Transcript Work and Kinetic Energy

Kinetic Energy, Work, Power,
and Potential Energy
8.01
W05D1
Today’s Reading Assignment:
W05D1
Young and Freedman: 6.1-6.4
Math Review Module: Scalar Product
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:
K 
1 2 1 2 1
1
2
2
mv f  mv0  m(vx,2 f  v 2y, f  vz,2 f )  m(vx,0
 v 2y,0  vz,0
)0
2
2
2
2
Concept Question: Work
and Kinetic Energy
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
(5) unsure.
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
Concept Question: Pushing
against a wall
The work done by the contact force
of the wall on the person as the
person moves away from the wall is
1. positive.
2. negative.
3. zero.
4. impossible to determine from
information given in question and
the figure.
Concept Question: Work
and Walking
When a person walks, the force of friction between the
floor and the person's feet accelerates the person
forward. The work done by the friction force is
1. positive.
2. negative.
3. zero.
Pushing a Stalled Car
Table Problem: Work Done by
Gravity Near the Surface of the
Earth
Consider an object of mass near the surface of
the earth falling directly towards the center of the
earth. The gravitational force between the object
and the earth is nearly constant. Suppose the
object starts from an initial point that is a distance
y0 from the surface of the earth and moves to a
final point a distance yf from the surface of the
earth. How much work does the gravitational
force do on the object as it falls?
Work done by Non-Constant
Force: 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
Concept Question: Work
due to Variable Force
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)
Table Problem: Work Done by
the Spring Force
Connect one end of a spring of length l0 with spring
constant k to an object resting on a smooth table and fix
the other end of the spring to a wall. Stretch the spring
until it has length l and release the object.
How much work does the spring do on the object as a
function of x = l - l0, the distance the spring has been
stretched or compressed?
Worked Example: Work Done
by Several Forces
A block of mass m slides along a horizontal
table with speed v0. At x = 0 it hits a spring with
spring constant k and begins to experience a
friction force. The coefficient of kinetic friction is
given by m. How far did the spring compress
when the block first momentarily comes to rest?
Recall integration formula
for acceleration with respect
to time
The x-component of the acceleration of an object
is the derivative of the x-component of the velocity
dvx
ax 
dt
Therefore the integral of x-component of the
acceleration with respect to time, is the xcomponent of the velocity

tf
t0
ax dt 

tf
t0
dvx
dt 
dt

vx , f
vx ,0
dvx  vx, f  vx,0
Integration formula for
acceleration with respect to
displacement
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
vx , f
1 2
2
2
a
dx

d
(1
/
2)v

(v

v
)
x
x, f
x,0
x0 x vx ,0
2


Multiply both sides by the mass of the object giving
integration formula

xf
x0
max dx 

vx , f
vx ,0

d (1 / 2)vx
2

1 2
1 2
 mvx, f  mvx,0  K
2
2
Work-Kinetic Energy Theorem
One Dimensional Motion
Substitute Newton’s Second Law (in one dimension)
Fx  max
in definition of work integral which then becomes
W

