Download Potential energy and conservation of energy

Chapter 8
Potential energy and conservation
of energy
I. Potential energy  Energy of configuration
II. Work and potential energy
III. Conservative / Non-conservative forces
IV. Determining potential energy values:
- Gravitational potential energy
- Elastic potential energy
V. Conservation of mechanical energy
VI. External work and thermal energy
VII. External forces and internal energy changes
I. Potential energy
Energy associated with the arrangement of a system of
objects that exert forces on one another.
Units: 1J
Examples:
- Gravitational potential energy: associated with the state of
separation between objects which can attract one another
via the gravitational force.
- Elastic potential energy: associated with the state of
compression/extension of an elastic object.
II. Work and potential energy
If tomato rises  gravitational force
transfers energy “from” tomato’s kinetic
energy “to” the gravitational potential
energy of the tomato-Earth system.
If tomato falls down  gravitational force transfers
energy “from” the gravitational potential energy “to” the
tomato’s kinetic energy.
System Example
– This system consists
of Earth and a book
– Do work on the
system by lifting the
book through dy
– The work done is
mgyb - mgya
Potential Energy
― The energy storage mechanism is called
potential energy
― A potential energy can only be associated
with specific types of forces
― Potential energy is always associated with a
system of two or more interacting objects
Gravitational Potential Energy
− Gravitational Potential Energy is associated
with an object at a given distance above
Earth’s surface
− Assume the object is in equilibrium and
moving at constant velocity
− The work done on the object is done by Fapp
and the upward displacement is
r  yˆj
Gravitational Potential Energy
−
W   Fapp   r
W  (mgˆj)   yb  ya  ˆj
W  mgyb  mgya
− The quantity mgy is identified as the
gravitational potential energy, Ug
Ug = mgy
− Units are joules (J)
Gravitational Potential Energy
− The gravitational potential energy depends
only on the vertical height of the object above
Earth’s surface
− In solving problems, you must choose a
reference configuration for which the
gravitational potential energy is set equal to
some reference value, normally zero
• The choice is arbitrary because you normally need
the difference in potential energy, which is
independent of the choice of reference
configuration
Conservation of Mechanical Energy
• The mechanical energy of a system is the
algebraic sum of the kinetic and potential
energies in the system
Emech = K + Ug
• The statement of Conservation of Mechanical
Energy for an isolated system is
Kf + Uf = Ki+ Ui
– An isolated system is one for which there are no
energy transfers across the boundary
Conservation of Mechanical Energy
• Look at the work done by the
book as it falls from some
height to a lower height
Won book = ΔKbook
• Also,
W = mgyb – mgya
• So,
ΔK = -ΔUg
Elastic Potential Energy
• Elastic Potential Energy is associated with a
spring
• The force the spring exerts (on a block, for
example) is
Fs = - kx
• The work done by an external applied force on a
spring-block system is
W = ½ kxf2 – ½ kxi2
– The work is equal to the difference between the initial
and final values of an expression related to the
configuration of the system
Elastic Potential Energy
• This expression is the
elastic potential energy:
Us = ½ kx2
• The elastic potential
energy can be thought
of as the energy stored
in the deformed spring
• The stored potential
energy can be
converted into kinetic
energy
Elastic Potential Energy
• The elastic potential energy stored in a spring is
zero whenever the spring is not deformed (U =
0 when x = 0)
– The energy is stored in the spring only when the
spring is stretched or compressed
• The elastic potential energy is a maximum when
the spring has reached its maximum extension
or compression
• The elastic potential energy is always positive
– x2 will always be positive
General:
- System of two or more objects.
- A force acts between a particle in the system and the rest of
the system.
- When system configuration changes  force does work on
the object (W1) transferring energy between KE of the object
and some other form of energy of the system.
- When the configuration change is reversed  force reverses
the energy transfer, doing W2.
Problem Solving Strategy – Conservation
of Mechanical Energy
• Define the isolated system and the initial and
final configuration of the system
– The system may include two or more interacting
particles
– The system may also include springs or other
structures in which elastic potential energy can be
stored
– Also include all components of the system that
exert forces on each other
Problem-Solving Strategy
• Identify the configuration for zero
potential energy
– Include both gravitational and elastic
potential energies
– If more than one force is acting within the
system, write an expression for the
potential energy associated with each force
Problem-Solving Strategy
• If friction or air resistance is present,
mechanical energy of the system is not
conserved
• Use energy with non-conservative
forces instead
Problem-Solving Strategy
• If the mechanical energy of the system
is conserved, write the total energy as
Ei = Ki + Ui for the initial configuration
Ef = Kf + Uf for the final configuration
• Since mechanical energy is conserved,
Ei = Ef
and you can solve for the unknown quantity
Conservation of Energy
Example 1 (Drop a Ball)
• Initial conditions:
Ei = Ki + Ui = mgh
– The ball is dropped, so Ki
=0
• The configuration for
zero potential energy is
the ground
• Conservation rules
applied at some point y
above the ground gives
½ mvf2 + mgy = mgh
Conservation of Energy
Example 2 (Pendulum)
• As the pendulum swings,
there is a continuous change
between potential and kinetic
energies
• At A, the energy is potential
• At B, all of the potential
energy at A is transformed
into kinetic energy
– Let zero potential energy
be at B
• At C, the kinetic energy has
been transformed back into
potential energy
Conservation of Energy
Example 3 (Spring Gun)
• Choose point A as the initial
point and C as the final point
EA = EC
KA + UgA + UsA = KC + UgC + UsB
½ kx2 = mgh
U  W
Also valid for elastic potential energy
Spring compression
fs
Spring extension
fs
Spring force does –W on block
 energy transfer from kinetic
energy of the block to potential
elastic energy of the spring.
Spring force does +W on block 
energy transfer from potential
energy of the spring to kinetic
energy of the block.
III. Conservative / Nonconservative forces
- If W1=W2 always  conservative force.
Examples: Gravitational force and spring force  associated
potential energies.
- If W1≠W2  nonconservative force.
Examples: Drag force, frictional force  KE transferred into
thermal energy. Non-reversible process.
- Thermal energy: Energy associated with the random
movement of atoms and molecules. This is not a potential
energy.
- Conservative force: The net work it does on a particle
moving around every closed path, from an initial point and then
back to that point is zero.
- The net work it does on a particle moving between two
points does not depend on the particle’s path.
Conservative force  Wab,1= Wab,2
Proof:
Wab,1+ Wba,2=0  Wab,1= -Wba,2
Wab,2= - Wba,2
 Wab,2= Wab,1
IV. Determining potential energy values
xf
xi
W   F ( x)dx  U
Force F is conservative
Gravitational potential energy:
U   (mg )dy  mg y   mg ( y f  yi )  mgy
yf
yi
yf
yi
Change in the gravitational potential energy
of the particle-Earth system.
U i  0, yi  0  U ( y )  mgy
Reference configuration
The gravitational potential energy associated with particleEarth system depends only on particle’s vertical position “y”
relative to the reference position y=0, not on the horizontal
position.
Elastic potential energy:
 
