Transcript Lecture 5: Energy

Chapter 5: Energy
Energy
 Energy
is present in a variety of forms:
mechanical, chemical, electromagnetic, nuclear, mass, etc.
 Energy
 The
 If
can be transformed from one from to another.
total amount of energy in the Universe never changes.
a collection of objects can exchange energy with
each other but not with the rest of the Universe (an
isolated system), the total energy of the system is
constant.
If one form of energy in an isolated system decreases,
another form of energy must increase.
 In this chapter, we focus on mechanical energy: kinetic
energy and potential energy.
Work

The work W done on an object by a constant force F
when the object is displaced by Dx by the force:
W  FDx
SI unit: joule (J) = newton-meter (N m) = kg m2/s2
• Work is a scalar quantity.
• If the force exerted on an
object is not in the same
direction as the displacement:
 
W  ( F cos  )Dx  F  Dx
component of
dot product
the force along
or
the direction of inner product
the displacement
Work

If an object is displaced vertical to the direction of
a force exerted, then no work is done.
W  ( F cos  )Dx  0 (  90)

If an object is displaced in opposite
direction to that of an exerted force,
the work done by the
force is negative (if F<Fg).
W  ( F cos  )Dx
  FDx (  180)
Work

Work and dissipative forces
• The friction force between two objects in contact and in relative
to each other always dissipate energy in complex ways.
• Friction is a complex process caused by numerous microscopic
interactions over the entire area of the surfaces in contact.
• The dissipated energy above is converted to heat and other forms
of energy.
• Frictional work is extremely important: without it Eskimos can’t
pull sled, cars can’t move, etc.
Work

Examples
• Example 5.1: Sledding through the Yukon
(a) How much work is done if =0?
m=50.0 kg
F= 1.20x102 N
W  FDx
Dx=5.00 m
2
 (1.20 10 J )(5.00 m)
 6.00 10 2 J
(b) How much work is done if =30o?
W  ( F cos  )Dx
 (1.20 10 2 J ) cos 30(5.00 m)
 5.20 10 2 J
Work

Examples
• Example 5.2: Sledding through the Yukon (with friction)
(a) How much work is done if =0?
m=50.0 kg
2N
F=
1.20x10
F

n

mg

0

n

mg
 y
Dx=5.00 m
W   f Dx    nDx
fric
k
k
   k mgDx  4.90  10 2 J
Wnet  Wapp  W fric  Wn  Wg
fk=0.200
 6.00  10 2 J  (4.90  10 2 J )
 0  0  1.10  10 2 J
(b) How much work is done if =30o?
F
y
 n  mg  0  n  mg
W fric   f k Dx    k nDx    k (mg  Fapp sin  )Dx  4.30 102 J
Wnet  Wapp  W fric  Wn  Wg  5.20 102 J  (4.30 102 J)  0  0  90.0 J
Kinetic Energy

Kinetic energy (energy associated with motion)
• Consider an object of mass m moving to the right under action of
a constant net force Fnet directed to the right.
(constant acceleration)
Wnet  Fnet Dx  (ma)Dx
v 2  v02  2aDx
v 2  v02
aDx 
2
Wnet
Wnet
 v 2  v02 

 m
 2 
1
1
 mv 2  mv02
2
2
 KE f  KEi  DKE
work-energy theorem
Define the kinetic energy KE as:
1 2
KE  mv
2
SI unit: J
Kinetic Energy

An example
• Example 5.3: Collision analysis
m=1.00x103 kg
vi = 35.0 m/s -> 0
=8.00x103 N
(a) The minimum necessary stopping distance?
1 2 1 2
1 2
Wnet  mv f  mvi   f k Dx  0  mvi
2
2
2
Dx  76.6 m
(b) If Dx=30.0 m what is the speed at impact?
Wnet  W fric   f k Dx 
v 2f  vi2 
1 2 1 2
mv f  mvi
2
2
2
f k Dx  745 m 2 / s 2  v f  27.3 m/s
m
Kinetic Energy

Conservative and non-conservative forces
• Two kinds of forces: conservative and non-conservative forces
• Conservative forces : gravity, electric force, spring force, etc.
 A force is conservative if the work it does moving an object
between two points is the same no matter what path is taken.
 It can be derived from “potential energy”.
• Non-conservative forces : friction, air drag, propulsive force, etc.
 In general dissipative – it tends to randomly disperse the energy
of bodies on which it acts.
 The dispersal of energy often takes the form of heat or sound.
 The work done by a non-conservative force depends on what
path of an object that it acts on is taken.
 It cannot be derived from “potential energy”.
• Work-energy theorem in terms of works by conservative and nonconservative force
Wnc  Wc  DKE
Gravitational Potential Energy

Gravitational work and potential energy
• Gravity is a conservative force and can be derived from a potential
energy.
Work done by gravity on the book:
Wg  FDy cos   mg ( yi  y f ) cos 0
 mg ( y f  yi )
 mgDy
v 2  v02  2 gDy  v 2  2 gDy
KEi  0
KE f 
1 2
mv  mgDy  DKE
2
Wnet  Wnc  Wg  DKE
Wnc  DKE  mg ( y f  yi )
Gravitational Potential Energy

