Physics 121,Lecture 13,Pg 1
Physics 121,Sections 9,10,11,and 12
Lecture 13
Today’s Topics:
? Homework 6,due after midterm week
? Chapter 6,Work and Energy
? Definition of work
? Work-kinetic energy theorem
? Potential energy
? Non-conservative forces
? Generalized work-kinetic energy theorem
? Examples
Physics 121,Lecture 13,Pg 2
Definition of Work:
Ingredients,Force ( F ),displacement ( ? r )
Work,W,of a constant force F
acting through a displacement ? r is:
W = F ? r cos ? ?
F
? rF
r
Physics 121,Lecture 13,Pg 3
Units:
N-m (Joule) Dyne-cm (erg)
= 10-7 J
BTU = 1054 J
calorie = 4.184 J
foot-lb = 1.356 J
eV = 1.6x10-19 J
cgs othermks
Force x Distance = Work
Newton x
[M][L] / [T]2
Meter = Joule
[L] [M][L]2 / [T]2
Physics 121,Lecture 13,Pg 4
Work Kinetic-Energy Theorem:
{Net Work done on object}
=
{change in kinetic energy of object}
KW net ??
12 KK ??
2122 mv21mv21 ??
Physics 121,Lecture 13,Pg 5
Work done by Variable Force,(1D)
? When the force was constant,
we wrote W = F?x
?area under F vs x plot:
F
x
Wg
?x
F(x)
x1 x2 ?x ??
W ? F ( x )? x?
Physics 121,Lecture 13,Pg 6
Springs
? A very common problem with a variable force is a spring,
? In this spring,the force gets greater as the spring is further
compressed,
? Hook’s Law,
FS = - k ?x
?x is the amount the spring is stretched or compressed
from it resting position.
F
?x
Active Figure
Physics 121,Lecture 13,Pg 7
1-D Variable Force Example,Spring
? For a spring we know that Fx = -kx (Hook’s law).
F(x) x2
x
x1
-kxrelaxed position
F= - k x1
F= - k x2
Physics 121,Lecture 13,Pg 8
Spring...
? The work done by the spring Ws during a displacement
from x1 to x2 is the area under the F(x) vs x plot
between x1 and x2.
Ws
F(x) x2
x
x1
-kxrelaxed position
Physics 121,Lecture 13,Pg 9
Spring...
??
W s ? ? 12 k x 22 ? x 12? ?
? The work done by the spring Ws during a
displacement from x1 to x2 is the area under the
F(x) vs x plot between x1 and x2.
x2 x1
F(x)
x
Ws
kx1
kx2
-kx
Ws = - 1/2 [ ( kx2) (x2) - (kx1) (x1) ]
Physics 121,Lecture 13,Pg 10
Lecture 13,ACT 1
Work & Energy
? A box sliding on a horizontal frictionless surface runs into a
fixed spring,compressing it a distance x from its relaxed
position while momentarily coming to rest.
?If the initial speed of the box were doubled and its mass
were halved,how far x’ would the spring compress?
x
(a) (b) (c)x'x = x2'x = x2'x =
Physics 121,Lecture 13,Pg 11
Problem,Spring pulls on mass.
? A spring (constant k) is stretched a distance d,and a mass m
is hooked to its end,The mass is released (from rest),What
is the speed of the mass when it returns to the relaxed
position if it slides without friction?
relaxed position
stretched position (at rest)
d
after release
back at relaxed position
vr
v
m
m
m
m
Physics 121,Lecture 13,Pg 12
Problem,Spring pulls on mass.
? First find the net work done on the mass during the motion
from x= d to x= 0 (only due to the spring):
stretched position (at rest)
d
relaxed position
vr
m
m
i
??
W s ? ? 12 k x 22 ? x 12? ? ? ? 12 k 0 2 ? d 2? ? ? 12 kd 2
Physics 121,Lecture 13,Pg 13
Problem,Spring pulls on mass.
? Now find the change in kinetic energy of the mass:
stretched position (at rest)
d
relaxed position
vr
m
m
i
??
? KE ? 12 mv 22 ? 12 mv 12 ? 12 mv r2
Physics 121,Lecture 13,Pg 14
Problem,Spring pulls on mass.
