Transcript Chapter 13 Equilibrium

Fundamentals of Physics
Mechanics
(Bilingual Teaching)
张昆实
School of Physical Science and Technology
Yangtze University
Chapter 7 Kinetic Energy and work
7-1
7-2
7-3
7-4
7-5
7-6
Energy
Work
Work and Kinetic Energy
Work Done by a Gravitational Force
Work Done by a Spring Force
Work Done by a General Variable
Force
7-7 Power
7-1 Energy
★Newton’s laws of motion allow us to
analyze many kinds of motion. However,
the analysis is often complicated,
requiring details about the motion that
we simply do not know.
There is another technique for analyzing
motion , which involves energy.
Kinetic energy K is energy associated
with the state of motion of an object.
K  mv
2
2
SI unit: joule(J), 1J  1kg  m / s
1
2
2
(7-1)
(7-2)
7-2 Work
work W is energy transferred to or from
an object by means of a force acting on
the object. Energy transferred to the
object is positive work, and energy
transferred from the object is negative
work.
You acceletate an object
its kinatic
energy increaced
the work done by
your force is positive;
You deceletate an object
its kinatic
energy decreaced
the work done by
your force is negative;
7-3 Work and Kinetic Energy
Finding an Expression for Work
A bead can slide along a frictionless
horizontal wire ( x axis), A constant force
directed at an angle  to the wire,
accelerates the bead along the wire.
Fx  max
(7-3)
v  v0  2ax d
2
1
2
2
(7-4)
mv  mv0  Fx d
2
1
2
W  Fx d
v0
2
(7-5)
(7-6)
Bead
start
v

F
wire
d
end
x
7-3 Work and Kinetic Energy
To calculate the work done on an object
by a force during a displacement, we use
only the force component along the object’s
displacement. The force component perpendicular to the displacement does zero work.
Fx  F cos 
Since
General form
W  Fx d  Fd cos 
Scalar (dot) product
W  Fd cos   F  d
(7-7)
(7-8)
Cautions: (1) constant force (magnitude
and direction) (2) particle-like object.
7-3 Work and Kinetic Energy
Signs for work
0    90 cos   0 W  0 positive work
90    180 cos   0 W  0 negative work
  90 cos   0 W  0 does’t do work
A force does positive work when it has a vector
component in the same direction as the
displacement, and it does negative work when it
has a vector component in the opposite direction.
It does zero work when it has no such vector
component.
7-3 Work and Kinetic Energy
Units for work
The unit for work is the same as the unit for energy
1J  kg  m / s  1N  m
2
2
(7-9)
net work
When two or more forces act on an object,
their net work is the sum of the individual
workes by the forces, which is also equal
to the work that would be done on the
object by the net force F of those forces.
net
7-3 Work and Kinetic Energy
Work-kinetic Energy Theorem
2
2
1
1
(7-5)
2 mv  2 mv0  Fx d
K  K f  Ki  W
change in the kinetic
energy of a particle
=
(7-10)
net work done on
the particle
K f  Ki  W
kinetic energy after
The net work is done
(7-11)
=
“Work-kinetic Energy Theorem”
P120 bottom; Problem 7-2
kinetic energy
Before the net work
+
the net
Work done
7-4 Work Done by a Gravitational Force
A tomato is thrown upward, the work done by the
grivatational force : Wg  mgd cos 
(7-12)
In the rising prosses
Wg  mgd cos180  mgd
(7-13)
minus sign: the grivatational force transfers
energy (mgd) from the object’s kinetic energy,
consistant with the slowing of the object.
In the falling down prosses
Wg  mgd cos0  mgd
(7-14)
plus sign: the grivatational force transfers
energy (mgd) to the object’s kinetic energy,
consistant with the speeding up of the object.
7-4 Work Done by a Gravitational Force
Work done in lifting and lowering an object
Lifting an object
Displacement: upward
Lifting force: positive work;
transfers energy to the object;
Gravitational force: negative
work; transfes energy from
the object.
Lowering an object. Displacement: downward
Lifting force: negative work; transfers energy
from the object;
Gravitational force: positive work; transfes
energy to the object.
7-4 Work Done by a Gravitational Force
Work done in lifting and lowering an object
The change  K in the kinetic
energy due to these two
energy transfers is
K  K f  Ki  Wa  Wg
If
(7-15)
K f  Ki ( 0)
Wa  Wg  0
Wa  Wg
(7-17)