k 2
U    (kx)dx 
x
2
xf
xi
xf
xi
1 2 1 2
 kx f  kxi
2
2
Change in the elastic potential energy of the spring-block
system.
Reference configuration  when the spring is at its relaxed
length and the block is at xi=0.
1 2
U i  0, xi  0  U ( x)  kx
2
Remember! Potential energy is always associated with a
system.
V. Conservation of mechanical energy
Mechanical energy of a system: Sum of the its potential (U)
and kinetic (K) energies.
Emec= U + K
Only conservative forces cause energy transfer
within the system.
The system is isolated from its environment  No external
force from an object outside the system causes energy
changes inside the system.
Assumptions:
W  K
W  U
K  U  0  ( K2  K1 )  (U 2  U1 )  0  K2  U 2  K1  U1
ΔEmec= ΔK + ΔU = 0
- In an isolated system where only conservative
forces cause energy changes, the kinetic energy and
potential energy can change, but their sum, the
mechanical energy of the system cannot change.
- When
the mechanical energy of a system is
conserved, we can relate the sum of kinetic energy
and potential energy at one instant to that at another
instant without considering the intermediate motion
and without finding the work done by the forces
involved.
A bead slides without friction around a loop-the-loop. The
bead is released from a height h = 3.50R.
(a) What is
its speed at point A? (b) How large is the normal force on
it if its mass is 5.00 g?
Emec= constant
y
x
Em ec  K  U  0
Potential energy curves
K 2  U 2  K1  U1
Finding the force analytically:
dU ( x)
U ( x)  W   F ( x)x  F ( x)  
(1D motion)
dx
- The force is the negative of the slope of the curve U(x)
versus x.
- The
particle’s kinetic energy is:
K(x) = Emec – U(x)
Turning point: a point
x at which the particle
reverses its motion
(K=0).
K always ≥0
(K=0.5mv2 ≥0 )
Examples:
x= x1 Emec= 5J=5J+K 
K=0
x<x1  Emec= 5J= >5J+K
 K<0  impossible
Equilibrium points:
where the slope of the U(x) curve is zero  F(x)=0
ΔU = -F(x) dx  ΔU/dx = -F(x)
ΔU(x)/dx = -F(x)
-F(x)  Slope
Emec,1
Emec,2
Equilibrium points
Emec,3
Example: x ≥ x5  Emec,1= 4J=4J+K  K=0 and also F=0
x5 neutral equilibrium
x2<x<x1, x5<x<x4  Emec,2= 3J= 3J+K  K=0  Turning points
x3  K=0, F=0  particle stationary  Unstable equilibrium
x4  Emec,3=1J=1J+K  K=0, F=0, it cannot move to x>x4 or
x<x4, since then K<0 
Stable equilibrium
VI. Work done on a system by an external force
Work is energy transfer “to” or
“from” a system by means of an
external force acting on that system.
When more than one force acts on
a system their net work is the energy
transferred to or from the system.
No Friction:
W = ΔEmec= ΔK+ ΔU  Ext. force
Remember!
Friction:
ΔEmec= ΔK+ ΔU = 0 only when:
- System isolated.
- No external forces act on a system.
- All internal forces are conservative.
F  f k  ma
v 2  v02  2ad  a  0.5(v 2  v02 ) / d
F  f k  ma
v  v  2ad  a  0.5(v  v ) / d
2
2
0
F  fk 
2
2
0
m 2 2
1
1
(v  v0 )  Fd  mv2  mv02  f k d
2d
2
2
W  Fd  K  f k d
General:
Example: Block sliding up a ramp.
W  Fd  Emec  f k d
A 15.7 kg block is dragged over a rough, horizontal
surface by a 72.2 N force acting at 21° above the horizontal.
The block is displaced 4.5 m, and the coefficient of kinetic
friction is 0.177. Find the work done on the block by (a) the
72.2 N force, (b) the normal force, and (c) the gravitational
force. (d) What is the increase in internal energy of the blocksurface system due to friction? (e) Find the total change in the
block's kinetic energy.
y
F
N F
21
Fy
fk
d
Fg
Fx
Thermal energy:
Eth  f k d
Friction due to cold welding
between two surfaces. As
the block slides over the
floor, the sliding causes
tearing and reforming of the
welds between the block and
the floor, which makes the
block-floor warmer.
Work done on a system by an external force, friction involved
W  Fd  Emec  Eth
VI. Conservation of energy
Total energy of a system = Emechanical + Ethermal + Einternal
- The total energy of a system can only change by amounts
of energy transferred “from” or “to” the system.
W  Emec  Eth  Eint
 Experimental law
-The total energy of an isolated system cannot change.
(There cannot be energy transfers to or from it).
Isolated system:
Emec  Eth  Eint  0
In an isolated system we can relate the total energy at one
instant to the total energy at another instant without
considering the energies at intermediate states.
VII. External forces and internal energy changes
Example: skater pushes herself away from a railing. There is a
force F on her from the railing that increases her kinetic energy.
One part of an object (skater’s arm)
does not move like the rest of body.
ii) Internal energy transfer (from one part
of the system to another) via the
external force F. Biochemical energy
from muscles transferred to kinetic
energy of the body.
i)
Change in system’s mechanical energy by an external force
WF ,ext  K  F (cos  )d
Non  isolated system 
K  U  WF ,ext  Fd cos 
Emec  Fd cos 
Eint  Em ec  0
Eint  Em ec   Fd cos 
Change in system’s
internal energy by a
external force
v 2  v02  2a x d (M )
Proof:
1
1
2
Mv  Mv02  Max d
2
2
K  Fd cos 
129. A massless rigid rod of length L has a ball of mass m attached to one end.
The other end is pivoted in such a way that the ball will move in a vertical
circle. First, assume that there is no friction at the pivot. The system is
launched downward from the horizontal position A with initial speed v0.
The ball just barely reaches point D and then stops.
D
y
A
L
C
x
v0
T
B
Fc
mg
7. A particle is attached between two identical springs on a horizontal
frictionless table. Both springs have spring constant k and are initially
unstressed. (a) If the particle is pulled a distance x along a direction
perpendicular to the initial configuration of the springs show that the force
exerted by the springs on the particle is

