Hi ! This Presentation would be really helpful to all of you...I hope you like it....Save it !!! Show It...Have Fun....

2. Work
The VERTICAL component of the force DOES
NOT cause the block to move the right. The energy
imparted to the box is evident by its motion to the
right. Therefore ONLY the HORIZONTAL
COMPONENT of the force actually creates energy
or WORK.
When the FORCE and DISPLACEMENT are in
the SAME DIRECTION you get a POSITIVE
WORK VALUE. The ANGLE between the force
and displacement is ZERO degrees.
When the FORCE and DISPLACEMENT are in
the OPPOSITE direction, yet still on the same axis,
you get a NEGATIVE WORK VALUE. IT simply
means that the force and displacement oppose each
other. The ANGLE between the force and
displacement in this case is 180 degrees.
When the FORCE and DISPLACEMENT are
PERPENDICULAR, you get NO WORK!!! The
ANGLE between the force and displacement in this
case is 90 degrees.

3. Scalar Dot Product?
  
W  F  x 
Fx cos
A dot product is basically a CONSTRAINT
on the formula. In this case it means that
F and x MUST be parallel. To ensure that
they are parallel we add the cosine on the
end.
FORCE
Displacement
  
W F x Fx
  
0 
; cos0 1
 

W 
Fx
cos



4.   
W  F  x 
Fx cos
FORCE
Displacement
  
W F x Fx
  
180 
; cos180 1
  
 
W F x
f
 
cos



5.   
W  F  x 
Fx cos
FORCE
Displacement
  
W  F  x 
Fx
 
90 ; cos90 0
W 0
J
cos





6. Work = The Scalar Dot Product between
Force and Displacement. So that means if
you apply a force on an object and it covers
a displacement you have supplied ENERGY
or done WORK on that object.
cos
 F  r 
F r 
Wdisplacement vector

r
  


7. Example
W  F 
r
W 
F r cos
A box of mass m = 2.0 kg is moving
over a frictional floor has a force
whose
magnitude is F = 25 N applied to it at
an angle of 30 degrees. The box is
observed to move 16 meters in the
horizontal direction before falling off
the table.
a) How much work does F do before
taking the plunge?
W  F 
r
W F r
W
cos
25 16 cos30


W 
346.4
Nm
W 
346.4
J


8. Example
What if we had done this in UNIT VECTOR
notation?
21.65ˆ 12.5 ˆ
F  i 
j
W F r F r
W
   
( ) ( )
x x y y
   
(21.65 16) (12.5 0)
W 
346.4
Nm
W 
346.4
J

9. Work-Energy Theorem
dx
 ( )

 
W m v dv
2 2 2
v o
v


  
2 2
)
2 2
| ) (
2
( |
2 2
o
v
v
mv mv
W
v v
m
v
W m
dv m v dv
dt
W m
o
o
 
  2
K mv
W  
K
2
Kinetic Energy 1
Kinetic energy is the ENERGY of MOTION.

10. Example
Suppose the woman in the figure above applies a 50 N force
to a 25-kg box at an angle of 30 degrees above the
horizontal. She manages to pull the box 5 meters.
a) Calculate the WORK done by the woman on the box
b) The speed of the box after 5 meters if the box started from
rest.
W Fx 

cos
(50)(5)cos30
 
W

W   KE 
mv
W 
v

v
2
2
(25)
2
1
2
1
216.5 J 4.16 m/s

11. Conservative Force
 Conservative Forces
◦ A force where the work it does is independent of
the path
A conservative force depends only on the position of the
particle, and is independent of its velocity or acceleration.
x
y
z
A
F B

12. Something is missing….
Consider a mass m that moves from position 1 ( y1)
to position 2 m,(y2), moving with a constant velocity.
How much work does gravity do on the body as it
executes the motion?
W F r Fr
gravity
  
W mg y y
gravity
 
W mg y
gravity
  
W U
gravity
 
cos
( ) cos180
1 2

 
W F r Fr
gravity
  
W mg y y
gravity
 
W mg y
gravity
  
W U
gravity
 
cos
( ) cos0
2 1

 
Suppose the mass was thrown UPWARD.
How much work does gravity do on the
body as it executes the motion?
In both cases, the negative
sign is supplied

13. The bottom line..
The amount of Work gravity does on a
body is PATH INDEPENDANT. Force
fields that act this way are
CONSERVATIVE FORCES
FIELDS. If the above is true, the amount
of work done on a body that moves
around a CLOSED PATH in the field will
always be ZERO
FRICTION is a non conservative force. By NON-CONSERVATIVE
we mean it DEPENDS on the PATH. If a
body slides up, and then back down an incline the total
work done by friction is NOT ZERO. When the direction of
motion reverses, so does the force and friction will do
NEGATIVE WORK in BOTH directions.

