# Physics 207: Lecture 2 Notes by HC120212201041

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```									            Physics 207, Lecture 13, Oct. 15
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, Oct. 22
 For Monday: Read all of Chapter 11

Physics 207: Lecture 13, Pg 1
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 2
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 3
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 4
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
h1
y=0
v
Physics 207: Lecture 13, Pg 5
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 6
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
Recall that “g” is the source of
Car has mass m the centripetal acceleration
U=mg2R
and N just goes to zero is
the limiting case.
h?                    Also recall the minimum speed
v
R       at the top is
gR
y=0
Physics 207: Lecture 13, Pg 7
Example
The Loop-the-Loop … again
Use E = K + U = constant
v

   mgh + 0 = mg 2R + ½ mv2                                gR
mgh = mg 2R + ½ mgR = 5/2 mgR

h = 5/2 R

h?
R

Physics 207: Lecture 13, Pg 8
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 9
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
..              Use E = K + U = constant

R=10 m                      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 10
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. Ball is dropped
2. Ball slides down a straight incline
3. 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 11
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 12
Example
Skateboard
   Now what is the normal force on the skate boarder?

N

m = 25 kg             60°
..                      mg
R=10 m
 S Fr = mar = m v2 / R
30°
= 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 13
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 14
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 15
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 16
Inelastic collision in 1-D: Example 1

What is the momentum of the bullet with speed v ?              
mv
 What is the initial energy of the system ?    1   1 2
mv  v  mv
2         2
 What is the final energy of the system ?     1
(m  M )V 2
2
 Is momentum conserved (yes)?          mv  M 0  (m  M )V
 Is energy conserved?     Examine Ebefore-Eafter
1       1               1      1      m     1          m
2
mv 2  [( m  M )V]V  mv 2  (mv)
2               2      2     mM
v  mv 2 1 
2
(
mM
)
v                       No!                           V

x
before                          after
Physics 207: Lecture 13, Pg 17
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 19
Home Exercise
Hooke’s Law

8m
9m

What is the spring constant “k” ?
Fspring
SF =  0 = Fs – mg = k Ds - mg
50 kg            Use k = mg/Ds = 5 N / 0.01 m
(A) 50 N/m   (B) 100 N/m (C) 400 N/m (D) 500 N/m
mg                                  Physics 207: Lecture 13, Pg 21
F-s relation for a single DNA molecule

Physics 207: Lecture 13, Pg 22
Measurement technique: optical tweezers

Physics 207: Lecture 13, Pg 23
Force vs. Energy for a Hooke’s Law spring
   F = - k (x – xequilibrium)
   F = ma = m dv/dt
= m (dv/dx dx/dt)                                                   m
= m dv/dx v
= mv dv/dx                          xf                   vf
   So - k (x – xequilibrium) dx = mv dv          ku du   mv dv
   Let u = x – xeq. & du = dx                  xi                   vi


1
2
ku | 
2 xf
xi
1
2
2 vf
mv vi    |

1
2
kx  kx 
2
f
1
2
2
i
1
2
2
mv f      
1
2
2
mvi

1
2
kx  mv  kx  mv
2
i
1
2
2
i
1
2
2
f
1
2
2
f
Physics 207: Lecture 13, Pg 24
Energy for a Hooke’s Law spring
1
2
kx  mv  kx  mv
2
i
1
2
2
i
1
2
2
f
1
2
2
f

   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 25
Energy diagrams
    In general:

Ball falling                         Spring/Mass system
Emech                                Emech
K                                            K

Energy
Energy

U
U

y                                   s

Physics 207: Lecture 13, Pg 26
Energy diagrams
Spring/Mass/Gravity system

spring
alone
Force

y

-mg                                                                      Ug
Emech                 K

Energy
spring &
m                 K
gravity
Uspring
UTotal

y
Physics 207: Lecture 13, Pg 27
Equilibrium
   Example
 Spring: Fx = 0 => dU / dx = 0 for x=0
The spring is in equilibrium position

   In general: dU / dx = 0  for ANY function establishes
equilibrium

U                          U

stable equilibrium       unstable equilibrium

Physics 207: Lecture 13, Pg 28
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 29
Physics 207, Lecture 13, Oct. 15

Assignment:
 HW6 due Wednesday
 For Monday: Read all of chapter 11

Physics 207: Lecture 13, Pg 30

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