Transcript Ch 10 Energy - APPhysicsCBrookstone

Chapter 10
Energy
What is Energy?
"It is important to realize that in
physics today, we have no
knowledge of what energy is. We do
not have a picture that energy
comes in little blobs of a definite
amount." -Richard Feynman
"Lectures on Physics"
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What is Energy? Con’t
Energy - A measure of being able to
do work…..–NASA.gov
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Matter and Energy
The combination of matter and
energy makes up the universe.
Matter is substance and energy
is the mover of substance.
-Paul Hewitt
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Definition
Energy is an abstract quantity
with the ability to effect physical
change in matter.
Energy is the substance from
which all things in the Universe
are made up.
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Forms of Energy
Two broad categories of energy:
Potential: Stored Energy
Chemical – Stored in chemical bonds
Mechanical – Stored in objects by tension
Nuclear – Stored in nucleus of atoms
Gravitational – Stored based on height and weight
Electrical – Stored in batteries
Kinetic: Energy due to motion
Radiant – electromagnetic
Thermal - Heat
Motion – Moving objects
Sound – motion of air and other media by sound waves
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Conservation of Energy
Energy Conservation during freefall:
v f  v i  2g  y f  y i 
2
2
v f  v i  2gy f  2gy i
2
2
v f  2gy f  v i  2gy i
2
m
2
2
v
2
f
 2gy f  v i  2gy i  
mv
2
2
u n its  k g
 
m
s
2
1
2
m v f  m gy f 
2
1
2
m v i  m gy i
2
m 

  kg 2  m  Nm
s 

m 
 m 

m g y u n its  k g  2  m   k g 2  m  N m
s 
s 

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Calculus Approach
Lets look at the freefall of a mass, m, from N2L and calculus:
F  ma  m
dv
dt
dy dv

mvy
dt dy
dv
dy
dv
dt
 mg
 vy
dv
dy
 mg
m v y dv   m gdy
m
1
2
vf
vi
v y dv   m g 
2
mvf
vf
vi

1
2
mvf 
2
1
2
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yf
dy
yi
1
2
m v i   m gy
2
m v f  m gy f 
2
1
2
yf
yi
  m g y f  m fy i
m v i  m gy i
2
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Kinetic Energy
K 
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mv
2
2
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Gravitational Potential Energy
Similar concept to kinetic energy except that the orientation is
vertical and the force is gravity. Think KE which has not been
created.
M
d
mg
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Mechanical Energy
1
2
m v  m gy f 
2
f
1
2
m v i  m gy i
2
Kf  Uf  Ki  Ui
K U  k
K   U
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Problem 1
A boy reaches out of a window and tosses a ball straight up with a speed
of 10 m/s. The ball is 20 m above the ground as he releases it. Use
energy to find:
a. The ball’s maximum height above the ground.
b. The ball’s speed as it passes the window on its way down.
c. The speed of impact on the ground.
Analysis: Use Kf + Uf = Ki + Ui to solve for y1, v2 and v3.
1
2
m v 0  m gy 0 
2
1
2
m v 1  m gy 1
2
a. At y1, v1 = 0; at y0, v0 = 10 and y0 = 0
1
2
m 10  0  0  m gy 1
y1 
2
100
2  9 .8 
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Problem 1 con’t
b. At y0, v0 = 10; at y2=0, find v2
1
2
v0  g 0 
2
1
2
v2  g 0
2
v0  v2
2
2
v 2  v 0  10
Speed up = speed down.
c. At y0 = 0, v0 = 10; At y3 = -20, find v3.
1
10  g  0  
2
1
v 3  g  20 
2
2
2
1

2
2  1 0  2 0  9 .8    v 3  2 2 .2
2

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m
s
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Energy Bar Charts
Bar chart modeling a rock thrown upward and returning to same elevation:
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Zero of Potential Energy
1
1
2
2
Our earlier calculus based derivation of
m v f  m gy f  m v i  m gy i
2
2
The following expression represented
the change in potential energy:

Uf
Ui
U  mg 
yf
dy
yi
The initial position was considered to be zero and resulted in U = mgyf
U will vary based on where the zero potential energy level is placed.
ΔU will always be the same regardless of the location of the zero level.
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Non-Freefall Ug
Horizontal surface: no change in gravitational potential energy
Slanted/undulating surface: define s-axis parallel to surface of movement.
 Fnet  s
 m as  m
 Fnet  s
 m
dv s
dt
dv s
dt
 m
 m g s in   m v s
dv s ds
ds dt
 mv s
dv s
ds
dv s
ds
 m g s in  d s  m v s d v s
 m g d y  m v sd v s
1
2
m v f  m gy f 
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1
2
m v i  m gy i
2
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Ballistic Pedulum
A 10 g bullet is fired into a 1200 g wood block hanging from a 150 cm long
string. The bullet embeds itself into the block and the block swings out
to an angle of 40°. What was the speed of the bullet?
Analysis: Two part problem: The
impact of the bullet with the
block is inelastic. Momentum is
conserved. After the collision
the block swings as a pendulum.
The sum of the kinetic and
potential energies before and
after do not change as the block
swings to its largest angle.
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Ballistic Pendulum con’t
Collision/Momentum calculations:
( m w  m B )v 1 x  m w  v 0 x
v 0 x B

mW  m B
mB
w
 m B v 0x
B
v 1x
If we can calculate v1x from swing/energy
relationships, we can calculate the speed of
the bullet.
1
2
m
 m B  v 2   mW  m B  gy 2
2
W

