Transcript Work, Power & Energy Explaining the Causes of Chapter 4

Work, Power & Energy
Chapter 4
Explaining the Causes of
Motion in a Different Way
Work
The product of force and the amount of
displacement along the line of action of
that force.
Work  Force  displacement
Units:
ft . lbs (horsepower)
Newton•meter (Joule) e
Work = F x d
To calculate work done on an object, we
need:
The Force
The average magnitude of the force
The direction of the force
The Displacement
The magnitude of the change of position
The direction of the change of position
Calculate Work
During the ascent phase of a rep of the
bench press, the lifter exerts an
average vertical force of 1000 N
against a barbell while the barbell
moves 0.8 m upward
How much work did the lifter do to the
barbell?
Calculate Work
Table of Variables:
Force = +1000 N
Displacement = +0.8 m
Force is positive due to pushing upward
Displacement is positive due to moving
upward
Calculate Work
Table of Variables:
Force = +1000 N
Displacement = +0.8 m
Select the equation and solve:
Work  Force  displaceme nt
Work   1000 N   0.8m 
Work  800 Nm  800 Joule  800 J
- & + Work
Positive work is performed when
the direction of the force and
the direction of motion are the
same
 ascent phase of the bench press
 Throwing a ball
 push off (upward) phase of a jump
Calculate Work
During the descent phase of a rep of
the bench press, the lifter exerts an
average vertical force of 1000 N
against a barbell while the barbell
moves 0.8 m downward
Calculate Work
Table of Variables
Force = +1000 N
Displacement = -0.8 m
Force is positive due to pushing upward
Displacement is negative due to movement
downward
Calculate Work
Table of Variables
Force = +1000 N
Displacement = -0.8 m
Select the equation and solve:
Work  Force  displaceme nt
Work   1000 N   0.8m 
Work  800 Nm  800 Joule  800 J
- & + Work
Positive work
Negative work is performed
when the direction of the force
and the direction of motion are
the opposite
 descent phase of the bench press
 catching
 landing phase of a jump
Contemplate
During negative work on the bar, what is
the dominant type of activity
(contraction) occurring in the muscles?
When positive work is being performed
on the bar?
EMG during the Bench Press
On elbow
180
90
Work performed climbing
stairs
 Work = Fd
 Force
 Subject weight
 From mass, ie 65 kg
 Displacement
 Height of each step
 Typical 8 inches (20cm)
 Work per step
 650N x 0.2 m = 130.0 Nm
 Multiply by the number of steps
Work on a stair stepper
Work = Fd
Force
 Push on the step
????
Displacement
 Step Height
8 inches
“Work” per step
 ???N x .203 m = ???Nm
Work on a cycle ergometer
Work = Fd
Force
 belt friction on the flywheel
mass (eg 3 kg)
Displacement
 revolution of the pedals
Monark: 6 m
“Work” per revolution
Work on a cycle ergometer
 Work = Fd
 Force
 belt friction on the flywheel
 mass (eg 3 kg)
 Displacement
 revolution of the pedals
 Monark: 6 m
 “Work” per revolution
 3kg x 6 m = 18 kgm
Similar principle for wheelchair
…and for handcycling
ergometer
Energy
 Energy (E) is defined as the capacity to do
work (scalar)
 Many forms
No more created, only converted
 chemical, sound, heat, nuclear, mechanical
 Kinetic Energy (KE):
 energy due to motion
 Potential Energy (PE):
 energy due to position or deformation
Kinetic Energy
Energy due to motion reflects
 the mass
 the velocity
of the object
KE = 1/2
2
mv
Kinetic Energy
Units: reflect the units of mass * v2
KE 
Units KE = Units work
KE 
KE 
KE 
KE 
1 2
mv
2
1
(kg)( m / s ) 2
2
1
kg  m  m / s / s
2
1
(kg  m / s / s )  m
2
1
Nm
2
Calculate Kinetic Energy
How much KE in a 5
ounce baseball (145 g)
thrown at 80 miles/hr
(35.8 m/s)?
