Tuesday June 14, 20 05 1 PHYS 1443-001, Summer I 2005 Dr. Andrew Brandt PHYS 1443 – Section 001 Lecture #8 Tuesday June 14, 2005 Dr. Andrew Brandt • Accelerated Frames • Work • Kinetic and Potential Energy
Jan 21, 2016
Tuesday June 14, 2005 1 PHYS 1443-001, Summer I 2005Dr. Andrew Brandt
PHYS 1443 – Section 001Lecture #8
Tuesday June 14, 2005Dr. Andrew Brandt
• Accelerated Frames• Work• Kinetic and Potential Energy
Tuesday June 14, 2005 2 PHYS 1443-001, Summer I 2005Dr. Andrew Brandt
Announcements• Test 2 Thursday 6/16• Homework:
– HW4 on ch5 due Tuesday 6/14 at 8pm– HW5 on ch 6 due Wednesday 6/15 at 6pm
Tuesday June 14, 2005 3 PHYS 1443-001, Summer I 2005Dr. Andrew Brandt
Motion in Accelerated FramesNewton’s laws are valid only when observations are made in an inertial frame of reference. What happens in a non-inertial frame?Fictitious forces are needed to apply Newton’s second law in an accelerated frame.
This force does not exist when the observations are made in an inertial reference frame.
What does this mean and why is this true?
Let’s consider a “free” ball inside a box that is moving under uniform circular motion.
We see that the box has a radial force exerted on it but none on the ball directly
How does this motion look like in an inertial frame (or frame outside a box)?
rF
rHow does this motion look like in the box?
The ball appears to experience a force that moves it in a curved path towards the wall of the box
Why?
According to Newton’s first law, the ball should continue in its original direction but since the box is turning, the ball feels like it is being pushed toward the wall relative to everything else in the box.
v
Tuesday June 14, 2005 4 PHYS 1443-001, Summer I 2005Dr. Andrew Brandt
Non-Inertial Frame
Example of Motion in Accelerated FramesA ball of mass m is hung by a cord to the ceiling of a boxcar that is moving with an acceleration a. What do the inertial observer at rest and the non-inertial observer traveling inside the car conclude? How do their conclusions differ?
m
This is how the ball looks like no matter which frame you are in.
F
Inertial Frame
How do the free-body diagrams look for two frames?
Fg=mgm
T
Fg=mgm
T
Ffic
ac
How are the motions interpreted in these two frames? Any differences?
For an inertial frame observer, the forces being exerted on the ball are only T and Fg. The acceleration of the ball is the same as that of the box car and is provided by the x component of the tension force.
F
In the non-inertial frame observer, the forces being exerted on the ball are T, Fg, and Ffic.
For some reason the ball is under a force, Ffic, that provides acceleration to the ball.
xF
yF
cos
mgT tangac
xF
yF
cos
mgT
sinTmaF ficfic
tanga fic While the mathematical expression of the acceleration of the ball is identical to that of inertial frame observer’s, the cause of the force is dramatically different.
TF g
xma cma sinT
mgT cos 0
ficg FTF
ficFT sin 0
mgT cos 0
Tuesday June 14, 2005 5 PHYS 1443-001, Summer I 2005Dr. Andrew Brandt
Scalar (Dot) Product of Two Vectors • Product of magnitude of the two vectors and the cosine of the
angle between them BA
• Operation is commutative cosBABA
• Operation follows distribution law of multiplication
CABA
• How does the scalar product look in terms of components?
kAjAiAA zyx
kBjBiBB zyx
kBjBiBkAjAiABA zyxzyx termscrosskkBAjjBAiiBA zzyyxx
kkjjii
ikkjji• Scalar products of Unit Vectors
zzyyxx BABABABA
cosBA
cosAB AB
CBA
1 0
Is this a vector or a scalar?
=0
Tuesday June 14, 2005 6 PHYS 1443-001, Summer I 2005Dr. Andrew Brandt
x
y
Work Done by a Constant ForceWork in physics is done only when a sum of forces exerted on an object causes motion of the object.
M
F Free Body
DiagramM
d gMF G
NF
F
Which force did the work?
W
Force F
How much work did it do?
What does this mean?Physical work is done only by the component of the force along the movement of the object.
Unit?Joule)(for J
mN
Work is energy transfer!!
dF cosFd
Tuesday June 14, 2005 7 PHYS 1443-001, Summer I 2005Dr. Andrew Brandt
Example of Work w/ Constant ForceA 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 it is displaced by 3.00m to the East.
Does work depend on mass of the object being worked on?
M
F
M
d
W
JW 13030cos00.30.50
Yes
Why don’t I see the mass term in the work at all then?
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.
dF cosdF
Tuesday June 14, 2005 8 PHYS 1443-001, Summer I 2005Dr. Andrew Brandt
Example of Work by Scalar ProductA 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.
d
F
b) Calculate the work done by the force F.
W
jiji 0.20.50.30.2 )(166100.20.30.50.2 Jjjii
Y
X
d F
Can you do this using the magnitudes and the angle between d and F?
cosdFdFW
22yx dd m6.30.30.2 22
22yx FF N4.50.20.5 22
dF
Tuesday June 14, 2005 9 PHYS 1443-001, Summer I 2005Dr. Andrew Brandt
Work Done by Varying Force• If the force depends on position of the object during the motion
– one must consider work in small segments of the position where the force can be considered constant
– Then add all the work-segments throughout the entire motion (xi xf)
W
f
i
x
xx xFW
0lim
f
i
x
xx
x
F x
( ) f
i
x
ixxW net F dx
– If more than one force is acting, the net work is done by the net force
In the limit where x0
An example of a force that depends on position is the spring force kxFs
The work done by the spring force is
W
f
i
x
xxF dx W
max
0
sxF dx
max
0
xkx dx
21
2kx
xF x
Tuesday June 14, 2005 10 PHYS 1443-001, Summer I 2005Dr. Andrew Brandt
Kinetic Energy and Work-Kinetic Energy Theorem• Some problems are hard to solve using Newton’s second law
– If forces acting on the object during the motion are complicated– Relate the work done on the object by the net force to the change of the
speed of the object
MF
M
d
vi vf
Suppose net force F was exerted on an object over a displacement d to increase its speed from vi to vf.
