Download Physics Talk 2.3

Newton’s Second Law
September 30, 2013
HW:
 Do Now:
 Copy LO and SC
 Agenda:
 Do Now
 LO and SC
 Investigate
 Physics Talk, Notes
 Active Physics Plus
 Learning Objective:
 Students use F=ma to solve
problems relating to Newton’s
Second Law of Motion
 Success Criteria:
 Identify the forces acting on an
object
 Determine when the forces on an
object are either balanced or
unbalanced
 Compare amounts of acceleration
semi-quantitatively
 Apply Newton’s Second Law of
Motion
 Apply the definition of the Newton
as a unit of force
WDYS/WDYT pg. 157
Notebook Set up
 Read Investigate pg. 157-159
 Set up Notebook for tomorrow
September 28, 2010
HW: Grade Report Signed, Missing
work due Friday
 Do Now:
 Read Investigate pg. 157-159
 Agenda:
 Do Now
 LO and SC
 Investigate
 Physics Talk, Notes
 Active Physics Plus
 Learning Objective:
 Students use F=ma to solve
problems relating to Newton’s
Second Law of Motion
 Success Criteria:
 Identify the forces acting on an
object
 Determine when the forces on an
object are either balanced or
unbalanced
 Compare amounts of acceleration
semi-quantitatively
 Apply Newton’s Second Law of
Motion
 Apply the definition of the Newton
as a unit of force
Investigate:
 #1-5 30 minutes
 6-7 10 minutes
Force
Mass
Acceleration
Exit Ticket (on a half sheet of
paper)
 Explain the relationship between Mass, Acceleration, and
Force.
 Hint: keep mass constant, explain what happens to force and
acceleration
 Hint: keep force constant, explain what happens to mass and
acceleration
 Hint: given a constant acceleration, how are mass and force
related?
 Vocab: Mass, Acceleration, Force, increase, decrease,
constant
September 29, 2010
HW: PTG 1-4, 6, 7, 9-11
 Do Now: How did
the mass on the car
change the force
needed to push it?
 Agenda:
 Do Now
 LO/SC
 Physics Talk
 PTG
 L.O. Students use F=ma to solve
problems relating to Newton’s
Second Law of Motion
 S.C.
 Identify the forces acting on an
object.
 Determine when the forces on an
object are either balanced or
unbalanced.
 Compare amounts of acceleration
semi-quantitatively.
 Apply Newton’s Second Law of
Motion.
 Apply the definition of the Newton
as a unit of force
In your notebook:
 Look at pg. 162
 Explain why force is measure in Newtons which is
defined as
 1N=1kg*m/s2
Physics Talk 2.3
 What is Newton’s
Second Law?
 Relationship between
force, mass, and
acceleration
 F=ma
Physics Talk 2.3
 What is the equation for
Newton’s Second Law?
F  m a
 What does each variable a = acceleration (m/s2)
represent?
F = force (Newton – N)
m = mass (kg)
Physics Talk 2.3
 What is a Newton?
 The Newton is the unit for
force. 1 N is the force
required to make on kg of
mass accelerate at 1m/s2
 1N = 1 kg*m/s2
 What causes
acceleration?
 Unbalanced forces
Physics Talk 2.3
 What are some examples
of Newton’s second law?
 Does Newton’s 2nd Law
ever stop working?
 If you push a small cart
with a large force, it will
accelerate a great deal. If
you use the same force
on a car, it will accelerate
less.
 No, there is always
acceleration, it just may
be too small to measure.
Sample Problem 1
 A tennis ball with mass
58g accelerates at
430m/s2 when it is
served. What is the
force responsible for
this acceleration?
 Given:
 m =58 g = 0.058 kg
 a = 430m/s2
 Unknown: Force
 Tool: F = ma
 Solution:
 F = 0.058kg*430m/s2
 F = 24.95 kg*m/s2
 F ≈ 25N
Sample Problem 1
 Could an identical force
accelerate a 5.0 kg
bowling ball at the same
rate?
 Given:
 F=25N
 m = 5.0kg
 Unknown: acceleration
 Tool: F=ma
 Solution:
 25N=5kg*a m/s2
 25N/5kg = a
Sample Problem 1
 Could an identical force
accelerate a 5.0 kg
bowling ball at the same
rate?
Solution:
25N=5kg*a m/s2
25N/5kg = a
25kg  m / s 2
a
5kg
5m / s 2  a
 No, an identical force
would not accelerate the
bowling ball at the same
rate.
Sample Problem 2
 A tennis racket hit a
sand-filled tennis ball
with a force of 4 N.
While the 275 g ball is in
contact with the racket,
what is its acceleration?
 Given:
 F=4N
 m=275g = 0.275 kg
 Unknown: acceleration
 Tool: F=ma
 Solution:
 4N=0.275kg*a m/s2
4kg  m / s 2
a
0.275kg
14.5m / s 2  a
Gravity, Mass, Weight, and
Newton’s Second Law
 What is the acceleration
 9.8 m/s2
due to gravity?
 What does this mean
about the force of
gravity?
 What is weight?
 If you drop a 1kg mass,
there is a force of 9.8N
acting on the object
 The vertical, downward
force exerted on a mass
as a result of gravity
Gravity, Mass, Weight, and
Newton’s Second Law
 How do you calculate an
object’s weight?
 What do the variables
mean?
Fgravity  m  a gravity
w  m g
w = weight
m = mass in kg
g = acceleration due to
gravity (9.8 m/s2)
September 30, 2010
HW: 2.3 PTG #1-4, 6, 7, 9-11, Due
Friday
 Do Now: Use the concept of F=ma to
explain why someone has different
weights on different planets.
 Agenda:
 Do Now
 LO/SC
 Physics Talk
 Vector Addition
 Learning Objectives:


Students use F=ma to solve
problems relating to Newton’s
Second Law of Motion
Explain the difference between
mass and weight
 Success Criteria:
 Identify the forces acting on an
object
 Determine when the forces on an
object are either balanced or
unbalanced
 Compare amounts of
acceleration semi-quantitatively
 Apply Newton’s Second Law of
Motion
 Apply the definition of the
Newton as a unit of force
 Describe weight as the force due
to gravity on an object
Balanced and Unbalanced Forces
 What is a free-body
diagram?
 When will an object
accelerate?
 A diagram showing the
forces acting on an
object
 Balanced force=no
acceleration
 Unbalanced
forces=acceleration
Free-Body Diagram
 Example of free body
diagram for
Free-Body Diagram
 Example of a free-body
diagram for a car moving
on the road at a constant
velocity.
Since the car is traveling at a
 Is the car accelerating?
constant speed, it is not
accelerating. This means that
the force of the road on the
tires is equal to the air
resistance and we have
balanced forces.
What do you think now?
 In your notebook: Pg. 170