law of acceleration
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THE PARADOX OF TIMEThe paradox of our time in history is that we
havetaller buildings, but shorter tempers;wider freeways, but narrower viewpoints;we spend more, but enjoy it less;
We have bigger houses, but smaller families;more conveniences, but less time;
We have more degrees, but less sense; more knowledge, but less judgment;
more experts, but more problems;more medicine, but less wellness
THE PARADOX OF TIME
We have been all the way to the moon and back, but have trouble crossing the street to meet the new neighbor.
We have conquered outer space, but not inner space; we have done larger things, but not better things.
We write more, but learn less; we plan more, but accomplish less, we’ve learned to rush, but not to wait; we have higher incomes but lower morals; we have more food, but less appeasement; we build more computers to hold more information to produce more copies than ever, but have less communication; we’ve become long in quantity, but short in quality.
THE PARADOX OF TIME
These are the times of fast foods and slow digestion; tall men and short character; steep profits and shallow relationships. These are the times of world peace, but domestic warfare; more leisure, but less fun; more kinds of food, but less nutrition.
These are the days of two incomes, but more divorces; of fancier houses, but broken homes. These are the days of quick trips, disposable diapers, throw away morality, one-night stands, overweight bodies, and pills that do everything from cheer to quiet to kill.
QUESTIONS
1. How would you describe the pacing of life today?
2. Do we need to accelerate our pace of life some more or do we need to slow down? Why?
3. What drives one to live life so fast?
THE LAW OF ACCELERATION
Newton’s Second Law
LET’S RECALL
Motion
Mass Weight
Force Acceleration
Position
ACCELERATION?
NET FORCE
MASS
ACCELERATION?NET FORCE
MASS
2 Objects of the same Mass
F
a2F
2a
FORCE OF THE SAME MAGNITUDE ON DIFFERENT MASSES
ACCELERATION?NET FORCE, Fnet
MASS
MASS
MASS
ACCELERATION?
NET FORCE, Fnet
F
F
a
½a
THE EXPERIMENT SET UP
Metal cartmass
Fixed Pulley
Slotted weight
Weight hanger
StringWooden plane
. . .
QUESTIONS
1. What represents the constant mass in procedure A? The mass of the metal cart.How about the varying force? The slotted weights.
2. How does the time of drop change as the accelerating force (slotted weights) is increased? The time of drops becomes shorter.
QUESTIONS
3. What represents the constant force in procedure B? The weight of the slotted weights.How about the varying mass? The mass of the metal cart.
4. How does the time of drop change as the accelerating force (slotted weights) is increased? The time of drop becomes greater or longer.
THE BIG QUESTION
How does the acceleration of the object (cart) related to the net force acting on the object (cart) and the mass of the object, and how can this relationship be expressed in a single equation?
a α Fnet
a α a α
How do we represent the 2nd law mathematically?
When m is constant
When Fnet is constant
Thus:
1m
Fnet
m
NOW, WE INSERT A PROPORTIONALITY CONSTANT, K, TO CREATE AN EQUATION RELATING ALL THREE VARIABLES:
a = k
a =
Fnet
mIf the units in both sides of the equation are consistent, with newtons for force, meters, kilograms and seconds for the acceleration and mass, the value of k is 1. Thus the final equation for Newton’s Second Law of Motion is
Fnet
mFnet = ma
OR
ILLUSTRATIVE EXAMPLE 1
A constant force of 20 N acts on a mass of (a) 2 kg, (b) 4 kg and (c) 6 kg. What are the resulting accelerations?
a1 = —— = ——— = 10 m/s2
F 20Nm1 2kg
a2 = —— = ——— = 5 m/s2 a3 = —— = ——— = 3.3 m/s2
F 20Nm2 4 kgF 20Nm3 6kg
ACCELERATION-MASS RELATIONSHIP
Mass in kg
Accele
rati
on
in
m
/s2
0 1 2 3 4 5 6 7
8
10
9
8
7
6
5
4
3
2
1
.
..
ILLUSTRATIVE EXAMPLE 2
A 4.0-kg mass is acted on by a resultant force of (a) 4.0N; (b) 8.0 N and (c) 12 N. What are the resulting accelerations?a1 = —— = ——— = 1.0 m/s2
F 4.0Nm1 4kg
a2 = —— = ——— = 2.0 m/s2 a3 = —— = ——— = 3.0 m/s2
F 8.0Nm2 4 kgF 12Nm3 4kg
ACCELERATION-FORCE RELATIONSHIP
Force in N
Accele
rati
on
in
m
/s2
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14
15 16
5
4
3
2
1
..
.
HOW CAN WE APPLY NEWTON’S 2ND LAW OF MOTION IN OUR LIFE?
How can we relate the second law in our own life?
If the net force is the same as the effort you exert in the different tasks you do everyday, how could the amount of effort you exert affect the success of completing your tasks?
CHRISTIAN LIVING EDUCATION:
And Jesus said to them, “Suppose one of you has a friend to whom he goes at midnight and says, “Friend lend me three loaves of bread, for a friend of mine has arrived at my house from a journey and I have nothing to offer him,” and he says in reply from within. “Do not bother me; the door has been locked and my children and I are already in bed. I cannot get up to give you anything.” I tell you, If he does not get up to give him the loaves because of their friendship, he will get up to give him whatever he needs because of his persistence.
- Luke 11: 5-8
LET’S CHECK WHAT YOU LEARNED.
True or False: (in ¼ sheet of paper)
Write T if the statement is true and F if it is not.
1. As the applied net force increases the acceleration of the object of constant mass decreases.
2. If the mass of the object is doubled and the
applied force is kept the same the acceleration produced will be halved.
3. The heavier the person the harder it is for him or her to change his or her state of motion.
4. An object can accelerate only if force is applied.
5. The law of acceleration applies only to objects
at rest.
COMPLETE THE STATEMENT:
Today, I discovered that ______. The most important thing I
learned in this lesson was______.
In this activity I find it difficult ______.
Newton’s Second Law of Motion: “Whenever an unbalanced force acts on a body, it produces in the direction of the force an acceleration that is directly proportional to the force and inversely proportional to the mass of the body.”
ΣF = + F1 + F2 + F3 + …Force (N) = mass (kg) x acceleration
(m/s2)Force (lb) = mass (slug) x acceleration (ft/s2)1 lb = 4.448 N 1 slug = 14.59 kg
Fnet = ma
SOLVE:
A 1000kg car moving north at 100 km/h brakes to a stop in 50 m. What are the magnitude and direction of the force?
Given: m = 1000kg; vi = 100 km/h = 27.8 m/s; ΔX = 50m; vf = 0 Find: Fnet = ?
Formula: Fnet = ma
a = ————
vf – vi
ΔtΔX 50 m
v 13.9 m/sΔt = —— = —————
Δt = 3.6 sa = ———————
0 – 27.8 m/s3.6 s
a = – 7.72 m/s2
or 7.72 m/s2, S
F = ma = (1000kg)(- 7.72 m/s2)F = 7720 N, South
Given: m = 1000kg; vi = 100 km/h = 27.8 m/s; ΔX = 50m; vf = 0 Find: Fnet = ?
Formula: Fnet = ma
ASSIGNMENT
Answer in your lecture notebook.
1. Does Newton’s second law agree with the first law of motion? Explain.
2. How can the concept of the second law of motion apply in safe driving?
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