1
m
2
 m B  v 1   mW  m B  gy 1
2
W
V2 = 0 and dividing by (mW + mB) gives:
1
2
 0   gy
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
1
2
v 1  gy 1
2
or v 1 
v1 
2gy 2
2  9 .8  0 .3 5 1  2 .6 2
m
s
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Restoring Forces
Restoring Force
Elastic
Equilibrium Length, L0
Displacement, Δs
Δs = L - L0
Fsp = k Δs
Spring constant, k
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Hooke’s Law
( Fs p ) s   k  s
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Hooke’s Law Problem
You need to make a spring scale for measuring mass. You want
each 1.0 cm length along the scale to correspond to a mass
difference of 100 g. What should be the value of the spring
constant?
F sp  k  x  m g .
k  m g /  x  (0 .1 0 0 k g)(9 .8 N /m )/(0 .0 1 0 m )  9 8 N /m
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Elastic Potential Energy
Is the force applied to the ball constant?
Describe the mechanical energy in
both situations; before and after.
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Elastic Potential Energy con’t
N2L for the ball is:
 Fnet  s
 m as  m
dv
dt
By Hooke’s law, (Fnet)s = -k(s – se), substituting gives:
m
dv
dt
  k  s  se 
Using the chain rule:
dv s
dt

dv s ds
ds dt
 vs
dv s
ds
Substituting gives:
m v sdv s   k  s  se  ds
Integrating from initial to final conditions:

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vs
vi
m v sd s 
1
2
mvf 
2
1
2
mvi  k 
2
sf
si
( s  s e )d s
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Elastic Potential Energy con’t
From previous slide:

vs
vi
k 
m v sd s 
sf
si
1
2
mvf 
2
( s  s e )d s  
1
2
1
2
mvi  k 
2
k  sf

2

1
2
sf
si
( s  s e )d s
k  si 
2
Substituting and rewriting gives:
1
2
mvf 
2
1
2
k  sf

2

1
2
mvi 
2
1
2
k  si 
2
½ mv2 is obviously the kinetic energy.
1
2
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k  s 
2
is the elastic potential energy, Us
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Elastic Potential Energy Problem
How far must you stretch a spring with k = 1000 N/m to store 200 J of
energy?
Elastic potential energy is defined as:
Us 
1
2
k ( s )
2
Solving for ∆s:
s 
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2U s /k 
2(2 0 0 J) / 1 0 0 0 N /m  0 .6 3 2 m
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Elastic Collisions
Perfectly Elastic Collision: A collision in which mechanical energy is
conserved.
Must conserve momentum and mechanical energy
Not possible where friction is involved
If only one object, m1, initially moving and all motion is along a line:
m 1  v fx  1  m 2  v fx  2  m 1  v ix  1
1
2
m 1  v fx  1 
2
1
2
m 2  v fx  2 
2
1
2
m 1  v ix  1
2
Solving the first equation for (vfx)1 and substituting into the second
2
gives:


m
2
2
m 1   v ix  1  2  v fx  2   m 2  v fx   m 1  v ix 
2
1
m1


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Elastic Collisions con’t
2


m
2
2
m 1   v ix  1  2  v fx  2   m 2  v fx   m 1  v ix 
2
1
m1


Squaring and simplifying gives:
 v fx  2


m2 
 1 
  v fx  2  2  v ix  1   0
m
1 


There are two solutions, (vfx)2 = 0 which is trivial and:
 v fx  2

2m 1
m1  m 2
 v ix  1
Substituting this into the momentum equation gives
 v fx  1

m1  m 2
m1  m 2
 v ix  1
an d
 v fx  2

2m 1
m1  m 2
 v ix  1
These allow us to compute the final velocity of each object in terms
of the initial velocity of m1 and the relative masses of each object
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Elastic Collision Problem
A 50 g marble moving at 2.0 m/s strikes a 20 g marble at rest. What is
the speed of each marble immediately after the collision?
Analysis: Expect that v1 will decrease and v2 will significantly
increase. Laws conservation of Momentum and ME will be observed.
(v f x ) 1 
(v f x ) 2 
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m1  m2
m1  m2
2m 1
m1  m2
(v i x ) 1 
(v i x ) 1 
50 g  20 g
50 g  20 g
2(5 0 g)
50 g  20 g
(2 .0 m / s)  0 .8 6 m / s
(2 .0 m / s)  2 .9 m / s
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Using Reference Frames
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Energy Diagrams
Energy Diagram: A graph showing a system’s potential energy and total
energy as a function of position.
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Energy Diagram for a Spring
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Generalized Energy Diagram
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Stable and Unstable Equilibrium
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