Monday, Oct. 13, 2008
Dr. Jae hoon Yu
• Work done by a Constant Force
• Work done by a Varying Force
• Work and Kinetic Energy Theorem
Today’s homework is homework #7, due 9pm, Monday, Oct. 22!!
Monday, Oct. 13, 2008

• Reading assignments
– CH. 8 – 5, 8 – 6 and 8 – 7
• No colloquium this week
• 2 nd term exam on Wednesday, Oct. 22
– Covers from Ch. 1 to what we cover up to this Wednesday, Oct.
15
– Time: 1 – 2:20pm in class
– Location: SH103
– Jason will conduct a summary session Monday, Oct. 20
• Please do NOT miss the exam
• Will have a mid-term grade discussion
– as soon as I am done with grading the exam and putting
PHYS 1443-002, Fall 2008
Dr. Jaehoon Yu
2

• Derive the formula for the gravitational in
 
R
E the center, inside of the Earth. (10 points)

• Compute the fractional magnitude of the gravitational acceleration 1km and 500km inside the surface of the Earth with respect to that on the surface. (6 points, 3 points each)
• Due at the beginning of the class Wednesday,
Oct. 15
Monday, Oct. 13, 2008 PHYS 1443-002, Fall 2008
Dr. Jaehoon Yu
3

A meaningful work in physics is done only when a sum of forces exerted on an object made a motion to the object.
y
M
F q
M
Free Body
Diagram d
Which force did the work?
How much work did it do?
W
Force

F
  r ur ur
Why?
 
Fd cos q
F
N q
F
G

M g
Unit?
F x

J (for Joule)
What does this mean?
Physical work is done only by the component of the force along the movement of the object.
Monday, Oct. 13, 2008 PHYS 1443-002, Fall 2008
Dr. Jaehoon Yu
Work is an energy transfer
4
!!

• Product of magnitude of the two vectors and the cosine of the angle between them u u
A B
 u r
A B cos q
• Operation is commutative ur ur ur ur
A B A B cos q  ur ur
B A cos q  ur ur

• Operation follows the distribution law of multiplication
• Scalar products of Unit Vectors i

 i
 ur
A

 ur
B
 ur
C



 j

 j

 k
 k

 1 ur ur
 i


 j
 ur ur
 
 j

 k

 k
 i

 0
• How does scalar product look in terms of components?
ur 
 
 
A A i x
A j y

A k z ur 
 
 
B B i x
B j y

B k z ur ur
  



 y
 

  x

 y

B i B j B k
 ur ur
A B




 x x
 
A B i i
 
A B j j y y z z
 
A B k k
A B
 x x
A B
 y y
A B z z
Monday, Oct. 13, 2008 PHYS 1443-002, Fall 2008
Dr. Jaehoon Yu
 cross terms
5
=0

A particle moving on the xy plane undergoes a displacement d=(2.0i+3.0j)m as a constant force F=(5.0i+2.0j) N acts on the particle.
Y d
F
X a) Calculate the magnitude of the displacement and that of the force.
ur d
 d x
2  d y
2 
   
2 
3 .
6 m ur
F

F x
2 
F y
2 
   
2 
5 .
4 N b) Calculate the work done by the force F.
W
 ur ur
 



2 .
0 i


3 .
0
 j







5 .
0 i


2 .
0 j





2 .
0

5 .
0 i


 i

3 .
0

2 .
0
 j

 j

10

6

16 ( J )
Can you do this using the magnitudes and the angle between d and F?
W
 ur ur
  ur ur
F d
Monday, Oct. 13, 2008 PHYS 1443-002, Fall 2008
Dr. Jaehoon Yu cos q
6

• If the force depends on the position of the object in motion,
→ one must consider the work in small segments of the displacement where the force can be considered constant
F x
 x
– Then add all the work-segments throughout the entire motion (x i
 x f
)
W
 x
 f
F x
  x In the limit where  x  0 lim x 0 x
 f
F x
   x i x f
F dx
 x x i x i
– If more than one force is acting, the net work done by the net force is
( )
  x f
 
F ix
 dx x i
One of the position dependent forces is the force by the spring
W
F
 s
 kx
The work done by the spring force is
W
 