? Now use work kinetic-energy theorem,
Wnet = WS = ? KE.
stretched position (at rest)
d
relaxed position
vr
m
m
i
1
2
2kd ?2
rmv21 m
kdv
r ?
Physics 121,Lecture 13,Pg 15
Problem,Spring pulls on mass.
? Now suppose there is a coefficient of friction ? between the
block and the floor
? The total work done on the block is now the sum of the work
done by the spring WS (same as before) and the work done
by friction Wf.
Wf = f,?r = - ? mg d
stretched position (at rest)
d
relaxed position
vr
m
m
i
f = ? mg
?r
Physics 121,Lecture 13,Pg 16
Problem,Spring pulls on mass.
? Again use Wnet = WS + Wf = ?KE
Wf = - ?mg d
stretched position (at rest)
d
relaxed position
vr
m
m
i
f = ?mg
?r
W kdS ? 12 2 2
rmv2
1K ??
2r2 mv21m gdkd21 ?? ?
??
v r ? km d 2 ? 2 ? gd
Physics 121,Lecture 13,Pg 17
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:
? A spring produces a conservative force:
??
W s ? ? 12 k x 22 ? x 12? ?
ymgW g ???
??
?
??
? ??
12
g R
1
R
1G M mW
Physics 121,Lecture 13,Pg 18
Conservative Forces:
? We have seen that the work done by a conservative force
does not depend on the path taken.
W1
The work done can be,reclaimed”.
W2
W1
W2
W1 = W2
WNET = W1 - W2 = 0
Therefore the work done in
a closed path is 0.
Physics 121,Lecture 13,Pg 19
Potential Energy
? For any conservative force F we can define a
potential energy function U in the following way:
?The work done by a conservative force is equal and
opposite to the change in the potential energy
function.
? This can be written as:
W = - ?U
?U = U2 - U1 = - W r1
r2 U2
U1
Physics 121,Lecture 13,Pg 20
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.
?y
m
Wg = -mg?y
j
Physics 121,Lecture 13,Pg 21
Gravitational Potential Energy
? So we see that the change in U near the earths surface is:
?U = - Wg = mg?y = mg(y2 -y1).
? So U = mgy + 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.
y1
m
Wg = -mg?y
j y2
Physics 121,Lecture 13,Pg 22
Potential Energy Recap:
? For any conservative force we can define a potential
energy function U such that:
? 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.
?U = U2 - U1 = - W
Physics 121,Lecture 13,Pg 23
Conservative Forces & Potential Energies
(stuff you should know):
Force
F
Work
W(1-2)
Change in P.E
?U = U2 - U1
P.E,function
U
Fg = -mg j
Fg = r
Fs = -kx
^
^
??
?
??
? ?
12 R
1
R
1G M m
? ?? ?12 22 12k x x
-mg(y2-y1) mg(y2-y1)
? ???
?
?
?
?
G M m
R R
1 1
2 1
? ?12 22 12k x x?
? G M mR 2
mgy + C
CRG Mm ??
Ckx ?221
Physics 121,Lecture 13,Pg 24
Lecture 13,ACT 2
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 121,Lecture 13,Pg 25
A Non-Conservative Force
Friction
? Looking down on an air-hockey table with no air,
Path 2
Path 1
For which path does friction do more work?
Physics 121,Lecture 13,Pg 26
A Non-Conservative Force
Path 2
Path 1
W1 = -?mg d1
W2 = -?mg d2
since d2 > d1,-W2 > -W1
Since |W2|>|W1| the puck
will be traveling slower at
the end of path 2,
Work done by a non-
conservative force takes
energy out of the system,
Physics 121,Lecture 13,Pg 27
Conservation of Energy
? If only conservative forces are present,the total energy
(sum of potential and kinetic energies) of a system is
conserved.
E = K + U
?E = ?K + ?U
= W + ?U
= W + (-W) = 0
using ?K = W
using ?U = -W
? Both K and U can change,but E = K + U remains constant.
E = K + U is constant !!! Active Figure
Physics 121,Lecture 13,Pg 28
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?
v
h1 h2
m
Physics 121,Lecture 13,Pg 29
Example,The simple pendulum.
