Transcript PPTX - University of Colorado Boulder

Announcements
• CAPA assignment #7 due Friday, October 7 at 10 PM
• CAPA assignment #8 is due Friday, October 14 at 10 PM
• Read Ch. 6 (Work & Energy): 6.1-6.6
• Next week in Section, Assignment #5: Work & Energy
Print and bring to your Lab Section.
• Midterm Exam #2 on Tuesday, October 11
Exam Room Assignments
Exam #2 Topics
A. Recall “Kinematical Equations” for 1D motion.
B. Ch. 3: Kinematics in 2D
3-4. Adding Vectors by Components
3-5 & 3-6. Projectile Motion & Solving
Problems
D. Ch. 5: Circular Motion; Gravity
5-1. Kinematics of Uniform Circular Motion
C. Ch. 4: Newton’s Laws of Motion
4-2. First Law
5-2. Dynamics of Uniform Circular Motion
4-3. Mass
5-6. Newton’s Law of Universal Gravitation
4-4. Second Law
5-7. Gravity Near Earth’s Surface
4-5. Third Law
5-8. Satellites and “Weightlessness”
4-6. Weight & the Normal Force
4-7. Solving Problems & Free-Body
5-9. Kepler’s Laws and Newton’s Synthesis.
Diagrams
4-8. Friction & Inclines
Exam #2 Topics
Remember: A large part of what you’ll be expected to do is to apply
Newton’s Laws. You can’t just memorize and apply formulas.
Steps for Linear Motion:
1. Draw a free-body diagram identifying and labeling all forces.
2. Choose a coordinate system – typically with the forces pointing in the coordinate
directions (x,y) as much as possible.
3. Write down Newton’s 2nd law in each coordinate direction (typically), summing
the forces. The equation perpendicular to the direction of motion often allows
you to find the Normal Force, which is needed to determine the force of friction.
Σ Fx = max
Σ Fy = may
(for rectilinear motion)
If a force is not in a coordinate direction, you must find its components in the
coordinate directions.
4. If there is more than one mass, then Newton’s 2nd law may be needed for each
mass.
Exam #2 Topics
Remember: A large part of what you’ll be expected to do is to apply
Newton’s Laws. You can’t just memorize and apply formulas.
Steps for Uniform Circular Motion:
1. Draw a free-body diagram identifying and labeling all forces.
2. Choose a coordinate system – one of the directions will point in the
radial direction. Others directions: tangent direction (T) or vertical (y).
3. Write down Newton’s 2nd law in each coordinate direction (typically),
summing the forces.
Σ FR = maR = mv2/r
Σ FT = maT (for uniform circular motion)
4. If there is more than one mass, then Newton’s 2nd law may be
needed for each mass.
Incomplete Gallery of Problems Involving Newton’s Laws:
F
w & w/0 friction
m
w & w/0 friction
F
Box 1
M
Box 2
mm
fc
M1
w & w/0 friction
+x
M2
M
m
T
w & w/0 gravity
Practice setting up the free-body diagram and Fnet = ma
Incomplete Gallery of Problems Involving
Newton’s Law of Universal Gravitation:
What is Energy?
All kinds of things come to mind…
Things in motion, collisions, electricity, stored energy
What is Energy?
Energy comes in many forms.
Energy can be converted from one form to
another.
• Kinetic energy (KE) - energy of motion
• Thermal energy - energy of atomic vibrations
• Potential energy (PE) - energy stored in configuration
- gravitational
- electrostatic
- elastic
- chemical
- nuclear
• Radiant (solar) energy – energy of light
• Mass energy – Einstein’s E=mc2
What is Energy?
Definition: Energy is the ability to do “Work”
Of course now we will need a physics definition of “Work”
Whenever work is being done, energy is being changed
from one form to another or being transferred from one
body to another.
The amount of work done on a system (or by a system) is
the change in energy of the system.
Energy is such a useful concept
because it is “Conserved”.
To understand energy,
its inter-conversion to different forms,
and its conservation,
we must first define some terms:
Work
Kinetic Energy (KE)
“Work-Energy Principle”
Potential Energy (PE)
Work
Consider an object moved by a force F
While the force is applied, If the object moves a distance d
in the direction of the force vector, then the work done is:
Force
|Dx| =d
Work  W  Force  Displaceme nt
Work must have units of Newtons x meters = (kg m/s2) x m = kg m2/s2
We also call this Joules – the SI unit for work and also energy!
Work
Force
What if the force vector
is at some angle with
respect to the
displacement vector?
|Dx| =d
Work  W  Fx  d  Fcos( ) d  F d
Component of force along the displacement
Clicker Question
Room Frequency BA
Atlas holds up the entire earth for 20 seconds.
What can you say about the work
done by Atlas?
A)An enormous amount of work
(many, many Joules)
B) 20 Joules
C) None
There is no displacement of the earth.
Thus, despite the very large force (Mg) needed to hold
up the earth, no work is done.
Note: Work is a scalar (not a vector), but it does have a sign.
When the component of force in the direction of displacement points
along the displacement, Work is positive.
When the force is perpendicular to the displacement, Work is zero.
When the component of force in the direction of displacement points
opposite to displacement, Work is negative.
Clicker Question
Room Frequency BA
You are at the gym and observe this
fellow lifting weights.
He lifts 10 kg at constant velocity
upwards a distance of 0.1 meters.
Flift
Mg
What are the forces on the barbell in a free body diagram?
What is true about the work done by the fellow on the
barbell (via Flift)?
A)Work > 0
B) Work < 0
C) Work = 0
What is true about the work done by the force of gravity
on the barbell?
A)Work > 0
B) Work < 0
C) Work = 0
Whenever you talk about “Work Done”,
you have to be careful to specify
which force does the work!
Consider an object displaced Δx = +1 m at a constant speed
on a rough table by a constant external force F = 10 N.
N
Work done by the external force =
Wext = +Fext Δx = (10 N)(1m) = +10 J
Work done by the friction force =
Fg=mg
Wfric = -Ffric Δx = -(10 N)(1m) = -10 J
Work done by the normal force =
WN = (0) Δx = 0 J
Work done by gravity
Wg = (0) Dx = 0 J
Work done by the net force = Wnet = (0) Δx = 0 J
Room Frequency BA
Clicker Question
A rock of mass m is twirled on a
string in a horizontal plane.
The work done by the tension in the string on the rock is
T
A) Positive
B) Negative
C) Zero
The work done by the tension force
is zero, because the force of the
tension in the string is perpendicular
to the direction of the displacement:
W = F cos 90°= 0
Kinetic Energy
The kinetic energy (KE) of an object of mass m
moving with speed v is:
1 2
KE  mv
2
• KE is the energy of motion.
• KE is always positive (KE >= 0).
• KE and Work are related by the “Work-Energy Principle”
• Units of KE
1 2
 kgm 
2
KE  mv  [kg][ m / s]   2 m   Nm  [Joules]
2
 s

 
A bowling ball of mass m = 7.7 kg (17 lbs) is
rolled down the lane with speed v = 7 m/s.
1 2 1
2
KE  mv  (7.7kg)( 7m / s)  190 J
2
2