Transcript PPT

Physics 211
Lecture 7
Today’s Concepts:
Work & Kinetic Energy
I need a refresher on how work works (pun intended).
F=ma
Let’s go back to just F=ma. Please?
r
r
Fnet  ma
Why are we doing this?
Simply because it has proven useful.
Why are we doing this?
Simply because it has proven useful.
The Dot Product
Work-Kinetic Energy Theorem
The work done by force F as it acts on an object that
moves between positions r1 and r2 is equal to the
change in the object’s kinetic energy:
W  K
r2
W   F  dl
r1
1 2
K  mv
2
Work-Kinetic Energy Theorem: 1-D Example
If the force is constant and the directions aren’t
changing then this is very simple to evaluate:
car
F
d
r2
W   F  dl  F  d
Demo: suitcase,
r1
Rope, scale
In this case =
Fd
since cos(0)=1
This is probably what you remember from High School.
ACT
A lighter car and a heavier van, each initially at rest, are
pushed with the same constant force F. After both
vehicles travel a distance d, which of the following
statements is true? (Ignore friction)
F
d
F
d
car
van
A) They will have the same velocity
B) They will have the same kinetic energy
C) They will have the same momentum
r2
F

dl


K

Derivation – not so important
Concept – very important
r1
r2
 F  dl
r1
A force pushing over some distance
will change the kinetic energy.
 K
q
W 
r2
 F  dl
r1
Work done by gravity near the Earth’s surface
mg
Work done by gravity near the Earth’s surface
WTOT  W1  W2  ...  WN
 mg  dl1  mg  dl2  ...  mg  dlN
dlN
dl1
mg
dl2
dy1
dl1
dx1
mg
Work done by gravity near the Earth’s surface
WTOT  W1  W2  ...  WN
 mg  dl1  mg  dl2  ...  mg  dlN
 mgdy1  mgdy2 ...  mgdyN
  mg y
dlN
y
dl1
mg
dl2
Wg  mg ( y final  yinitial )
Work-Kinetic Energy Theorem
If there are several forces acting then W is the work
done by the net (total) force:
WNET  K
 W1  W2  ...
You can just add up the
work done by each force
WNET  WTOT
Checkpoint
Three objects having the same mass begin
at the same height, and all move down the
same vertical distance H. One falls straight
down, one slides down a frictionless
inclined plane, and one swings on the end
of a string.
In which case
does the object
have the biggest
net work done
on it by all forces
during its
motion?
H
Free Fall
A) Free Fall
Frictionless incline
B) Incline
C) String
String
D) All the same
Checkpoint Comments
Three objects having the same mass begin
at the same height, and all move down the
same vertical distance H. One falls straight
down, one slides down a frictionless
inclined plane, and one swings on the end
of a string.
mgh doesn’t matter about path taken
Because the string has the largest displacement it has the largest total work
done. Be careful, it’s the VERTICAL displacement where work is done by gravity.
w = mgy. Since the height is the same for all the objects, the work done is the
same in all cases.
Since all three objects have same mass and fall down same height, all the work
should be same.
I’m actually not too sure. I feel like they all would have the same amount of work
because they are at the same height; have the same downward acceleration. But they
do NOT have the same downward acceleration!
Because the object has the most work. I don’t know what this means.
Not sure to tell you the truth, the prelecture slide said the normal force doesn’t do
anything so I’m assuming they are the same for all examples. WHY is no work done
by the normal force? The normal force is perpendicular to the motion and the dot
product is zero for perpendicular vectors.
ACT
Three objects having the same mass begin at the same
height, and all move down the same vertical distance H.
One falls straight down, one slides down a frictionless
inclined plane, and one swings on the end of a string.
What is the relationship between their speeds when they
reach the bottom?
H
Free Fall
Frictionless incline
A) vf > vi > vp
B) vf > vp > vi
String
C) vf = vp = vi
Checkpoint / ACT
YES!
Will you do more examples in lecture?
H
Free Fall
A) vf > vi > vp
Frictionless incline
B) vf > vp > vi
C) vf = vp = vi
Demo: rolling balls on tracks
String
Checkpoint / ACT
Will you do harder examples in lecture?
H
Free Fall
A) vf > vi > vp
Frictionless incline
B) vf > vp > vi
Only gravity will do work: Wg =  K
String
C) vf = vp = vi
Wg = mgH
K =
1/
2
mv2
2
v  2 gH
v f  vi  v p  2 gH
Checkpoint / ACT
A car drives up a hill with constant speed. Which statement
best describes the total work WTOT done on the car by
all forces as it moves up the hill?
A) WTOT = 0
B) WTOT > 0
C) WTOT < 0
About 25% got this right…
Checkpoint
A car drives up a hill with constant speed. Which statement
best describes the total work WTOT done on the car by
all forces as it moves up the hill?
A) WTOT = 0
B) WTOT > 0
C) WTOT < 0
ACT
A car drives up a hill with constant speed. How does
the kinetic energy of the car change as it moves up
the hill?
A) It increases
B) It stays the same
C) It decreases
Checkpoint
A car drives up a hill with constant speed.
Which statement best describes the
total work WTOT done on the car by all
forces as it moves up the hill?
A) WTOT = 0
B) WTOT > 0
C) WTOT < 0
A) since the car moves with constant speed, then its change in kinetic
energy is 0, which means that total work done is 0
B) It will take a lot of work to move the truck up the hill-- a positive value.
C) The normal force is always parallel to the direction of motion so it does
no work but the gravitational force is negative because it acts in a
direction opposite the direction of motion.
Checkpoint
A box sits on the horizontal bed of a moving truck. Static
friction between the box and the truck keeps the box
from sliding around as the truck drives.
S
a
The work done on the box by the static frictional force as the
truck moves a distance D is:
Only 40% got this right…
A) Positive
B) Negative
C) Zero
Could we go over the directions of work and go over the friction problem with the boxes?
I've tried every number combination available and frankly, the frustration has caused me to
break numerous pencils in my dorm room.
Checkpoint
A box sits on the horizontal bed of a moving truck. Static
friction between the box and the truck keeps the box
from sliding around as the truck drives.
S
a
The work done on the box by the static frictional force as the
truck moves a distance D is:
A) Positive
B) Negative
C) Zero
From last time:
A box sits on the horizontal bed of a moving truck. Static
friction between the box and the truck keeps the box from
sliding around as the truck drives.
S
a
If the truck moves with constant acceleration to the left as
shown, which of the following diagrams best describes the
static frictional force acting on the box:
A
B
C
Checkpoint
F
S
a
D
The work done on the box by the static frictional
force as the truck moves a distance D is:
A) Positive
B) Negative
C) Zero
A) It is positive because the force is in the same direction as the
displacement.
B) the static friction opposes the motion therefor work is negative
C) No work is done since the box is not moving in relation to the truck.
Checkpoint Comments
F
S
a
D
The work done on the box by the static frictional
force as the truck moves a distance D is:
The work done on the box must be opposite to the work done by the truck as it
accelerates.
the frictional force points in the negative direction. Be careful! The friction force and the
displacement point in the SAME direction.
The net force acted on the box is zero. What force acts to the RIGHT?
Total Work done is the change in Kinetic Energy. Since the box does not move its kinetic
energy is zero and the work done must also be zero. But the box does move; the box, along
with the truck, is even accelerating.
The static frictional force and box’s movement are all same direction.
The static frictional force works in the positive direction. Ooops! Work does NOT have
a direction. That’s important.
The work is done in the direction of acceleration, which is the same direction as the
truck. Since the velocity of the box is increasing, work is being done.
The box isn’t moving. If the box doesn’t move, it will have to fall out the back door
because the truck is certainly moving.
Work done by a Spring
ACT
A box attached at rest to a spring at its equilibrium length. You
now push the box with your hand so that the spring is compressed
a distance D, and you hold the box at rest in this new location.
D
During this motion, the spring does:
A) Positive Work
B) Negative Work
C) Zero work
ACT
A box attached at rest to a spring at its equilibrium length. You
now push the box with your hand so that the spring is compressed
a distance D, and you hold the box at rest in this new location.
D
During this motion, your hand does:
A) Positive Work
B) Negative Work
C) Zero work
ACT
A box attached at rest to a spring at its equilibrium length. You
now push the box with your hand so that the spring is compressed
a distance D, and you hold the box at rest in this new location.
D
During this motion, the total work done on the box is:
A) Positive
B) Negative
C) Zero
Work done by gravity
r2
GM e m  GM e m  GM m  1  1 
W   F (r )  dr  
dr

