The
and


Stuck in traffic? Can’t make to be in time in 9:00 am
Phys250 class?
What about the ability to fly in your own car?
o o o o
Dutch design engineering firm has just developed a three-wheeled vehicle that travels both on ground and in air, via a set of unfolding helicopter blades.
The PALV (personal air and land vehicle), powered by a rotary engine, has a top ground speed of 125 mph
(120 mph in the air) and can get between 60 and 70 miles per gallon of conventional gasoline. It can take off at close range, and can land vertically.
We will se how this project will be developed…
What is the thrust (forward force on the PALV) developed by the PALV Rotary engine with power output 213 hp when the vehicle is airborne and traveling in air horizontally at 120 mph?
http://www.sparkdesign.nl


Runner with mass m runs up the stairs to the top of 443-m-tall Sears
Tower. To lift herself there in 15 minutes (900 s), what must be her average power output in watts? Kilowatts? Horsepower?


Treat the runner as a particle of mass m.
Let’s find first how much work she must do against the gravity to lift herself at height h.


W
P av
 mgh

( 50 kg )( 9 .
8 m s
2
)( 443 m )

2 .
17

10
5
J

W
 t

2 .
17

10
5
J
900 s

241 W

0 .
241 kW

0 .
323 hp
Another way: calculate average upward force and then multiply by upward velocity
Upward force here is vertical, average vertical component of velocity is (443m) / (900s) = 0.492 m/s
P av

F v av

( mg ) v av

( 50 kg )( 9 .
8 m s
2
)( 0 .
492 s m )

241 W



Energy associated with position is called potential energy
If elevation for which the gravitational potential energy is chosen to be zero has been selected then the expression for the gravitational potential energy as a function of position y is given by
U grav
 mgy

Gravitational potential energy U grav is associated with the work done by the gravitational force according to
W grav

U
1

U
2
 
( U
2

U
1
)
  
U

Non



Work done by the conservative force only depends on the initial and final positions, and doesn’t depend on the path
Runner: gravitational force is conservative

From point 1 to point 2, same work

The work done by a conservative force has these properties:
 It can always be expressed as the difference between the initial and final values of a
potential energy function:  U = -W.
 It is reversible.
 It is independent of the path of the body and depends only on the starting and ending points.
 When the initial and final points are the same
(closed loop), the total work is zero.
All forces which do not satisfy these properties are non-conservative forces .

 When this guy is in midair, only gravity does work on him (air resistance can be neglected)
 Mechanical energy (sum of kinetic and gravitational potential energy) is conserved
 E = K + U = const

Moving up
K decreases
U increases
Moving down
K increases
U decreases
 When this guy is in midair, only gravity does work on him (air resistance can be neglected)
 Mechanical energy (sum of kinetic and gravitational potential energy) is conserved
 E = K + U = const
W = m g

Near the Earth
F = mg
W = x x f i

F dx
  x x f i
mg dx
W = mg x f x i
= mg (x f
x i
)
Away from the Earth
F = G m
1 m
2 x
2
W = x x i f 
F dx
  x x f i
G m
1 m
2 x
2 dx
W = G m
1 m
2
1 x x i x f
= G m
1 m
2


1 x f
1
x i



They disappeared at the boundary between the Cretaceous and
Tertiary periods (C-T boundary)
W hy ?
Luis Alvarez (
1911 – 1988
)
~ Nobel Prize winner in Physics ~ suggested an asteroid impact might be responsible

Alvarez calculated the asteroid would need to be 10 km across and would leave a crater
150 km in diameter
A crater off the Yucatan peninsula of Mexico has been identified as a possible impact site. Research on this crater has shown it is the result of a
extra-terrestrial impact.

Many asteroids and comets that cross the
Earth’s path originate in the Oort cloud.
This is a dense ring of asteroids that surrounds our solar system
Most asteroids that hit the Earth originate in the inner Oort cloud that extends from
40 to 10,000 times the radius of the earth’s orbit from the sun.