? 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.
v
h1 h2
y
y=0
Physics 121,Lecture 13,Pg 30
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
y=h1
Physics 121,Lecture 13,Pg 31
Example,The simple pendulum.
? 1/2mv2 will be maximum at the bottom of the swing.
? So at y = 0 1/2mv2 = mgh1 v2 = 2gh1
v
h1
y
y=h1
v gh? 2 1
y=0
Physics 121,Lecture 13,Pg 32
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=h2
y=0
Physics 121,Lecture 13,Pg 33
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
Physics 121,Lecture 13,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?
d
M
m
v
v
Physics 121,Lecture 13,Pg 35
Example,Airtrack & Glider
? Energy is conserved since all forces are conservative.
? Choose initial configuration to have PE=0.
?KE = -?PE
d
M
m
v
? ?12 2m M v mg d? ?
Physics 121,Lecture 13,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.
h
d
v1
v2
Physics 121,Lecture 13,Pg 37
Problem,Hotwheel...
? Energy is conserved,so ?E = 0 ?KE = - ?PE
? Moving down a distance d,?PE = -mgd,?KE = 1/2mv12
? Solving for the speed:
hd v1
v gd1 2?
Physics 121,Lecture 13,Pg 38
Problem,Hotwheel...
? At the end,we are a distance d-h below our starting point.
? ?PE = -mg(d-h),?KE = 1/2mv22
? Solving for the speed:
hd
v2
? ?v g d h2 2? ?
d-h
Physics 121,Lecture 13,Pg 39
? What speed will skateboarder reach at bottom of the hill?
R=3 m
..
m = 25 kg
Initial,K1 = 0 U1 = mgR
Final,K2 = 1/2 mv2 U2 = 0
Conservation of
Total Energy,
Lecture 13,Example
Skateboard
K1 + U1 = K2 + U2
0 + mgR = 1/2mv2 + 0
v = (2gR)1/2
v ~ (2 x10m/s2 x 3m)1/2
v ~ 8 m/s (~16mph) !
Does NOT depend on the mass !
..
Physics 121,Lecture 13,Pg 40
Non-conservative Forces,
? If the work done does not depend on the path taken,the
force involved is said to be conservative.
? If the work done does depend on the path taken,the force
involved is said to be non-conservative.
? An example of a non-conservative force is friction:
? 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 121,Lecture 13,Pg 41
Non-conservative Forces,Friction
? Suppose you are pushing a box across a flat floor,The mass of
the box is m and the kinetic coefficient of friction is ?.
? The work done in pushing it a distance D is given by:
Wf = Ff D cos ?= -?mgD.
D
Ff = -?mg
Physics 121,Lecture 13,Pg 42
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.
A
B
path 1
path 2
Physics 121,Lecture 13,Pg 43
Generalized Work Energy Theorem:
? Suppose FNET = FC + FNC (sum of conservative and non-
conservative forces).
? The total work done is,WTOT = WC + WNC
? The Work Kinetic-Energy theorem says that,WTOT = ?K.
?WTOT = WC + WNC = ?K
? WNC = ?K - WC
? But WC = -?U
So WNC = ?K + ?U = ?E
Physics 121,Lecture 13,Pg 44
Generalized Work Energy Theorem:
? The change in total energy of a system is equal to the work
done on it by non-conservative forces,E of system not
conserved !
?If all the forces are conservative,we know that energy of
the system is conserved,?K + ?U = ?E = 0 which says that
WNC = 0,which makes sense.
?If some non-conservative force (like friction) does work,
energy of the system will not be conserved and WNC = ?E,
which also makes sense.
WNC = ?K + ?U = ?E
Physics 121,Lecture 13,Pg 45
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 kinetic coefficient of friction between the
block and the track is ?,
?How far,x,does the block go along the bottom
portion of the track before stopping?
x
d ?
Physics 121,Lecture 13,Pg 46
Problem,Block Sliding with Friction...
? Using WNC = ?K + ?U
? As before,?U= -mgd
? WNC = work done by friction = -?mgx.
? ?K = 0 since the block starts out and ends up at
rest.
? WNC = ?U -?mgx = -mgd
x = d / ?
x
d ?
