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Physics : Lecture 09
(Chapter 8 Halliday)
 Conservative forces & potential energy
 Conservation of “total mechanical energy”
 Example: pendulum
 Non-conservative forces
 friction
 General work/energy theorem
 Example problem
Conservative Forces:
 In general, if the work done does not depend on the path
taken (only depends the initial and final distances between
objects), the force involved is said to be conservative.


 Gravity is a conservative force: Wg  GMm 1  1 


 R2
 Gravity near the Earth’s surface:
R1 
Wg  mgy
 A spring produces a conservative force: W   1 k x 2  x 2 
s
2
1
2
Conservative Forces:
 We have seen that the work done by a conservative force
does not depend on the path taken.
W2
W 1 = W2

W1
Therefore the work done in a
closed path is 0.
W2
WNET = W1 - W2
= W1 - W1 = 0
W1
Potential Energy
 For any conservative force F we can define a
potential energy function U in the following way:
W =
 F.dr = -U
 The work done by a conservative force is equal and opposite to
the change in the potential energy function.
 This can be written as:
U = U2 - U1 = -W = -
F.dr
r2
r2
r1
r1
U1
U2
Gravitational Potential Energy
 We have seen that the work done by gravity near the
Earth’s surface when an object of mass m is lifted a
distance y is
Wg = -mg y
 The change in potential energy of this object is therefore:
U = -Wg = mg y
j
m
y
Wg = -mg y
Gravitational Potential Energy
 So we see that the change in U near the Earth’s surface is:
U = -Wg = mg y = mg(y2 -y1).
 So U = mg y + U0 where U0 is an arbitrary constant.
 Having an arbitrary constant U0 is equivalent to saying that we
can choose the y location where U = 0 to be anywhere we
want to.
j
m
y2
y1
Wg = -mg y
Conservative Forces:
 We have seen that the work done by gravity does not depend
on the path taken.
m
 1
1 

Wg  GMm


R
R
 2
1 
R2
R1
M
m
h
Wg = -mgh
Work & Energy
 A rock is dropped from a distance RE above the
surface of the earth, and is observed to have kinetic
energy K1 when it hits the ground. An identical rock
is dropped from twice the height (2RE) above the
earth’s surface and has kinetic energy K2 when it hits.
RE is the radius of the earth.
(a)
2
 What
(b)
(c)
is K2 / K1?
2RE
3
2
4
3
RE
RE
Solution
 Since energy is conserved, K = WG.
 1
1 
WG = GMm
 
 R2 R1 
 1
1 
ΔK = c 
 
 R2 R1 
Where c = GMm is the
same for both rocks
2RE
RE
RE
 1
1 
ΔK = c 
 
 R2 R1 
Solution
 For the first

 1
1 
1 1


K
=
c


c


rock: 1  RE 2RE 
2 RE
 1
1 
2 1
  c  

3 RE
 RE 3RE 
For the second rock: K 2 = c 
So:
2
K2 3 4
= 
K1 1 3
2
2RE
RE
RE
Conservative Forces:
 We have seen that the work done by a conservative
force does not depend on the path taken.
W2
W 1 = W2

Therefore the work done in a
closed path is 0.
W1
W2
WNET = W1 - W2
= W1 - W1 = 0
W1
Conservative Forces
 The pictures below show force vectors at different points in space for two
forces. Which one is conservative ?
(a) 1
(b) 2
y
(c) both
y
x
(1)
x
(2)
Lecture 11, Act 2
Solution
 Consider the work done by force when moving along different paths in each
case:
WA = WB
(1)
WA > WB
(2)
 In fact, you could make money on type (2) if it
ever existed:
 Work done by this force in a “round trip” is >
0!
 Free kinetic energy!!
WNET = 10 J = K
W=0
W = 15 J
Note: NO REAL
FORCES OF THIS
TYPE EXIST, SO
FAR AS WE KNOW
W = -5 J
W=0
Elastic Potential Energy
Energy stored in a stretched or compressed spring.
Work Done By Spring
(Elastic) Force:
Fe = -kx
kx2
We  
2
Choosing Uo = 0 for x = 0 (unstretched position), we have:
U e  We
kx2
Ue 
2
Potential Energy Recap:
 For any conservative force we can define a potential energy
function U such that:
U = U2 - U1 = -W = -