Gravitational work and potential energy
• Gravity is a conservative force and can be derived from a potential
energy.
Wnc  DKE  mg ( y f  yi )
• Let’s define the gravitational potential energy of a system consisting
of an object of mass m located near the surface of Earth and Earth
as:
y : the vertical position of the mass to
PE  mgy
a reference point ( often at y=0 )
g : the acceleration of gravity
SI unit: J
Wnc  DKE  DPE
where
DPE  PE f  PEi
 mg ( y f  yi )
Gravitational Potential Energy

Reference levels for gravitational potential energy
• As far as the gravitational potential is concerned, the important
quantity is not y (vertical coordinate) but the difference Dy between
two positions.
• You are free to choose a reference point at any level (but usually
at y=0).
yi
yf
Gravitational Potential Energy

Gravity and the conservation of mechanical energy
• When a physical quantity is conserved the numeric value of
the quantity remains the same throughout the physical process.
• When there is no non-conservative force involved,
Wnc  DKE  DPE  0
KEi  PEi  KE f  PE f
• Define the total mechanical energy as:
E  KE  PE
• The total mechanical energy is conserved.
Ei  E f
1 2
1
mvi  mgyi  mv2f  mgy f
2
2
• In general, in any isolated system of objects interacting only
conservative forces, the total mechanical energy of the system
remains the same at all times.
Gravitational Potential Energy

Examples
• Example 5.5: Platform diver
(a) Find the diver’s speed at y=5.00 m.
KEi  PEi  KE f  PE f
1 2
1 2
mvi  mgyi  mv f  mgy f
2
2
0  gyi 
1 2
v f  gy f
2
v f  2 g ( yi  y f )  9.90 m/s
(b) Find the diver’s speed at y=0.0 m.
1 2
0  mgyi  mv f  0
2
v f  2 gyi  14.0 m/s
Gravitational Potential Energy

Examples
• Example 5.8: Hit the ski slopes
(a) Find the skier’s speed at the bottom (B).
KEi  PEi  KE f  PE f
1 2
1 2
mvi  mgyi  mv f  mgy f
2
2
0  gyi 
fk  0
1 2
v f  gy f
2
v f  2 g ( yi  y f )  19.8 m/s
(b) Find the distance traveled on
the horizontal rough surface.
Wnet   f k d  DKE 
d
vB2
2 k g
 95.2 m
1 2 1 2
1
mvC  mvB    k mgd   mvB2
2
2
2
f k  0.210
Spring Potential Energy

Spring and Hooke’s law
• Force exerted by a spring Fs
Fs  kx Hooke’s law
If x > 0, Fs <0 Fs to the left
If x < 0, Fs >0 F to the right
s
k : a constant of proportionality
called spring constant.
SI unit : N/m
• The spring always exerts its
force in a direction opposite
the displacement of its end
and tries to restore the attached
object to its original position.
Restoring force
x>0
Fs
Spring Potential Energy

Potential due to a spring
• The spring Fs is associated with elastic potential energy.
Between xi -1/2Dx and xi+1/2Dx the work
exerted by the spring is approximately: Wi  Fs ,i Dx  Fs ,i ( x / N )
Between x=0 and x, the total work exerted
-Fs
by the spring is approximately:
Ws  lim N  i 1Ws ,i
N
 lim N  i 1 Fs ,i Dx
N
 lim N  i 1 ( areai )
N
width = Dx
xi-1/2Dx xi+1/2Dx
1 2
 Fs , N x   kx
2
In general when the spring is stretched -Fi
from xi to xf, the work done by the spring
is:
1
1

-Ws,i= areai xi xi+1
Ws   kx2f  kxi2 
2
2

x
Spring Potential Energy

Potential due to a spring (cont’d)
• The energy-work theorem including a spring and gravity
1 2 1 2
Wnc   kx f  lxi   DKE  DPEg
2 
2
1
elastic potential energy
PEs  kx2
2
Wnc  ( KE f  KEi )  ( PEgf PEgi )  ( PEsf  PEsi )
( KE  PEg  PEs ) i  ( KE  PEg  PEs ) f
Extended form of conservation of mechanical energy
Spring Potential Energy
Examples

• Example 5.9: A horizontal spring
(a) Find the speed at x=0 without friction. m=5.00 kg
k=4.00x102 N/m
1 2 1 2 1 2 1 2
mvi  kxi  mv f  kx f
xi=0.0500 m
2
2
2
2
vi  0, x f  0
k 2
k
xi  v 2f  v f 
xi
m
m
 0.447 m/s
(b) Find the speed at x=xi/2.
2
i
kx2f
kx
 v 2f 
m
m
k 2
vf 
( xi  x 2f )  0.387 m/s
m
k=0
Spring Potential Energy