The work on the object by the net force F is
cos 0W F d ma d ma d ur ur
tvvd if 2
1 a
W
Displacement Acceleration
W
Work 21
2KE mv
Kinetic Energy
Work The work done by the net force causes a change of object’s kinetic energy.
dma
tvv
t
vvm if
if 2
1 22
2
1
2
1if mvmv
t
vv if
22
2
1
2
1if mvmv if KEKE KE
Work-Kinetic Energy Theorem
Tuesday June 14, 2005 11 PHYS 1443-001, Summer I 2005Dr. Andrew Brandt
Example of Work-KE TheoremA 6.0kg block initially at rest is pulled to the 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.
Work done by the force F is
From the work-kinetic energy theorem, we know
Since the initial speed is 0, the above equation becomes:
M F M
d
vi=0 vf
22
2
1
2
1if mvmvW
2
2
1fmvW
fv Solving the equation for vf, we obtain 2W
m
2 363.5 /
6.0m s
W cosF d ur ur
12 3.0cos 0 36 J
Tuesday June 14, 2005 12 PHYS 1443-001, Summer I 2005Dr. Andrew Brandt
Work and Energy Involving Kinetic Friction• How will friction affect the work that can be done by a
force?– Does static friction matter?
M M
d
vf
Frictional force Ffr works on the object to slow it down
The work on the object by friction Ffr is
frW
The final kinetic energy of an object, taking into account its initial kinetic energy, friction force and other source of work, is
Ffr
KE
fKE
t=0, KEi Friction,Engine work
t=T, KEf
No, since there must be motion for work to occur.
cos 180frF d frF d frF d
iKE W frF d
Reduces it, of course!
vi
Tuesday June 14, 2005 13 PHYS 1443-001, Summer I 2005Dr. Andrew Brandt
Example of Work with FrictionA 6.0kg block initially at rest is pulled to the East along a horizontal surface with a coefficient of kinetic friction k=0.15 by a constant horizontal force of 12N. Find the speed of the block after it has moved 3.0m.
Work done by the force F is
Thus the total work is
M F M
d=3.0m
vf
fv
Work done by friction Fk is
Fk
kkW F d ur ur
g
J26180cos0.38.90.615.0
W
Using work-kinetic energy theorem and the fact that initial speed is 0, we obtain
W Solving the equation
for vf, we obtain
cosdmgk
kF WW )(102636 J
F kW W 21
2 fmv 2W
m
2 101.8 /
6.0m s
coskF d uur ur
FW cosF d ur ur
12 3.0cos 0 36 J vi=0
What’s another way to solvethis problem?
Tuesday June 14, 2005 14 PHYS 1443-001, Summer I 2005Dr. Andrew Brandt
Kinetic Energy at High SpeedThe laws of Newtonian mechanics are no longer valid for objects moving at a speed close to that of light, c. It must be generalized for these special cases. Theory of relativity. The kinetic energy must be
modified to reflect the fact that the object is moving very high speed.
The speed of an object cannot be faster than light in vacuum. Have not seen any particle that goes faster than light, yet.However this equation must satisfy the Newtonian expression for smaller speeds!!
11
12
2
cv
mcK
What does this expression tell you?
1...8
3
2
111
1
142
2
2
2
c
v
c
vmc
cv
mcK
22
22
2
2
1
2
11
2
11 mv
c
vmc
c
vmcK
Tuesday June 14, 2005 15 PHYS 1443-001, Summer I 2005Dr. Andrew Brandt
Potential EnergyEnergy associated with a system of objects Stored energy which has the potential or the possibility to do work or to convert stored E to kinetic energy
What does this mean?
In order to describe potential energy, U, a system must be defined.
What are some other forms of energy?
The concept of potential energy can only be used under the special class of forces called, conservative forces which results in principle of conservation of mechanical energy.
Mechanical Energy
Biological EnergyElectromagnetic
EnergyNuclear Energy
Chemical Energy
ME
These different types of energies are stored in the universe in many different forms!!!
If one takes into account ALL forms of energy, the total energy in the entire universe is conserved. It just transforms from one form to another.
i iKE PE f fKE PE
Tuesday June 14, 2005 16 PHYS 1443-001, Summer I 2005Dr. Andrew Brandt
Gravitational Potential Energy
When an object is falling, a gravitational force, Mg, performs work on the object, increasing its kinetic energy. The potential energy of an object at a height y (its the potential to work ) is expressed as
m
yf
m
mgyigU
What does this mean?
gW The work performed on the object by the gravitational force as the brick goes from yi
to yf is:
mgyU g mgy
i fmgy mgy
Work by the gravitational force as the brick goes from yi to yf is the negative of the change in the system’s potential energy
Potential energy was lost in order for gravitational force to increase the brick’s kinetic energy.
cosgF y ur ur
gF yur ur
gU i fU U
gF y ur ur
Potential energy given to an object by the gravitational field due to its height above the surface of the Earth