PHYS 1443 – Section 001
Lecture #8
Monday, June 12, 2006
Dr. Jaehoon Yu
•
•
•
Monday, June 12, 2006
Quiz Solutions
Work done by a varying force
Work and Kinetic Energy Theorem
PHYS 1443-001, Summer 2006
Dr. Jaehoon Yu
1
Announcements
• Quiz result
– Average: 3.8/8
• 46/100
• Quiz 1: 76
– Top score: 7/8
• Mid-term exam
– 8:00 – 10am, this Thursday, June 15, in class
– CH 1 – 8
Monday, June 12, 2006
PHYS 1443-001, Summer 2006
Dr. Jaehoon Yu
2
Work Done by Varying Force
• If the force depends on the position of the object in motion
– one must consider work in small segments of the position where the force
can be considered constant
W  Fx  x
– Then add all work-segments throughout the entire motion (xi xf)
xf
W   Fx  x
xf
lim  Fx  x 
In the limit where x0
x 0
xi
xi

xf
xi
Fx dx  W
– If more than one force is acting, the net work is done by the net force
W (net ) 
   F  dx
xf
ix
xi
One of the forces depends on position is force by a spring Fs kx
The work done by the spring force is
Hooke’s Law
1 2
Fs dx   x   kx  dx  kx
W
 xmax
max
2
0
Monday, June 12, 2006
0
PHYS 1443-001, Summer 2006
Dr. Jaehoon Yu
3
Kinetic Energy and Work-Kinetic Energy Theorem
• Some problems are hard to solve using Newton’s second law
– If forces exerting on the object during the motion are so complicated
– Relate the work done on the object by the net force to the change of the
speed of the object
M
SF
Suppose net force SF was exerted on an object for
displacement d to increase its speed from vi to vf.
M
The work on the object by the net force SF is
r r
d
W   F  d   ma  d cos0   ma  d
1
v f  vi
d

v

v
t
f
i
Displacement
Acceleration a 
2
t
  v f  vi  1
1 2 1 2 Kinetic
1 2


W

mv

mv


m
v

v
t



ma
d

 

f
i
f
i
Work

KE  mv
2
2
Energy
  t  2
2
vi
vf

Work W 


1 2 1 2
mv f  mvi  KE f  KEi  KE
2
2
Monday, June 12, 2006

Work done by the net force causes
change of object’s kinetic energy.
PHYS 1443-001, Summer 2006
Dr. Jaehoon YuWork-Kinetic Energy
Theorem
4
Example of Work-KE Theorem
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
F
M
vi=0
vf
Work done by the force F is
W  F  d  F d cos  12  3.0cos0  36  J 
d
1 2 1 2
From the work-kinetic energy theorem, we know W  mv f  mvi
2
2
1 2
Since initial speed is 0, the above equation becomes W  mv f
2
Solving the equation for vf, we obtain
Monday, June 12, 2006
vf 
PHYS 1443-001, Summer 2006
Dr. Jaehoon Yu
2W
2  36

 3.5m / s
m
6.0
5
Work and Energy Involving Kinetic Friction
• What do you think the work looks like if there is friction?
– Static friction does not matter! Why? It isn’t there when the object is moving.
– Then which friction matters? Kinetic Friction
Ffr
M
M
vi
vf
d
Friction force Ffr works on the object to slow down
The work on the object by the friction Ffr is
W fr  Ffr d cos 180   F fr d KE   F fr d
The final kinetic energy of an object, taking into account its initial kinetic
energy, friction force and other source of work, is
KE f  KEi W  F fr d
t=0, KEi
Monday, June 12, 2006
Friction,
Engine work
PHYS 1443-001, Summer 2006
Dr. Jaehoon Yu
t=T, KEf
6
Example of Work Under Friction
A 6.0kg block initially at rest is pulled to East along a horizontal surface with coefficient
of kinetic friction mk=0.15 by a constant horizontal force of 12N. Find the speed of the
block after it has moved 3.0m.
Fk M
F
vi=0
Work done by the force F is
r r
WF  F d cos  12  3.0cos 0  36  J 
M
vf
d=3.0m
Work done by friction Fk is
Thus the total work is
r r
r r
Wk  Fk gd  Fk d cos 
mk mg d cos
 0.15  6.0  9.8  3.0 cos180  26J 
W  WF  Wk  36  26  10( J )
Using work-kinetic energy theorem and the fact that initial speed is 0, we obtain
1 2
W  WF  Wk  mv f
2
Monday, June 12, 2006
Solving the equation
for vf, we obtain
vf 
PHYS 1443-001, Summer 2006
Dr. Jaehoon Yu
2W
2 10

 1.8m / s
m
6.0
7
Work and Kinetic Energy
Work in physics is done only when the sum of forces
exerted on an object made a motion to the object.
What does this mean?
However much tired your arms feel, if you were just
holding an object without moving it you have not
done any physical work to the object.
Mathematically, work is written in a product of magnitudes
of the net force vector, the magnitude of the displacement
vector and the angle between them.
W


ur ur
F i gd 

ur
Fi

ur
d cos 
Kinetic Energy is the energy associated with motion and capacity to perform work. Work
causes change of energy after the completion Work-Kinetic energy theorem
1 2
K  mv
2
Monday, June 12, 2006
W  K f  Ki  K
PHYS 1443-001, Summer 2006
Dr. Jaehoon Yu
Nm=Joule
8