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Energy
Readings: Chapter 10
1
Free-Fall motion
v  v  2 g( yi  y f )
2
fy
2
iy
1 2
1 2
v fy  gy f  viy  gyi
2
2
This is the conservation law for free fall
motion: the quantity
1 2
v y  gy
2
has the same value before and after the
motion.
2
Free-Fall Motion
v  v  2 g( yi  y f )
2
fy
2
iy
v fx  vix
vi
Then
1 2 1 2
1 2 1 2
v fx  v fy  gy f  vix  viy  gyi
2
2
2
2
1 2
1 2
v f  gy f  vi  gyi
2
2
vf
Conservation law: the quantity
1 2
v  gy
2
has the same value before and after the
motion.
3
y
yi
Frictionless surface: acceleration
a  g sin 
a
Motion with constant acceleration:
s
yf
v 2f  vi2  2as
yi  y f
s
sin 

v  v  2 g sin 
2
f
2
i
Conservation law: the quantity
1 2
v  gy
2
yi  y f
sin 
 2 gyi  2 gy
1 2
1 2
v f  gy f  vi  gyi
2
2
has the same value before and after the
motion.
4
y
Conservation law: the quantity
yi
1 2
v  gy
2
has the same value at the initial and the
final points
yf
In all cases the velocity at the
final point is the same
(FRICTIONLESS MOTION)
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1 2
v  gy
2
Conservation law:
or
1
mv 2  mgy
2
1
K  mv 2
2
Kinetic Energy – energy of motion
U g  mgy
Gravitational Potential Energy – energy of position
Emech  K  U g
Mechanical Energy
Conservation law of mechanical energy (without
friction):
Emech  K  U  constant
The units of energy is Joule:
m2
J  kg 2
s
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7
8
Zero of potential energy
y
y
yi ,1
yi ,2
K i  U g ,i ,1  K f ,1  U g , f ,1
K i  U g ,i ,2  K f ,2  U g , f ,2
0
K f ,1  K i  (U g ,i ,1  U g , f ,1 )
y f ,1
0
y f ,2
K f ,2  K i  (U g ,i ,2  U g , f ,2 )
U g ,1  (U g ,i ,2  U g , f ,2 )  mg( yi ,2  y f ,2 ) 
 mg( yi ,1  y f ,1 )  (U g ,i ,1  U g , f ,1 )  U g ,2
Only the change of potential energy has the physical meaning
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K i  U g ,i  K f  U g , f
1
K i  mv02
2
U g ,i  mgy0
U g , f  mgy1  0
1
K f  mv12
2
1
1
2
mv1  mv02  mgy0
2
2
v1  v02  2 gy0  4  100  10.2m / s
10
Restoring Force: Hooke’s Law
Hooke’s Law:
Fsp   k s
spring constant
11
Restoring Force: Hooke’s Law
Fsp   k s
The sign of a restoring force is
always opposite to the sign of
displacement
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Restoring Force: Elastic Potential Energy
Fsp   k s
1
U s  k (s )2
2
The elastic potential energy is
always positive
13
Restoring Force: Elastic Potential Energy
1
U s  k (s )2
2
1
1
2
K  U s  mv  k (s )2  constant
2
2
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Restoring Force: Elastic Potential Energy
1
1
2
K  U s  mv  k (s )2  constant
2
2
1
U s  k (s )2
2
15
Law of Conservation of Mechanical Energy
1
1
2
K  U g  U s  mv  mgy  k (s )2  constant
2
2
16

m2v fx ,2  m2vix ,2   m1v fx ,1  m1vix ,1
m1vix ,1  m2vix ,2  m1v fx ,1  m2 v fx ,2
pix ,1  pix ,2  p fx ,1  p fx ,2
pix ,total  p fx ,total
The law of conservation of momentum
17

Perfectly Elastic Collisions
During the collision the kinetic energy will be transformed into elastic
energy and then elastic energy will transformed back into kinetic
energy
Perfectly Elastic Collision: Mechanical Energy is Conserved (no
internal friction)
A
B
v A,i
A
v A, f
v B ,i
vB, f
1
1
2
m Av A,i  m B v B2 ,i 
2
2
1
1
2
 m Av A, f  m B v B2 , f
2
2
B
Perfectly inelastic collision:
A collision in which the two objects stick together and move with a
common final velocity – NO CONSERVATION OF MECHANICAL
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ENERGY
Perfectly Elastic Collisions
Conservation of Momentum and Conservation of Energy
isolated system
A
B
v A,i
A
v A, f
v B ,i
vB, f
B
m Av A,i  m B v B ,i  m A v A , f  m B v B , f
1
1
1
1
2
2
2
m Av A,i  mB vB ,i  m Av A, f  mB vB2 , f
2
2
2
2
Now we have enough equations to find the final velocities.
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Example:
v B ,i  0
v A,i
vB, f
v A, f
m Av A,i  m Av A, f  m B v B , f
1
1
1
m Av A2 ,i  m Av A2 , f  mB v B2 , f
2
2
2
v A, f
m A  mB

v A,i
m A  mB
vB , f 
2m A
v A,i
m A  mB
vB , f is always positive, v A, f can be positive (if m A  mB ) or
negative (if
v A, f  0
v B , f  v A, i
m A  mB )
if
m A  mB
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