Download Energy, Kinetic Energy, Work, Dot Product, and

Energy, Kinetic Energy, Work,
Dot Product, and Power
8.01
W08D1 Fall 2006
Forms of Energy
• kinetic energy
• gravitational energy
• elastic energy
• thermal energy
• electrical energy
• chemical energy
• electromagnetic energy
• nuclear energy
• mass energy
Energy Transformations
• Falling water releases stored ‘gravitational potential energy’
turning into a ‘kinetic energy’ of motion.
• Human beings transform the stored chemical energy of food
into catabolic energy
• Burning gasoline in car engines converts ‘chemical energy’
stored in the atomic bonds of the constituent atoms of
gasoline into heat
• Stretching or compressing a spring stores ‘elastic potential
energy’ that can be released as kinetic energy
Energy Conservation
•
Energy is always conserved
N
 E
i 1
i
 E1  E2  ...  0
• It is converted from one form into another, from an “initial
state” to a “final state”
E  Efinal  Einitial
• Energy can also be transferred from a system to its
surroundings
Esystem  Esurroundings  0
Kinetic Energy
• Scalar quantity (reference frame dependent)
1 2
K  mv  0
2
• SI unit is joule:
1J  1kg  m2 /s 2
• Change in kinetic energy:
1 2 1 2 1
1
2
2
2
2
2
K  mv f  mv0  m(vx, f  v y, f  vz, f )  m(vx,0
 v 2y,0  vz,0
)0
2
2
2
2
Concept Question
Compared to the amount of energy required to
accelerate a car from rest to 10 mph (miles per
hour), the amount of energy required to accelerate
the same car from 10 mph to 20 mph is
(1) the same
(2) twice as much
(3) three times as much
(4) four times as much
Kinematics: Recall
Integration Formula
The x-component of the acceleration of an object
is the derivative of the x-component of the velocity
dv x
ax 
dt
Therefore the integral of x-component of the
acceleration with respect to time, is the xcomponent of the velocity
tf
t f dv
vx , f
x
t0 ax dt  t0 dt dt  vx ,0 dvx  vx, f  vx,0
Multiply both sides by the mass of the object

tf
t0
max dt  mvx, f  mvx,0  px
Kinematics: An Integral
Theorem for One Dimensional
Motion
The integral of x-component of the acceleration
with respect to the displacement of an object, is
given by
xf
x f dv
xf
vx , f
dx
x
x0 ax dx  x0 dt dx  x0 dvx dt  vx ,0 vx dvx

xf
x0
ax dx 

vx , f
vx ,0

d (1 / 2)vx
2

1 2
2
 (vx, f  vx,0
)
2
Multiply both sides by the mass of the object

xf
x0
max dx 

vx , f
vx ,0

d (1 / 2)vx
2

1 2
1 2
 mvx, f  mvx,0  K
2
2
Newton’s Second Law: An
Integral Theorem for One
Dimensional Motion
Newton’s Second Law If F is not constant, then (in
one dimension),
Fx  max
Therefore the integral theorem coupled with
Newton’s Second Law becomes

xf
x0
Fx dx  K
Work Done by a Constant
Force for One Dimensional
Motion
Definition:
The work W done by a constant force with an x-component,
Fx, in displacing an object by x is equal to the xcomponent of the force times the displacement:
W  Fx x
Work done by Non-Constant
Force for One Dimensional
Motion
(Infinitesimal) work is a scalar
Wi  (Fx )i xi
Add up these scalar quantities to get the total work as area under graph
of Fx vs x :
iN
iN
i 1
i 1
W   Wi   ( Fx )i xi
As N   and xi  0
W  lim
iN
 ( F ) x
N 
xi 0 i 1
x i
i
x x f


x  x0
Fx dx
Work-Kinetic Energy
Theorem for One
Dimensional Motion
Work-energy theorem is the statement that the work done by
a force in displacing an object is equal to the change in
kinetic energy of the object
W  K
Concept Question
Consider two carts, of masses m and 2m, at rest
on an air track. If you push one cart for 3 s and
then the other for the same length of time, exerting
equal force on each, the kinetic energy of the light
cart is
(1) larger than
(2) equal to
(3) smaller than
the kinetic energy of the heavy car.
Concept Question
A particle starts from rest at x = 0 and moves to x = L
under the action of a variable force F(x), which is
shown in the figure. What is the particle's kinetic
energy at x=L/2 and at x=L?
(1) (Fmax)(L/2), (Fmax)(L)
(2) (Fmax)(L/4), 0
(3) (Fmax)(L), 0
(4) (Fmax)(L/4), (Fmax)(L/2)
(5) (Fmax)(L/2), (Fmax)(L/4)
Concept Question
When a person walks, the force of friction
between the floor and the person's feet
accelerates the person forward. The floor
does
(1) Positive work on the person
(2) Negative work on the person
(3) No work on the person
Power
• The average power of an applied force is
the rate of doing work
W Fapplied, x x
P