e 
2

r
r
r
r

2
1 
r
r
1
r1
r2
1
r2
ACT
In Case 1 we send an object from the surface of the earth to a height
above the earth surface equal to one earth radius.
In Case 2 we start the same object a height of one earth radius above
the surface of the earth and we send it infinitely far away.
In which case is the magnitude of the work done by the Earth’s
gravity on the object biggest?
A) Case 1
B) Case 2
C) They are the same
1 1
W  GM e m   
 r2 r1 
ACT Solution
Case 1:
 1
1  GM e m
W  GM e m 

   2R
E
 2 RE RE 
Same!
1

GM e m
1
Case 2: W  GM e m  
 
2 RE
  2 RE 
RE
2 RE
Comments and Questions
I don’t really get this prelecture at all... So I guess I need help on most of it. Has
today’s lecture helped? Does it help to “rewind” the prelecture?
Could you go over the work done my gravity again please. I did not get the purpose of
the negative sign in the w= - mgy equation. If you make a diagram where y1 is lower than
y2, then y2 – y1 is positive or upward and the force mg is negative or downward. The work
done by the force of gravity is then negative or W = - mg (y2 – y1). Then – for
convenience – we make the equation look “pretty” by choosing to set y1 = 0 and then
relabel y2 as simply y so that W = - mg (y – 0) or W = - mgy .
I am having a hard time conceptualizing what work actually is. Also, I do not
understand why work is only the change in kinetic energy disregarding changes in potential
energy. Great observation! Hang on for two more days! We’ll add potential energy
tomorrow. Total Work is the change in Kinetic Energy by definition. Part of that total work
will be renamed Potential Energy. Just hang on.
The sign of the work done is a little confusing. It would be great to have some clarity
on that. Also, work done is only due to kinetic energy change or also potential energy. The
apple’s case was a little confusing because there is a change in potential energy and so
there should be positive work, why zero?? What’s “potential energy”? ; — ) Hang on for
two more days. We’ll relabel part of the total work as potential energy tomorrow.