Download Newton`s Second Law

Newton’s Second Law
The Mathematical One
What is the relationship?
Force
Force
Mass
Acceleration
Mass

Constant

Constant
Acceleration

Constant

Constant

Constant

Constant
Think of a race car!
Force is generated by the engine
Acceleration is how fast it increases it’s speed off the line
Student Race…
• Let’s hit the hall!
What are the relationships?
• Force is proportional to acceleration
▫ If F  then a  (Mass constant)
• Force is proportional to mass
▫ If F  then m  (Acceleration constant)
• Acceleration is inversely proportional to mass
▫ If a  then m  (Force constant)
Scenario 1:
• Talk to your partner about how these
relationships apply to:
You are late for school and your mom pulls up
behind a transport truck. What would the impact
be if it was: Empty? Full?
Scenario 2:
• Talk to your partner about how these
relationships apply to:
You have 2 cars that have the same mass – one car
has the engine of a Ford Focus, the other has the
engine of a Ford Mustang. Assume both overall
masses are the same
Newton’s Second Law
• When a net force acts on an object, the object
accelerates in the direction of the net force.
FNET = ma
Unit Analysis: 1N = 1 kg۰m/s2
Example # 1:
• A student is bowling with her friends. She gives a
7.0 kg bowling ball an acceleration of 5.0 m/s2
[forward]. Calculate the net force she exerted on the
ball.
ā = 5.0 m/s2
m = 7.0 kg
Fnet = mā
= (7.0 kg)(5.0 m/s2 [forward])
= 35 N [forward]
Example 2:
A motorcycle and driver have a combined mass of
280 kg. They accelerate from 7.0 m/s [E] to 34
m/s [E] in 4.2 s. What is the net force on the
motorcycle and driver?
m = 280 kg
v1 = 7.0 m/s [E]
v2 = 34 m/s [E]
Δt = 4.2 s
Solve for ā first
ā = v2 – v1
Δt
= 34 m/s [E] – 7.0 m/s [E]
4.2 s
= 6.4 m/s2 [E]
Fnet =ma
= (280 kg)(6.4 m/s2 [E])
= 1.8 x 103 N [E]
Newton’s Second Law and
Motion
Day 2 – The Real Math
Equations
2 2
 
v2  v1  2aavd

2
 
aav (t )
d  v2 t 
2
 
v2  v1

aav 
t

 
aav (t ) 2
d  v1t 
2
 (v1  v2 )t
d 
2
Motion


FNET  ma
BOTH!!!
Forces


FNETY   FY


FNETX   FX
Example 1:
A rocket ship has engines that can produce a force
of 1.8x105N [up]. If the ship has a mass of
1.3x104kg.
a) What will the acceleration of the ship be?
b) What will it’s speed be after 10s?
Example 1:
A driver approaches an intersection at a velocity of
14 m/s [forward] when the light turns amber. The
driver applies the brakes to get the maximum
stopping force. The car has a mass of 1500 kg,
and the force of friction between the tires and the
road is 1.1 x 104N. Ignoring the driver's reaction
time calculate:
a.) the maximum deceleration of the car
b.) the minimum stopping time
c.) the minimum stopping distance
Example 2:
An elevator, including passengers, has a mass of
600 kg. When leaving the first floor, it accelerates
upward at 2.0 m/s2. What force is the cable
exerting on the elevator?