Physics 121,Lecture 14,Pg 1
Physics 121,Sections 9,10,11,and 12
Lecture 14
Today’s Topics:
? Homework 6,
? Chap,6,# 6,12,20,24,29,38,52,57,78,and 83.
? Midterm 1,solutions
? Chapter 6,Work and Energy
? Review,Kinetic and Potential energy
? Non-conservative forces
? Generalized work-kinetic energy theorem
? Power
Physics 121,Lecture 14,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 14,Pg 3
Work Kinetic-Energy Theorem:
{Net Work done on object}
=
{change in kinetic energy of object}
KW net ??
12 KK ??
2122 mv21mv21 ??
Physics 121,Lecture 14,Pg 4
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 14,Pg 5
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 14,Pg 6
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 14,Pg 7
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 14,Pg 8
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 14,Pg 9
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 14,Pg 10
? 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 14,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 14,Pg 11
? What would be the speed if instead the skateboarder jumps
to the ground on the other side?
..
..
R=3 m
KINEMATICS:
* v = a t
* h = 1/2 at2
2h = v2/a
v = (2 h a)1/2 = (2 R g)1/2
the same magnitude as before !
and independent of mass
Lecture 14,Example
Skateboard
Physics 121,Lecture 14,Pg 12
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 14,Pg 13
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 14,Pg 14
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 14,Pg 15
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 14,Pg 16
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 14,Pg 17
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 14,Pg 18
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 14,Pg 19
Lecture 14,ACT 1
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 14,Pg 20
? 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 14,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 14,Pg 21
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 14,Pg 22
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 14,Pg 23
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 14,Pg 24
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 14,Pg 25
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 14,Pg 26
Lecture 14,ACT 2
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 14,Pg 27
Recap of today’s lecture
? Homework 6,
?Chap,6,# 6,12,20,24,29,38,52,57,78,and 83.
? Midterm 1,solutions
? Chapter 6,Work and Energy
?Review,Kinetic and Potential energy
?Non-conservative forces
?Generalized work-kinetic energy theorem
?Power
Physics 121,Sections 9,10,11,and 12
Lecture 14
Today’s Topics:
? Homework 6,
? Chap,6,# 6,12,20,24,29,38,52,57,78,and 83.
? Midterm 1,solutions
? Chapter 6,Work and Energy
? Review,Kinetic and Potential energy
? Non-conservative forces
? Generalized work-kinetic energy theorem
? Power
Physics 121,Lecture 14,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 14,Pg 3
Work Kinetic-Energy Theorem:
{Net Work done on object}
=
{change in kinetic energy of object}
KW net ??
12 KK ??
2122 mv21mv21 ??
Physics 121,Lecture 14,Pg 4
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 14,Pg 5
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 14,Pg 6
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 14,Pg 7
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 14,Pg 8
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 14,Pg 9
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 14,Pg 10
? 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 14,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 14,Pg 11
? What would be the speed if instead the skateboarder jumps
to the ground on the other side?
..
..
R=3 m
KINEMATICS:
* v = a t
* h = 1/2 at2
2h = v2/a
v = (2 h a)1/2 = (2 R g)1/2
the same magnitude as before !
and independent of mass
Lecture 14,Example
Skateboard
Physics 121,Lecture 14,Pg 12
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 14,Pg 13
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 14,Pg 14
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 14,Pg 15
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 14,Pg 16
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 14,Pg 17
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 14,Pg 18
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 14,Pg 19
Lecture 14,ACT 1
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 14,Pg 20
? 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 14,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 14,Pg 21
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 14,Pg 22
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 14,Pg 23
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 14,Pg 24
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 14,Pg 25
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 14,Pg 26
Lecture 14,ACT 2
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 14,Pg 27
Recap of today’s lecture
? Homework 6,
?Chap,6,# 6,12,20,24,29,38,52,57,78,and 83.
? Midterm 1,solutions
? Chapter 6,Work and Energy
?Review,Kinetic and Potential energy
?Non-conservative forces
?Generalized work-kinetic energy theorem
?Power
