Transcript No Slide Title

Chapter 6
Work and kinetic energy
In chapters 4 and 5 we studied Newton’s laws of motion and
applied them to various situations.
In chapter 6 we shall introduce a new approach that makes
the solution of mechanics problem easier
The following concepts will be introduced
• Kinetic energy (symbol: K )
• Work (symbol: W )
We will also introduce and use the work-energy theorem
(6-1)
The scalar product of two vectors (also known as "dot" product)
A  B  AB cos
Note 1:
A  B  B A
Note 2: A  B  0 when   90
The dot product in terms of vector components
A  Ax i  Ay j  Az k
B  Bx i  By j  Bz k
A  B  Ax Bx  Ay By  Az Bz
(6-2)
Work-energy theorem
(6-3)
Consider the motion of an object of mass m along the x-axis
from point 1 (coordinate x1) to point 2 (coordinate x2). A
constant net force Fnet acts on the object. The velocity at
points 1 and 2 is v1 and v2 , respectively
a
v1
.
O
.
1
x1
2a( x2  x1 )  v22  v12
m
Fnet
v2
.
2
x2
x-axis
(third equation of kinematics)
Fnet
a
Substitute a in the equation above 
m
2
2
Fnet
mv
mv
2
( x2  x1 )  v22  v12  Fnet ( x2  x1 )  2  1
m
2
2
v1
.
O
.
m
1
x1
x
Fnet
v2
(6-4)
.
2
x2
x-axis
mv22 mv12
Fnet ( x2  x1 ) 

Work-energy theorem x2  x1  x
2
2
mv22 mv12
Fnet x 

We define as work (symbol W)
2
2
performed by Fnet during the motion from point 1 to point 2:
Wnet  Fnet x
Units: Nm  Joule (J)
We define as the kinetic energy of a moving object
kg  m 2
Units:
 Nm  J
2
s
The work-energy theorem can be written as :
mv 2
K
2
Wnet  K
Algebraic sign for Wnet = Fnetx (all possible scenarios) (6-5)
I
II
III
IV
.
O
.
O
.
O
.
O
.
1
.
2
.
1
.
2
m
Fnet
x
m
x
Fnet
.
2
Fnet
.
x-axis
1
x-axis
m
Fnet m
x
x
.
x > 0
Fnet > 0
Wnet > 0
x < 0
Fnet > 0
Wnet < 0
x-axis
x > 0
Fnet < 0
Wnet < 0
x-axis
x < 0
Fnet < 0
Wnet > 0
2
.
1
Caution
If there are more than one forces (F1 , F2, F3 , …FN) acting on
the moving object we calculate the net work Wnet as follows:
We first determine the work each force performs:
W1 = F1 x , W2 = F2 x , W3 = F3 x , …, WN = FN x
Wnet is simply the sum of all the terms above i.e.
Wnet = W1 + W2 + W3 + …+ WN
Note: The work-energy theorem Wnet = K applies for the
net work. All forces acting on the object for which we are
applying the theorem must be included
(6-6)
Example (6-1) page 146 A car of mass m = 1200 kg falls a
vertical distance h = 24 m starting from rest. Find the velocity
v2 of the car before it hits the water.
.1
Fnet  mg
Wnet  Fnet y  (-mg )(-h)  mgh
y Wnet  1200  9.8  24  2.8 105 J
mg
Wnet  K
y
h
.2
water
(6-7)
v2
(work-energy theorem)
mv22
mv12
K 
v1  0
2
2
mv22

 Wnet  v2  2 gh
2
v2  2  9.8  24  21.7 m/s
(6-8)
m
Work-energy theorem when we have
motion of an object of mass m in a plane
from point 1 to point 2 under the
action of a constant force F
Displacement =  r
Work-energy theorem: Wnet  K
Below we give the general definition of W
W  F  r  F r cos
mv22 mv12
K 

v1 and v2 are the velocities of the object
2
2
at point 1 and point 2, respectively
1
m
mg
r
An object of mass m is thrown with an
initial speed v1 off a tall building. Find
the object’s speed after it has fallen a
vertical distance h
Wnet
h
2
L
mv22 mv12
 K 

