Transcript Document

Chapter 7
Work and Energy
Alternative description of dynamics
Energy
Integration over space Scalar
Momentum Integration over time
Vector
Significance: They are conserved
Valuable when the forces are not detailed
Wide range including Relativity & Quantum M.
Energy is useful in all areas of physics and in
other science as well
2
Work done by constant force
Accomplished by action of force over a distance
Work done by a constant force
W  Fd cos or W  F  d
When W = 0?
F d
F

d
Example1: An object moves as r  2ti  t 2 j ,
and there is a force F  5i  6k acting on it,
what is the work done by this force in 0~2s?
r  4i  4 j  W  F  r  20J
3
Work done by varying force
b
Divide the path into short intervals
F
During each interval, force ~ constant
Work
W  Fi  li
F
a
Total work W   Fi  li
li  0 Sum → Integral
b
W   F  dl
a
General definition of work
4
Work in component form
b
b
W   F  dl   F  dl
F
a
L
F
a
This is called a line integral.
dl
F  Fx i  Fy j  Fz k , dl  dxi  dyj  dzk
Work can be written as
b
b
b
a
a
a
W   Fx dx   Fy dy   Fz dz
Or:
dl  dl , F  dl  Ftan dl
b
 W   Ftan dl
a
5
Work depends on path
b
b
a
a
W   F  dl   Ftan dl
Work depends on the specific path from a to b
Moves on horizontal ground,
3 different paths from a to b,
work by friction is different!
b
W    Ndl    N S
1
a
2
b
3
a
Friction is called a nonconservative force
6
Work done by gravity
y
a
Gravity of object: G  mg
b
hb
a
ha
ha
W   Fy dy    mgdy
  mg  h
 L
mg
C
hb
b
o
x
Work done by gravity is path independent.
Determined only by the initial and final position.
Gravity is called a conservative force.
All constant forces are conservative.
7
Lose weight by climbing
Example2: Bob (80kg) wants to lose weight by
climbing mountains. How many calories he loses at
least when he goes from A to B? ( 1cal=4.184J )
Solution: The work done by gravity is
W   mgh   80  9.8  1500   1.18  10 J
6
So Bob loses energy 1.18  10 6 J
or 2.81  10 5 cal  281kcal
B
h=1500m
That is about 31g fat.
A
8
Stretching a spring
Example3: Someone slowly pulls a spring from
unstretched to make the object m leave the ground,
calculate the work done by this person.
Solution: force to stretch a spring: F = kx
x
x
1 2
 W   Fdx   kxdx  kx
0
0
2
How to make m leave the ground?
F
k
F  mg  x  mg / k
Work :
1 2 ( mg ) 2
W  kx 
2
2k
m
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Kinetic Energy
Energy: one of most important concepts in science.
There are many types of energy, it can be usually
regarded as “the ability to do work”.
Moving objects can do work → have energy
The energy of motion is called kinetic energy
Object starts from rest under constant net force:
v2 1 2
 mv
Wnet  Fnet d  mad  ma
2a 2
Translational kinetic energy Ek  1 mv 2
2
10
Work-energy principle
For varying force, 3 dimensional motion:
b
Wnet   Fnet  dl  a Ftan dl
b
a
dv 

 Ftan  m 
dt 

b
b
dv
dl
  m dl   m dv   mvdv
a
a
a
dt
dt
1 2 1 2
 mvb  mva  Ek
2
2
b
Work-energy principle
The net work done on an object is equal to the
change in its kinetic energy.
11
Work in an elliptical motion
Example4: A particle m moves under the equation
r  a cos  ti  b sin  tj , where a, b,  are positive
constants. Determine the net work during t=0~/2.
Solution1: The net force is
2
d r
F  m 2  m 2 r  m 2  xi  yj 
dt
x1
y1
Net work W   Fx dx   Fy dy
x0
0
y0
b
   m xdx    m 2 ydy
a
2
0
1
1
2 2
 m a  m 2 b 2
2
2
12
Example4: A particle m moves under the equation
r  a cos  ti  b sin  tj , where a, b,  are positive
constants. Determine the net work during t=0~/2.
Solution2: Using work-energy principle
dr
v
  a sin  ti   b cos  tj
dt
t=0: v0= b; t=/2: v = a
Net work equals to the change of kinetic energy:
1 2 1
1
1
2
2 2
2 2
W  mv  mv0  m a  m b
2
2
2
2
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Work done by friction
Example5: An object moves along a semi-circular
wall on smooth horizontal plane. Known: v0 and µ,
determine the work done by friction from A to B.
v2
Solution: Radial force N  m
R
A
dv
Tangential force   N  m
µ
v0
dt
v 2 dv dv d v dv
v



 
  
.
R dt d dt R d
o
dv
 
R
   d  v  v0 e
v
1
1
1 2 2 
2
2
B
W  mv  mv0  mv0 (e
 1)
2
2
2


14
Kinetic energy of spring
Example6: Considering the mass of spring ms (k, L),
it is attached by a mass M, determine kinetic energy
of the spring when M is moving with velocity v.
Solution: take dx of the spring, the mass is ms·dx/L
If the spring deforms uniformly, then vdx = v·x/L
2
x
2
1  dx  x  1
x
dEk   ms  v   ms v 2 3 dx
2  L  L  2
L
dx
L
1  ms  2
E k   dE k    v
0
2 3 
M
Effective mass
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*Forms of energy
There are 5 main forms of energy
Mechanical
Thermal / heat
Electromagnetic
Chemical
Nuclear
(Fission & Fusion)
Chain
reaction
*Nuclear bomb
“Little boy” & “Fat man”
“If the radiance of a thousand
suns were to burst into the
sky, that would be like the
splendor of the Mighty One.”
The effects of nuclear bomb:
Heat, shock wave and radiation (γ ray and neutrons)