Physics 111: Lecture 11
Today’s Agenda







Review
Work done by variable force in 3-D
 Newton’s gravitational force
Conservative forces & potential energy
Conservation of “total mechanical energy”
 Example: pendulum
Non-conservative forces
 friction
General work/energy theorem
Example problem
Physics 111: Lecture 11, Pg 1
Work by variable force in 3-D:

Work dWF of a force F acting
through an infinitesimal
displacement r is:
F
r
.
dW = F r

The work of a big displacement through a variable force will
be the integral of a set of infinitesimal displacements:

.
WTOT = F r
Physics 111: Lecture 11, Pg 2
Work by variable force in 3-D:
Newton’s Gravitational Force

Work dWg done on an object by gravity in a displacement dr is
given by:
.
.
^
^
dWg = Fg dr = (-GMm / R2 r)
(dR r ^+ Rd)
dWg = (-GMm / R2) dR
.
.
^
(since r ^ = 0, r ^r =^ 1)
^
dR
Rd
Fg

dr
m
r^
d
R
M
Physics 111: Lecture 11, Pg 3
Work by variable force in 3-D:
Newton’s Gravitational Force

Integrate dWg to find the total work done by gravity in a “big”
displacement:

R2

R2
Wg = dWg = (-GMm / R2) dR = GMm (1/R2 - 1/R1)
R1
R1
Fg(R2)
m
R2
Fg(R1)
R1
M
Physics 111: Lecture 11, Pg 4
Work by variable force in 3-D:
Newton’s Gravitational Force

Work done depends only on R1 and R2, not on the path taken.
 1
1 
Wg  GMm

R

 2 R1 
m
R2
R1
M
Physics 111: Lecture 11, Pg 5
Lecture 11, Act 1
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.
What is K2 / K1?
(a)
2
2RE
(b)
(c)
3
2
4
3
RE
RE
Physics 111: Lecture 11, Pg 6
Lecture 11, Act 1
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
Physics 111: Lecture 11, Pg 7
 1
1 
ΔK = c 
 
 R2 R1 
Lecture 11, Act 1
Solution
 1
1 
1 1
  c  

2 RE
 RE 2RE 

For the first rock: K1 = c 

For the second rock: K 2 = c 
 1
1 
2 1
  c  

3 RE
 RE 3RE 
So:
2
K2 3 4
= 
K1 1 3
2
2RE
RE
RE
Physics 111: Lecture 11, Pg 8
Newton’s Gravitational Force
Near the Earth’s Surface:

Suppose R1 = RE and R2 = RE + y
 R2  R1 
 RE  y   RE  
 GM 




Wg  GMm
 GMm
 m 2 y
 RR 
 R  y R  
R 
 1 2 


 E 
E
E
 GM 
but we have learned that  2   g
 RE 
So:
Wg = -mgy
m
RE+ y
RE
M
Physics 111: Lecture 11, Pg 9
Conservative Forces:

We have seen that the work done by gravity does not
depend on the path taken.
m
 1
1 

Wg  GMm

R

 2 R1 
R2
R1
M
m
h
Wg = -mgh
Physics 111: Lecture 11, Pg 10
Conservative Forces:

In general, if the work done does not depend on the path
taken, the force involved is said to be conservative.

Gravity is a conservative force:

Gravity near the Earth’s surface: Wg  mgy

A spring produces a conservative force:
 1
1 

Wg  GMm

R

R
 2
1 
Ws  
1
k x22  x12 
2
Physics 111: Lecture 11, Pg 11
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 = 0
W1

The work done can be “reclaimed.”
Physics 111: Lecture 11, Pg 12
Lecture 11, Act 2
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)
Physics 111: Lecture 11, Pg 13
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)
Physics 111: Lecture 11, Pg 14
Lecture 11, Act 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
W = -5 J
W=0
Physics 111: Lecture 11, Pg 15
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.
r2

U2
This can be written as:
F.dr
r2
U = U2 - U1 = -W = -
r1
r1
U1
Physics 111: Lecture 11, Pg 16
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
Physics 111: Lecture 11, Pg 17
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
Physics 111: Lecture 11, Pg 18
Potential Energy Recap:

