12/1 do now – on a new sheet1. A 10 kilogram block moves down a 30° incline at

constant velocity.

A. What is the magnitude of the component of the gravity force that is parallel to the incline?

B. What is the friction force acting on the block?

2. A 10 kilogram block is pulled by a 60 N force at 60° above the horizontal at constant velocity along the horizontal surface.

A. What is the magnitude of the component of the 60 N force that is parallel to the horizontal surface?

B. What is the friction force acting on the block?

assignment

• Castle learning – due tomorrow

• Due– Force review packet – count as a quiz– Egg drop project

• Progress report due 12/12 – make corrections!!!

Circular Motion and Satellite Motion - Chapter Outline

Lesson 1: Motion Characteristics for Circular Motion

Lesson 2: Applications of Circular Motion

Lesson 3: Universal Gravitation

Lesson 4: Satellite Motion

Chapter test is on Friday 12/12

Objective - Lesson 1: Motion Characteristics for Circular Motion

1. Speed and Velocity

2. Acceleration

3. The Centripetal Force Requirement

4. Mathematics of Circular Motion

Uniform circular motion• Uniform circular motion is the motion of an

object in a circle with a constant or uniform speed. The velocity is changing because the direction of motion is changing.

• R is radius of the circle• 2πR is the circumference of the circle• T is time to make one lap around the circle, also

known as period.

Speed is constant

T

R

t

dvavg

2

The average speed and the radius of the circle are directly proportional.

Velocity is changingThe Direction of the Velocity Vector

The direction of the velocity vector at every instant is in a direction tangent to the circle.

Acceleration• An object moving in uniform circular motion is moving in a

circle with a uniform or constant speed. The velocity vector is constant in magnitude but changing in direction, which is tangent to the circle..

• Since the velocity is changing. The object is accelerating.

vi represents the initial velocity

vf represents the final velocity

t represents the time interval

t

vv

t

va if

Direction of the Acceleration Vector

• The velocity change is directed towards the center of the circle.

• The acceleration is in the same direction as this velocity change. The acceleration is directed towards the center of the circle.

The Centripetal Force

• According to Newton's second law of motion, an object which experiences an acceleration requires a net force.

• The direction of the net force is in the same direction as the acceleration. Since acceleration is directed toward the center, the net force must be toward the center. This net force is referred to as the centripetal force. The word centripetal means center seeking.

• The Centripetal Force is the NET force.

Inertia, Force and Acceleration

Centrifugal force is a fictitious force.

Centripetal force is the reason objects moves in a circle

As a car makes a turn in a circle, centripetal force is provided by friction.

As a bucket of water tied to a string and spun in a circle, centripetal force is provided by the tension acting upon the bucket.

As the moon orbits the Earth in a circle, centripetal force is provided by gravity acting upon the moon.

• An object in uniform circular motion moves at ____________ speed. Its velocity is ___________ to the circle and its acceleration is directed toward the ___________ of the circle. The object experiences ____________ which is directed in the same direction as the acceleration, toward the _________ of the circle. This net force is called ______________ force. The fact that the centripetal force is directed perpendicular to the tangential velocity means that the force can alter the direction of the object's velocity vector without altering its magnitude.

Summary

Check Your Understanding• A tube is been placed upon the table and shaped into a

three-quarters circle. A golf ball is pushed into the tube at one end at high speed. The ball rolls through the tube and exits at the opposite end. Describe the path of the golf ball as it exits the tube.

The ball will move along a path which is tangent to the spiral at the point where it exits the tube. At that point, the ball will no longer curve or spiral, but rather travel in a straight line in the tangential direction.

example• The initial and final speed of a ball at two different points in

time is shown below. The direction of the ball is indicated by the arrow. For each case, indicate if there is an acceleration. Explain why or why not. Indicate the direction of the acceleration.

a.

b.

c.

No, no change in velocity

yes, change in velocity, right

yes, change in velocity, left

d.yes, change in velocity, left

example• Identify the three controls on an

automobile which allow the car to be accelerated.

The accelerator allows the car to increase speed. The brake pedal allows the car to decrease the speed. And the steering wheel allows the car to change direction.

example

1. Which vector below represents the direction of the velocity vector when the object is located at point B on the circle?

2. Which vector below represents the direction of the acceleration vector when the object is located at point C on the circle?

3. Which vector below represents the direction of the force vector when the object is located at point A on the circle?

An object is moving in a clockwise direction around a circle at constant speed.

d

b

d

Class work

• Packet pp. #1-8

• Packet is due Friday

12/2 do now

• The diagram shows a 5.00-kilogram block at rest on a horizontal, frictionless table. Which diagram best represents the force exerted on the block by the table? Explain.

