Transcript Slide 1

Chapters 7, 8
Energy
Energy
• What is energy?
• Energy - is a fundamental, basic notion in physics
• Energy is a scalar, describing state of an object or a
system
• Description of a system in ‘energy language’ is
equivalent to a description in ‘force language’
• Energy approach is more general and more effective
than the force approach
• Equations of motion of an object (system) can be
derived from the energy equations
Scalar product of two vectors
• The result of the scalar (dot) multiplication of two
vectors is a scalar
 
A  B  AB cos
• Scalar products of unit vectors
iˆ  iˆ  11cos 0  1
ˆj  ˆj  1
kˆ  kˆ  1
iˆ  ˆj  1 1 cos 90   0
iˆ  kˆ  0
ˆj  kˆ  0
Scalar product of two vectors
• The result of the scalar (dot) multiplication of two
vectors is a scalar
 
A  B  AB cos
• Scalar product via unit vectors
 
A  B  ( Axiˆ  Ay ˆj  Az kˆ)(Bxiˆ  By ˆj  Bz kˆ)
 
A  B  Ax Bx  Ay By  Az Bz
Some calculus
• In 1D case
dx
v
dt
 v2 
d  
dv dv dx vdv
2


a


dx dt
dt
dx
dx
m a  Fnet
 m v2 

d 
2 


dx
 m v2 

Fnet dx  d 
 2 
Some calculus
• In 1D case
 m v2 

Fnet dx  d 
 2 
• In 3D case, similar derivations yield

 m v2 

  d K 
Fnet  dr  d 
 2 
• K – kinetic energy
m v2
K
2
Kinetic energy
•K
= mv2/2
• SI unit: kg*m2/s2 = J (Joule)
James Prescott Joule
(1818 - 1889)
• Kinetic energy describes object’s ‘state of motion’
• Kinetic energy is a scalar
Chapter 7
Problem 31
A 3.00-kg object has a velocity of (6.00^i – 2.00^j) m/s. (a) What is its kinetic
energy at this moment? (b) What is the net work done on the object if its
velocity changes to (800^i + 4.00^j) m/s?
Work–kinetic energy theorem

 m v2 

  dK
Fnet  dr  d 
 2 
 
rf

K f  Ki   Fnet  dr  Wnet
ri
• Wnet – work (net)
• Work is a scalar
• Work is equal to the change in kinetic energy, i.e.
work is required to produce a change in kinetic
energy
• Work is done on the object by a force
Work: graphical representation
xf
W   Fx dx
xi
• 1D case: Graphically - work is the area under the
curve F(x)
Net work vs. net force
• We can consider a system, with several forces
acting on it
• Each force acting on the system, considered
separately, produces its own work

rf
Wk  
ri
• Since
 
Fk  dr


Fnet   Fk (vector sum)
k
Wnet  Wk (scalar sum)
k
Work done by a constant force
• If a force is constant

rf
W  
ri
 
F  dr
 rf   
 F   dr  F  r
r
i
• If the displacement and the constant force are not
parallel
 
W  F  r  Fr cos  Fd cos
Work done by a spring force
• Hooke’s law in 1D
Fs  kx
• From the work–kinetic energy theorem
xf
xf
xi
xi
Ws   Fs dx    kxdx
2
i
kx2f
kx


2
2
Work done by the gravitational force
• Gravity force is ~ constant near the surface of the
Earth
Wg  mgdcos
• If the displacement is vertically up
Wg  mgdcos180  m gd
• In this case the gravity force does a negative work
(against the direction of motion)
Lifting an object
• We apply a force F to lift an object
• Force F does a positive work Wa
• The net work done
Wnet  K  K f  Ki  Wa  Wg
• If in the initial and final states the object is at rest,
then the net work done is zero, and the work done by
the force F is
Wa  Wg  mgd
Power
• Average power
Pavg
W

t
• Instantaneous power – the rate of doing work
dW
P
dt
• SI unit: J/s = kg*m2/s3 = W (Watt)
James Watt
(1736-1819)
Power of a constant force
• In the case of a constant force
 