For any conservative force we can define a potential
energy function U such that:
F.dr
S2
U = U2 - U1 = -W = -
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.
Physics 111: Lecture 11, Pg 19
Conservative Forces & Potential Energies
(stuff you should know):
Work
W(1-2)
Force
F
^
Fg = -mg j
Fg = 
GMm ^
r
2
R
Fs = -kx
-mg(y2-y1)
 1
1 

GMm

R

 2 R1 

1
k x22  x12 
2
Change in P.E
U = U2 - U1
mg(y2-y1)
P.E. function
U
mgy + C
 1
GMm
1 


C
 GMm

R

R
 2 R1 
1
k x22  x12 
2
1 2
kx  C
2
Physics 111: Lecture 11, Pg 20
Lecture 11, Act 3
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)
Physics 111: Lecture 11, Pg 21
Lecture 11, Act 3
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)
Physics 111: Lecture 11, Pg 22
Lecture 11, Act 3
Solution
1
2
2
2
The potential energy stored in (1) is 2  kd  kd
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)
Physics 111: Lecture 11, Pg 23
Conservation of Energy

If only conservative forces are present, the total kinetic
plus potential energy of a system is conserved.
E=K+U
E = K + U
= W + U
= W + (-W) = 0
using K = W
using U = -W
E = K + U is constant!!!

Both K and U can change, but E = K + U remains constant.
Physics 111: Lecture 11, Pg 24
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
Physics 111: Lecture 11, Pg 25
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
Physics 111: Lecture 11, Pg 26
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
Physics 111: Lecture 11, Pg 27
Example: The simple pendulum
 1/2mv2

will be maximum at the bottom of the swing.
1/ mv2 = mgh
So at y = 0
v2 = 2gh1
2
1
v 
2 gh1
y
y = h1
y=0
h1
v
Physics 111: Lecture 11, Pg 28
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
Physics 111: Lecture 11, Pg 29
Example: The simple pendulum

Bowling
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
Physics 111: Lecture 11, Pg 30
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
Physics 111: Lecture 11, Pg 31
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
Physics 111: Lecture 11, Pg 32
Example: The simple pendulum
 1/2mv2

will be maximum at the bottom of the swing.
1/ mv2 = mgh
So at y = -h1
v2 = 2gh1
2
1
v 
2 gh1
y
Same as before!
y=0
y = -h1
h1
v
Physics 111: Lecture 11, Pg 33
Example: The simple pendulum


Galileo’s
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!
Physics 111: Lecture 11, Pg 34
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
Physics 111: Lecture 11, Pg 35
Example: Airtrack & Glider


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
v
M
m
d
Physics 111: Lecture 11, Pg 36
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
Physics 111: Lecture 11, Pg 37
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
Physics 111: Lecture 11, Pg 38
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
Physics 111: Lecture 11, Pg 39
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!
Physics 111: Lecture 11, Pg 40
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
Physics 111: Lecture 11, Pg 41
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
Physics 111: Lecture 11, Pg 42
Generalized Work/Energy Theorem:

Suppose FNET = FC + FNC (sum of conservative and nonconservative 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 = E
Physics 111: Lecture 11, Pg 43
Generalized Work/Energy Theorem:
WNC = K + U = E

The change in kinetic+potential energy of a system is equal
to the work done on it by non-conservative forces. E=K+U
of system not conserved!
If all the forces are conservative, we know that K+U
energy is conserved: K + U = E = 0 which says that
WNC = 0, which makes sense.
If some non-conservative force (like friction) does work,
K+U energy will not be conserved and WNC = E, which
also makes sense.
Physics 111: Lecture 11, Pg 44
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
Physics 111: Lecture 11, Pg 45
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
Physics 111: Lecture 11, Pg 46
Recap of today’s lecture

Work done by variable force in 3-D
(Text: 6-1)
 Newton’s gravitational force
(Text: 11-2)
Conservative Forces & Potential energy
(Text: 6-4)
Conservation of “Total Mechanical Energy”
(Text: 7-1)
 Examples: pendulum, airtrack, Hotwheel car
Non-conservative forces
(Text: 6-4)
 friction
General work/energy theorem
(Text: 7-2)
Example problem

Look at Textbook problems





Chapter 7: # 9, 21, 27, 51, 77
Physics 111: Lecture 11, Pg 47