A. B. C. D.

assignment

• Castle learning – due tomorrow

• Packet due Friday

• Progress report due 12/12 – make corrections!!!

• What do we know about

1.Speed

2.Velocity

3.Acceleration

4.Centripetal force

5.Centrifugal force

Uniform circular motion review

questions

• The centripetal force acting on the space shuttle as it orbits Earth is equal to the shuttle’s

A.inertia

B.momentum

C.velocity

D.weight

Mathematics of Circular Motion

T

Rvavg

2

R

va

2

amFnet R

vmFnet

2

2

24

T

RmFnet

m: mass

v: speed

R: radius

T: period

Relationships between quantities

vavg and R are direct: R double, v double

vavg and T are inverse: T double, v halves

ac and v are direct squared: v double, ac quadruple.

Fc and v are direct squared: v double, Fc quadruple.

ac and R are inverse: R double, ac halves.

Fc and R are inverse: R double, Fc halves.

Fc and m are direct: m double, Fc doubles.

ac and m are not related: m double, ac unchanged.

R

vmFc

2

R

vac

2

T

Rvavg

2

example• A car going around a curve is acted upon

by a centripetal force, F. If the speed of the car were twice as great, the centripetal force necessary to keep it moving in the same path would be

1. F 2. 2F            3. F/2 4. 4F  

F = m v2 / r

F in directly proportional to v2

F in increased by 22

example• Anna Litical is practicing a centripetal force

demonstration at home. She fills a bucket with water, ties it to a strong rope, and spins it in a circle. Anna spins the bucket when it is half-full of water and when it is quarter-full of water. In which case is more force required to spin the bucket in a circle?

It will require more force to accelerate a half-full bucket of water compared to a quarter-full bucket. According to the equation Fnet = m•v2 / R, force and mass are directly proportional. So the greater the mass, the greater the force. 

Example • The diagram shows a 5.0-kilogram cart

traveling clockwise in a horizontal circle of radius 2.0 meters at a constant speed of 4.0 meters per second. If the mass of the cart was doubled, the magnitude of the centripetal acceleration of the cart would be

1. doubled 2. halved 3. unchanged 4. quadrupled

ac = v2 / R

example• A 60.-kilogram adult and 30.-kilogram child are passengers on

a rotor ride at an amusement park. When the rotating hollow cylinder reaches a certain constant speed, v, the floor moves downward. Both passengers stay "pinned" against the wall of the rotor, as shown in the diagram. The magnitude of the frictional force between the adult and the wall of the spinning rotor is F. What is the magnitude of the frictional force between the child and the wall of the spinning rotor?

1. F

2. 2F

3. ½ F

4. ¼ F        

 

Ff = µFNorm

FNorm = Fnet = mv2/R

The child is half as massive as the adult, therefore FNorm of the child is half of that of the adult and its Ff is half of that of friction of the adult also.

example• Two masses, A and B, move in circular paths as

shown in the diagram. The centripetal acceleration of mass A, compared to that of mass B, is

1. the same

2. twice as great

3. one-half as great

4. four times as great

F = m v2 / r

Class work

• Work on packet

12/3 do now

• In the diagram, as angle θ between the lawnmower handle and the horizontal increases, the horizontal component of F

A. decreases

B. increases

C. remains the same

assignment

• Castle learning – due tomorrow

• Packet due Friday

• Progress report due 12/12 – make corrections!!!

Solving problems using Equations

• Identify the given and unknown.• Choose an equation • Substitute number with units• Answer with units

T

Rvavg

2

R

vac

2

R

vmFc

2

Example

• A 900-kg car moving at 10 m/s takes a turn around a circle with a radius of 25.0 m. Determine the acceleration and the net force acting upon the car.

Given: = 900 kg; v = 10.0 m/s R = 25.0 mFind: a = ? Fnet = ?

a = v2 / Ra = (10.0 m/s)2 / (25.0 m)a = (100 m2/s2) / (25.0 m)a = 4 m/s2

Fnet = m • aFnet = (900 kg) • (4 m/s2)Fnet = 3600 N

example• A 95-kg halfback makes a turn on the football field. The

halfback sweeps out a path which is a portion of a circle with a radius of 12-meters. The halfback makes a quarter of a turn around the circle in 2.1 seconds. Determine the speed, acceleration and net force acting upon the halfback.