Physics 121,Lecture 13,Pg 47
Lecture 13,ACT 3
Stones and Friction
? I throw a stone into the air,While in flight it feels
the force of gravity and the frictional force of the
air resistance,The time the stone takes to reach
the top of its flight path (i.e,go up) is,
A) larger than
B) equal to
C) less than
The time it takes to return from the top (i.e,go
down).
Physics 121,Lecture 13,Pg 48
? Let’s now suppose that the surface is not frictionless and
the same skateboarder reach the speed of 7.0 m/s at
bottom of the hill,What was the work done by friction on
the skateboarder?
R=3 m
..
m = 25 kg
Conservation of
Total Energy,
Lecture 13,Example
Skateboard
K1 + U1 = K2 + U2
Wf + 0 + mgR = 1/2mv2 + 0
Wf = 1/2mv 2 - mgR
Wf = (1/2 x25 kg x (7.0 m/s2)2 -
- 25 kg x 10m/s2 3 m)
Wf = 613 - 735 J = - 122 J
Total mechanical energy decreased by 122 J !
..
Wf +
Physics 121,Lecture 13,Pg 49
Work & Power:
? Two cars go up a hill,a BMW Z3 and me in my old Mazda
GLC,Both have the same mass,
? Assuming identical friction,both engines do the same
amount of work to get up the hill.
? Are the cars essentially the same?
? NO,The Z3 gets up the hill quicker
? It has a more powerful engine.
Physics 121,Lecture 13,Pg 50
Work & Power:
? Power is the rate at which work is done.
t
WP
?? dt
dWP ?
Instantaneous
Power:
Average
Power:
? A person of mass 80.0 kg walks up to 3rd floor (12.0m),If
he/she climbs in 20.0 sec what is the average power used,
W = mg h
W = 80.0kg 9.8m/s2 12.0 m = 9408 J
P = W / ?t
P = W / ?t = 9408 J / 20.0s = 470 W
Simple Example 1,
Units (SI) are
Watts (W):
1 W = 1 J / 1s
Physics 121,Lecture 13,Pg 51
Power
? We have seen that W = F cos ? ?r
F
?r
v
t
WP
??
? Average power:
? Units of power,J/sec = Nm/sec = Watts
?
? If the force does not depend on time:
??
P ? W? t ? F c o s ? ? r? t ? F c o s ? v
Physics 121,Lecture 13,Pg 52
Power,example
? A 2000 kg trolley is pulled up
a 30 degree hill at 20 mi/hr
by a winch at the top of the
hill,How much power is the
winch providing?
? The power is P = F cos ? v = T cos ? v
? Since the trolley is not accelerating,the net force on it
must be zero,In the x direction:
?T - mg sin ? = 0
?T = mg sin ?
?
v
mg
T
winch
x
y
Physics 121,Lecture 13,Pg 53
Power,example
? P = Tv
since T is parallel to v
? So P = mgv sin ?
v = 20 mi/hr = 8.93 m/s
g = 9.8 m/s2
m = 2000 kg
sin? = sin(30o) = 0.5
P = m g v sin ? =
(2000 kg)(9.8 m/s2)(8.93 m/s)(0.5) = 88,000,W
?
v
mg
T
winch
x
y
Physics 121,Lecture 13,Pg 54
Lecture 13,ACT 4
Power for Circular Motion
? I swing a sling shot over my head,The tension in the rope
keeps the shot moving in a circle,How much power must
be provided by me,through the rope tension,to keep the
shot in circular motion?
Note that Rope Length = 1m
Shot Mass = 1 kg
Angular frequency = 2 rads/sec v
A) 16 J/s B) 8 J/s C) 4 J/s D) 0
Physics 121,Lecture 13,Pg 55
Lecture 13,ACT 4
Power for Circular Motion
? Note that the string expends no power,i.e,does
no work,Makes sense?
? By the work – kinetic energy theorem,work done
equals change in kinetic energy.
