Lecture 13

Goals
 Introduce concepts of Kinetic and Potential energy
 Develop Energy diagrams
 Relate Potential energy to the external net force
 Discuss Energy Transfer and Energy Conservation
 Define and introduce power (energy per time)
Physics 201: Lecture 13, Pg 1
Conservative vs. Non-Conservative forces
For a spring one can perform negative work but
then reverse this process and recover all of this
energy.
 A compressed spring has the ability to do work
 For a Hooke’s law spring the work done is
independent of path
 The spring is said to be a conservative force

In the case of friction there is no immediate way to
back transfer the energy of motion
 In this case the work done can be shown to be
dependent on path
 Friction is said to be a non-conservative force

Physics 201: Lecture 13, Pg 2
Hidden energy is Potential Energy (U)
For the compressed spring the energy is “hidden”
but still has the ability to do work (i.e., allow for
energy transfer)
 This kind of “energy” is called “Potential Energy”

U ( x) spring  k ( x  xeq )
1
2

2
The gravitation force, if constant, has the same
properties.
U ( y ) gravity  mg ( y  y0 )
Physics 201: Lecture 13, Pg 3
Mechanical Energy (Kinetic + Potential)
If only “conservative” forces, then total mechanical energy
(potential U plus kinetic K energy) of a system is conserved
For an object in a gravitational “field”

½ m vyi2 + mgyi = ½ m vyf2 + mgyf = constant
K ≡ ½ mv2
U ≡ mgy
Emech = K + U
Emech = K + U = constant

K and U may change, but Emech = K + U remains a fixed value.
Emech is called “mechanical energy”
Physics 201: Lecture 13, Pg 4
Example of a conservative system:
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 201: Lecture 13, Pg 5
Example: The simple pendulum.
 What is the maximum speed of the mass and where
does this happen ?
E = K + U = constant and so K is maximum when U is
a minimum.
y
y=h1
y=
0
Physics 201: Lecture 13, Pg 6
Example: The simple pendulum.
 What is the maximum speed of the mass and where
does this happen ?
E = K + U = constant and so K is maximum when U is
a minimum
E = mgh1 at top
E = mgh1 = ½ mv2 at bottom of the swing
y
y=h1
y=0
h1
v
Physics 201: Lecture 13, Pg 7
Example: The simple pendulum.
To what height h2 does it rise on the other side?
E = K + U = constant and so when U is maximum
again (when K = 0) it will be at its highest point.
E = mgh1 = mgh2 or h1 = h2
y
y=h1=h2
y=0
Physics 201: Lecture 13, Pg 8
Potential Energy, Energy Transfer and Path
A ball of mass m, initially at rest, is released and follows
three difference paths. All surfaces are frictionless
1. The ball is dropped
2. The ball slides down a straight incline
3. The ball slides down a curved incline
After traveling a vertical distance h, how do the three speeds
compare?

1
2
3
h
(A) 1 > 2 > 3
(B) 3 > 2 > 1
(C) 3 = 2 = 1 (D) Can’t tell
Physics 201: Lecture 13, Pg 9
Energy diagrams

In general:
Ball falling
Spring/Mass system
Emech
Emech
K
U
Energy
Energy
K
0
y
U
0
u = x - xeq
Physics 201: Lecture 13, Pg 10
Conservative Forces & 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.
Uf
rf

This can be written as:
U = Uf - Ui = - W = -
rf
rF • dr
i
ri
Ui
Physics 201: Lecture 13, Pg 11
Conservative Forces and Potential Energy

So we can also describe work and changes in
potential energy (for conservative forces)
U = - W

Recalling (if 1D)
W = Fx x

Combining these two,
U = - Fx x

Letting small quantities go to infinitesimals,
dU = - Fx dx

Or,
Fx = -dU / dx
Physics 201: Lecture 13, Pg 12
Equilibrium

Example
 Spring: Fx = 0 => dU / dx = 0 for x=xeq
The spring is in equilibrium position

In general: dU / dx = 0  for ANY function establishes
equilibrium
U
stable equilibrium
U
unstable equilibrium
Physics 201: Lecture 13, Pg 13
Chapter 8 Conservation of Energy
Esystem   T (energy tr ansfer)
Examples of energy transfer
 Work (by conservative or non-conservative forces)
 Heat (thermal energy transfer)
Esystem  K  U  Eint
A “system” depends on situation
 Example: A can with internal non-conservative forces
In general, if just one external force acting on a system

K  W
Physics 201: Lecture 13, Pg 14
“Mechanical” Energy of a System…agai n
Emech  K  U

Isolated system without non-conservative forces
Emech  K  U  0
K f  Ki  U f  U i  0
K f  U f  Ki  U i  constant
Physics 201: Lecture 13, Pg 15
Example
The Loop-the-Loop … again



To complete the loop the loop, how high do we have to let
the release the car?
Condition for completing the loop the loop: Circular motion
at the top of the loop (ac = v2 / R)
Exploit the fact that E = U + K = constant ! (frictionless)
Ub=mgh
U=mg2R
h?
y=0(A) 2R
U=0
Recall that “g” is the source of
Car has mass m the centripetal acceleration
and N just goes to zero is
the limiting case.
Also recall the minimum speed
at the top is
R
(B) 3R
(C) 5/2 R
(D)
23/2
R
v
gR
Physics 201: Lecture 13, Pg 16
Example
The Loop-the-Loop … again