S2
.
F dr
S1
 The potential energy function U is always defined only
up to an additive constant.
 You can choose the location where U = 0 to be anywhere
convenient.
Conservative Forces & Potential
Energies
Work
W(1-2)
Force
F
^
Fg = -mg j
Fg = 
GMm ^
r
2
R
Fs = -kx
Change in P.E
U = U2 - U1
-mg(y2-y1)
mg(y2-y1)
 1
1 

GMm


R
R
 2
1 
 1
1 

 GMm


 R2 R1 
1
 k x22  x12 
2
1
k x22  x12 
2
P.E. function
U
mgy + C

GMm
C
R
1 2
kx  C
2
(R is the center-to-center distance, x is the spring stretch)
Potential Energy
 All springs and masses are identical. (Gravity acts down).
 Which of the systems below has the most potential energy stored in its spring(s),
relative to the relaxed position?
(a) 1
(b) 2
(c) same
(1)
(2)
Solution
 The displacement of (1) from equilibrium will be half of that of (2) (each spring
exerts half of the force needed to balance mg)
0
d
2d
(1)
(2)
Solution
1
The potential energy stored in (1) is 2  kd2  kd2
2
The potential energy stored in (2) is
1
2
k2d  2kd2
2
The spring P.E. is
twice as big in (2) !
0
d
2d
(1)
(2)
Potential Energy Summary
Can be defined only in a conservative field of forces.
Mathematical definition:
2
ds
1
F
 
dU  W  F  ds
Potential Energy
Potential Energy and Field Force
For a conservative force in one dimension:
dU = - W = - Fxdx
Therefore, the force can be expressed as the negative derivative
of the potential energy (- gradient of potential).
The field force will act in the direction of the drop of
potential energy with distance.
Let’s illustrate for a spring-mass system:
dU
d kx2
F x 
 (
)  k 2 x / 2  kx
dx
dx 2
Potential Energy and Equilibrium
Let’s illustrate for a spring and mass system:
First lets note that the slope is the derivative of the function.
The force being the negative derivative of the potential energy
U, is going to be also equal to the negative of the slope.
At x = 0, Fx = 0 and the
mass is in equilibrium,
slope = 0.
Stable
Equilibrium
Curve has a minimum
If x>0, slope>0, F<0
If x<0, slope<0, F>0
In either case the force is in the direction that will accelerate the
object towards lower potential energy.
A slight displaced from the equilibrium position x = 0, gives rise to
a force directed back toward x = 0.
Unstable Equilibrium
Curve has a maximum at x = 0 [example
skier (Avram) at the top of the hill]
F
If x>0, slope<0, F>0
If x<0, slope>0, F<0
Again the force moves the
object toward lower
potential energy, but away
from the equilibrium
position.
Neutral Equilibrium
Near x = 0 the region is flat.
Small displacements
near this location won’t
produce any net force,
and hence the particle is
still in equilibrium
Exercise1
dU
F r 
dr
dU/dr = slope <0
Fr>0
Ex.1
ΔK=-ΔU
mv2/2 = 3Uo-Uo = 2Uo
4U o
Uo
v
2
m
m
dU
F r 
dr
=0

U 03U 0 2U o

2ro  ro
r0
Ex.1
2Uo/ro
Ex. 2
Ex. 2
Area =( 4Fo + Fo) x ro/2 =
5/2 (Foro) = 2.5 Foro= W
ΔK = mv2/2 = 2.5Foro
v = √5Foro/m
o
 1
k
1
 1
W   2 dr  k     k  [
 ] 
r
 r  ro
 2ro ro 
ro
2 ro
F = k/r2
4 Fo 
k
2