ˆ
L
F  2kx 1 
i
2
2

x  L 
(b) Determine the amount of work done by this force in moving the
particle from x = A to x = 0; (c) show that the potential energy of the
system is:

U (x )  kx  2kL L  x  L
2
2
2

61. In the figure below, a block slides along a path that is
without friction until the block reaches the section of length
L=0.75m, which begins at height h=2m. In that section, the
coefficient of kinetic friction is 0.4. The block passes through
point A with a speed of
8m/s. Does it reach point B
N
(where the section of
f
C
friction ends)? If so, what is
mg
the speed there and if not,
what greatest height above
point A does it reach?
101. A 3kg sloth hangs 3m above the ground. (a) What is the
gravitational potential energy of the sloth-Earth system if we
take the reference point y=0 to be at the ground? If the sloth
drops to the ground and air drag on it is assumed to be
negligible, what are (b) the kinetic energy and (c) the speed
of the sloth just before it reaches the ground?
130. A metal tool is sharpen by being held against the rim of
a wheel on a grinding machine by a force of 180N. The
frictional forces between the rim and the tool grind small
pieces of the tool. The wheel has a radius of 20cm and
rotates at 2.5 rev/s. The coefficient of kinetic friction between
the wheel and the tool is 0.32. At what rate is energy being
transferred from the motor driving the wheel and the tool to
the kinetic energy of the material thrown from the tool?
v
F=180N
82. A block with a kinetic energy of 30J is about to collide with a spring at its
relaxed length. As the block compresses the spring, a frictional force between
the block and floor acts on the block. The figure below gives the kinetic energy
of the block (K(x)) and the potential energy of the spring (U(x)) as a function of
the position x of the block, as the spring is compressed. What is the increase in
thermal energy of the block and the floor when (a) the block reaches position
0.1 m and (b) the spring reaches its maximum compression?
N
f
mg