xf
x0
Fx dx 

xf
x0
max dx
Apply integration formula to get work-kinetic energy theorem
W

xf
x0
Fx dx 

xf
x0
max dx  K
Concept Question: WorkEnergy
An object is dropped to the earth from a height of 10m.
Which of the following sketches best represent the kinetic
energy of the object as it approaches the earth (neglect
friction)?
1.
2.
3.
4.
5.
a
b
c
d
e
Concept Question
Two objects are pushed on a frictionless surface from a starting line to a
finish line with equal constant forces. One object is four times as massive
as the other. Both objects are initially at rest. Which of the following
statements is true when the objects reach the finish line?
1. The kinetic energies of the two objects are equal.
2. Object of mass 4m has the greater kinetic energy.
3. Object of mass m has the greater kinetic energy.
4. Not information is given to decide.
Worked Example: Work-Energy
Theorem for Inverse Square
Gravitational Force
Consider a magnetic rail gun that shoots an
object of mass m radially away from the surface
of the earth (mass me). When the object leaves
the rail gun it is at a distance ri from the center of
the earth moving with speed vi . What speed of
the object as a function of distance from the
center of the earth?
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 
Work and the Dot Product
Dot Product
A scalar quantity
Magnitude:
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
A  B  A (cos  ) B  A B
A  B  A (cos  ) B  A B
Dot Product of Unit Vectors in
Cartesian Coordinates
For unit vectors
We have
φ
i, φ
j
ˆi  ˆi  | ˆi || ˆi | cos(0)  1
ˆi  ˆj | ˆi || ˆj |cos( /2)  0
Generally:
φ
iφ
i φ
j φ
j  kφ kφ  1
φ
iφ
j φ
i  kφ  φ
j kφ  0
Example: Dot Product in Cartesian
Coordinates
r
φ
A  Ax φ
i  Ay φ
j  Az k,
r
B  Bx φ
i  By φ
j  Bz kφ
Then
r r
φ
φ  (B φ
φ
A  B  ( Ax φ
i  Ay φ
j  Az k)
i

B
j

B
k)
x
y
z
 Ax φ
i  Bx φ
i  Ax φ
i  By φ
j  Ax φ
i  Bz kφ
 Ay φ
j  Bx φ
i  Ay φ
j  By φ
j  Ay φ
j  Bz kφ
 Az kφ  Bx φ
i  Az kφ  By φ
j  Az kφ  Bz kφ
 Ax Bx  Ay By  Az Bz
Kinetic Energy and Dot
Product
Velocity
r
v  vx φ
i  vyφ
j  vz kφ
Kinetic Energy:
1 r r
1
K  m( v  v)  m(vx2  v 2y  vz2 )  0
2
2
Change in kinetic energy:
1 2 1 2 1 r r
1 r r
K  mv f  mv0  m( v f  v f )  m( v 0  v 0 )
2
2
2
2
1
1
2
2
2
2
2
m(vx, f  v y, f  vz, f )  m(vx,0
 v 2y,0  vz,0
)
2
2
Work Done by a Constant
Force
Definition: Work
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:
W  F  r  F r cos   F cos  r  F 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
Concept Question: Work
and Gravity 1
A ball is given an initial horizontal velocity and allowed to
fall under the influence of gravity near the surface of the
earth, as shown below. The work done by the force of
gravity on the ball is:
(1) positive
(2) zero
(3) negative
Worked Example: Work
Done by a Constant Force in
Two Dimensions
Let the force exerted on an object be
F  F ˆi  F ˆj
x
y
Fx  F cos 
Fy  F sin 
Displacement: r  x ˆi
W  F  r  F x cos 
 ( Fx ˆi  Fy ˆj)  (x ˆi )  Fx x
Table Problem: Work Constant
Forces and Dot Product
An object of mass m, starting from rest,
slides down an inclined plane of length s.
The plane is inclined by an angle of θ to
the ground. The coefficient of kinetic
friction is μ.
a) Use the dot product definition of work to calculate the work done
by the normal force, the gravitational force, and the friction force
as the object displaces a distance s down the inclined plane.
b) For each force, is the work done by the force positive or
negative?
c) What is the sum of the work done by the three forces? Is this
positive or negative?
Concept Question: Work
and inverse square gravity
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 Done Along an Arbitrary
Path
Work done by force for
small displacement
r
r
Wi  Fi  ri
Work done by force along
path from A to B
r
r
 lim  Fi  ri 
i N
WAB
N r
ri 0 i1