? K = 1/2 mv2,thus since v doesn’t change,neither
does K,
? A force perpendicular to the direction of motion
does not change speed,v,and does no work.
v
T
Physics 121,Lecture 13,Pg 56
Recap of today’s lecture
? Homework 6,due after midterm week
? Chapter 6,Work and Energy
?Definition of work
?Work-kinetic energy theorem
?Potential energy
?Non-conservative forces
?Generalized work-kinetic energy theorem
?Examples
Physics 121,Sections 9,10,11,and 12
Lecture 13
Today’s Topics:
? Homework 6,due after midterm week
? Chapter 6,Work and Energy
? Definition of work
? Work-kinetic energy theorem
? Potential energy
? Non-conservative forces
? Generalized work-kinetic energy theorem
? Examples
Physics 121,Lecture 13,Pg 2
Definition of Work:
Ingredients,Force ( F ),displacement ( ? r )
Work,W,of a constant force F
acting through a displacement ? r is:
W = F ? r cos ? ?
F
? rF
r
Physics 121,Lecture 13,Pg 3
Units:
N-m (Joule) Dyne-cm (erg)
= 10-7 J
BTU = 1054 J
calorie = 4.184 J
foot-lb = 1.356 J
eV = 1.6x10-19 J
cgs othermks
Force x Distance = Work
Newton x
[M][L] / [T]2
Meter = Joule
[L] [M][L]2 / [T]2
Physics 121,Lecture 13,Pg 4
Work Kinetic-Energy Theorem:
{Net Work done on object}
=
{change in kinetic energy of object}
KW net ??
12 KK ??
2122 mv21mv21 ??
Physics 121,Lecture 13,Pg 5
Work done by Variable Force,(1D)
? When the force was constant,
we wrote W = F?x
?area under F vs x plot:
F
x
Wg
?x
F(x)
x1 x2 ?x ??
W ? F ( x )? x?
Physics 121,Lecture 13,Pg 6
Springs
? A very common problem with a variable force is a spring,
? In this spring,the force gets greater as the spring is further
compressed,
? Hook’s Law,
FS = - k ?x
?x is the amount the spring is stretched or compressed
from it resting position.
F
?x
Active Figure
Physics 121,Lecture 13,Pg 7
1-D Variable Force Example,Spring
? For a spring we know that Fx = -kx (Hook’s law).
F(x) x2
x
x1
-kxrelaxed position
F= - k x1
F= - k x2
Physics 121,Lecture 13,Pg 8
Spring...
? The work done by the spring Ws during a displacement
from x1 to x2 is the area under the F(x) vs x plot
between x1 and x2.
Ws
F(x) x2
x
x1
-kxrelaxed position
Physics 121,Lecture 13,Pg 9
Spring...
??
W s ? ? 12 k x 22 ? x 12? ?
? The work done by the spring Ws during a
displacement from x1 to x2 is the area under the
F(x) vs x plot between x1 and x2.
x2 x1
F(x)
x
Ws
kx1
kx2
-kx
Ws = - 1/2 [ ( kx2) (x2) - (kx1) (x1) ]
Physics 121,Lecture 13,Pg 10
Lecture 13,ACT 1
Work & Energy
? A box sliding on a horizontal frictionless surface runs into a
fixed spring,compressing it a distance x from its relaxed
position while momentarily coming to rest.
?If the initial speed of the box were doubled and its mass
were halved,how far x’ would the spring compress?
x
(a) (b) (c)x'x = x2'x = x2'x =
Physics 121,Lecture 13,Pg 11
Problem,Spring pulls on mass.
? A spring (constant k) is stretched a distance d,and a mass m
is hooked to its end,The mass is released (from rest),What
is the speed of the mass when it returns to the relaxed
position if it slides without friction?
relaxed position
stretched position (at rest)
d
after release
back at relaxed position
vr
v
m
m
m
m
Physics 121,Lecture 13,Pg 12
Problem,Spring pulls on mass.
? First find the net work done on the mass during the motion
from x= d to x= 0 (only due to the spring):
stretched position (at rest)
d
relaxed position
vr
m
m
i
??
W s ? ? 12 k x 22 ? x 12? ? ? ? 12 k 0 2 ? d 2? ? ? 12 kd 2
Physics 121,Lecture 13,Pg 13
Problem,Spring pulls on mass.
? Now find the change in kinetic energy of the mass:
stretched position (at rest)
d
relaxed position
vr
m
m
i
??
? KE ? 12 mv 22 ? 12 mv 12 ? 12 mv r2
Physics 121,Lecture 13,Pg 14
Problem,Spring pulls on mass.