Given: m = 95.0 kg; R = 12.0 m; Traveled 1/4 of the circumference in 2.1 s Find: v = ? a = ? Fnet = ?

v = d / t

v = (1/4•2•π•12.0m) /(2.1s)

v = 8.97 m/s

a = v2 / R

a = (8.97 m/s)2/(12.0 m)

a = 6.71 m/s2

Fnet = m*aFnet = (95.0 kg)*(6.71 m/s2)Fnet = 637 N

example• Determine the centripetal force acting upon a 40-kg child

who makes 10 revolutions around the Cliffhanger in 29.3 seconds. The radius of the barrel is 2.90 meters.

Given: m = 40 kg; R = 2.90 m; T = 2.93 s (since 10 cycles takes 29.3 s).Find: Fnet

Step 1: find speed: v = (2 • π • R) / T = 6.22 m/s.

Step 2: find the acceleration: a = v2 / R = (6.22 m/s)2/(2.90 m) = 13.3 m/s2

Step 3: find net force: Fnet = m • a = (40 kg)(13.3 m/s/s) = 532 N.

12/4 do now

• The diagram shows a sled and rider sliding down a snow-covered hill that makes an angle of 30.° with the horizontal. Which vector best represents the direction of the normal force, FN, exerted by the hill on the sled?

A B C D

assignment

• Castle learning – due tomorrow

• Packet due tomorrow

• Progress report due 12/12 – make corrections!!!

Question• A go-cart travels around a flat, horizontal, circular track

with a radius of 25 meters. The mass of the go-cart with the rider is 200. kilograms. The magnitude of the maximum centripetal force exerted by the track on the go-cart is 1200. newtons. Which change would increase the maximum speed at which the go-cart could travel without sliding off this track?

A. Decrease the coefficient of friction between the go-cart and the track.

B. Decrease the radius of the track.

C. Increase the radius of the track.

D. Increase the mass of the go-cart. T

Rvavg

2

R

vac

2

R

vmFc

2

Lesson 2: Applications of Circular Motion

1. Newton's Second law - Revisited

2. Amusement Park Physics

Where Fnet is the sum (the resultant) of all forces acting on the object.

Newton's second law is used in combination of circular motion equations to analyze a variety of physical situations.

Note: centripetal force is the net force!

Newton's Second Law - Revisited

Steps in solving problems involving forces

1. Drawing Free-Body Diagrams

2. Determining the Net Force from Knowledge of Individual Force Values

3. Determining Acceleration from Net Force Or Determining Individual Force Values from Knowledge of the Acceleration

When a car is moving in a horizontal circle on a level surface, there is a net force (centripetal force) acting on it, the net force is FRICTION, which is directed toward the center of the circle.

Case 1: car driven in a circle

example• A 945-kg car makes a 180-degree turn with a speed of 10.0 m/s.

The radius of the circle through which the car is turning is 25.0 m. Determine the force of friction and the coefficient of friction acting upon the car.

Given: m = 945 kg; v = 10.0 m/s; R = 25.0 m Find: Ffrict = ? μ = ?

Ff = μFNorm; FNorm = mg;

Ff = μmg

3780 N = μ(9270 N);

μ = 0.41

Ff = Fnet = m*v2/RFf = (945 kg)*(10m/s)2/25mFf = 3780 N

Ff

FNorm

Fgrav

example• The coefficient of friction acting upon a 945-kg car is 0.850.

The car is making a 180-degree turn around a curve with a radius of 35.0 m. Determine the maximum speed with which the car can make the turn.

Given: m = 945 kg; μ = 0.85; R = 35.0 mFind: v = ? (the minimum speed would be the speed achieved with the given friction coefficient)

Ff = μFN = μFN ;

Ff = (0.85)(9270N) = 7880 N

7880 N = (945 kg)(v2) / (35.0 m); v = 17.1 m/s

Ff = Fnet

Fnet = m*v2/R

Ff

FNorm

Fgrav

Case 2: a swing bucket

In a swing bucket, the net force (Fc) is the combinations of gravity and tension.

At top:

Fnet = Ftens + Fg

Fnet = mv2/R

At bottom:

Fnet = Ftens - Fg

Fnet = mv2/R

• A 1.50-kg bucket of water is tied by a rope and whirled in a circle with a radius of 1.00 m. At the top of the circular loop, the speed of the bucket is 4.00 m/s. Determine the acceleration, the net force and the individual force values when the bucket is at the top of the circular loop.

m = 1.5 kga = ________ m/s/sFnet = _________ N

 

Fgrav = mg = 14.7 N

a = v2 / R = 16 m/s2

Fnet = ma = 24 N, down

Fnet = Fgrav + Ftens

24 N = 14.7 N + Ften

Ftens = 24 N - 14.7 N = 9.3 N

Example

• A 1.50-kg bucket of water is tied by a rope and whirled in a circle with a radius of 1.00 m. At the bottom of the circular loop, the speed of the bucket is 6.00 m/s. Determine the acceleration, the net force and the individual force values when the bucket is at the bottom of the circular loop.