2
2
 r  Li  h j
Fnet  mg j
A  B  Ax Bx  Ay By
Wnet  Fnet   r  (mg )(h)  mgh
mv22 mv12
K 

 mgh
2
2
(6-9)
v2  v12  2 gh
Solve for v 2
(6-10)
Consider the
motion of an
object from
point xo to point
xf along the xaxis under the
action of a force
F(x) that is not
constant.
xf
In this case the work W performed by F is given by:
W
 F ( x)dx
xo
W  Area under the F versus x curve from x o to x f
Work performed by a spring (spring constant = k) as it is
stretched from its relaxed length by L
L
x 
L2
W   F ( x)dx   (kx)dx  k  xdx  k    k
2
 2 0
0
0
0
L
L
L
2
W = shaded area in the F versus x plot
(6-11)
The most general case is when the force F changes both
magnitude and direction from point to point. In this case the
work W performed by F as it moves an object from point A to
point B along a given path. W depends on rA, rB and the path.
W is calculated as follows:
(6-12)
• Divide the path into segments r
• Calculate the work W = F•r for each element
• Sum all the contributions and take the limit as r  0
rB
The limit of the sum gives: W   F  d r
rA
This type of integral is known as "line intergal"
(6-13)
Work performed during uniform circular motion.
The net force F points towards the center C of the orbit
(centripetal force). For the path segment ds the work dW is:
dW  F  d s  Fds cos(90)  0
W   dW  0
Conclusion:
C
.
No work is done on an
object that undergoes
uniform circular motion
(6-14)
y
A
Classification of forces
1
2
B
3
Work W performed by a force
F as it moves an object along
one of the three paths from
point A to point B:
rB
O
x
W   F  dr
rA
A force is called “conservative” if W does not depend on the
path but only on the coordinates of the start and finish points.
In this case: W1 = W2 = W3
A force is called “non-conservative” if W depends not only
on the coordinates of the start and finish points but on the path
as well. In this case W1  W2  W3
(6-15)
y
A
(6-16)
1
C
B
2
D
x
O
If the force F is conservative
than W along any closed path
is zero. This statement can
be used as an alternative
definition of a conservative
force
Consider a closed path ACBDA. This can be divided into two
different paths that take us from point A to point B. Path 1
(ACB), and path 2 (ADB). WACB = WADB
A
WACBDA  WACB  WBDA
WBDA   F  d r (along path  2 )
B
A
B
WBDA   F  d r    F  d r  WADB
B
A
 WACBDA  WACB  WADB  0
Example of a conservative force: The gravitational force
We shall prove that the work gone by Fg along path 1 and path
2 is the same.
Path 1: W1 = Fg•r1
W1 = mgLcos(90-)
cos (90 - ) = sin
Path 1
r1

W1 = mgLsin
Path 2: W2 = WAC + WCB
WAC = mghcos0 = mgh
h
= Lsin  WAC = mgLsin
Path 2
WCB = mgLcos(90) = 0

W2 = mgLsin = W1
(6-17)
(6-18)
.
A
.
B
Example of a non-conservative force: friction f
We shall calculate the work the work done by friction as it
moves the cup along a closed path that starts at point A and
ends at point A. During the trip we apply a force F = -f so
that the net force on the cup, and thus its acceleration a is zero
r1
A
f1
.
r2
f2
.
B
(6-19)
f1 = f2 = kmg
Note: friction opposes motion
W = WAB + WBA
WAB = f1 xmaxcos(180) = - f1xmax = -kmgxmax
WBA = f2 xmaxcos(180) = - f2xmax = -kmgxmax 
W = -kmgxmax -kmgxmax = -2kmgxmax  0
Power Consider a force F that moves an object of mass m
from point x to point x + dx in time dt
F
.
O
.
t
m
x
(6-20)
t+dt
.
x+dx
x-axis
dx
Power (symbol P) is defined as the rate at which F performs
work.
dW
P
dt
Units: Joule/second  Watt (W)
dW Fdx
dx
dW  Fdx  P 

F
 Fv
dt
dt
dt
Note: The equation P = Fv is valid when F and the
displacement dx are parallel
y (6-21)
P 
dr
v
path
O
F  Fx i  Fy j
In general the force F and
the displacement dr are
not parallel. In this case
the power P is given by
the equation:
F
P  F  v  Fv cos
x
,
d r  dxi  dy j
dW  F  d r  Fx dx  Fy dy
dW Fx dx  Fy dy
P

 Fx vx  Fy v y  F  v
dt
dt

Commonly used practical (non-SI) Units
P = W/t

W = Pt
Power: The horsepower (symbol: hp) is the average power
that a horse can generate
1 hp = 746 W
Work: The kilowatt-hour (symbol: kWh) is the work
produced by a machine of power P = 1 kW = 1000 W in a time
interval t = 1 hour = 3600 s
W = Pt
 1 kWh = 10003000 Ws = 3.6106 J
(6-22)