february 21, 2014 day 2

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February 21, 2014 Day 2 February 21, 2014 Day 2 Science Starters Sheet 1. Please have these Items on your desk. 2- Science Starter: List 3 “Physics” words. http://www.youtube.com/watch?v=B DpxSLv89Y8&list=PLB6B38A6C0540EE B6 Agenda Lab Write Up

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1. Please have these Items on your desk. Science Starters Sheet. Lab Write Up. Agenda. February 21, 2014 Day 2. 2- Science Starter: List 3 “Physics” words. http://www.youtube.com/watch?v=BDpxSLv89Y8&list=PLB6B38A6C0540EEB6. Continuation of notes. 2 nd Law. F = m x a. - PowerPoint PPT Presentation

TRANSCRIPT

Page 1: February 21, 2014     Day 2

February 21, 2014 Day 2February 21, 2014 Day 2

ScienceStarters Sheet

1. Please have theseItems on your desk.

2- Science Starter:

List 3 “Physics” words.

http://www.youtube.com/watch?v=BDpxSLv89Y8&list=PLB6B38A6C0540EEB6

AgendaLab

WriteUp

Page 2: February 21, 2014     Day 2

Continuation of notes.Continuation of notes.

Page 3: February 21, 2014     Day 2

22ndnd Law Law

Page 4: February 21, 2014     Day 2

22ndnd Law Law

The net force of an object is The net force of an object is equal to the product of its mass equal to the product of its mass and acceleration, or F=ma.and acceleration, or F=ma.

Page 5: February 21, 2014     Day 2

22ndnd Law Law

When mass is in kilograms and acceleration is When mass is in kilograms and acceleration is in m/s/s, the unit of force is in newtons (N).in m/s/s, the unit of force is in newtons (N).

One newton is equal to the force required to One newton is equal to the force required to accelerate one kilogram of mass at one accelerate one kilogram of mass at one meter/second/second.meter/second/second.

Page 6: February 21, 2014     Day 2

22ndnd Law (F = m x a) Law (F = m x a)

How much force is needed to accelerate a 1400 kilogram car 2 meters per second/per second?

Write the formulaWrite the formula F = m x a Fill in given numbers and unitsFill in given numbers and units F = 1400 kg x 2 meters per second/second Solve for the unknownSolve for the unknown 2800 kg-meters/second/second or 2800 N

Page 7: February 21, 2014     Day 2

Newton’s 2nd Law proves that different masses accelerate to the earth at the same rate, but with different forces.

• We know that objects with different masses accelerate to the ground at the same rate.

• However, because of the 2nd Law we know that they don’t hit the ground with the same force.

F = maF = ma

98 N = 10 kg x 9.8 m/s/s98 N = 10 kg x 9.8 m/s/s

F = maF = ma

9.8 N = 1 kg x 9.8 9.8 N = 1 kg x 9.8 m/s/sm/s/s

Page 8: February 21, 2014     Day 2

Check Your UnderstandingCheck Your Understanding

1. 1. What acceleration will result when a 12 N net force applied to a 3 kg What acceleration will result when a 12 N net force applied to a 3 kg object? A 6 kg object?object? A 6 kg object?

   2. A net force of 16 N causes a mass to accelerate at a rate of 5 m/s2. A net force of 16 N causes a mass to accelerate at a rate of 5 m/s22. .

Determine the mass.Determine the mass.

3. How much force is needed to accelerate a 66 kg skier 1 m/sec/sec?3. How much force is needed to accelerate a 66 kg skier 1 m/sec/sec?

4. What is the force on a 1000 kg elevator that is falling freely at 9.8 4. What is the force on a 1000 kg elevator that is falling freely at 9.8 m/sec/sec?m/sec/sec?

Page 9: February 21, 2014     Day 2

Check Your UnderstandingCheck Your Understanding

1. What acceleration will result when a 12 N net force applied to a 3 kg object? 1. What acceleration will result when a 12 N net force applied to a 3 kg object? 12N/3kg = 4m/sec/sec12N/3kg = 4m/sec/sec

   2. A net force of 16 N causes a mass to accelerate at a rate of 5 m/s2. A net force of 16 N causes a mass to accelerate at a rate of 5 m/s22. Determine the . Determine the

mass.mass. m=f/a = 16N/5 m/sec/secm=f/a = 16N/5 m/sec/sec = 3.2kg= 3.2kg

   3. How much force is needed to accelerate a 66 kg skier 1 m/sec/sec?3. How much force is needed to accelerate a 66 kg skier 1 m/sec/sec? f= mxa =66kg/1m/sec/secf= mxa =66kg/1m/sec/sec = 66N= 66N

4. What is the force on a 1000 kg elevator that is falling freely at 9.8 m/sec/sec?4. What is the force on a 1000 kg elevator that is falling freely at 9.8 m/sec/sec? f=mxa =1000kg x 9.8m/sec/sec =9800 Nf=mxa =1000kg x 9.8m/sec/sec =9800 N

Page 10: February 21, 2014     Day 2

33rdrd Law Law

For every action, there is an For every action, there is an equal and opposite reactionequal and opposite reaction..