Calculate Kinetic Energy
Table of Variables
Mass = 145 g  0.145 kg
Velocity = 35.8 m/s
Calculate Kinetic Energy
Table of Variables
Mass = 145 g  0.145 kg
Velocity = 35.8 m/s
Select the equation and solve:
KE = ½ m v2
KE = ½ (0.145 kg)(35.8 m/s)2
KE = ½ (0.145 kg)(1281.54 m/s/s)
KE = ½ (185.8 kg m/s/s)
KE = 92.9 kg m/s/s, or 92.9 Nm, or 92.9J
Calculate Kinetic Energy
How much KE possessed by
a 150 pound female
volleyball player moving
downward at 3.2 m/s after
a block?
Calculate Kinetic Energy
Table of Variables
 150 lbs = 68.18 kg of mass
 -3.2 m/s
Select the equation and solve:
KE = ½ m v2
 KE = ½ (68.18 kg)(-3.2 m/s)2
 KE = ½ (68.18 kg)(10.24 m/s/s)
 KE = ½ (698.16 kg m/s/s)
 KE = 349.08 Nm or J
Calculate Kinetic Energy
Compare KE possessed by:
 a 220 pound (100 kg) running back
moving forward at 4.0 m/s
 a 385 pound (175 kg) lineman moving
forward at 3.75 m/s
Bonus: calculate the momentum
of each player
Calculate Kinetic Energy
Table of Variables
m = 100 Kg
v = 4.0 m/s
Select the equation
and solve:
KE = ½ m v2
KE = ½ (100 kg)(4.0
m/s)2
KE = 800 Nm or J
Table of Variables
m = 175 kg
v = 3.75 m/s
Select the equation
and solve:
KE = ½ m v2
KE = ½ (175)(3.75)2
KE = 1230 Nm or J
Calculate Momentum
Momentum = mass times velocity
Player 1 = 100 kg * 4.0 m/s
Player 1 = 400 kg m/s
Player 2 = 175 * 3.75 m/s
Player 2 = 656.25
Potential Energy
Two forms of PE:
Gravitational PE:
energy due to an object’s position
relative to the earth
Strain PE:
due to the deformation of an
object
Gravitational PE
Affected by the object’s
 weight
mg
 elevation (height) above reference point
 ground or some other surface
h
GPE = mgh
Units = Nm or J (why?)
Calculate GPE
How much gravitational
potential energy in a 45 kg
gymnast when she is 4m
above the mat of the
trampoline?
Take a look at the energetics of a roller coaster
Calculate GPE
How much gravitational potential energy
in a 45 kg gymnast when she is 4m
above the mat of the trampoline?
Trampoline mat is 1.25 m
above the ground
Calculate GPE
GPE relative to mat
Table of Variables
m = 45 kg
g = -9.81 m/s/s
h=4m
PE = mgh
PE = 45kg * -9.81
m/s/s * 4 m
PE = - 1765.8 J
More on this
GPE relative to ground
Table of Variables
m = 45 kg
g = -9.81 m/s/s
h = 5.25 m
PE = mgh
PE = 45m * -9.81
m/s/s * 5.25 m
PE = 2317.6 J
Conversion of KE to GPE and
GPE to KE and KE to GPE and …
Strain PE
Affected by the object’s
amount of deformation
 greater deformation = greater SE
 x2 = change in length or deformation of
the object from its undeformed position
stiffness
 resistance to being deformed
 k = stiffness or spring constant of material
SE = 1/2 kx2
Strain Energy
When a fiberglass vaulting pole bends,
strain energy is stored in the bent pole
Pole vault explosion
Strain Energy
When a fiberglass vaulting pole bends,
strain energy is stored in the bent pole
Bungee jumping
Strain Energy
When a fiberglass vaulting pole bends,
strain energy is stored in the bent pole
Bungee jumping
Hockey sticks
Strain Energy
 When a fiberglass vaulting pole bends, strain
energy is stored in the bent pole
 Bungee jumping
 When a tendon/ligament/muscle is stretched,
strain energy is stored in the elongated
elastin fibers (Fukunaga et al, 2001, ref#5332)
 k = 10000 n /m
tendon in walking
x = 0.007 m (7 mm), Achilles
 When a floor/shoe sole is deformed, energy
is stored in the material
Plyometrics
Work - Energy Relationship
The work done by an external force
acting on an object causes a change in
the mechanical energy of the object
Fd  Energy
Fd  KE  PE
1
2
Fd  mv f  vi   mg (rf  ri )
2
Work - Energy Relationship
The work done by an external force
acting on an object causes a change in
the mechanical energy of the object
 Bench press ascent phase
initial position = 0.75 m; velocity = 0
final position = 1.50 m; velocity = 0
m = 100 kg
g = -10 m/s/s
What work was performed on the bar by lifter?