14. Nonconservative Forces
 Examples of Nonconservative forces
◦ Friction
◦ Air resistance
◦ Tension
◦ Normal force
◦ Propulsion force of things like rocket
engine
 Each of these forces depends on the
path

15. Potential Energy
mg
h
W Fx F mg x h
 cos  ;

0,cos0 1
 
W  mgh 
PE



Since this man is lifting the
package upward at a CONSTANT
SPEED, the kinetic energy is NOT
CHANGING. Therefore the work
that he does goes into what is
called the ENERGY OF POSITION
or POTENTIAL ENERGY.
All potential energy is considering
to be energy that is STORED!

16. Conservation of Energy
A B C D

17. In Figure A, a pendulum is
released from rest at some
height above the ground
position.
It has only potential energy.
In Figure C, a pendulum is at
the ground position and
moving with a maximum
velocity.
It has only kinetic energy.
In Figure B, a pendulum is still
above the ground position, yet
it is also moving.
It has BOTH potential energy
and kinetic energy.
In Figure D, the pendulum has
reached the same height
above the ground position as
A.
It has only potential energy.

18. Hooke’s Law
 If a ‘spring’ is bent, stretched or compressed from
its equilibrium position, then it will exert a
restoring force proportional to the amount it is
bent, stretched or compressed.
 Fs = kx (Hooke’s Law)
 Related to Hooke’s Law, the work required to bend
a spring is also a function of k and x.
 Ws = (½ k)x2 = amount of stored strain energy
◦ K is the stiffness of the spring. Bigger means Stiffer
◦ x is the amount the spring is bent from equilibrium

19. Based on Hooke’s Law
 A force is required to bend a spring
 The more force applied the bigger the bend
 The more bend, the more strain energy stored
Force (N)
Relationship between F and x
slope of the line = k
x (m)
Area under curve is
the amount of stored
strain energy
Stored strain energy = ½ k x2

20. Hooke’s Law from a Graphical Point of
View
Suppose we had the following data:
x(m) Force(N)
0 0
0.1 12
0.2 24
0.3 36
0.4 48
0.5 60
0.6 72
F 
kx

F
 Slope of a F vs. x graph
k
x
k
s
s
Force vs. Displacement y = 120x + 1E-14
R2 = 1
80
70
60
50
40
30
20
10
0
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7
Displacement(Meters)
Force(Newtons)
k =120 N/m

21. ELASTIC POTENTIAL ENERGY
Recall that the force of an elastic spring is F = ks. It
is important to realize that the potential energy of a
spring, while it looks similar, is a different formula.
Ve (where ‘e’ denotes an elastic
spring) has the distance “s”
raised to a power (the result of
an integration) or
2
1
V ks e 
2
Notice that the potential
function Ve always yields
positive energy.

22. Elastic potential energy
2
W  F ( x ) dx 
( kx )
dx
 
W  kx dx 
k xdx
   
0 0
0
2
2
| 1
2
|
( )
W U kx
x
W k
spring
x
x
x
x
x
x
   

Elastic “potential” energy is a fitting term as
springs STORE energy when there are
elongated or compressed.

23. Conservation of Energy in
Springs

24. Example
A slingshot consists of a light leather cup, containing a
stone, that is pulled back against 2 rubber bands. It takes a
force of 30 N to stretch the bands 1.0 cm (a) What is the
potential energy stored in the bands when a 50.0 g stone
is placed in the cup and pulled back 0.20 m from the
equilibrium position? (b) With what speed does it leave the
slingshot?
a F kx k k
) 30 (0.01) 3000 N/m
s
2
b U kx k
s
c E E U K
U mv v

  
  
 
 
v
s
B A s
2 2
(0.050)
2
1
2
1
)
0.5( )(.20)
2
) 1
300 J
109.54 m/s