Lecture 13
Goals:
• Chapter 10
•
 Understand the relationship between motion and energy
 Define Potential Energy in a Hooke’s Law spring
 Develop and exploit conservation of energy principle
in problem solving
Chapter 11
 Understand the relationship between force,
displacement and work
Assignment:
 HW6 due Wednesday, Feb. 11
 For Thursday: Read all of Chapter 11
Physics 207: Lecture 13, Pg 1
Energy
-mg Dy= ½ m (vy2 - vy02 )
-mg (yf – yi) = ½ m ( vyf2 -vyi2 )
A relationship between y-displacement and change in the
y-speed
Rearranging to give initial on the left and final on the right
½ m vyi2 + mgyi = ½ m vyf2 + mgyf
We now define mgy as the “gravitational potential energy”
Physics 207: Lecture 13, Pg 2
Energy

Notice that if we only consider gravity as the external force then
the x and z velocities remain constant

To
½ m vyi2 + mgyi = ½ m vyf2 + mgyf

Add
½ m vxi2 + ½ m vzi2
and
½ m vxf2 + ½ m vzf2
½ m vi2 + mgyi = ½ m vf2 + mgyf

where
vi2 = vxi2 +vyi2 + vzi2
½ m v2 terms are defined to be kinetic energies
(A scalar quantity of motion)
Physics 207: Lecture 13, Pg 3
Energy

If only “conservative” forces are present, the total energy
(sum of potential, U, and kinetic energies, K) of a system is
conserved
For an object in a gravitational “field”
½ m vyi2 + mgyi = ½ m vyf2 + mgyf
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 207: 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 207: 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 207: 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 207: 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 207: Lecture 13, Pg 8
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)
Use fact that E = U + K = constant !
Ub=mgh
U=mg2R
h?
y=0(A) 2R
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 207: Lecture 13, Pg 9
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 207: Lecture 13, Pg 10
Example
Skateboard



What speed will the skateboarder reach halfway down the hill
if there is no friction and the skateboarder starts at rest?
Assume we can treat the skateboarder as a “point”
Assume zero of gravitational U is at bottom of the hill
m = 25 kg
..
R=10 m
30°
R=10 m
y=0
Physics 207: Lecture 13, Pg 11
Example
Skateboard



What speed will the skateboarder reach halfway down the hill
if there is no friction and the skateboarder starts at rest?
Assume we can treat the skateboarder as “point”
Assume zero of gravitational U is at bottom of the hill
m = 25 kg
..
R=10 m
Use E = K + U = constant
Ebefore = Eafter
0 + m g R = ½ mv2 + mgR (1-sin 30°)
30°
mgR/2 = ½ mv2
R=10 m
gR = v2  v= (gR)½
v = (10 x 10)½ = 10 m/s

Physics 207: Lecture 13, Pg 12
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 207: Lecture 13, Pg 13
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 207: Lecture 13, Pg 14
Example
Skateboard

What is the normal force on the skate boarder?
N
m = 25 kg
60°
..
mg
R=10 m
30°
R=10 m
Physics 207: Lecture 13, Pg 15
Example
Skateboard

Now what is the normal force on the skate boarder?
N
m = 25 kg
60°
..
R=10 m
30°
mg
 S Fr = mar = m v2 / R
= N – mg cos 60°
R=10 m
N = m v2 /R + mg cos 60°
N = 25 100 / 10 + 25 10 (0.87)
N = 250 + 220 =470 Newtons
Physics 207: Lecture 13, Pg 16
Elastic vs. Inelastic Collisions

A collision is said to be elastic when energy as well as
momentum is conserved before and after the collision.
Kbefore = Kafter
 Carts colliding with a perfect spring, billiard balls, etc.
vi
Physics 207: Lecture 13, Pg 17
Elastic vs. Inelastic Collisions

A collision is said to be inelastic when energy is not
conserved before and after the collision, but momentum is
conserved.
Kbefore  Kafter
 Car crashes, collisions where objects stick together, etc.
Physics 207: Lecture 13, Pg 18
Inelastic collision in 1-D: Example 1