k

4
F
r
o
o
ro2
ΔK = mv2/2 = 2Foro
4 Fo ro2
k

 2 Fo ro
2r o
2ro
v = √4Foro/m
2r
Ex.2
1 1 
k
1
U  U  0  W    2 dr  k    k   
r
 r  ro
 r ro 
ro
r
Gravity
r
Conservation of Energy
 If only conservative forces are present, the total kinetic plus potential energy of a
system is conserved,
i.e. the total “mechanical energy” is conserved.
 (note: E=Emechanical throughout this discussion)
E=K+U
 using K = W
E = K + U
 using U = -W
= W + U
= W + (-W) = 0 E = K + U is constant!!!
 Both K and U can change, but E = K + U remains constant.
 But we’ll see that if non-conservative forces act then energy can be dissipated into
other modes (thermal,sound)
Example: The simple pendulum
 Suppose we release a mass m from rest a distance h1 above its
lowest possible point.
 What is the maximum speed of the mass and where
does this happen?
 To what height h2 does it rise on the other side?
m
h1
h2
v
Example: The simple pendulum
 Kinetic+potential energy is conserved since gravity is a
conservative force (E = K + U is constant)
 Choose y = 0 at the bottom of the swing,
and U = 0 at y = 0 (arbitrary choice)
E = 1/2mv2 + mgy
y
y=0
h1
h2
v
Example: The simple pendulum
 E = 1/2mv2 + mgy.
 Initially, y = h1 and v = 0, so E = mgh1.
 Since E = mgh1 initially, E = mgh1 always since energy is conserved.
y
y=0
Example: The simple pendulum
 1/2mv2 will be maximum at the bottom of the swing.
 So at y = 0
1/
2 = mgh
mv
2
1
v 
2 gh1
y
y = h1
y=0
h1
v
v2 = 2gh1
Example: The simple pendulum
 Since E = mgh1 = 1/2mv2 + mgy it is clear that the
maximum height on the other side will be at y = h1 = h2
and v = 0.
 The ball returns to its original height.
y
y = h1 = h 2
y=0
Example: The simple pendulum
 The ball will oscillate back and forth. The limits on its height
and speed are a consequence of the sharing of energy
between K and U.
E = 1/2mv2 + mgy = K + U = constant.
y
Example: The simple pendulum
 We can also solve this by choosing y = 0 to be at the original
position of the mass, and U = 0 at y = 0.
E = 1/2mv2 + mgy.
y
y=0
h1
h2
v
Example: The simple pendulum
 E = 1/2mv2 + mgy.
 Initially, y = 0 and v = 0, so E = 0.
 Since E = 0 initially, E = 0 always since energy is
conserved.
y
y=0
Example: The simple pendulum
 1/2mv2 will be maximum at the bottom of the swing.
 So at y = -h1
1/
2mv
2
v2 = 2gh1
= mgh1
v 
2 gh1
y
Same as before!
y=0
y = -h1
h1
v
Example: The simple pendulum
 Since 1/2mv2 - mgh = 0 it is clear that the maximum height on
the other side will be at y = 0 and v = 0.
 The ball returns to its original height.
y
y=0
Same as before!
Example: Airtrack & Glider
 A glider of mass M is initially at rest on a horizontal
frictionless track. A mass m is attached to it with a massless
string hung over a massless pulley as shown. What is the
speed v of M after m has fallen a distance d ?
v
M
m
d
v
Example: Airtrack & Glider
 Kinetic+potential energy is conserved since all forces are conservative.
 Choose initial configuration to have U=0.
K = -U
1
m  M v 2  mgd
2
M
m
d
v
Problem: Hotwheel
 A toy car slides on the frictionless track shown below. It
starts at rest, drops a distance d, moves horizontally at speed
v1, rises a distance h, and ends up moving horizontally with
speed v2.
 Find v1 and v2.
v2
d
v1
h
Problem: Hotwheel...
 K+U energy is conserved, so E = 0
K = - U
 Moving down a distance d, U = -mgd, K = 1/2mv12
 Solving for the speed:
v1 
2 gd
d
v1
h
Problem: Hotwheel...
 At the end, we are a distance d - h below our starting point.
 U = -mg(d - h), K = 1/2mv22
 Solving for the speed:
v2 
2 g d  h 
d-h
d
v2
h
Non-conservative Forces:
 If the work done does not depend on the path taken, the
force is said to be conservative.
 If the work done does depend on the path taken, the force is
said to be non-conservative.
 An example of a non-conservative force is friction.
 When pushing a box across the floor, the amount of work that is
done by friction depends on the path taken.