What is GPE at the start & end of the press?
Work - Energy Relationship









What work was performed on the bar by lifter?
Fd =  KE +  PE
Fd = ½ m(vf –vi)2 + mgh
Fd = 100kg * - 10 m/s/s * 0.75 m
Fd = 750 J
W = Fd
W = 100 kg * .75m
W = 75 kg m
W = 75 kg m (10) = 750 J
Work - Energy Relationship
 What is GPE at the start & end of the press?
 End (ascent)
 PE = mgh
 PE = 100 kg * -10 m/s/s * 1.50 m
 PE = 1500 J
 Start (ascent)
 PE = 100 kg * -10 m/s/s * 0.75m
 PE = 750 J
Work - Energy Relationship
Of critical importance
Sport and exercise =  velocity
 increasing and decreasing kinetic energy of a
body
Fd  Energy
Fd  KE  PE
1
2
Fd  mv f  vi   mg (rvf  rvi )
2
 similar to the impulse-momentum relationship
Ft = m (vf-vi)
Work - Energy Relationship
If more work is done, greater energy
 greater average force
 greater displacement
Ex. Shot put technique (121-122).
If displacement is restricted, average
force is __________ ? (increased/decreased)
 “giving” with the ball
 landing hard vs soft
Power
The rate of doing work
 Work = Fd
Power  Work / time
Power  Fd / t
Power  Force  velocity
Units: Fd/s = J/s = watt
Calculate & compare power
During the ascent phase of a rep of the
bench press, two lifters each exert an
average vertical force of 1000 N
against a barbell while the barbell
moves 0.8 m upward
Lifter A: 0.50 seconds
Lifter B: 0.75 seconds
Calculate & compare power
Lifter A
Table of Variables
F = 1000 N
d = 0.8 m
t = 0.50 s
Fd
t
1000 N  0.8m
Power 
0.50 s
800 J
Power 
 1600 w
0.50 s
Power 
Lifter B
Power on a cycle ergometer




Work = Fd
Force: 3kg
Displacement: 6m /rev
“Work” per revolution
 3kg x 6 m = 18 kgm
 60 rev/min
" Power" Fd / t
" Power" Fd  rev / min
" Power"  18kgm  60 / min
" Power" 1080kgm / min
Power on a cycle ergometer




Work = Fd
Force: 3kg
Displacement: 6m /rev
“Work” per revolution
 3kg x 6 m = 18 kgm
 60 rev/min
" Power" Fd / t
" Power" Fd  rev / min
" Power"  18kgm  60 / min
" Power" 1080kgm / min
1 Watt = 6.12 kgm/min
Compare “power” in typical
stair stepping
 Work = Fd
 Force: Push on the step
 constant setting
 Displacement
 Step Height: 5” vs 10”
 0.127 m vs 0.254 m
 step rate
 56.9 /min vs 28.8 /min
Time per step
 60s/step rate
Thesis data from Nikki Gegel
and Michelle Molnar
Compare “power” in typical
stair stepping
 Work = Fd
 Force: Push on the step
 constant setting
 Displacement
 Step Height: 5” vs 10”
 0.127 m vs 0.254 m
 step rate
 56.9 /min vs 28.8 /min
Power  F  v
Power5inch  F  (.127m / 1.05s)
Power10inch  F  (.254m / 2.08s)
Compare “power” in typical
stair stepping
 Work = Fd
 Force: Push on the step
 constant setting
 Displacement
 Step Height: 5” vs 10”
 0.127 m vs 0.254 m
 step rate
 56.9 /min vs 28.8 /min
Power  F  v
Power5inch  F  0.121m / s
Power10inch  F  0.122m / s
Results: VO2 similar fast/short
steps vs slow/deep steps