 Fapplied,x vx
t
t
• SI units of power: Watts
1W  1J/s  1kg  m /s
2
• Instantaneous power
3

W
x 
P  lim
 Fapplied,x  lim
 Fapplied,x vx

t0 t
 t0 t 
Dot Product
A scalar quantity
Magnitude:
r r
r r
A  B  A B cos
The dot product can be positive, zero, or negative
Two types of projections: the dot product is the parallel
component of one vector with respect to the second vector
times the magnitude of the second vector
r r
r
r
r
A  B  A (cos ) B  AP B
r r
r
r
r
A  B  A (cos ) B  A BP
Dot Product Properties
AB  BA
cA  B  c( A  B)
( A  B)  C  A  C  B  C
Dot Product in Cartesian
Coordinates
With unit vectors ˆi, ˆj and kˆ
ˆi  ˆi  ˆj  ˆj  kˆ  kˆ  1
ˆi  ˆj  ˆi  kˆ  ˆj  kˆ  0
ˆi  ˆi  | ˆi || ˆi | cos(0)  1
ˆi  ˆj | ˆi || ˆj |cos( /2)  0
Example:
ˆ,
A  Axˆi  Ayˆj  Az k
ˆ
B  Bxˆi  Byˆj  Bz k
A  B  Ax Bx  Ay By  Az Bz
Work Done by a Constant
Force
Definition: Work
r
The work done by a constant force F on an object is equal to the
component of the force in the direction of the displacement times the
magnitude of the displacement:
r r r r
r
r
r
W  F  r  F r cos  F cos r  FP r
Note that the component of the force in the direction of the
displacement can be positive, zero, or negative so the work may be
positive, zero, or negative
Work as a Dot Product
Let the force exerted on an object be
F  F ˆi  F ˆj
x
y
Fx  F cos 
Fy  F sin 
r
Displacement: r  x φ
i
r r
W  F  r  Fx cos 
 (Fx φ
i  Fy φ
j)  (x φ
i)  Fx x
Concept Question: Work
A ball is given an initial horizontal velocity and allowed to
fall under the influence of gravity, as shown below. The
work done by the force of gravity on the ball is:
(1) positive
(2) zero
(3) negative
Concept Question: Work
A comet is speeding along a hyperbolic orbit toward the Sun.
While the comet is moving away from the Sun, the work
done by the Sun on the comet is:
(1) positive
(2) zero
(3) negative
Work in Three Dimensions
Let the force acting on an object be given by
r
F  Fx φ
i  Fy φ
j  Fz kφ
The displacement vector for an infinitesimal displacement is
r
dr  dx φ
i  dy φ
j  dy kφ
The work done by the force for this infinitesimal displacement is
r r
φ  (dx φ
φ
dW  F  dr  ( Fx φ
i  Fy φ
j  Fz k)
i  dy φ
j  dy k)
dW  Fx dx  Fy dy  Fz dz
By Newton’s Second Law
dW  max dx  may dy  maz dz
Three Integral Formulas for
Three- Dimensional Motion
Recall

xf
x0
max dx 

dvx
m
dx 
dt
xf
x0

xf
x0
dx
m dvx 
dt

vx , f
vx ,0
mvx dvx 
1 2
1 2
mvx, f  mvx,0
2
2
Repeat the argument for the y-direction

yf
y0
1 2
1 2
ma y dx  mv y, f  mv y,0
2
2
And for the z-direction

zf
z0
1 2
1 2
maz dx  mvz, f  mvz,0
2
2
Adding these three results gives the work done in moving an
object from A to B
B
1
1
2
2
2
2
2
2
(ma
dx

ma
dy

ma
dz)dx

m(v

v

v
)

m(v

v

v
)
y
z
x, f
y, f
z, f
x,0
y,0
z,0
A x
2
2
1 2 1 2
 mv f  mv0  K
2
2
Work-Energy Theorem in
Three-Dimensions
Recall

B
A
(max dx  ma y dy  maz dz)dx  K
From Newton’s Second Law the infinitesimal work is
r r
dW  max dx  may dy  maz dz  F  dr
Therefore
r r 1 2 1 2
A dW  A F  dr  2 mv f  2 mv0
B
B
Work-Energy Theorem
W  K
Appendix: Work Done Along an
Arbitrary Path
r
r
Wi  Fi  ri
r
r
W  lim  Fi  ri 
i N
N r
ri 0 i1

B
A
r r
F  dr
Appendix: Line Integrals
• force vector
r
F  Fx φi  Fy φj  Fz kφ
• line element
ˆ
dr  dxˆi  dy ˆj  dz k



ˆ  dxˆi  dy ˆj  dz k
ˆ  F dx  F dy  F dz
F  dr  Fx ˆi  Fy ˆj  Fz k
x
y
z
r
• total work: Denote A by r0 and B
B
r r
W   F  dr 
A
r r
r  rf

r r
r  r0
r r
r  rf
Fx dx 

r r
r  r0
r
by r f
r r
r  rf
Fy dy 

r r
r  r0
Fz dz