Work done is proportional to the length of the path!
Energy dissipation: e.g. sliding
friction

As the parts scrape by each other
they start small-scale vibrations,
which transfer kinetic and potential
energy into atomic motions
The atoms’ vibrations go
back and forththey have energy,
but no average momentum.
Non-conservative Forces: Friction
 Suppose you are pushing a box across a flat floor. The mass
of the box is m and the coefficient of kinetic friction is k.
 The work done in pushing it a distance D is given by:
Wf = Ff • D = -kmgD.
Ff = -kmg
D
Non-conservative Forces: Friction
 Since the force is constant in magnitude and opposite in
direction to the displacement, the work done in pushing the
box through an arbitrary path of length L is just
Wf = -mgL.
 Clearly, the work done depends on the path taken.
 Wpath 2 >Wpath 1
B
path 1
path 2
A
Generalized Work/Energy Theorem:
 Suppose FNET = FC + FNC (sum of conservative and non-
conservative forces).
 The total work done is: WNET =WC + WNC
 The Work/Kinetic Energy theorem says that: WNET = K.
 WNET =WC +WNC = K

WNC = K -WC
 But WC = -U
So
WNC = K + U = Emechanical
Generalized Work/Energy Theorem:
WNC = K + U = Emechanical
 The change in kinetic+potential energy of a system is equal to the work
done on it by non-conservative forces.
conserved!
Emechanical =K+U of system not
 If all the forces are conservative, we know that K+U energy is
conserved: K + U = Emechanical = 0 which says that WNC = 0, which
makes sense.
 If some non-conservative force (like friction, a “push” or a “pull”) does
work, K+U energy will not be conserved and WNC = E, which also
makes sense.
Problem: Block Sliding with Friction
 A block slides down a frictionless ramp. Suppose the horizontal (bottom) portion
of the track is rough, such that the coefficient of kinetic friction between the
block and the track is k.
 How far, x, does the block go along the bottom portion of the track before
stopping?
d
k
x
Problem: Block Sliding with Friction...
 Using WNC = K + U
 As before, U = -mgd
 WNC = work done by friction = -kmgx.
 K = 0 since the block starts out and ends up at rest.
 WNC = U
-kmgx = -mgd
x = d / k
d
k
x
Recap of today’s lecture
 Conservative forces & potential energy - review
 Conservation of “Total Mechanical Energy”
 Examples: pendulum, airtrack, Hotwheel car
 Non-conservative forces
 friction
 General work/energy theorem
 Example problem
 Look at Textbook problems
7.3.1. Which one of the following choices represents the largest
kinetic energy?
a) Mars is moving in its orbit around the Sun.
b) A cyclist is racing in the annual Tour de France bicycle race.
c) A leaf falls from a tree.
d) A cheetah runs at its maximum speed to catch a fleeing zebra.
e) An oil tanker sails through the Panama Canal.
7.3.1. Which one of the following choices represents the largest
kinetic energy?
a) Mars is moving in its orbit around the Sun.
b) A cyclist is racing in the annual Tour de France bicycle race.
c) A leaf falls from a tree.
d) A cheetah runs at its maximum speed to catch a fleeing zebra.
e) An oil tanker sails through the Panama Canal.
7.3.3. Ignoring friction effects, the amount of energy required to
accelerate a car from rest to a speed v is E. The energy is delivered
to the car by burning gasoline. What additional amount of energy
is required to accelerate the car to a speed 2v?
a) 0.5E
b) E
c) 2E
d) 3E
e) 4E
7.3.3. Ignoring friction effects, the amount of energy required to
accelerate a car from rest to a speed v is E. The energy is delivered
to the car by burning gasoline. What additional amount of energy
is required to accelerate the car to a speed 2v?
a) 0.5E
b) E
c) 2E
d) 3E
e) 4E
7.5.1. A block is in contact with a rough surface as shown in the drawing. The block has a
rope attached to one side. Someone pulls the rope with a force , which is represented by
the vector in the drawing. The force is directed at an angle  with respect to the
horizontal direction. The magnitude of is equal to two times the magnitude of the
frictional force, which is designated f. For what value of  is the net work on the block
equal to zero joules?
a) 0
b) 30
c) 45
d) 60
e) Net work will be done in the object for all values of .
7.5.1. A block is in contact with a rough surface as shown in the drawing. The block has a