m = 1.5 kga = ________ m/s/sFnet = _________ N

 

Fgrav = m • g = 14.7 N

a = v2 / R = 36 m/s2

Fnet = ma = 54 N, up

Fnet = Ften - Fgrav

54 N = Ften – 14.7 N

Ften = 54 N +14.7 N = 68.7 N

Example

Case 3: Roller Coasters and Amusement Park Physics

In a roller coaster, the centripetal force is provided by the combination of normal force and gravity.

example• Anna Litical is riding on The Demon at Great America. Anna

experiences a downwards acceleration of 15.6 m/s2 at the top of the loop and an upwards acceleration of 26.3 m/s2 at the bottom of the loop. Use Newton's second law to determine the normal force acting upon Anna's 864 kg roller coaster car at top and at bottom of the loop.

Given:

m = 864 kg

atop = 15.6 m/s2 , down

abottom = 26.3 m/s2 , up

Fgrav = m•g = 8476 N.

Find:

Fnorm at top and bottom

top of Loop

• Fnet = m * a

Fnet = (864 kg)∙(15.6 m/s2 ) = 13 478 N, downward

Fnet = Fnorm + Fgrav

13478 N = Fnorm + 8475 N

Fnorm = 5002 N

Bottom of Loop

• Fnet = m * a

Fnet = (864 kg) * (26.3 m/s2 ) = 22 723 N, upward

Fnet = Fnorm + (-Fgrav)

22723 N = Fnorm + (-8476 N)

Fnorm = 31199 N

Class work

• Work on packet

• lab

12/5 do now

• A 10.-kilogram box, sliding to the right across a rough horizontal floor, accelerates at -2.0 meters per second2 due to the force of friction.

A. Calculate the magnitude of the net force acting on the box.

B. Calculate the coefficient of kinetic friction between the box and the floor.

assignment

• Castle learning – due tomorrow

• Packet due Monday

• Progress report due 12/12 – make corrections!!!

Lesson 3: Universal Gravitation

1. What is Gravity?

2. The Apple, the Moon, and the Inverse Square Law

3. Newton's Law of Universal Gravitation

4. Cavendish and the Value of G

5. The Value of g

What is Gravity?

• We know that gravity is a force and we represent it by the symbol Fg. It causes an acceleration of all objects around it. The acceleration is referred as the acceleration of gravity. On and near Earth's surface, the value for the acceleration of gravity (g) is approximately 9.81 m/s/s. It is the same acceleration value for all objects, regardless of their mass (and assuming that the only significant force is gravity).

The Apple, the Moon, and the Inverse Square Law

• It was known that the moon orbits Earth and Earth orbits the sun. Since both the moon and Earth moves in a circular path, they are accelerating. But what is the force that causes these accelerations?

• According to legend, a breakthrough came to Newton at age 24 in an apple orchard in England.

• Newton's was able to relate the cause for heavenly motion (the orbit of the moon about the earth) to the cause for Earthly motion (the falling of an apple to the Earth). This connection led him to his notion of universal gravitation.

• It was known that earthbound objects (such as falling apples) accelerate towards the earth at a rate of 9.81 m/s2. And it was also known that the moon accelerated towards the earth at a rate of 0.00272 m/s2.

The difference in the accelerations between the moon and the apple is due to the difference in distance from the center of the earth to the two objects.

The force of gravity follows an inverse square law.

Do now

• On assigned sheet: page 77 #163-166

• Homework – packet pp. #13-15

• Any two bodies in the universe attract each other with a force that is directly proportional to the product of their masses and inversely proportional to the square of the distance between them.

Newton’s law of Universal Gravitation

G is gravitational constant. G = 6.67x10-11N∙m2/kg2

m1 is the mass of object 1m2 is the mass of object 2r is the distance between the objects’ centers

F1 and F2 are action/reaction forces.

• The relationship between the force of gravity (Fg) between

any two objects and the distance (d) that separates their centers is inverse squared.

• If the separation distance is increased by a factor of 2, then the force of gravity is decreased by a factor of four (22). And if the separation distance is increased by a factor of 3, then the force of gravity is decreased by a factor of nine (32).

Gravity and distance - Inverse Square Law

r

Fg

Check Your Understanding1 . Suppose that two objects attract each other with a gravitational force of

16 units. If the distance between the two objects is doubled, what is the new force of attraction between the two objects?