Page 11: February 21, 2014     Day 2

33rdrd Law Law

According to Newton, According to Newton, whenever objects A and whenever objects A and B interact with each B interact with each other, they exert forces other, they exert forces upon each other. When upon each other. When you sit in your chair, you sit in your chair, your body exerts a your body exerts a downward force on the downward force on the chair and the chair chair and the chair exerts an upward force exerts an upward force on your body. on your body.

Page 12: February 21, 2014     Day 2

33rdrd Law Law

There are two forces There are two forces resulting from this resulting from this interaction - a force on interaction - a force on the chair and a force on the chair and a force on your body. These two your body. These two forces are called forces are called actionaction and and reactionreaction forces. forces.

Page 13: February 21, 2014     Day 2

Newton’s 3rd Law in NatureNewton’s 3rd Law in Nature Consider the propulsion of a Consider the propulsion of a

fish through the water. A fish fish through the water. A fish uses its fins to push water uses its fins to push water backwards. In turn, the backwards. In turn, the water water reactsreacts by pushing the by pushing the fish forwards, propelling the fish forwards, propelling the fish through the water.fish through the water.

The size of the force on the The size of the force on the water equals the size of the water equals the size of the force on the fish; the force on the fish; the direction of the force on the direction of the force on the water (backwards) is water (backwards) is opposite the direction of the opposite the direction of the force on the fish (forwards).force on the fish (forwards).

Page 14: February 21, 2014     Day 2

33rdrd Law Law

Flying gracefully Flying gracefully through the air, through the air, birds depend on birds depend on Newton’s third Newton’s third law of motion. As law of motion. As the birds push the birds push down on the air down on the air with their wings, with their wings, the air pushes the air pushes their wings up their wings up and gives them and gives them lift.lift.

Page 15: February 21, 2014     Day 2

Consider the flying motion of birds. A bird flies by Consider the flying motion of birds. A bird flies by use of its wings. The wings of a bird push air use of its wings. The wings of a bird push air downwards. In turn, the air reacts by pushing the bird downwards. In turn, the air reacts by pushing the bird upwards. upwards.

The size of the force on the air equals the size of the The size of the force on the air equals the size of the force on the bird; the direction of the force on the air force on the bird; the direction of the force on the air (downwards) is opposite the direction of the force on (downwards) is opposite the direction of the force on the bird (upwards).the bird (upwards).

Action-reaction force pairs make it possible for birds Action-reaction force pairs make it possible for birds to fly.to fly.

Page 16: February 21, 2014     Day 2
Page 17: February 21, 2014     Day 2

Other examples of Newton’s Other examples of Newton’s Third LawThird Law

The baseball forces the The baseball forces the bat to the left (an bat to the left (an action); the bat forces action); the bat forces the ball to the right (the the ball to the right (the reaction). reaction).

Page 18: February 21, 2014     Day 2

33rdrd Law Law

Consider the motion of Consider the motion of a car on the way to a car on the way to school. A car is school. A car is equipped with wheels equipped with wheels which spin backwards. which spin backwards. As the wheels spin As the wheels spin backwards, they grip the backwards, they grip the road and push the road road and push the road backwardsbackwards..

Page 19: February 21, 2014     Day 2

33rdrd Law LawThe reaction of a rocket The reaction of a rocket is an application of the is an application of the third law of motion. third law of motion. Various fuels are burned Various fuels are burned in the engine, producing in the engine, producing hot gases. hot gases.

The hot gases push The hot gases push against the inside tube of against the inside tube of the rocket and escape out the rocket and escape out the bottom of the tube. the bottom of the tube. As the gases move As the gases move downward, the rocket downward, the rocket moves in the opposite moves in the opposite direction.direction.

Page 20: February 21, 2014     Day 2

Now complete the summary Now complete the summary portion of your notes :o)portion of your notes :o)