? Now use work kinetic-energy theorem,
Wnet = WS = ? KE.
stretched position (at rest)
d
relaxed position
vr
m
m
i
1
2
2kd ?2
rmv21 m
kdv
r ?
Physics 121,Lecture 13,Pg 15
Problem,Spring pulls on mass.
? Now suppose there is a coefficient of friction ? between the
block and the floor
? The total work done on the block is now the sum of the work
done by the spring WS (same as before) and the work done
by friction Wf.
Wf = f,?r = - ? mg d
stretched position (at rest)
d
relaxed position
vr
m
m
i
f = ? mg
?r
Physics 121,Lecture 13,Pg 16
Problem,Spring pulls on mass.
? Again use Wnet = WS + Wf = ?KE
Wf = - ?mg d
stretched position (at rest)
d
relaxed position
vr
m
m
i
f = ?mg
?r
W kdS ? 12 2 2
rmv2
1K ??
2r2 mv21m gdkd21 ?? ?
??
v r ? km d 2 ? 2 ? gd
Physics 121,Lecture 13,Pg 17
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:
? A spring produces a conservative force:
??
W s ? ? 12 k x 22 ? x 12? ?
ymgW g ???
??
?
??
? ??
12
g R
1
R
1G M mW
Physics 121,Lecture 13,Pg 18
Conservative Forces:
? We have seen that the work done by a conservative force
does not depend on the path taken.
W1
The work done can be,reclaimed”.
W2
W1
W2
W1 = W2
WNET = W1 - W2 = 0
Therefore the work done in
a closed path is 0.
Physics 121,Lecture 13,Pg 19
Potential Energy
? For any conservative force F we can define a
potential energy function U in the following way:
?The work done by a conservative force is equal and
opposite to the change in the potential energy
function.
? This can be written as:
W = - ?U
?U = U2 - U1 = - W r1
r2 U2
U1
Physics 121,Lecture 13,Pg 20
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.
?y
m
Wg = -mg?y
j
Physics 121,Lecture 13,Pg 21
Gravitational Potential Energy
? So we see that the change in U near the earths surface is:
?U = - Wg = mg?y = mg(y2 -y1).
? So U = mgy + 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.
y1
m
Wg = -mg?y
j y2
Physics 121,Lecture 13,Pg 22
Potential Energy Recap:
? For any conservative force we can define a potential
energy function U such that:
? 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.
?U = U2 - U1 = - W
Physics 121,Lecture 13,Pg 23
Conservative Forces & Potential Energies
(stuff you should know):
Force
F
Work
W(1-2)
Change in P.E
?U = U2 - U1
P.E,function
U
Fg = -mg j
Fg = r
Fs = -kx
^
^
??
?
??
? ?
12 R
1
R
1G M m
? ?? ?12 22 12k x x
-mg(y2-y1) mg(y2-y1)
? ???
?
?
?
?
G M m
R R
1 1
2 1
? ?12 22 12k x x?
? G M mR 2
mgy + C
CRG Mm ??
Ckx ?221
Physics 121,Lecture 13,Pg 24
Lecture 13,ACT 2
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 121,Lecture 13,Pg 25
A Non-Conservative Force
Friction
? Looking down on an air-hockey table with no air,
Path 2
Path 1
For which path does friction do more work?
Physics 121,Lecture 13,Pg 26
A Non-Conservative Force
Path 2
Path 1
W1 = -?mg d1
W2 = -?mg d2
since d2 > d1,-W2 > -W1
Since |W2|>|W1| the puck
will be traveling slower at
the end of path 2,
Work done by a non-
conservative force takes
energy out of the system,
Physics 121,Lecture 13,Pg 27
Conservation of Energy
? If only conservative forces are present,the total energy
(sum of potential and kinetic energies) of a system is
conserved.
E = K + U
?E = ?K + ?U
= W + ?U
= W + (-W) = 0
using ?K = W
using ?U = -W
? Both K and U can change,but E = K + U remains constant.
E = K + U is constant !!! Active Figure
Physics 121,Lecture 13,Pg 28
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?
v
h1 h2
m
Physics 121,Lecture 13,Pg 29
Example,The simple pendulum.
? 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.
v
h1 h2
y
y=0
Physics 121,Lecture 13,Pg 30
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
y=h1
Physics 121,Lecture 13,Pg 31
Example,The simple pendulum.
