Chapters 6, 7
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1cos 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
ma  Fnet
 mv2 

d 
2 


dx
 mv2 

Fnet dx  d 
 2 
Some calculus
• In 1D case
 mv2 

Fnet dx  d 
 2 
• In 3D case, similar derivations yield

 mv2 

  d K 
Fnet  dr  d 
 2 
• K – kinetic energy
mv 2
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
Work-kinetic energy theorem

 mv2 

  dK
Fnet  dr  d 
 2 
 
rf

K f  K i   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  lim
xi
x 0
xf
 F x
x
xi
• 1D case: Graphically - work is the area under the
curve Fx(x)
Chapter 6
Problem 52
A force with magnitude F = a√x acts in the x-direction, where a = 9.5 N/m1/2.
Calculate the work this force does as it acts on an object moving from (a) x = 0
to x = 3.0 m; (b) 3.0 m to 6.0 m; and (c) 6.0 m to 9.0 m.
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
• Since

rf
Wnet  
ri

Fnet

Fk  dr
Wk  
 ri
  Fk (vector sum)
k

 


r
r
 

f 

f

Fnet  dr     Fk   dr     Fk  dr 
ri
r

 k

k  i
Wnet  Wk ( scalar sum)
k
Work done by a constant force
• If a force is constant

rf
W  
ri
 
 rf   
F  dr  F   dr  F  r
ri
• If the displacement and the constant force are not
parallel
 
W  F  r  Fr cos   Fd cos 
Work done by a constant force
 
W  F  r  Fr cos   Fd cos 
Work done by a spring force
• Hooke’s law in 1D
Fs  kx
• From the definition of work
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  mgd cos 
• If the displacement is vertically up
Wg  mgd cos 180   mgd
• 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  K i  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)
 
 
F  dr
dW
 F v

P
dt
dt
P  Fv cos
James Watt
(1736-1819)
Chapter 6
Problem 36
A 75-kg long-jumper takes 3.1 s to reach a prejump speed of 10 m/s. What’s his
power output?
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  U i  W    F ( x)dx
U ( x)    F ( x)dx  C
xi
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
Gravitational potential energy
• For an upward direction the y axis
yf
U ( y )    (mg )dy  mgy f  mgyi  mgy
yi
U g ( y )  mgy
Gravitational potential energy
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
Chapter 7
Problem 37
A particle moves along the x-axis under the influence of a force F = ax2 + b,
where a and b are constants. Find its potential energy as a function of position,
taking U = 0 at x = 0.
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  U i
K f  K i  (U f  U i )
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
Conservation of mechanical energy
• From the work-kinetic energy theorem
K  Wnet
• When both conservative a nonconservative forces
do work on the object
Wnet  Wc  Wnc  U  Wnc  K
K  U  Wnc
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
Chapter 7
Problem 53
A spring of constant k = 340 N/m is used to launch a 1.5-kg block along a
horizontal surface whose coefficient of sliding friction is 0.27. If the spring is
compressed 18 cm, how far does the block slide?
Conservation of
mechanical
energy:
pendulum
Potential energy curve
dU ( x)
F ( x)  
dx
Potential energy curve: equilibrium
points
Neutral equilibrium
Unstable equilibrium
Stable equilibrium
Questions?
Answers to the even-numbered problems
Chapter 6
Problem 14:
9.6 × 106 J
Answers to the even-numbered problems
Chapter 6
Problem 40:
The hair dryer consumes more energy.
Answers to the even-numbered problems
Chapter 6
Problem 50:
360 J
Answers to the even-numbered problems
Chapter 7
Problem 14:
(a) 7.0 MJ
(b) 1.0 MJ
Answers to the even-numbered problems
Chapter 7
Problem 24:
(a) ± 4.9 m/s
(b) ± 7.0 m/s
(c) ≈ 11 m
Answers to the even-numbered problems
Chapter 7
Problem 38:
95 m