B
A
r r
F  dr
Work-Energy Theorem in
Three-Dimensions
As you will show in the problem set, the one dimensional
work-kinetic energy theorem generalizes to three dimensions
r B
r r B r r B d vr r B
r dr
r r
  F  d r   ma  d r   m
 dr   mdv 
  mdv  v
dt
dt A
A
A
A
A
B
WAB
1 2 1 2
 mv B  mv A  K B  K A
2
2
 K
Work: Path Dependent Line
Integral
Work done by force along path from A to B
r
r
 lim  Fi  ri 
i N
WAB
N r
ri 0 i1

B
A
r r
F  dr
In order to calculate the line integral, in principle,
requires a knowledge of the path. However we will
consider an important class of forces in which the work
line integral is independent of the path and only
depends on the starting and end points
Conservative Forces
Definition: Conservative Force If the work done by
a force in moving an object from point A to point B is
independent of the path (1 or 2),
B
Wc   Fc  dr
(path independent)
A
then the force is called a conservative force which
we denote by Fc . Then the work done only depends
on the location of the points A and B.
Example: Gravitational Force
Consider the motion of an object under the influence
of a gravitational force near the surface of the earth
The work done by gravity depends only on the
change in the vertical position
Wg  Fg y   mg y
Potential Energy Difference
Definition: Potential Energy Difference between the
points A and B associated with a conservative force Fc
is the negative of the work done by the conservative
force in moving the body along any path connecting
the points A and B.
B
U    Fc  dr   Wc
A
Potential Energy Differnece:
Constant Gravity
Force:
r
r
F  mg  F y φ
j  mg φ
j
Work:
W  Fy y  mgy
Potential Energy:

U  W  mgy  mg y f  y0
Choice of Zero Point: Choose y0  0 and
choose U ( y0 )  0 . Then
Potential Energy:
U ( y)  mgy

Worked Example: Change in
Potential Energy for Inverse
Square Gravitational Force
Consider an object of mass m1 moving towards
the sun (mass m2). Initially the object is at a
distance r0 from the center of the sun. The object
moves to a final distance rf from the center of the
sun. For the object-sun system, what is the
change in potential during this motion?
Worked Example Solution:
Inverse Square Gravity
Force:
Work done:
Fm1 ,m2
rf
Gm1m2

rˆ
2
r
rf
Gm1m2
 Gm1m2 
W   F  dr    
 dr 
2
r
r

r0
r0 
Potential Energy
Change:
Zero Point:
Potential Energy
Function
rf
r0
1 1
 Gm1m2   
r r 
0 
 f
 1 1
U  W  Gm1m2   
 r f r0 
U (r0  )  0
Gm1m2
U (r) = 
r
Table Problem: Change in
Potential Energy Spring Force
Connect one end of a spring of length l0 with
spring constant k to an object resting on a
smooth table and fix the other end of the spring
to a wall. Stretch the spring until it has length l
and release the object. Consider the objectspring as the system. When the spring returns to
its equilibrium length what is the change in
potential energy of the system?
Potential Energy Difference:
Spring Force
Force:
F  Fx ˆi  kxˆi
xxf
Work done:
W
 
x  x0

1
kx dx   k x 2f  x02
2

Potential Energy
Change:
1
U  W  k x 2f  x02
2
Zero Point:
U (x  0)  0
Potential Energy
1 2
U (x)  kx
2


Work-Energy Theorem:
Conservative Forces
The work done by the force in moving an object
from A to B is equal to the change in kinetic energy
r r 1 2 1 2
W   F  d r  mvB  mv A  K
A
2
2
B
When the only forces acting on the object are
conservative forcesr
r
F  Fc
then the change in potential energy is
U  W
Therefore
U  K
Table Problem : Asteroid
about Sun
An asteroid of mass m is in a non-circular closed orbit
about the sun. Initially it is a distance ri from the sun, with
speed vi. What is the change in the kinetic energy of the
asteroid when it is a distance is rf, from the sun?
Next Reading Assignment:
W05D2
Young and Freedman: 7.1-7.5,12.3
Experiment 3: Energy Transformations
44