? 1/2mv2 will be maximum at the bottom of the swing.
? So at y = 0 1/2mv2 = mgh1 v2 = 2gh1
v
h1
y
y=h1
v gh? 2 1
y=0
Physics 121,Lecture 13,Pg 32
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=h2
y=0
Physics 121,Lecture 13,Pg 33
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
Physics 121,Lecture 13,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?
d
M
m
v
v
Physics 121,Lecture 13,Pg 35
Example,Airtrack & Glider
? Energy is conserved since all forces are conservative.
? Choose initial configuration to have PE=0.
?KE = -?PE
d
M
m
v
? ?12 2m M v mg d? ?
Physics 121,Lecture 13,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.
h
d
v1
v2
Physics 121,Lecture 13,Pg 37
Problem,Hotwheel...
? Energy is conserved,so ?E = 0 ?KE = - ?PE
? Moving down a distance d,?PE = -mgd,?KE = 1/2mv12
? Solving for the speed:
hd v1
v gd1 2?
Physics 121,Lecture 13,Pg 38
Problem,Hotwheel...
? At the end,we are a distance d-h below our starting point.
? ?PE = -mg(d-h),?KE = 1/2mv22
? Solving for the speed:
hd
v2
? ?v g d h2 2? ?
d-h
Physics 121,Lecture 13,Pg 39
? What speed will skateboarder reach at bottom of the hill?
R=3 m
..
m = 25 kg
Initial,K1 = 0 U1 = mgR
Final,K2 = 1/2 mv2 U2 = 0
Conservation of
Total Energy,
Lecture 13,Example
Skateboard
K1 + U1 = K2 + U2
0 + mgR = 1/2mv2 + 0
v = (2gR)1/2
v ~ (2 x10m/s2 x 3m)1/2
v ~ 8 m/s (~16mph) !
Does NOT depend on the mass !
..
Physics 121,Lecture 13,Pg 40
Non-conservative Forces,
? If the work done does not depend on the path taken,the
force involved is said to be conservative.
? If the work done does depend on the path taken,the force
involved is said to be non-conservative.
? An example of a non-conservative force is friction:
? 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 121,Lecture 13,Pg 41
Non-conservative Forces,Friction
? Suppose you are pushing a box across a flat floor,The mass of
the box is m and the kinetic coefficient of friction is ?.
? The work done in pushing it a distance D is given by:
Wf = Ff D cos ?= -?mgD.
D
Ff = -?mg
Physics 121,Lecture 13,Pg 42
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.
A
B
path 1
path 2
Physics 121,Lecture 13,Pg 43
Generalized Work Energy Theorem:
? Suppose FNET = FC + FNC (sum of conservative and non-
conservative forces).
? The total work done is,WTOT = WC + WNC
? The Work Kinetic-Energy theorem says that,WTOT = ?K.
?WTOT = WC + WNC = ?K
? WNC = ?K - WC
? But WC = -?U
So WNC = ?K + ?U = ?E
Physics 121,Lecture 13,Pg 44
Generalized Work Energy Theorem:
? The change in total energy of a system is equal to the work
done on it by non-conservative forces,E of system not
conserved !
?If all the forces are conservative,we know that energy of
the system is conserved,?K + ?U = ?E = 0 which says that
WNC = 0,which makes sense.
?If some non-conservative force (like friction) does work,
energy of the system will not be conserved and WNC = ?E,
which also makes sense.
WNC = ?K + ?U = ?E
Physics 121,Lecture 13,Pg 45
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 kinetic coefficient of friction between the
block and the track is ?,
?How far,x,does the block go along the bottom
portion of the track before stopping?
x
d ?
Physics 121,Lecture 13,Pg 46
Problem,Block Sliding with Friction...
? Using WNC = ?K + ?U
? As before,?U= -mgd
? WNC = work done by friction = -?mgx.
? ?K = 0 since the block starts out and ends up at
rest.
? WNC = ?U -?mgx = -mgd
x = d / ?
x
d ?
