Download Physics 121C Mechanics

Physics 114C - Mechanics
Lecture 16 (Walker: Ch. 7.1-2)
Work & Energy
February 3, 2012
John G. Cramer
Professor Emeritus, Department of Physics
B451 PAB
[email protected]
Announcements
 Because the last two problems of HW#4 are covered in
today’s lecture, I moved the due date for HW#4 to 11:59 PM
on Friday, February 3 (tonight). HW#5 is still due at 11:59
PM on Thursday, February 9. HW#6 is due at 11:59 PM on
Thursday, February 16.
 Register your clicker.
 We will have Exam 2 on Friday, February 10. It will cover
Chapters 5-8 and will be similar to Exam 1 in its structure.
There will again be assigned seating. If you have not already
done so and would like to request a left-handed seat, righthanded aisle seat, or front row seat, E-mail your request to
me ASAP.
February 3, 2012
Physics 114A - Lecture 16
2/28
Lecture Schedule (Part 2)
Physics 114A - Introduction to Mechanics - Winter-2012
Lecture: Professor John G. Cramer
Textbook: Physics, Vol. 1 (UW Edition), James S. Walker
Week
4
5
6
Date
L#
23-Jan-12
E1
24-Jan-12
10 Newton's Laws
14
29
5-1 to 5-4
26-Jan-12
11 Common Forces
11
26
5-5 to 5-7
27-Jan-12
12 Free Body Diagrams
-
24
30-Jan-12
13 Friction
9
27
6-1
31-Jan-12
14 Strings & Springs
12
29
6-2 to 6-4
2-Feb-12
15 Circular Motion
5
30
6-5
3-Feb-12
16 Work & Energy
11
23
7-1 to 7-2
6-Feb-12
17 Work & Power
7
25
7-3 to 7-4
7-Feb-12
18 Potential Energy
10
26
8-1 to 8-2
9-Feb-12
19 Energy Conservation I
16
18
8-3 to 8-5
10-Feb-12
E2
February 3, 2012
Lecture Topic
Pages Slides
Reading
HW Due
Lab
EXAM 1 - Chapters 1-4
-
1-D Dynamics
HW3
Newton's Laws
Tension
HW4
Work-energy
HW5
We are here.
EXAM 2 - Chapters 5-8
Physics 114A - Lecture 16
3/28
Circular Orbits (1)
Thought Experiment:
On an airless planet,
cannon balls are shot from a
cannon mounted on a tower
ar increasing muzzle
velocities, and go farther
and farther as the velocity
is increased.
What limits their range?
February 3, 2012
Physics 114A - Lecture 16
4/28
Circular Orbits (2)
w  (mg , toward center)
Fnet
a
 ( g , toward center)
m
(vorbit ) 2
ar 
 g , so vorbit  rg
r
w  (mg , toward center)
vorbit  rg  (6.37 106 m)(9.80 m/s 2 )  7,900 m/s  16,000 mph
February 3, 2012
Physics 114A - Lecture 16
5/28
Example: A Satellite’s Motion
A satellite moves at constant speed in a circular
orbit about the center of the Earth and near the
surface of the Earth. If the magnitude of its
acceleration is g = 9.81 m/s2 and the Earth’s
radius is 6,370 km, find:
(a) its speed v; and
(b) the time T required for one complete
revolution.
v2
acp   g
r
v  rg  (6,370 103 m)(9.81 m/s 2 )  7.91103 m/s  17, 700 mi/h
T  2 r / v  2 (6,370 103 m) /(7.91103 m/s)  5,060 s  84.3 min
Note: if you dug a tunnel directly through the center of the Earth and
dropped in a subway car, its round-trip transit time would also be 84.3 minutes.
February 3, 2012
Physics 114A - Lecture 16
6/28
Work Done by a Constant Force
The definition of work, when the force is
parallel to the displacement:
(7-1)
SI work unit:
newton-meter (N·m) = joule, J
February 3, 2012
Physics 114A - Lecture 16
7/28
Typical Work
February 3, 2012
Physics 114A - Lecture 16
8/28
Work for Force at an Angle
If the force is at an angle to the displacement:
(7-3)
Only the horizontal component of the force
does any work (horizontal displacement).