Use E = K + U = constant
mgh + 0 = mg 2R + ½ mv2
mgh = mg 2R + ½ mgR = 5/2 mgR
v
gR
h = 5/2 R
h?
R
Physics 201: Lecture 13, Pg 17
With non-conservative forces

Mechanical energy is not conserved.
Emech  K  U  WNC  WC
Physics 201: Lecture 13, Pg 18
An experiment
Two blocks are connected on the table as shown. The table
has a kinetic friction coefficient of mk. The masses start at
rest and m1 falls a distance d. How fast is m2 going?
N
T
T
Mass 1
m2
f
k
S Fy = m1ay = T – m1g
Mass 2
m1
m2g
S Fx = m2ax = -T + fk = -T + mk N
m1g
S Fy = 0 = N – m2g
| ay | = | ay | = a =(mkm2 - m1) / (m1 + m2)
2ad = v2 =2(mkm2 - m1) g / (m1 + m2)
K= - mkm2gd – Td + Td + m1gd = ½ m1v2+ ½ m2v2
v2 =2(mkm2 - m1) g / (m1 + m2)
Physics 201: Lecture 13, Pg 19
Energy conservation for a Hooke’s Law spring
U si  K i  U sf  K f

Associate ½ kx2 with the
“potential energy” of the spring
1
2

m
kx  mv  kx  mv
2
i
1
2
2
i
1
2
2
f
1
2
2
f
Ideal Hooke’s Law springs are conservative so
the mechanical energy is constant
Physics 201: Lecture 13, Pg 20
Energy (with spring & gravity)
1
h
2
0
-x
3
mass: m
Given m, g, h & k,

how much does the spring compress?
Emech = constant (only conservative forces)

At 1: y1 = h ; v1y = 0 At 2: y2 = 0 ; v2y = ? At 3: y3 = -x ; v3 = 0

Em1 = Ug1 + Us1 + K1 = mgh + 0 + 0
Em2 = Ug2 + Us2 + K2 = 0 + 0 + ½ mv2
Em3 = Ug3 + Us3 + K3 = -mgx + ½ kx2 + 0 Physics 201: Lecture 13, Pg 21


Energy (with spring & gravity)
1
h
0
-x







2
3
mass: m
Given m, g, h & k, how much does the
spring compress?
Emech = constant (only conservative forces)
At 1: y1 = h ; v1y = 0 At 2: y2 = 0 ; v2y = ? At 3: y3 = -x ; v3 = 0
Em1 = Ug1 + Us1 + K1 = mgh + 0 + 0
Em2 = Ug2 + Us2 + K2 = 0 + 0 + ½ mv2
Em3 = Ug3 + Us3 + K3 = -mgx + ½ kx2 + 0
Given m, g, h & k, how much does the spring compress?
Em1 = Em3 = mgh = -mgx + ½ kx2  Solve ½ kx2 – mgx +mgh = 0
Physics 201: Lecture 13, Pg 22
Energy (with spring & gravity)
1
h
0
-x







2
3
mass: m
Given m, g, h & k, how much does the
spring compress?
Emech = constant (only conservative forces)
At 1: y1 = h ; v1y = 0 At 2: y2 = 0 ; v2y = ? At 3: y3 = -x ; v3 = 0
Em1 = Ug1 + Us1 + K1 = mgh + 0 + 0
Em2 = Ug2 + Us2 + K2 = 0 + 0 + ½ mv2
Em3 = Ug3 + Us3 + K3 = -mgx + ½ kx2 + 0
Given m, g, h & k, how much does the spring compress?
Em1 = Em3 = mgh = -mgx + ½ kx2  Solve ½ kx2 – mgx +mgh = 0
Physics 201: Lecture 13, Pg 23
Energy (with spring & gravity)
1
h
mass: m
2
3
0
-x

When is the child’s speed greatest?
(A) At y1 (top of jump)
(B) Between y1 & y2
(C) At y2 (child first contacts spring)
(D) Between y2 & y3
(E) At y3 (maximum spring compression)
Physics 201: Lecture 13, Pg 24
Work & Power:





Two cars go up a hill, a Corvette and a ordinary Chevy
Malibu. Both cars 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 Corvette can get up the hill quicker
It has a more powerful engine.
Physics 201: Lecture 13, Pg 25
Work & Power:

Power is the rate at which work is done.
Average Power is,

Instantaneous Power is,

If force constant in 1D, W= F x = F (v0 t + ½ at2)

and

dW  dr  
P
 F   F v
dt
dt
Pavg
W

t
dE
P
dt
P = W / t = F (v0 + at)
1 W = 1 J / 1s
Physics 201: Lecture 13, Pg 26
Exercise
Work & Power


Starting from rest, a car drives up a hill at constant
acceleration and then suddenly stops at the top.
The instantaneous power delivered by the engine during this
drive looks like which of the following,
A.
B.
C.
Top
Middle
Bottom
time
time
time
Physics 201: Lecture 13, Pg 27
Exercise
Work & Power

P = dW / dt and W = F d = (m mg cos q  mg sin q) d
and d = ½ a t2 (constant accelation)
So W = F ½ a t2  P = F a t = F v

(A)
time

(B)
time

(C)
time
Physics 201: Lecture 13, Pg 28
Work & Power:

Power is the rate at which work is done.
Pavg
W

t
Example:



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.
Pavg = F h / t = mgh / t = 80.0 x 9.80 x 12.0 / 20.0 W
P = 470. W
Physics 201: Lecture 13, Pg 29
Recap

Read through Chap 9.1-9.4
Physics 201: Lecture 13, Pg 30