Physics 121,Lecture 13,Pg 47
Lecture 13,ACT 3
Stones and Friction
? I throw a stone into the air,While in flight it feels
the force of gravity and the frictional force of the
air resistance,The time the stone takes to reach
the top of its flight path (i.e,go up) is,
A) larger than
B) equal to
C) less than
The time it takes to return from the top (i.e,go
down).
Physics 121,Lecture 13,Pg 48
? Let’s now suppose that the surface is not frictionless and
the same skateboarder reach the speed of 7.0 m/s at
bottom of the hill,What was the work done by friction on
the skateboarder?
R=3 m
..
m = 25 kg
Conservation of
Total Energy,
Lecture 13,Example
Skateboard
K1 + U1 = K2 + U2
Wf + 0 + mgR = 1/2mv2 + 0
Wf = 1/2mv 2 - mgR
Wf = (1/2 x25 kg x (7.0 m/s2)2 -
- 25 kg x 10m/s2 3 m)
Wf = 613 - 735 J = - 122 J
Total mechanical energy decreased by 122 J !
..
Wf +
Physics 121,Lecture 13,Pg 49
Work & Power:
? Two cars go up a hill,a BMW Z3 and me in my old Mazda
GLC,Both have the same mass,
? Assuming identical friction,both engines do the same
amount of work to get up the hill.
? Are the cars essentially the same?
? NO,The Z3 gets up the hill quicker
? It has a more powerful engine.
Physics 121,Lecture 13,Pg 50
Work & Power:
? Power is the rate at which work is done.
t
WP
?? dt
dWP ?
Instantaneous
Power:
Average
Power:
? A person of mass 80.0 kg walks up to 3rd floor (12.0m),If
he/she climbs in 20.0 sec what is the average power used,
W = mg h
W = 80.0kg 9.8m/s2 12.0 m = 9408 J
P = W / ?t
P = W / ?t = 9408 J / 20.0s = 470 W
Simple Example 1,
Units (SI) are
Watts (W):
1 W = 1 J / 1s
Physics 121,Lecture 13,Pg 51
Power
? We have seen that W = F cos ? ?r
F
?r
v
t
WP
??
? Average power:
? Units of power,J/sec = Nm/sec = Watts
?
? If the force does not depend on time:
??
P ? W? t ? F c o s ? ? r? t ? F c o s ? v
Physics 121,Lecture 13,Pg 52
Power,example
? A 2000 kg trolley is pulled up
a 30 degree hill at 20 mi/hr
by a winch at the top of the
hill,How much power is the
winch providing?
? The power is P = F cos ? v = T cos ? v
? Since the trolley is not accelerating,the net force on it
must be zero,In the x direction:
?T - mg sin ? = 0
?T = mg sin ?
?
v
mg
T
winch
x
y
Physics 121,Lecture 13,Pg 53
Power,example
? P = Tv
since T is parallel to v
? So P = mgv sin ?
v = 20 mi/hr = 8.93 m/s
g = 9.8 m/s2
m = 2000 kg
sin? = sin(30o) = 0.5
P = m g v sin ? =
(2000 kg)(9.8 m/s2)(8.93 m/s)(0.5) = 88,000,W
?
v
mg
T
winch
x
y
Physics 121,Lecture 13,Pg 54
Lecture 13,ACT 4
Power for Circular Motion
? I swing a sling shot over my head,The tension in the rope
keeps the shot moving in a circle,How much power must
be provided by me,through the rope tension,to keep the
shot in circular motion?
Note that Rope Length = 1m
Shot Mass = 1 kg
Angular frequency = 2 rads/sec v
A) 16 J/s B) 8 J/s C) 4 J/s D) 0
Physics 121,Lecture 13,Pg 55
Lecture 13,ACT 4
Power for Circular Motion
? Note that the string expends no power,i.e,does
no work,Makes sense?
? By the work – kinetic energy theorem,work done
equals change in kinetic energy.
? K = 1/2 mv2,thus since v doesn’t change,neither
does K,
? A force perpendicular to the direction of motion
does not change speed,v,and does no work.
v
T
Physics 121,Lecture 13,Pg 56
Recap of today’s lecture
? Homework 6,due after midterm week
? Chapter 6,Work and Energy
?Definition of work
?Work-kinetic energy theorem
?Potential energy
?Non-conservative forces
?Generalized work-kinetic energy theorem
?Examples
