
this can be solved to find the this can be solved to find the takeoff velocity (final velocity takeoff velocity (final velocity of countermovement phase)of countermovement phase)
vvtakeofftakeoff = = J/m since vJ/m since vii = 0 = 0
Knowing vKnowing vtakeofftakeoff allows you to compute jump height allows you to compute jump height
Knowing vKnowing vtakeofftakeoff allows you to allows you to
compute jump heightcompute jump height
vvtakeofftakeoff
vvtoptop = 0 m/s = 0 m/s
vvff22 = v = vii
22 + 2ad + 2ad
wherewhere vvff = v = vtoptop
vvii = v = vtakeofftakeoff
jump heightjump height
SO - jump height is increased bySO - jump height is increased byincreasing the total net increasing the total net impulseimpulse… … not just the net forcenot just the net force
Conservation of Momentum
Momentum = mass x velocity
Total momentumbefore = total momentumafter
for example - if there are 2 objects in the systemm1v1 + m2v2 = m1u1 + m2u2
v = velocity beforeu = velocity aftersubscripts represent object number
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.
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 jointThe net force acting across a jointe.g. when standing, the thigh exerts a downward forcee.g. when standing, the thigh exerts a downward force
on the shank, conversely, the shank exerts anon the shank, conversely, the shank exerts anupward force on the thigh of equal magnitudeupward force on the thigh of equal magnitude
NOTE: this JRF does not includeNOTE: this JRF does not includethe forces exerted on the jointthe forces exerted on the jointby the muscle crossing theby the muscle crossing thejoint. The force that includesjoint. The force that includesall of the forces crossing the jointall of the forces crossing the jointis known as the is known as the bone-on-bonebone-on-boneforceforce
Contact ForcesContact Forces
• 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.
Block Pulling Force
Friction Force
Normal Force(weight of the block)
Contact ForcesContact Forces
Normal Force = force perpendicular to surface
FN
When surface is horizontal the perpendicular direction is vertical so the normal force is simply the weight of the object
Weight
Normal Force
When surface is inclined the perpendicular direction is NOT vertical so the normal force is only a component of the object’s weight.
• At some point the pulling force will be great enough so that the friction force cannot prevent movement.
Fri
ctio
n F
orce
Applied Force
Static
Dynamic
impending motion
maximal static friction force
Contact ForcesContact Forces
• The coefficient of static friction is expressed as:
where s= coefficient of static friction
Fnormal = normal force
Fmax = maximal static friction force
Friction
normals F
Fmax
•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 s 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.
where d= coefficient of dynamic friction
N = normal force
Ffriction = force resisting motion
Normal
frictiond F
F
Contact ForcesContact Forces
Friction
• It has been found experimentally that d< s.
• d depends of the relative speed of the surfaces.
• At speeds from 1 cm/s to several m/s, d is approximately constant.
Contact ForcesContact Forces
Why is it easier to pull a desk than push it?
Ph
v
WW
R = W+v
When you push you usually have a downwardcomponent of force -- so the normalforce is increased and therefore thefrictional force is increased.
Contact ForcesContact Forces
P
h
v
WW
R = W-v
When you pull you usually have an upwardcomponent of force -- so the normalforce is reduced and therefore the frictional force is reduced.
Contact ForcesContact Forces
• 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.
Fluid Resistance
Contact ForcesContact Forces
• 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%.
Surface and Form Drag
Contact ForcesContact Forces
Van Ingen Schenau (1982) reported a 10% reduction in surface drag when a speed skater wears a smooth body suit.
Surface Drag
Contact ForcesContact Forces
Form drag occurs when air is driven past an object and is diverted outward creating a low pressure region behind the object.
high pressure
low pressure
Form Drag
Contact ForcesContact Forces
The orientation of the object will affect the frontal area and will play an important role in the amount of form drag.
Low form drag
High form drag
Form Drag
Contact ForcesContact Forces
frontal area .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.
Contact ForcesContact Forces
At low velocities laminar flow occurs. The boundary layer remains attached to the surface. During separated flow the boundary layerseparates toward theback of the object and alow pressure regionis formed. During fully turbulent flow the boundary layer becomes turbulent and the size of the pocket is decreased.
laminar
separated
fully turbulent
Flow Type
Contact ForcesContact Forces
Factors Affecting Flow Type
• size
• shape
• surface roughness
• viscosity of the fluid
• flow velocity
Contact ForcesContact Forces
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.
Airfoil
Bernoulli’s Principle, 1738Contact ForcesContact Forces
direction of movement
Lift always acts perpendicular to drag.
Fdrag = 1/2(CdAv2)
Flift = 1/2(ClAv2)
Fair resistance
Fdrag
Flift
Lift
Contact ForcesContact 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 - 10oContact ForcesContact Forces
Angle of Attack Lift Drag Lift
(degrees) (N) (N) Drag
0 0.00 1.17 0.0010 4.33 1.50 2.8920 10.64 4.13 2.5825 12.83 5.79 2.2227 13.80 6.88 2.0128 13.80 7.41 1.8629 11.01 7.94 1.3930 11.21 8.18 1.3735 10.12 8.74 1.1640 8.50 9.55 0.8945 8.90 11.13 0.8050 8.62 12.22 0.7160 6.88 14.98 0.4670 4.77 16.43 0.2980 2.55 16.88 0.1590 0.00 17.73 0.00
Adapted from Aerodynamic Factors Which Influence Discus Flight, Ganslen.
Lift to drag ratios for the discus.
Contact ForcesContact Forces
Rotating objects can also create a pressure difference.
low pressurezone
high pressurezone
Direction of air flow
Rotating Objects
Magnus Effect, 1852Contact ForcesContact Forces
low pressurezone
high pressurezone
intended direction of flight
actual directionof flight
Contact ForcesContact Forces
The golf club imparts backspin on the golf ball andincreases the length of the drive.
