#### Transcript Physics 207: Lecture 2 Notes

```Until recently…
Physics 207: Lecture 14, Pg 1
Lecture 14
Goals:
• Chapter 10
•
Understand the relationship between motion and energy
 Define Potential Energy in a Hooke’s Law spring
 Exploit conservation of energy principle in problem solving
 Understand spring potential energies
 Use energy diagrams
Chapter 11
 Understand the relationship between force, displacement
and work
Assignment:
 HW6 due Tuesday Oct. 26th
 For Monday: Read Ch. 11
Physics 207: Lecture 14, Pg 2
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 squared
Rearranging to give initial on the left and final on the right
½ m vyi2 + mgyi = ½ m vyf2 + mgyf
We now define mgy = U as the “gravitational potential energy”
Physics 207: Lecture 14, Pg 3
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

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 = K terms are defined to be kinetic energies
(A scalar quantity of motion)
Physics 207: Lecture 14, Pg 4
 What
is the SI unit of energy
A erg
B dyne
CNs
D calorie
E Joule
Physics 207: Lecture 14, Pg 5
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 14, Pg 6
Inelastic collision in 1-D: Example 1

mv
What is the momentum of the bullet with speed v ?
1   1 2
mv  v  mv
2
2
1
 What is the final energy of the system ?
(m  M )V 2
2
 Is momentum conserved (yes)?
mv  M 0  (m  M )V
 What is the initial energy of the system ?
Examine Ebefore-Eafter
1
1
1
1
m
1
m
mv 2  [( m  M )V]V  mv 2  (mv)
v  mv 2 1 
2
2
2
2
mM
2
mM
(
v
No!
before
)
V
after
x
 Is energy conserved?
Physics 207: Lecture 14, Pg 7
Elastic vs. Inelastic Collisions


A collision is said to be inelastic when “mechanical” energy (
K +U ) is not conserved before and after the collision.
How, if no net Force then momentum will be conserved.
Kbefore  Kafter
 E.g. car crashes on ice: Collisions where objects stick
together

A collision is said to be elastic when both energy & momentum
are conserved before and after the collision.
Kbefore = Kafter
 Carts colliding with a perfect spring, billiard balls, etc.
Physics 207: Lecture 14, Pg 8
Inelastic collision:
Where did that energy go?
Physics 207: Lecture 14, Pg 9
Kinetic & Potential energies

Kinetic energy, K = ½ mv2, is defined to be the
large scale collective motion of one or a set of
masses

Potential energy, U, is defined to be the “hidden”
energy in an object which, in principle, can be
converted back to kinetic energy

Mechanical energy, EMech, is defined to be the sum
of U and K.
Physics 207: Lecture 14, Pg 10
Energy
If only “conservative” forces are present, the total mechanical
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 14, Pg 11
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 14, Pg 12
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 14, Pg 13
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 14, Pg 14
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 14, Pg 15
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 14, Pg 16
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 14, Pg 17
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 207: Lecture 14, Pg 18
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 14, Pg 19
Variable force devices: Hooke’s Law Springs

Springs are everywhere, (probe microscopes, DNA, an
effective interaction between atoms)
Rest or equilibrium position
F
Du


In this spring, the magnitude of the force increases as the
spring is further compressed (a displacement).
Hooke’s Law,
Fs = - k Du
Ds is the amount the spring is stretched or compressed
from it resting position.
Physics 207: Lecture 14, Pg 25
Exercise 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 14, Pg 27
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 Du - mg
Use k = mg/Du = 500 N / 1.0 m
(B) 100 N/m (C) 400 N/m (D) 500 N/m
mg
Physics 207: Lecture 14, Pg 28
Force (N)
F vs. Dx relation for a foot arch:
Displacement (mm)
Physics 207: Lecture 14, Pg 29
F-Dx relation for a single DNA molecule
Physics 207: Lecture 14, Pg 30
Measurement technique: optical tweezers
Physics 207: Lecture 14, Pg 31
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 
uf
vf
ui
vi
 ku du   mv dv
1

2
ku | 
2 uf
ui
ku  ku 
1
2
ku  mv  ku  mv
2
f

1
2
m
2
i
1
2
2
i
1
2
1
2
2
f
2
f
1
2
2
i
1
2
1
2
2 vf
mv vi
2
mv f
|

1
2
2
mvi
Physics 207: Lecture 14, Pg 32
Energy for a Hooke’s Law spring
1
2

ku  mv  ku  mv
2
i
1
2
2
i
1
2
2
f
Associate ½ ku2 with the
“potential energy” of the spring
1
2
2
f
m
U si  K i  U sf  K f

Ideal Hooke’s Law springs are conservative so
the mechanical energy is constant
Physics 207: Lecture 14, Pg 33
Energy diagrams

In general:
Ball falling
Spring/Mass system
Emech
Emech
K
U
Energy
Energy
K
0
y
U
0
u = x - xeq
Physics 207: Lecture 14, Pg 34
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 14, Pg 36
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 207: Lecture 14, Pg 38
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 207: Lecture 14, Pg 39
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 207: Lecture 14, Pg 40
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 207: Lecture 14, Pg 41
Energy (with spring & gravity)
1
h
2
mg
3 kx
0
-x
When is the child’s speed greatest? (D) Between y2 & y3
 A: Calc. soln. Find v vs. spring displacement then maximize
(i.e., take derivative and then set to zero)
 B: Physics: As long as Fgravity > Fspring then speed is increasing
Find where Fgravity- Fspring= 0  -mg = kxVmax or xVmax = -mg / k
So mgh = Ug23 + Us23 + K23 = mg (-mg/k) + ½ k(-mg/k)2 + ½ mv2

 2gh = 2(-mg2/k) + mg2/k + v2  2gh + mg2/k = vmax2
Physics 207: Lecture 14, Pg 42
Recap
Assignment:
 HW6 due Tuesday Oct. 26th
 For Monday: Read Ch. 11
Physics 207: Lecture 14, Pg 45
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