Transcript Work produced by the net force

Work
Readings: Chapter 11
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Newton’s Second Law: Net Force is zero – acceleration is zero –
velocity is constant – kinetic energy is constant
Fnet  0  a  0  v  const 
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K  mv 2  const
2
Newton’s Second Law: If Net Force is not zero then: What is the
relation between the change of Kinetic energy and the Net Force?
v
Fnet
a
Where
v
Fnet  ma  m
t
v
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Fnet vt  m
vt  mvv  mv 2
t
2
1
Fnet s  mv 2  K
2
s  vt
- displacement
2
A
B
v
Fnet
a
s  xB  x A
If
Fnet  const
then
1
1
2
Fnet ( xB  x A )  K  mv B  mv A2
2
2
This equation gives the relation between the displacement and
velocity. It is the same equation as
v B2  v A2  2a ( x B  x A )
for the motion with constant acceleration (constant net force)
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A
B
v
a
Fnet
s  xB  x A
If
Fnet  const
then we have integral
xB
1
1
2
2
F
dx


K

mv

mv
B
A
x net
2
2
A
If the net force depends also on the velocity then the relation becomes more
complicated
xB
Fnet s
or
F
net
dx
is the Work done by net force
xA
W 
xB
F
net
xA
dx
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Then Work is
W  Fs s  F cos s
This combination is called the
dot product (or scalar product)
of two vectors
F coss  F  s
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Dot Product
a
Dot product is a scalar:

a  b  ab cos
b
If we know the components of the vectors
then the dot product can be calculated as
a  b  a x bx  a y by
y
ay
ax
by
If vectors are orthogonal then dot product is zero
  90  cos   0
0
a b  0
a
bx
x
b
a
900
b
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a  b  ab cos
Dot Product: properties
(ca )  b  c(a  b )  cab cos 
a
(a1  a2 )  b  a1  b  a2  b
Dot product is positive if
  900
cos  0
Dot product is negative if
  900
cos  0
The magnitude of
angle  is 600
a
is 5, the magnitude of
What is the dot product of
a
and
b

b
is 2, the
b
1
a  b  5  2cos 60  5  2   5
2
0
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Work produced by a force, acting on an object, is the dot product
of the force and displacement
final
point
W  F  s
W 

initial
point
Work has the same units as the energy:
F  s
kg  m
kg  m 2
1J  1 N  m  1 2 m  1
s
s2
Example: What is the work produced by the tension force?
T  10 N
s  100m
  450
W  T  s  T s cos  
s
 10  100cos 450 
 707 N  m  707J8
K  K final  K initial  W  F  s
W  F  s  F s  0
(a  0)
K final  K initial
v f  vi
W  F  s  0
K final  K initial
(a  0)
v f  vi
W  F  s  0
K final  K initial
(a  0)
v f  vi
W  F  s  0
K final  K initial
(a  0)
v f  vi
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Work produced by a NET FORCE is equal to a change of KINETIC
ENERGY (it follows from the second Newton’s law)
Wnet  Fnet  s  K  K final  K initial
The NET FORCE is equal to (vector) sum of all forces acting on the
object
Fnet  F1  F2  ...
then
Wnet  ( F1  F2  ...)  s  F1  s  F2  s  ...  W1  W2  ...  K
so the sum of the works produced by all forces acting on the object is
equal to the change of kinetic energy,
W1  W2  ...  K
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Example: What is the change of kinetic energy? Or what is the
final velocity of the block if the initial velocity is 5 m/s? The mass
of the block is 5 kg.
Forces:
normal force
n
gravitational force
tension
w
T  10 N
s
n
w
T
  450
s  100m
Work produced by the net force is equal to the change of kinetic energy.
Wnet
1
1
2
 Fnet  s  K  mv f  mvi2
2
2
Since
Fnet  n  w  T
then
Wnet  Wn  Ww  WT  n  s  w  s  T  s  T s cos
Wn  n  s  0
Ww  w  s  0
WT  T  s  T s cos
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Example: What is the change of kinetic energy? Or what is the
final velocity of the block if the initial velocity is 5 m/s? The mass
of the block is 5 kg.
Forces:
normal force
n
gravitational force
tension
w
T  10 N
s
n
w
T
  450
1
1
1
2
2
mv f  mvi  Wnet  mvi2  T s cos 
2
2
2
T cos 
2
2
v f  vi  2
s
m
s  100m
This problem can be also solved by finding acceleration (this is the
motion with constant acceleration). Then
v  v  2as
2
f
2
i
T cos 
a
m
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Example: Free fall motion.
Forces:
yi
gravitational force
w
yf
s
w
Work produced by the net force (gravitational
force) is equal to the change of kinetic energy.
Wnet
1
1
2
 Fnet  s  K  mv f  mvi2
2
2
Wnet  Ww  w  s  mgs  mg( yi  y f ) 
then
1
1
mv 2f  mvi2
2
2
1
1
mgyi  mvi2  mgy f  mv 2f
2
2
Conservation of mechanical energy
Work produced by gravitational force can be written
as the change of gravitational potential energy
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Not for all forces the work can be written as the difference between
the potential energy between two points. Only if the force does not
depend on velocity then we can introduce potential energy as
W  F  s  U initial  U final
Potential energy depends only on the position of the object
It means also that the work done
by the force does not depend on
the trajectory (path) of the object:
W path1  W path 2  U A  U B
Such forces are called conservative forces
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Conservative Forces
Example 1: Gravitational Force
U  mgh
Example 2: Elastic Force
xf
W 

