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Laws of Motion

Laws of Motion

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Laws of Motion. Chapter 11. Newton ’ s First Law. an object at rest remains at rest and an object in motion maintains its velocity (stays in motion) unless it experiences an unbalanced force (outside force). Aka Law of Inertia. INERTIA. tendency of an object to resist being moved or - PowerPoint PPT Presentation

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Page 1: Laws of Motion

Laws of Motion

Page 2: Laws of Motion

Newton’s First Law

an object at rest remains at rest and an object in motion maintains its velocity

(stays in motion) unless it experiences an unbalanced force (outside force).

Aka Law of Inertia

Chapter 11

Page 3: Laws of Motion

INERTIA tendency of an object to resist being

moved or if object is moving, to resist a change in

speed or direction until an outside force acts on object.

All objects resist changes in motion related to an object’s mass. Objects with small mass have less inertia

than objects with large mass

Page 4: Laws of Motion

INERTIA EXAMPLE

Seat belts and car seats provide protection.

• Because of inertia, you slide toward the side of a car when the driver makes a sharp turn.

• When the car you are riding in comes to a stop, your seat belt and the friction between you and the seat stop your forward motion.

Chapter 11

Page 5: Laws of Motion

Newton’s Second Law

unbalanced force acting on an object equals object’s mass times its acceleration.

Force = mass acceleration F = ma

SI Unit: newtons (N) 1 N = 1 kg 1 m/s2 m

F

a

Page 6: Laws of Motion

Newton’s Second LawChapter 11

Page 7: Laws of Motion

FORCEZookeepers lift a stretcher that holds a sedated lion. The total mass of the lion and stretcher is 175 kg, and the lion’s upward acceleration is 0.657 m/s2. What is the unbalanced force necessary to produce this acceleration of the lion and the stretcher?

GIVEN:

F = ?

m= 175 kg

a = 0.657 m/s2

WORK:

F = ma

F = (175 kg) (0.657 m/s2)

F = 115 N

mF

a

Page 8: Laws of Motion

FORCEWhat is the net force necessary for a 1.6 x 103

kg automobile to accelerate forward at 2.0 m/s2?

GIVEN:

F = ?

m= 1.6 x 103 kg

a = 2.0 m/s2

WORK:

F = ma

F = (1.6 x 103 kg) (2.0 m/s2)

F = 3.2 x 103 N

mF

a

Page 9: Laws of Motion

FORCEA baseball accelerates downward at 9.8 m/s2. If the gravitational force is the only force acting on the baseball and is 1.4 N, what is the baseball’s mass?

GIVEN:

F = 1.4 N

m= ?

a = 9.8 m/s2

WORK:

m = F/a

m = 1.4 N / 9.8 m/s2)

m = 0.14 kg

mF

a

Page 10: Laws of Motion

FORCEA sailboat and its crew have a combined mass of 655 kg. Ignoring frictional forces, if the sailboat experiences a net force of 895 N pushing it forward, what is the sailboat’s acceleration?

GIVEN:

F = 895 N

m= 655 kg

a = ?

WORK:

a = F/m

a = 895 N / 655 kg

F = 1.37 m/s2

mF

a

Page 11: Laws of Motion

GRAVITY

Page 12: Laws of Motion

Law of Universal Gravitation All objects in universe attract each other through

gravitational force Sir Isaac Newton (1642–1727) Universal Gravitation Equation

F G

m1m

2

d 2

• m1 and m2 are masses of two objects• d distance between two objects• G constant

Chapter 11

Page 13: Laws of Motion

Law of Universal Gravitation

All matter is affected by gravity.• Two objects, whether large or small,

always have a gravitational force between them.

Gravitational force increases as mass increases.

Gravitational force decreases as distance increases.

Page 14: Laws of Motion

Free Fall Acceleration

motion of a body when only force of gravity is acting on body.

constant near Earth’s surface. g =9.8 m/s2

Page 15: Laws of Motion

Weight equal to mass times free-fall acceleration. weight = mass free-fall acceleration

w = mg

Weight is different from mass.Mass • measure of amount of matter in an object.Weight • gravitational force an object experiences

because of its mass.• influences shape

Page 16: Laws of Motion

Terminal Velocity constant velocity of a falling object when force of

air resistance is equal in magnitude and opposite in direction to force of gravity

Page 17: Laws of Motion

Two Motions Cause Orbiting

Page 18: Laws of Motion

Projectile Motion and Gravity curved path an object follows when thrown,

launched, or otherwise projected near surface of Earth.

applies to objects that are moving in two dimensions under the influence of gravity.

has two components horizontal and vertical. two components are independent.

Chapter 11

Page 19: Laws of Motion

Projectile MotionChapter 11

Page 20: Laws of Motion

Newton’s Third Law

for every action force, there is an equal and opposite reaction force.

Forces always occur in action-reaction pairs.

Action-reaction force pairs are equal in size and opposite in direction.

Chapter 11

Page 21: Laws of Motion

Action and Reaction Forces

Force pairs do not act on the same object.

Equal forces don’t always have equal effects.

example, the action force of Earth pulling on an object and causing it to fall is much more obvious than the equal and opposite reaction force of the falling object pulling on Earth.

Chapter 11

Page 22: Laws of Motion

Momentum

quantity defined as product of the mass and velocity of an object.

momentum = mass velocity

p = mv

Moving objects have momentum.• the more mass an object has, the greater its momentum is.• faster an object is moving, the greater its momentum is.• When you force an object to change its motion, you force

it to change its momentum.

Chapter 11

Page 23: Laws of Motion

MomentumCalculate the momentum of a 6.00 kg bowling ball moving at 10.0 m/s down the alley toward the pins.

GIVEN:

p = ?

m= 6.00 kg

v = 10 m/s

WORK:

p = mv

p = (6.00 kg) (10 m/s)

p = 60 kg x m/s down the alley

mp

v

Page 24: Laws of Motion

MomentumA 75 kg speed skater moving forward at 16 m/s. Calculate the momentum.

GIVEN:

p = ?

m= 75 kg

v = 16 m/s

WORK:

p = mv

p = (75 kg) (16 m/s)

p = 1200 kg x m/s forward

mp

v

Page 25: Laws of Motion

MomentumA 135 kg ostrich running north at 16.2 m/s. Calculate the momentum.

GIVEN:

p = ?

m= 135 kg

v = 16.2 m/s

WORK:

p = mv

p = (135 kg) (16.2 m/s)

p = 2190 kg x m/s north

mp

v

Page 26: Laws of Motion

MomentumCalculate the velocity of a 0.8 kg kitten with a momentum of 5 kg x m/s forward.

GIVEN:

p = 5 kg x m/s forward

m= 0.8 kg

v = ?

WORK:

v = p/m

v = 5 kg x m/s / 0.8 kg

v = 6.3 m/s forward

mp

v

Page 27: Laws of Motion

Law of Conservation of Momentum• total amount of momentum in an isolated system

is conserved.• When a moving object hits a second object, some

or all of the momentum of first object is transferred to second object.

• Momentum can be transferred in collisions, but total momentum before and after a collision is same.

• rocket propulsion.• Cars collided • Billiards

Chapter 11