February 3, 2012
Physics 114A - Lecture 16
9/28
Work Summary
Energy is transferred from
person to spring as the person
stretches the spring. This is
“work”.
W  F x
Work = 0
SI Units for work:
W  Fx x  F cos  x
February 3, 2012
1 joule = 1 J = 1 N·m
1 electron-volt = 1 eV = 1.602 x 10-19 J
Physics 114A - Lecture 16
10/28
Work Done by a Constant Force
The work can also be written as the dot
product of the force F and the displacement d:
February 3, 2012
Physics 114A - Lecture 16
11/28
Negative and Positive Work
The work done may be positive, zero, or
negative, depending on the angle between the
force and the displacement:
February 3, 2012
Physics 114A - Lecture 16
12/28
Perpendicular Force and Work
A car is traveling on a curved
highway. The force due to friction fs
points toward the center of the
circular path.
How much work does the frictional
force do on the car?
Zero!
General Result: A force that is
everywhere perpendicular to the
motion does no work.
February 3, 2012
Physics 114A - Lecture 16
13/24
Work on a System with
Many Forces
Wtotal  F1x x1  F2 x x2  F3 x x3 
Model the system as a particle  a single x
Wtotal  F1x x  F2 x x  F3 x x 
 ( F1x  F2 x  F3 x 
February 3, 2012
)x  Fnet x x
Physics 114A - Lecture 16
14/24
Work Done by a Constant Force
If there is more than one force acting on an
object, we can find the work done by each force,
and also the work done by the net force:
(7-5)
February 3, 2012
Physics 114A - Lecture 16
15/24
Example: Pulling a Suitcase
A rope inclined upward at 45o pulls
a suitcase through the airport. The
tension on the rope is 20 N.
How much work does the tension
do, if the suitcase is pulled 100 m?
W  T (x) cos 
W  (20 N)(100 m) cos 45  1410 J
Note that the same work could have been done by a tension of just
14.1 N by pulling in the horizontal direction.
February 3, 2012
Physics 114A - Lecture 16
16/28
Gravitational Work
In lifting an object of weight mg by a height h, the
person doing the lifting does an amount of work W = mgh.
If the object is subsequently allowed to fall a
distance h, gravity does work W = mgh on the object.
W  mgh
February 3, 2012
Physics 114A - Lecture 16
17/24
Example: Loading with a Crane
A 3,000 kg truck is to be loaded onto a ship by a
crane that exerts an upward force of 31 kN on the
truck. This force, which is large enough to
overcome the gravitational force and keep the truck
moving upward, is applied over a distance of 2.0 m.
(a) Find the work done on the truck by the crane.
(b) Find the work done on the truck by gravity.
(c) Find the net work done on the truck.
Wapp  Fapp y y  (31 kN)(2.0 m)  62 kJ
Wg  mg y y  (3000 kg)(9.81 m/s2 )(2.0 m)  58.9 kJ
Wnet  Wapp  Wg  (62.0 kJ)  (58.9 kJ)  3.1 kJ
February 3, 2012
Physics 114A - Lecture 16
18/28
Positive & Negative
Gravitational Work
When positive work is done on an object, its
speed increases; when negative work is done, its
speed decreases.
February 3, 2012
Physics 114A - Lecture 16
19/28
Kinetic Energy &
The Work-Energy Theorem
v  v  2ax  mv  mv  2F x
2
f
2
i
2
f
2
i
After algebraic manipulations of the equations
of motion, we find:
Therefore, we define the kinetic energy:
(7-6)
February 3, 2012
Physics 114A - Lecture 16
20/28
Kinetic Energy &
The Work-Energy Theorem
Work-Energy Theorem: The
total work done on an object is
equal to its change in kinetic
energy.
(7-7)
February 3, 2012
Physics 114A - Lecture 16
21/28
Clicker Question 1
Car 1 has twice the mass of Car 2, but
they both have the same kinetic energy.