Assume an asteroid started at rest in the middle of the inner Oort cloud (~5000 R
E-S
)
Assume it is acted on primarily by the Sun
Assume mass ~10 16 kg (10-km-rock)

9
W = G m s m a


1 x f
1
x i


= (6.672

10
11 N
 m
2 kg
2
)(1.99

10
30 kg)(10
16 kg)


1
1.5x10
11 m
-
1
7.5
x 10
14 m
= 8.9
x 10
24
J
9
v =
2 K
= m
2 (8.9

10
24
J)
10
16 kg
= 42,100 m s
Over 80,000 MPH !

B
1
60º
A
3
2
30º
AB sin 30

 sin
AC
60


BC sin 90

C

Elastic


When you compress a spring:


If there is no friction, spring moves back
Kinetic energy has been “stored” in the
elastic deformation of the spring
Rubber-band slingshot: the same principle

Work is done on the rubber band by the force that stretches it


That work is stored in the rubber band until you let it go
You let it go, the rubber gives kinetic energy to the projectile

Elastic body: if it returns to its original shape and size after being deformed

Elastic
Equilibrium
Spring is stretched
It does negative work on block
Spring relaxes
It does positive work on block
Spring is compressed
Positive work on block
Block moves from one position x
1 to another position x
2
: how much work does the elastic (spring) force do on the block?

Elastic

Work done ON a spring to move one end from elongation x
1 to a different elongation x
2

When we stretch the spring, we do positive work on the spring

When we relax the spring, work done
on the spring is negative


Work done BY the spring

From N3L: quantities of work are negatives of each other
Thus, work W el done by the spring

We can express the work done BY the spring in terms of a given quantity at the beginning and end of the displacement
W

1
2 kx
2
2

1
2 kx
1
2
W el
U


1
2 kx
1
2 
1
2 kx
2
2
1 kx
2
2
Elastic potential energy

U

1
2 kx
2





The graph of elastic potential energy for ideal spring is a parabola
For extension of spring, x>0
For compression, x<0
Elastic potential energy U is NEVER negative!
In terms of the change of potential energy:
W el

U
1

U
2
  
U


1
2 kx
1
2 
1
2 kx
2
2

W el

U
1

U
2
  
U

1
2 kx
1
2 
1
2 kx
2
2

When a stretched spring is stretched greater, W in the spring el is negative and
U increases: greater amount of elastic potential energy is stored

When a stretched spring relaxes, x decreases, W el is positive and U decreases: spring loses its elastic potential energy

More spring compressed OR stretched, greater its elastic potential energy


Work – Energy Theorem: W tot
=K
2
-K
1
, no matter what kind of forces are acting on the body. Thus:
W tot

W el

U
2

U
1
W tot

U
1

U
2

K
2

K
1

K
1

U
1

K
2

U
2
1
2 mv
1
2 
1
2 kx
1
2 
1
2 mv
2
2 
1
2 kx
2
2
If only elastic force does work


Total mechanical energy E (the sum of elastic potential energy and kinetic energy) is conserved
Ideal spring is frictionless and massless

If spring has a mass, it also has kinetic energy



Your car has a mass of 1.2 ton or more
Suspension spring has a mass of few kg
So we can neglect spring’s mass in study of how the car bounces on its suspension
E

K

U


If forces other than elastic force also do work on the body, the total work is
W tot

W el

W other

K
2

K
1

K
1

U
1

W other

K
2

U
2 elastic force + other forces
1 mv
1
2 
1 kx
1
2 
W other

1 mv
2
2 
1 kx
2
2
2 2 2 2
The work done by all forces other than the elastic force equals the change in the total mechanical energy E of the system, where U is the elastic potential energy:
E

K

U



“System” is made up of the body of mass m and the spring of force constant k
When W other
When W other is positive, E increases is negative, E decreases



Both Gravitational Potential Energy and Elastic Potential Energy


Spring with a body is hanged vertically
Bungee jumper
U
1 and U
2 then are initial and final values of the total potential energy
U
1

U grav , 1

U el , 1
U
2

U grav , 2

U el , 2
K
1

U grav , 1

U el , 1

W other

K
2

U grav , 2

U el , 2

The work done by all forces other than the gravitational force or elastic force equals the change in the total mechanical energy E=K+U of the system, where U is the sum of the gravitational potential energy and the elastic potential energy

Force





We have studied in detail two specific conservative forces, gravitational force and elastic force.
We have seen there is a definite relationship between a conservative force and the corresponding potential energy function.
The force on a mass in a uniform gravitational field is
F y
= - mg. The corresponding potential energy function is U(y) = mgy.
The force exerted on a body by a spring of force constant k is F x
= - kx. The corresponding potential energy function is U s
(x) = (1/2)kx 2 .
In some situations, you are given an expression for potential energy as a function of position and have to find corresponding force.