Contact ForcesContact Forces
Depth of Dimple Carry Length of Drive(mm) (m) (m)
0.05 107 1340.10 171 1940.15 194 2120.20 204 2180.25 218 2390.30 206 219
From The Mechanics of Sport, E. Bade.
Dimples on a golf ball increase the velocity of the boundary layer and can dramatically influence the length of a drive.
Contact ForcesContact 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 ForcesContact Forces
Estimated Terminal Speeds of Selected Spheres
KC DD
2
g
W8
Vg
KT
CD: coefficient of drag: fluid densityD: sphere diameterW: weight of sphereVT: terminal speed
Adapted from SportScience by Peter J. Brancazio.
Weight Diameter K Terminal VBall (N) (cm) (Drag Factor) (m/s)
16-lb shot 71.27 1.86 0.00014 145.28Baseball 1.43 1.14 0.0016 42.47Golf ball 0.45 0.66 0.0018 40.23Basketball 5.84 3.73 0.007 20.12Ping-Pong ball 0.03 0.58 0.04 8.94
Contact ForcesContact Forces
Centripetal vs. Centrifugal Force
• Centripetal force (center seeking force) = mass xcentripetal acceleration
• Centrifugal force (center fleeing force) -- reaction to the centripetal force; applied to the other body
rmr
vmmaF t
cc2
2
Consider Newton’s second law of motion:F = 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.
Exampleyou make a right turn in your caryou 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
youFcpFcf
Forces occurring Forces occurring along a curved pathalong a curved path
Pressure
• localized effect of a force being applied to an area of a certain surface
W
Rn
x x x x xx x x x xx x x x x
Bird’s eye viewx = pt of application for a force
Pressure Force
Area
P F
As
if W = 100 lb, A = 500 in2
P lb
inlb in psi 100
50002 02
22. / .
Special ForceSpecial ForceApplicationApplication
40 in2 =0.0258 m2
10 in2 = 0.00645 m2
W = 110 lb *4.45 N/lb = 489.5 N
What is the avg. pressurewhen standing on one foot?
kPa 9.75m 00645.0
N 5.489P
2 kPa 0.19
m 0258.0
N 5.489P
2
In SI Units?
102 54
64 5 0 0062
2
2 2incm
incm m
.. .
Special ForceSpecial ForceApplicationApplication
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 ForceSpecial ForceApplicationApplication
• 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.
direction of motion
Special ForceSpecial ForceApplicationApplication
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 = E
• Energy is measured in Joules.
Special ForceSpecial ForceApplicationApplication
• 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 ForceSpecial ForceApplicationApplication
• Kinetic Energy (KE) - energy due to motion
– 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?
KE mv unit kg m s J 12
2 2 2: /
KE mv kg ms J 1
212
70 14 68602 2( ) ( )
Kinetic Energy
Special ForceSpecial ForceApplicationApplication
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 ForceSpecial ForceApplicationApplication
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 ForceSpecial ForceApplicationApplication
Conservation of Energy• Energy can not be created nor destroyed, it can
only change forms (von Helmholtz, 1847).
Energy is often transferredbetween kinetic, potential and strain.
PE = mgh
KE =.5mv2
Special ForceSpecial ForceApplicationApplication
• 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
KE = maxPE = 0
KE = 0PE = max
KE = 0PE = max 1 release 3
2
.5m
Special ForceSpecial ForceApplicationApplication
1.0
1.5
2.0
2.5
3.0 29.4
24.5
19.6
14.7
9.8
0
3.1
4.4
5.4
6.3 19.6
14.7
9.8
4.9
0
Ht (m) PE (J) v (m/s) KE (J)
time Special ForceSpecial ForceApplicationApplication
Conversion of PE to KE and vice-versaStart 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
• 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.
PE = mgh1
PE = mgh2
W = E
Work-Energy Relationship
Special ForceSpecial ForceApplicationApplication
• 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.
PE1 = (300)0 = 0 J
PE2 = (300)1 = 300 J
PE = mgh1
PE = mgh2
W = E
Special ForceSpecial ForceApplicationApplication
W = TE = PE+ KE+ SE
Work-Energy Relationship
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 = KEF * d = ½ m(v)
So the stopping distance (d) is proportional to the square of velocityd 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.
Dawson and Taylor, 1973
• 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?
Special ForceSpecial ForceApplicationApplication
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 ForceSpecial ForceApplicationApplication
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 ForceSpecial ForceApplicationApplication
Objects typically deform during an impact
This is referred to as the period of deformation
This is followed by a period of restitutionthus the name coefficient of restitution
Coefficient of Restitution
• e = -
• e = h
h
drop
bounce
Power• Power is the rate of performing work.
P =
P =
P = F *
P = F*v
t
W
t
s*F
t
s
Special ForceSpecial ForceApplicationApplication
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 ForceSpecial ForceApplicationApplication
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 ForceSpecial ForceApplicationApplication
Percent of Stance
-20
-15
-10
-5
0
5
10
15
0 10 20 30 40 50 60 70 80 90 100
Pow
er (
Wat
ts·k
g-1)
EnergyAbsorption
EnergyGeneration
Knee Power During Running
Special ForceSpecial ForceApplicationApplication
Power Examplem = 100 kg, g = 9.8 m/s2, h = 2 mW = mgh = 100 (9.8) 2 = 1960 J
now add timeCase 1: raise the barbell slowly -- t = 5 s
W392s 5
J 1960t
WP
Case 2: raise the barbell quickly -- t = 1.5 s
W 1307s 5.1J 1960
tWP
Special ForceSpecial ForceApplicationApplication