rope attached to one side. Someone pulls the rope with a force , which is represented by
the vector in the drawing. The force is directed at an angle  with respect to the
horizontal direction. The magnitude of is equal to two times the magnitude of the
frictional force, which is designated f. For what value of  is the net work on the block
equal to zero joules?
a) 0
b) 30
c) 45
d) 60
e) Net work will be done in the object for all values of .
7.5.4. Determine the amount of work done in firing a 2.0-kg projectile
with an initial speed of 50 m/s. Neglect any effects due to air
resistance.
a) 900 J
b) 1600 J
c) 2500 J
d) 4900 J
e) This cannot be determined without knowing the launch angle.
7.5.4. Determine the amount of work done in firing a 2.0-kg projectile
with an initial speed of 50 m/s. Neglect any effects due to air
resistance.
a) 900 J
b) 1600 J
c) 2500 J
d) 4900 J
e) This cannot be determined without knowing the launch angle.
7.6.1. An elevator supported by a single cable descends a shaft at a
constant speed. The only forces acting on the elevator are the
tension in the cable and the gravitational force. Which one of the
following statements is true?
a) The work done by the tension force is zero joules.
b) The net work done by the two forces is zero joules.
c) The work done by the gravitational force is zero joules.
d) The magnitude of the work done by the gravitational force is larger
than that done by the tension force.
e) The magnitude of the work done by the tension force is larger than
that done by the gravitational force.
7.6.1. An elevator supported by a single cable descends a shaft at a
constant speed. The only forces acting on the elevator are the
tension in the cable and the gravitational force. Which one of the
following statements is true?
a) The work done by the tension force is zero joules.
b) The net work done by the two forces is zero joules.
c) The work done by the gravitational force is zero joules.
d) The magnitude of the work done by the gravitational force is larger
than that done by the tension force.
e) The magnitude of the work done by the tension force is larger than
that done by the gravitational force.
7.6.2. A mountain climber pulls a supply pack up the side of a
mountain at constant speed. Which one of the following
statements concerning this situation is false?
a) The net work done by all the forces acting on the pack is zero
joules.
b) The work done on the pack by the normal force of the mountain is
zero joules.
c) The work done on the pack by gravity is zero joules.
d) The gravitational potential energy of the pack is increasing.
e) The climber does "positive" work in pulling the pack up the
mountain.
7.6.2. A mountain climber pulls a supply pack up the side of a
mountain at constant speed. Which one of the following
statements concerning this situation is false?
a) The net work done by all the forces acting on the pack is zero
joules.
b) The work done on the pack by the normal force of the mountain is
zero joules.
c) The work done on the pack by gravity is zero joules.
d) The gravitational potential energy of the pack is increasing.
e) The climber does "positive" work in pulling the pack up the
mountain.
7.6.3. Two balls of equal size are dropped from the same height from the
roof of a building. One ball has twice the mass of the other. When the
balls reach the ground, how do the kinetic energies of the two balls
compare?
a) The lighter one has one fourth as much kinetic energy as the other does.
b) The lighter one has one half as much kinetic energy as the other does.
c) The lighter one has the same kinetic energy as the other does.
d) The lighter one has twice as much kinetic energy as the other does.
e) The lighter one has four times as much kinetic energy as the other does.
7.6.3. Two balls of equal size are dropped from the same height from the
roof of a building. One ball has twice the mass of the other. When the
balls reach the ground, how do the kinetic energies of the two balls
compare?
a) The lighter one has one fourth as much kinetic energy as the other does.
b) The lighter one has one half as much kinetic energy as the other does.
c) The lighter one has the same kinetic energy as the other does.
d) The lighter one has twice as much kinetic energy as the other does.
