PHYS 1443 – Section 003
Lecture #12
Wednesday, Oct. 6, 2004
Dr. Jaehoon Yu
1.
2.
3.
4.
5.
Work done by a constant force
Scalar Product of Vectors
Work done by a varying force
Work and Kinetic Energy Theorem
Potential Energy
Homework #7 due at 1pm next Wednesday, Oct. 13!!
Wednesday, Oct. 6, 2004
PHYS 1443-003, Fall 2004
Dr. Jaehoon Yu
1
Work Done by a Constant Force
Work in physics is done only when a sum of forces
exerted on an object made a motion to the object.
F
M
y
q
M
FN
Free Body
Diagram
F
q
d
x
FG  M g
Which force did the work? Force F
How much work did it do? W 
What does this mean?
Wednesday, Oct. 6, 2004
 F  d  Fd cosq
Unit?
Nm
 J (for Joule)
Physical work is done only by the component of
the force along the movement of the object.
PHYS 1443-003, Fall 2004
Dr. Jaehoon Yu
Work is an energy transfer!!
2
Example of Work w/ Constant Force
A man cleaning a floor pulls a vacuum cleaner with a force of magnitude F=50.0N at
an angle of 30.0o with East. Calculate the work done by the force on the vacuum
cleaner as the vacuum cleaner is displaced by 3.00m to East.
F
M
30o
W
M
 F  d   F  d cosq
W  50.0  3.00  cos 30  130 J
d
Does work depend on mass of the object being worked on?
Why don’t I see the mass
term in the work at all then?
Wednesday, Oct. 6, 2004
Yes
It is reflected in the force. If the object has smaller
mass, its would take less force to move it the same
distance as the heavier object. So it would take less
work. Which makes perfect sense, doesn’t it?
PHYS 1443-003, Fall 2004
Dr. Jaehoon Yu
3
Scalar Product of Two Vectors
• Product of magnitude of the two vectors and the cosine of the
angle between them
A B  A B cos q
• Operation is commutative A B  A B cos q  B A cos q  B A


• Operation follows distribution
A BC  A B A C
law of multiplication
 
 
 
 
 
 
• Scalar products of Unit Vectors i  i  j j  k  k  1 i  j  j k  k  i  0
• How does scalar product look in terms of components?



A  Ax i  Ay j  Az k



B  Bx i  By j  Bz k



C  Cx i  C y j  Cz k
 
 
 






 
 

A B   Ax i  Ay j  Az k   Bx i  By j  Bz k    Ax Bx i i  Ay By j j  Az Bz k k  cross terms





A  B  Ax Bx  Ay By  Az Bz
Wednesday, Oct. 6, 2004
PHYS 1443-003, Fall 2004
Dr. Jaehoon Yu
=0
4
Example of Work by Scalar Product
A particle moving in 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.
a) Calculate the magnitude of the displacement
and that of the force.
Y
d
F
X
d  d x2  d y2 
2.02  3.02  3.6m
F  Fx2  Fy2  5.02  2.02  5.4 N
b) Calculate the work done by the force F.
W  F d 


 
  
  2.0  5.0 i  i  3.0  2.0 j j  10  6  16( J )
 2.0 i  3.0 j    5.0 i  2.0 j 

 

Can you do this using the magnitudes and the angle between d and F?
W  F  d  F d cosq
Wednesday, Oct. 6, 2004
PHYS 1443-003, Fall 2004
Dr. Jaehoon Yu
5
Work Done by Varying Force
• If the force depends on position of the object through the 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 )  
xf
xi
  F  dx
ix
One of the forces depends on position is force by a spring
The work done by the spring force is
Fs  kx
1 2
Fs dx   x   kx  dx  kx
W
 xmax
max
2
0
Wednesday, Oct. 6, 2004
0
PHYS 1443-003, Fall 2004
Dr. Jaehoon Yu
6
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
ur ur
d
W   F  d   ma  d cos 0   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
Wednesday, Oct. 6, 2004

The work done by the net force caused
change of object’s kinetic energy.
PHYS 1443-003, Fall 2004
Dr. Jaehoon Yu Work-Kinetic Energy
Theorem
7
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 q  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
Wednesday, Oct. 6, 2004
vf 
PHYS 1443-003, Fall 2004
Dr. Jaehoon Yu
2W
2  36

 3.5m / s
m
6.0
8
Work and Energy Involving Kinetic Friction
• What do you think the work looks like if there is friction?
– Why doesn’t static friction matter?
Ffr
M
M
vi
vf
d
Because it isn’t there while
the object is moving.
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
Wednesday, Oct. 6, 2004
PHYS 1443-003, Fall 2004
Dr. Jaehoon Yu
Friction,
Engine work
t=T, KEf
9
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
ur ur
WF  F d cos q  12  3.0cos 0  36  J 
M
vf
d=3.0m
Work done by friction Fk is
Thus the total work is
uur ur
ur ur
Wk  F k gd  Fk d cos q 
mk mg d cos q
 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
Wednesday, Oct. 6, 2004
Solving the equation
for vf, we obtain
PHYS 1443-003, Fall 2004
Dr. Jaehoon Yu
vf 
2W
2 10

 1.8m / s
m
6.0
10