(7-16)
Wa  mgd cos 
The angle between
(Work in lifting and lowering;K f  Ki )
Fgand d .
Wa  mgd
Up:
Down: Wa  mgd
7-5 Work Done by a Spring Force
The spring force
Fig(a): A block is attached to a
spring in equilibrium (neither
compessed nor stretched).
X axis along the spring
relaxed state
stretched
Fig(b): when stretched to right
The spring pulls the block to the
Left (restoring force)
Fig(c): when compressed to left compressed
The spring pulls the block to the
right (restoring force)
The spring force
(Hooke’s law) (7-20)
(variable force)
K: spring constant
F  kd
F   kx
(Hooke’s law) (7-21)
7-5 Work Done by a Spring Force
The work done by a spring force
Simplifying assumptions:springmassless; ideal
spring(obeys Hooke’s law); contact frictionless.
Ws  Fj x (7-22)
x  0 Sumintegration
xf
xf
xi
xi
Ws   Fdx   kxdx

(7-23)
1 2 1 2 (7-24)
kxi  kx f
(7-25)
2
2
( work by a spring force )
'
F

F
Calculus metheod
o
F
x
P
segments
dW
O
xf
dx
xi x
x
7-5 Work Done by a Spring Force
The work done by a spring force
Work Ws is positive if the block ends up closer to
the relaxed position (x=0) than it was initially. It is
negative if the block ends up father away from x=0.
It is zero if the block ends up at the same distance
from x=0.
(7-25)
If
xi  0
and
xf  x
1 2
Ws   kx
2
'
F

F
1 2 1 2
Ws  kxi  kx f
2
2
o
x
P
then
( work by a spring force )
(7-26)
x
7-5 Work Done by a Spring Force
The work done by an applied force
Suppose we keep applying a force Fa on the
block, our force does work Wa , the spring force
does work Ws on the block.

F
The change  K in
the kinetic energy
of the block
K  K f  Ki  Wa  Ws
o
x
(7-27)
If the block is stationary before and after the
displacement, then K  K  0
f
i
Wa  Ws
(7-28)
Fa
P
x
7-5 Work Done by a Spring Force
The work done by an applied force
Wa  Ws
(7-28)
If a block that is attached to a spring is stationary
before and after a displacement, then the work
done on it by the applied force displacing it is the
negative of the work done on it by the spring
force.

F
o
x
Sample problem 7-8 : P128
Fa
P
x
7-6 Work Done by a General Variable Force
One-dimensional Analysis
the same method as the calculation of the
work done by a spring force (calculus).
Wj  Fj ,avg x
W  Wj  Fj ,avg x
W  lim Fj ,avg x
x 0
xf
W   F ( x)dx
xi
( Work:
variable force )
(7-29)
(7-30)
(7-31)
(7-32)
7-6 Work Done by a General Variable Force
Three-dimensional Analysis
A three-dimensional force acts on a body
(7-33)
F  Fx i  Fy j  Fz k
Simplifications: Fx only depends on x and so on
Incremental displacement
dr  dxi  dyj  dzk
The work dW done by
(7-34)
F during dr is
dW  F  dr  Fx dx  Fy dy  Fz dz
(7-35)
rf
xf
yf
zf
ri
xi
yi
zi
W   dW   Fx dx   Fy dy   Fz dz
(7-36)
7-6 Work Done by a General Variable Force
work-kinetic energy theorem with a variable
force
Let us prove:
xf
W   F ( x)dx  K
xi
xf
xf
xi
xi
( Work: variable force ) (7-32)
W   F ( x)dx   madx
dv
dv
madx  m dx  m vdx  mvdv
dt
dx
(7-37)
(7-38)
(7-40)
From the “chain rule” of calculus
dv dv dx dv

 v
dt dx dt dx
(7-39)
7-6 Work Done by a General Variable Force
work-kinetic energy theorem with a variable
force
Let us prove:
xf
xf
W   F ( x)dx   madx
xi
xi
dv
madx  m vdx  mvdv
dx
(7-37)
(7-40)
Substituting Eq. 7-40 into Eq. 7-37:
xf
vf
vf
xi
vi
vi
W   madx   mvdv  m vdv
 mv f  mvi
1
2
2
1
2
W  K f  Ki  K
2
work-kinetic energy theorem
(7-41)
7-7 Power
Power : The power due to a force is the rate
at which that force does work on an object.
If the force does work W during a time interval
t , the average power due to the force over
that time interval is
W
Pavg 
Instantaneous power is the
instantaneous rate of doing
work
If the direction of a force F is
at an angle  to the direction
of travel of the object, the
Instantaneous power is
t
dW
P
dt
(7-42)
(7-43)
P  Fv cos  F  v
(7-44)