Examples
• Example 5.9: A horizontal spring (cont’d)
(c) Find the speed at x=0 with friction
1 2 1 2 1 2 1 2
W fric  mv f  mvi  kx f  kxi
2
2
2
2
1 2 1 2
  k nxi  mv f  kxi
2
2
1 2 1 2
mv f  kxi   k nxi
2
2
k 2
vf 
xi  2  k gxi
m
 0.230 m/s
m=5.00 kg
k=4.00x102 N/m
xi=0.0500 m
k= 0.150
Spring Potential Energy

Examples
• Example 5.10 : Circus acrobat
What is the max. compression
of the spring d?
( KE  PEg  PEs )i
 ( KE  PEg  PEs ) f
1
0  mg (h  d )  0  0  0  kd 2
2
d 2  (0.123 m) d - 0.245 m 2  0
d  0.560 m
m=50.0 kg
h =2.00 m
k = 8.00 x 103 N/m
Spring Potential Energy

Examples
• Example 5.11 : A block projected up a frictionless incline
(a) Find the max. distance
d the block travels up the
incline.
1 2
1 2
mvi  mgyi  kxi
2
2
1 2
1 2
 mv f  mgy f  kx f
2
2
m=0.500 kg
xi=10.0 cm
k=625 N/m
=30.0o


kxi2 / 2
1 2
kxi  mgh  mgd sin   d 
 1.28 m
2
mg sin 
(b) Find the velocity at hafl height h/2.
1 2 1 2
k 2
1 
kxi  mv f  mg  h   xi  v 2f  gh
2
2
m
2 
k 2
vf 
xi  gh  2.50 m/s
m
Spring Potential Energy

Systems and energy conservation
• Work-energy theorem
Wnc  Wc  DKE
• Consider changes in potential
Wnc  DKE  DPE  ( KE f  KEi )  ( PE f  PEi )
 ( KE f  PE f )  ( KEi  PEi )
 E f  Ei
The work done on a system by all non-conservative forces is
equal to the change in mechanical energy of the system.
If the mechanical energy is changing, it has to be going somewhere.
The energy either leaves the system and goes into the surrounding
environment, or stays in the system and is converted into nonmechanical form(s).
Systems and Energy Conservation

Forms of energy
• Forms of energy stored
kinetic, potential, internal energy
• Forms of energy transfer between a non-isolated system and its
environment
 Mechanical work :
transfers energy to a system by displacing it with a force.
 Heat : transfers energy through microscopic collisions between
atoms or molecules.
 Mechanical waves :
transfers energy by creating a disturbance that propagates
through a medium (air etc.).
 Electrical transmission :
transfers energy through electric currents.
 Electromagnetic radiation :
transfers energy in the form of electromagnetic waves such
as light, microwaves, and radio waves.
Systems and Energy Conservation

Energy conservation
• Principle of energy conservation:
Energy cannot be created or destroyed, only transferred from
one form to another.
• The principle of conservation of energy is not only true in physics
but also in other fields such as biology, chemistry, etc.
Power

Power
• The rate at which energy is transferred is important in the design
and use of practical devices such as electrical appliances and engines.
• If an external force is applied to an object and if the work done by this
force is W in time interval Dt, then the average power delivered to the
object during this interval is the work done divided by the time interval:
W
P
Dt
F Dx

 Fv
Dt
P  Fv
SI unit : watt (W) = J/s = 1 kg m2/s3
W=FDt
More general definition
U.S. customary unit : 1 hp = 550 ft lb/s
= 746 W
1 kWh = (103 W)(3600 s) = 3.60 x 105 J
Power

Examples
• Example 5.12 : Power delivered by an elevator
What is the min. power to lift the
elevator with the max. load?


F  ma

 

T  f  Mg  0
T  f  Mg  0
T  f  Mg  2.16 104 N
P  Fv  6.48 10 4 W
 64.8 kW  86.9 hp
M=1.00x103 kg
m=8.00x102 kg
f =4.00x103 N
v = 3.00 m/s
Power

Examples
• Example 5.14 : Speedboat power
How much power would a 1.00x103 kg speed boat need to go from
rest to 20.0 m/s in 5.00 s, assuming the water exerts a constant drag
force of magnitude fd=5.00x102 N and the acceleration is constant?
1
1
Wnet  DKE  mv 2f  mvi2
2
2
1
Wengine  Wdrag  Wengine  f d Dx  mv 2f
2
v f  at  vi  v f  at  a  4.00 m/s 2
v 2f  vi2  2aDx  50.0 m
W fric   f d Dx  2.50  10 4 J
1 2
Wengine  mv f  f d Dx  2.25  105 J
2
Wengine
P
 4.50  10 4 W  60.3 hp
Dt
Power

Energy and power in a vertical jump
• Center of mass (CM)
The point in an object at which all the may be considered to be
concentrated.
h=0.40 m depth of crouch
• Stationary jump
Dt=0.25 s time for extension
Two phases:
m=68 kg
(1) Extension, (2) free flight
PEi  KEi  PE f  KE f
2
vCM
1 2
mvCM  mgH  H 
2
2g
vCM  2v  2h / Dt  3.2 m/s
H  0.52 m
1 2
KE  mvCM
 3.5  10 2 J
2
KE
P
 1.4  103 W
Dt