Force
Consider motion along a straight line, with coordinate x


F x
(x) is the x-component of force as function of x
U(x) is the potential energy as function of x

Work done by conservative force equals the negative of the change
 U in potential energy:
W
  
U


For infinitesimal displacement this displacement is ~ F
F x
( x
In the limit  x  0:
)
 x
  
U
 x, the work done by force F that the force will vary just a little)
F x
( x )
 


U x x
(x) during x
(x)  x (suppose that this interval is so small
F x
( x )
  dU ( x ) dx
Force from potential energy, one dimension

Force
F x
( x )
  dU ( x ) dx
Force from potential energy, one dimension





In regions where U(x) changes most rapidly with x (i.e. where dU(x)/dx is
large) the greatest amount of work is done during the displacement, and it corresponds to a large force magnitude
When F x
(x) is in positive x-direction, U(x) decreases with increasing x
Thus, F x
(x) and U(x) have opposite sign
Thus, the force is proportional to the negative slope of the potential
energy function
The physical meaning: conservative force always acts to push the system toward lower potential energy

Force

Lets verify if this expression correctly gives the gravitational force and the elastic force when using the gravitational potential energy and the elastic potential energy:
U ( y )
 mgy F y
( x )
  dU ( y ) dy
  d dy
 mgy

  mg


U ( x )

1
2 kx
2 F x
( x )
  dU ( x ) dx
  d dx
1
2 kx
2
The gravitational potential energy is linearly related to the elevation (i.e. constant slope) and the force is constant.
The elastic potential energy varies quadratically with position.
The force varies in a linearly.
  kx

Potential Energy

Force
3D



Conservative force in three dimensions has components F x
, F y
, and F z

Each component may be function of coordinates x, y, z
Potential energy change  U is the function of coordinates as well
When particle moves a small distance
~constant. It does NOT depend on F y
 x in x-direction, the force F and F z x is because these components of force are perpendicular to the displacement and do NO work
F x
 

U
 x
F y
 


U y
F z
 


U z

 x y


0
0
F x
  dU dx
F y
  dU dy
F z
  dU dz
 z

0
Force from potential energy, three dimensions

F
 

 dU dx i
ˆ  dU dy
ˆ j
 dU dz k
ˆ


 
F
  
U

Energy Diagrams


In situations where a particle moves in one-dimension only under influence of a single conservative force it is very useful to study the graph of the potential energy as a function of position U(x)
At any point on a graph of U(x), the
force can be calculated as the negative of the slope of the potential
energy function

F x
= - dU/dx

Example: Glider on an air track



Spring exerts a force F x
=-kx
Potential energy function U(x)
Limits of the motion are the points where U curve intersects the horizontal line representing the total mechanical energy E




Any point where the force is zero is called equilibrium point
These are the "critical points" on the graph of U(x):


Points on the graph that are local minima correspond to
"stable equilibria" since the force on particle tends to push it back toward the equilibrium point.
Points on the graph that are local maxima correspond to
"unstable equilibria" since force on particle tends to push it back toward the equilibrium point.

Points on the graph that are inflection points correspond to
"neutral equilibria".
If the total mechanical energy is known, then the potential energy graph can be used to determine the speed at any point since E = K +
U is constant (i.e. use K = E – U and then find speed)

Energy Diagrams

y x y


R sin( q
)
R ( 1
 cos( q
)
50
40 q
U
 mgy
30
20 x E = K + U

=
1
2
m v
2
+ m g y
10
0
A Pendulum
-10
-4 -3 -2 -1 0 1 2 3 4 x (m)
Turning Points