0 x max
F dx s
 

0 x max
   
1
2
2 kx max
Hooke’s Law
Monday, Oct. 13, 2008 PHYS 1443-002, Fall 2008
Dr. Jaehoon Yu
7

Kinetic Energy and Work-Kinetic Energy Theorem
• Some problems are hard to solve using Newton’s second law
– If forces exerting on an object during the motion are complicated
M
– Relate the work done on the object by the net force to the change of the speed of the object
S F
M
Suppose net force S F was exerted on an object for displacement d to increase its speed from v i to v f
. v i v f d
Displacement
Work W
   d
 d

1
2


 m

 v f t
 v i
 v




 f
The work on the object by the net force S F is
W

1
2
 v i

  t
 v


 f v i t
1
2 d r
 
Acceleration mv f
2 
1
2 mv i
2 a cos 0

Kinetic
Energy v f
 t
 

KE
 v i
1
2 mv
2
Work
W

1 mv f
2 
2
Monday, Oct. 13, 2008
1
2 mv i
2
Haven’t we seen this equation somewhere?

KE f

KE i
 
KE
PHYS 1443-002, Fall 2008
Dr. Jaehoon Yu
Work done by the net force causes change of the object’s kinetic energy.

A 6.0kg block initially at rest is pulled to East along a horizontal, frictionless surface by a constant horizontal force of 12N. Find the speed of the block after it has moved 3.0m.
M v i
=0
F v f
M
W
Work done by the force F is
 ur ur
  ur ur
F d cos q   d
From the work-kinetic energy theorem, we know W

1
2 mv
2 f

1
2 mv i
2
Since initial speed is 0, the above equation becomes W

1
2 mv
2 f
 
Solving the equation for v f
, we obtain v
 f
2 W m 6.0

Monday, Oct. 13, 2008 PHYS 1443-002, Fall 2008
Dr. Jaehoon Yu
9

F fr
Work and Energy Involving Kinetic Friction
• What do you think the work looks like if there is friction?
– Static friction does not matter! Why?
– Then which friction matters?
It isn’t there when the object is moving.
Kinetic Friction
M v i d v f
M
Friction force F fr works on the object to slow down
The work on the object by the friction F fr
W
 fr
F d fr
   
F d fr

KE
  is
F d fr
The negative sign means that the work is done on the friction!!
The final kinetic energy of an object, taking into account its initial kinetic energy, friction force and other source of work, is
KE
 f
KE i
 
W

F d fr
Monday, Oct. 13, 2008 t=0, KE i
PHYS 1443-002, Fall 2008
Dr. Jaehoon Yu
Friction,
Engine work t=T, KE f
10

A 6.0kg block initially at rest is pulled to East along a horizontal surface with coefficient of kinetic friction m k
=0.15 by a constant horizontal force of 12N. Find the speed of the block after it has moved 3.0m.
F k
M
F
M v i
=0 v f d=3.0m
Work done by friction F k is
Thus the total work is
Work done by the force F is
W

F r
F d r cos q 
12 3.0cos 0

36
 
W k
 r g r
F d k

 r r
F d k cos q  m k r mg d
0 .
15

6 .
0

9 .
8

3 .
0 cos 180
 cos q

26
 
W

W
F

W k

36

26

10 ( J )
Using work-kinetic energy theorem and the fact that initial speed is 0, we obtain
W

W
F

W k

1
2 mv
2 f
Solving the equation for v f
, we obtain
Monday, Oct. 13, 2008 PHYS 1443-002, Fall 2008
Dr. Jaehoon Yu v
 f
2 W m

6.0

11

A meaningful work in physics is done only when the sum of the forces exerted on an object made a motion to the object.
However much tired your arms feel, if you were
What does this mean?
just holding an object without moving it you have not done any physical work to the object.
Mathematically, the work is written as the product of magnitudes of the net force vector, the magnitude of the displacement vector and the angle between them.
W



  i
  g ur d ur d
 cos q
Kinetic Energy is the energy associated with the motion and capacity to perform work.
Work causes change of energy after the completion  Work-Kinetic energy theorem
K

1
2 mv
2

W

K f
K i

K Nm=Joule
Monday, Oct. 13, 2008 PHYS 1443-002, Fall 2008
Dr. Jaehoon Yu
12