 

d ( F  r )
dW
dr
 F v

P
F
dt
dt
dt
P  Fv cos 
Chapter 8
Problem 32
A 650-kg elevator starts from rest. It moves upward for 3.00 s with constant
acceleration until it reaches its cruising speed of 1.75 in/s. (a) What is the
average power of the elevator motor during this time interval? (b) How does
this power compare with the motor power when the elevator moves at its
cruising speed?
Conservative forces
• The net work done by a conservative force on a
particle moving around any closed path is zero
Wab,1  Wba, 2  0
Wab,1  Wba,2
Wab, 2  Wba, 2
Wab,1  Wab, 2
• The net work done by a conservative force on a
particle moving between two points does not depend
on the path taken by the particle
Conservative forces: examples
• Gravity force
 mghup  mghdown  0
• Spring force

2
kxright
2

2
kxleft
2
0
Potential energy
• For conservative forces we introduce a definition of
potential energy U
U  W
• The change in potential energy of an object is being
defined as being equal to the negative of the work
done by conservative forces on the object
• Potential energy is associated with the arrangement
of the system subject to conservative forces
Potential energy
• For 1D case
xf
U  U f  Ui  W   F ( x)dx
xi
U ( x)    F ( x)dx  C
dU ( x)
F ( x)  
dx
• A conservative force is associated with a potential
energy
• There is a freedom in defining a potential energy:
adding or subtracting a constant does not change the
force
• In 3D F ( x, y, z )   U ( x, y, z ) iˆ  U ( x, y, z ) ˆj  U ( x, y, z ) kˆ
x
y
z
Chapter 7
Problem 44
A single conservative force acting on a particle varies F = (– Ax + Bx2) ^i N,
where A and B are constants and x is in meters. (a) Calculate the potential
energy function U(x) associated with this force, taking U = 0 at x = 0. (b) Find
the change in potential energy and the change in kinetic energy of the system
as the particle moves from x = 2.00 m to x = 3.00 m.
Gravitational potential energy
• For an upward direction the y axis
yf
U ( y)   (mg)dy  mgyf  mgyi  mgy
yi
U g ( y)  mgy
Elastic potential energy
• For a spring obeying the Hooke’s law
U ( x)  
xf
xi
kx2f
kxi2
(kx)dx 

2
2
kx
U s ( x) 
2
2
Internal energy
• The energy associated with an object’s temperature
is called its internal energy, Eint
• In this example, the friction does work and
increases the internal energy of the surface
Conservation of mechanical energy
• Mechanical energy of an object is
Emec  K  U
• When a conservative force does work on the object
K  W U  W K  U
K f  U f  Ki  Ui
K f  Ki  (U f Ui )
Emec, f  Emec,i
• In an isolated system, where only conservative
forces cause energy changes, the kinetic and
potential energies can change, but the mechanical
energy cannot change
Work done by an external force
• Work is transferred to or from the system by means
of an external force acting on that system
W  K  U  Eint
• The total energy of a system can change only by
amounts of energy that are transferred to or from the
system
• Power of energy transfer, average and intantaneous
Pavg
E

t
dE
P
dt
Conservation of
mechanical
energy:
pendulum
Potential energy curve
dU ( x)
F ( x)  
dx
Potential energy curve: equilibrium
points
Neutral equilibrium
Unstable equilibrium
Stable equilibrium
Chapter 8
Problem 55
A 10.0-kg block is released from point A. The track is frictionless except for the
portion between points B and C, which has a length of 6.00 m. The block
travels down the track, hits a spring of force constant 2250 N/m, and
compresses the spring 0.300 m from its equilibrium position before coming to
rest momentarily. Determine the coefficient of kinetic friction between block
and the rough surface between B and C.
Answers to the even-numbered problems
Chapter 7
Problem 2:
(a) 3.28 × 10−2 J
(b) - 3.28 × 10−2 J
Answers to the even-numbered problems
Chapter 7
Problem 10:
16.0
Answers to the even-numbered problems
Chapter 7
Problem 46:
(7−9x2y)ˆi−3x3ˆj
Answers to the even-numbered problems
Chapter 8
Problem 14:
(a) 0.791 m/s
(b) 0.531 m/s
Answers to the even-numbered problems
Chapter 8
Problem 28:
8.01 W
Answers to the even-numbered problems
Chapter 8
Problem 34:
194 m
Answers to the even-numbered problems
Chapter 8
Problem 50:
(a) 0.588 J
(b) 0.588 J
(c) 2.42 m/s
(d) UC = 0.392 J, KC = 0.196 J