xi
xf
Fdx 

(  kx )dx   k
xi
2
x
2
xf

xi
x 2f
xi2
k
k
 U initial  U final
2
2
x2
Uk
2
Nonconservative (dissipative) forces
Example: Friction
Direction of friction force is always opposite to the direction of velocity, so
the friction force depends on velocity.
The work done by friction force depends on path.
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If we know force we can find the corresponding potential energy
W  F  s  U  (U final  U initial )
We assume that the potential (for example)
at point A is 0, then the potential energy at
point B is
B
B
A
A
U B  U B   F  ds    F  ds
U
s
mg y
Fy  
  mg
y
If we know potential we can find the force as
Example: Gravitational force:
Example: Elastic force:
U  mgy
x2
Uk
2
The exact relation between
U ( x, y, z )
F


the force and potential:
x
x
Fx  
Fs  
k  x 2 / 2
x
y
  kx
U ( x16
, y, z )
U ( x , y , z )
Fz  
Fy  
y
z
Nonconservative (dissipative) forces
Example 1: Friction
Direction of friction force is always opposite to the direction of
velocity, so the friction force depends on velocity.
The work done by friction force depends on path.
W path1   f k s1  f k s2   f k ( s1  s2 )
fk
path 1
B
s2
s1
fk
path 2
s3
fk
A
Example 2: External Force
W path 2   f k s3
s3  ( s1  s2 )
W path1  W path 2
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The work done by the net force is equal to the change of kinetic
energy.
The net force is the sum of conservative forces (gravitational force,
elastic force, …) and nonconservative force (friction force ..). The
work done by conservative forces can be written as the change of
potential energy. Then
Wnet  Fnet  s 
 Fconservative  s  Fnonconservative  s 
 U initial  U final  Fnonconservative  s  K final  K initial
Fnonconservative  s  (U final  K final )  (U initial  K initial ) 
 Emech, final  Emech,initial
Work done by nonconservative forces is equal to the change of
mechanical energy of the system (sum of kinetic energy,
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gravitational potential energy, elastic potential energy, and so on).
Example
vinitial  0
h
v final  0
Work done by friction force is equal to the change of mechanical
energy:
W friction   f k  ds  Emech, final  Emech,initial   mgh
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Nonconservative forces: friction (dissipative) forces and external
forces.
Friction force decreases the mechanical energy of the system but
increases the TEMPERATURE of the system – increases thermal
energy of the system. Then
F friction  s  Ethermal ,initial  Ethermal , final
Fnonconservative  s  Fexternal  s  Ethermal ,initial  Ethermal , final 
 Emech, final  Emech,initial
Fexternal  s  ( Emech, final  Ethermal , final )  ( Emech,initial  Ethermal ,initial )
Work done by external forces is equal to the change of the total
energy of the system (mechanical + thermal)
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Work done by external forces is equal to the change of the total
energy of the system (mechanical + thermal)
Thermal energy is also a mechanical energy – this is the energy of
motion of atoms or molecules inside the objects.
Fexternal  s  ( Emech, final  Ethermal , final )  ( Emech,initial  Ethermal ,initial )
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