e) The lighter one has four times as much kinetic energy as the other does.
7.6.4. Consider the box in the drawing. We can slide the box up the frictionless incline from
point A and to point C or we can slide it along the frictionless horizontal surface from
point A to point B and then lift it to point C. How does the work done on the box along
path A-C,WAC, compare to the work done on the box along the two step path A-B-C,
WABC?
a) WABC is much greater than WAC.
b) WABC is slightly greater than WAC.
c) WABC is much less than WAC.
d) WABC is slight less than WAC.
e) The work done in both cases is the same.
7.6.4. Consider the box in the drawing. We can slide the box up the frictionless incline from
point A and to point C or we can slide it along the frictionless horizontal surface from
point A to point B and then lift it to point C. How does the work done on the box along
path A-C,WAC, compare to the work done on the box along the two step path A-B-C,
WABC?
a) WABC is much greater than WAC.
b) WABC is slightly greater than WAC.
c) WABC is much less than WAC.
d) WABC is slight less than WAC.
e) The work done in both cases is the same.
7.7.1. Block A has a mass m and block B has a mass 2m. Block A is pressed
against a spring to compress the spring by a distance x. It is then
released such that the block eventually separates from the spring and it
slides across a surface where the friction coefficient is µk. The same
process is applied to block B. Which one of the following statements
concerning the distance that each block slides before stopping is correct?
a) Block A slides one-fourth the distance that block B slides.
b) Block A slides one-half the distance that block B slides.
c) Block A slides the same distance that block B slides.
d) Block A slides twice the distance that block B slides.
e) Block A slides four times the distance that block B slides.
7.7.1. Block A has a mass m and block B has a mass 2m. Block A is pressed
against a spring to compress the spring by a distance x. It is then
released such that the block eventually separates from the spring and it
slides across a surface where the friction coefficient is µk. The same
process is applied to block B. Which one of the following statements
concerning the distance that each block slides before stopping is correct?
a) Block A slides one-fourth the distance that block B slides.
b) Block A slides one-half the distance that block B slides.
c) Block A slides the same distance that block B slides.
d) Block A slides twice the distance that block B slides.
e) Block A slides four times the distance that block B slides.
7.7.2. In designing a spring loaded cannon, determine the spring
constant required to launch a 2.0 kg ball with an initial speed of
1.2 m/s from a position where the spring is displaced 0.15 m from
its equilibrium position.
a) 16 N/m
b) 32 N/m
c) 64 N/m
d) 130 N/m
e) 180 N/m
7.7.2. In designing a spring loaded cannon, determine the spring
constant required to launch a 2.0 kg ball with an initial speed of
1.2 m/s from a position where the spring is displaced 0.15 m from
its equilibrium position.
a) 16 N/m
b) 32 N/m
c) 64 N/m
d) 130 N/m
e) 180 N/m
7.8.1. A 12 500-kg truck is accelerated from rest by a net force that
decreases linearly with distance traveled. The graph shows this
force. Using the information provided and work-energy methods,
determine the approximate speed of the truck when the force is
removed.
a) 8.41 m/s
b) 12.5 m/s
c) 17.7 m/s
d) 25.0 m/s
e) 35.4 m/s
7.8.1. A 12 500-kg truck is accelerated from rest by a net force that
decreases linearly with distance traveled. The graph shows this
force. Using the information provided and work-energy methods,
determine the approximate speed of the truck when the force is
removed.
a) 8.41 m/s
b) 12.5 m/s
c) 17.7 m/s
d) 25.0 m/s
e) 35.4 m/s
7.8.2. A net force given by F  12 N  (1.4 x N) î is applied to an object
that is initially at rest. What is the change in the object’s kinetic
energy as it moves from x1 = 0.50 m to x2 = 2.50 m?
a) 14 J
b) 17 J
c) 21 J
d) 24 J
e) 28 J
7.8.2. A net force given by F  12 N  (1.4 x N) î is applied to an object
that is initially at rest. What is the change in the object’s kinetic
energy as it moves from x1 = 0.50 m to x2 = 2.50 m?
a) 14 J
b) 17 J
c) 21 J
d) 24 J
e) 28 J