If the speed of Car 1 is v, approximately
what is the speed of Car 2?
a) 0.50 v
February 3, 2012
b) 0.707 v
c) v
d) 1.414 v
Physics 114A - Lecture 16
e) 2.00 v
22/28
Problem Solving Strategy
Picture: The way you choose the +y direction or the +x direction can help
you to easily solve a problem that involves work and kinetic energy.
Solve:
1. Draw the particle first at its initial position and second at its final
position. For convenience, the object can be represented as a dot or box.
Label the initial and final positions of the object.
2. Put one or more coordinate axes on the drawing.
3. Draw arrows for the initial and final velocities, and label them
appropriately.
4. On the initial-position drawing of the particle, place a labeled vector for
each force acting on it.
5. Calculate the total work done on the particle by the forces and equate
this total to the change in the particle’s kinetic energy.
Check: Make sure you pay attention to negative signs during your
calculations. For example, values for work done can be positive or negative,
depending on the direction of the displacement relative to the direction of
the force. Kinetic energy values, however, are always positive.
February 3, 2012
Physics 114A - Lecture 16
23/28
Example: A Dogsled Race
During your winter break, you enter a
“dogsled” race across a frozen lake, in which
the sleds are pulled by students instead of
dogs. To get started, you pull the sled (mass
80 kg) with a force of 180 N at 40° above the
horizontal. The sled moves x = 5.0 m,
starting from rest. Assume that there is no
friction.
(a) Find the work you do.
(b) Find the final speed of your sled.
Wtotal  Wyou  Fx x  F cos x
 (180 N)(cos 40)(5.0 m)  689 J
Wtotal  12 mv2f  12 mvi2  12 mv2f
February 3, 2012
2Wtotal
v 
m
2
f
vf 
Physics 114A - Lecture 16
2Wtotal
2(689 J)

 4.15 m/s
m
(80 kg)
24/28
Example: Work and Kinetic
Energy in a Rocket Launch
A 150,000 kg rocket is launched straight
up. The rocket engine generates a thrust of
4.0 x 106 N.
What is the rocket’s speed at a height of
500 m? (Ignore air resistance and mass loss
due to burned fuel.)
Wthrust  Fthrust (y )  (4.0  106 N)(500 m)  2.0  109 J
Wgrav  w(y)  mg (y)  (1.5  104 kg)(9.80 m/s2 )(500 m)  0.74  109 J
K  12 mv2  0  Wthrust  Wgrav  1.26  109 J
February 3, 2012
Physics 114A - Lecture 16
v
2K
 129.6 m/s
m
25/28
Example: Pushing a Puck
A 500 g ice hockey puck slides across
frictionless ice with an initial speed of
2.0 m/s. A compressed air gun is used to
exert a continuous force of 1.0 N on the
puck to slow it down as it moves 0.50 m.
The air gun is aimed at the front edge of
the puck, with the compressed air flow
30o below the horizontal.
What is the puck’s final speed?
W  F (r ) cos   (1.0 N)(0.5 m) cos150  0.433 J
K  12 mv12  12 mv02  W
v1  v0 2 
February 3, 2012
2W
2(0.433 J)
 (2.0 m/s) 2 
 1.51 m/s
m
(0.5 kg)
Physics 114A - Lecture 16
26/28
Example: Work on an Electron
In a television picture tube, electrons are accelerated
by an electron gun. The force that accelerates the
electron is an electric force due to the electric field
in the gun. An electron is accelerated from rest by an
electron gun to an energy of 2.5 keV (2,500 eV) over a distance
of 2.5 cm. (1 eV = 1.60 x 10-19 J)
Find the force on the electron, assuming that it is constant and in the
direction of the electron’s motion.
Fx x  K f  Ki
Wtotal  K
Fx 
K f  Ki
x
(2,500 ev)(1.6 1019 J/eV)  0

(0.025 m)
 1.6 1014 N
February 3, 2012
Physics 114A - Lecture 16
27/28
End of Lecture 16
 Before Monday, read Walker Chapter
7.3-4
 Homework Assignments #4 should be
submitted using the Tycho system by
11:59 PM on Friday, February 3 (Tonight!)
 Register your clicker, using the “Clicker”
link on the Physics 114A Syllabus page.
February 3, 2012
Physics 114A - Lecture 16
28/28