A block of mass M is initially at rest on a frictionless
horizontal surface. A bullet of mass m is fired at the block
with a muzzle velocity (speed) v. The bullet lodges in the
block, and the block ends up with a speed V.
 What is the initial energy of the system ?
 What is the final energy of the system ?
 Is energy conserved?
x
v
V
before
after
Physics 207: Lecture 13, Pg 19
Inelastic collision in 1-D: Example 1

mv
What is the momentum of the bullet with speed v ?
1
 
2
mv  v  mv
2
2
1
2
(m  M )V
2
1
 What is the initial energy of the system ?
 What is the final energy of the system ?
 Is momentum conserved (yes)?
mv  M 0  (m  M )V
 Is energy conserved? Examine Ebefore-Eafter
2
mv 
2
1
[( m  M )V]V 
2
1
2
v
mv 
2
1
( mv)
2
m
mM
v
1
mv
2
No!
before
2
(1 
m
mM
)
V
after
x
1
Physics 207: Lecture 13, Pg 20
Variable force devices: Hooke’s Law Springs

Springs are everywhere, (probe microscopes, DNA, an
effective interaction between atoms)
Rest or equilibrium position
F
Ds


In this spring, the magnitude of the force increases as the
spring is further compressed (a displacement).
Hooke’s Law,
Fs = - k Ds
Ds is the amount the spring is stretched or compressed
from it resting position.
Physics 207: Lecture 13, Pg 22
Exercise 2
Hooke’s Law
8m
9m
What is the spring constant “k” ?
50 kg
(A) 50 N/m
(B) 100 N/m (C) 400 N/m (D) 500 N/m
Physics 207: Lecture 13, Pg 23
Exercise 2
Hooke’s Law
8m
9m
What is the spring constant “k” ?
Fspring
50 kg
(A) 50 N/m
SF =
0 = Fs – mg = k Ds - mg
Use k = mg/Ds = 500 N / 1.0 m
(B) 100 N/m (C) 400 N/m (D) 500 N/m
mg
Physics 207: Lecture 13, Pg 24
Force (N)
F-s relation for a foot arch:
Displacement (mm)
Physics 207: Lecture 13, Pg 25
Force vs. Energy for a Hooke’s Law spring




F = - k (x – xequilibrium)
F = ma = m dv/dt
= m (dv/dx dx/dt)
= m dv/dx v
= mv dv/dx
So - k (x – xequilibrium) dx = mv dv
Let u = x – xeq. & du = dx 
m
xf
 ku du   mv dv
xi


1
2
1
2
vf
1
2
vi
ku |x  2 mv |v
2
xf
1
2 vf
i
i
kx f  2 kxi  2 mv f  2 mvi
2
1
2
1
2
1
kxi  2 mvi  2 kx f  2 mv f
2
1
2
1
2
1
2
Physics 207: Lecture 13, Pg 26
2
Energy for a Hooke’s Law spring
kx

mv

kx

mv
i
i
f
f
2
2
2
2
1

2
1
2
1
2
1
2
Associate ½ kx2 with the
“potential energy” of the spring
m
U si  K i  U sf  K f

Hooke’s Law springs are conservative so the
mechanical energy is constant
Physics 207: Lecture 13, Pg 27
Energy diagrams

In general:
Ball falling
Spring/Mass system
Emech
Emech
K
U
y
Energy
Energy
K
U
s
Physics 207: Lecture 13, Pg 28
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 207: Lecture 13, Pg 30
Comment on Energy Conservation

We have seen that the total kinetic energy of a system
undergoing an inelastic collision is not conserved.
 Mechanical energy is lost:
 Heat (friction)
 Bending of metal and deformation

Kinetic energy is not conserved by these non-conservative
forces occurring during the collision !

Momentum along a specific direction is conserved when
there are no external forces acting in this direction.
 In general, easier to satisfy conservation of momentum
than energy conservation.
Physics 207: Lecture 13, Pg 31
Lecture 13
Assignment:
 HW6 due Wednesday 2/11
 For Monday: Read all of chapter 11
Physics 207: Lecture 13, Pg 32