   2. Suppose that two objects attract each other with a gravitational force of

16 units. If the distance between the two objects is tripled, then what is the new force of attraction between the two objects?

   3. Suppose that two objects attract each other with a gravitational force of

16 units. If the distance between the two objects is reduced in half, then what is the new force of attraction between the two objects?

   4. Suppose that two objects attract each other with a gravitational force of

16 units. If the distance between the two objects is reduced by a factor of 5, then what is the new force of attraction between the two objects?

16/22 = 4 units

16/32 = 1.8 units

16 x 22 = 64 units

16 x 52 = 400 units

Relationship between quantities

Example• Determine the force of gravitational attraction between

the earth (m = 5.98 x 1024 kg) and a 70-kg physics student if the student is standing at sea level, a distance of 6.38 x 106 m from earth's center.

Fgrav = mg = (70 kg)(9.8 m/s2) = 686 N

The Universality of Gravity

Mass of Object 1

(kg)

Mass of Object 2(kg)

Separation Distance

(m)

Force of Gravity

(N)Student70 kg

Earth5.98 x1024 kg

6.60 x 106 m(low-height orbit)

Student 70 kg

Physics Student70 kg

1 m

Student 70 kg

Physics Book1 kg

1 m

Student 70 kg

Jupiter 1.901 x 1027 kg

6.98 x 107 m (on surface)

 

686 N

3.27 • 10-7 N

4.67 • 10-9 N

1822 N

Gravity is a force pulling together all matter. The more matter, the more gravity.

Example• The weight of an object was determined at five different

distances from the center of Earth. The results are shown in the table below. Position A represents results for the object at the surface of Earth. What is the approximate mass of the object?

100 kg

The Value of g

22

r

mmGF Earth

g

2r

mGg Earth

The acceleration of gravity is dependent upon the mass of the earth (approx. 5.98x1024 kg) and the distance (r) from the center of the earth.

gmFg 2

22

2 r

mmGgm Earth

G and g have different values!!!

g is inversely proportional to the distance squared – inverse square law.

LocationDistance from

Earth's center (m)Value of g

m/s2

Earth's surface 6.38 x 106 m 9.8

1000 km above surface

7.38 x 106 m

5000 km above surface

1.14 x 107 m

10000 km above surface

1.64 x 107 m

50000 km above surface

5.64 x 107 m

g is different at different location

7.33

3.08

1.49

0.13

Gravity is a field force

• Gravitational field – a region in space where an object experiences a gravitational force. Every mass is surrounded by a gravitational field.

The strength of gravitational field is inversely related to the distance from Earth. It is stronger near Earth and decreases as the distance from the Earth increases.

Gravitational field strength is a vector quantity (g)

• At any point in a gravitational field, gravitational field strength, g, equals the force per unit mass at that point.

– g (gravitational field strength) = 9.81 N/kg (near earth)– g (gravitational acceleration) = 9.81 m/s2

or 2r

mGg Earth

m

Fg g

The gravitational field lines are directed toward the center of Earth.

12/10 do now

• On assigned sheet: #171-178

• Homework questions?

• Homework: packet pp. 17, 19-20

Questions• Earth’s mass is approximately 81 times the

mass of the Moon. If Earth exerts a gravitational force of magnitude F on the Moon, the magnitude of the gravitational force of the Moon on Earth is

A.F

B.F/81

C.9F

D.81F

Lesson 4: Satellite Motion

– Circular Motion Principles for Satellites – Mathematics of Satellite Motion – Weightlessness in Orbit

Circular Motion Principles for Satellites

• A satellite is any object that is orbiting the earth, sun or other massive body. Every satellite's motion is governed by the same physics principles and described by the same mathematical equations.

Velocity, Acceleration and Force Vectors

• The motion of an orbiting satellite can be described by the same motion characteristics as any object in circular motion. – The velocity of the satellite would be directed

tangent to the circle at every point along its path. – The acceleration of the satellite would be directed

towards the center of the circle - towards the central body that it is orbiting.

– And this acceleration is caused by a net force that is directed inwards in the same direction as the acceleration. This centripetal force is supplied by gravity - the force that universally acts at a distance between any two objects that have mass.

Mathematics of Satellite Motion

• If the satellite moves in circular motion, then the net centripetal force is provided by the gravitational force acting upon this orbiting satellite.

2r

mmGF satEarth

g

r

vmF sat

c

2

2

2

r

mmG

r

vm satEarthsat

gc FF

r

GMv Earth 2r

MGa central

r

GMv central

Both the speed and acceleration of satellite depends only on the mass of the central body about which the satellite orbits, and the distance between the satellite and the center of the central body. It has nothing to do with mass of the satellite.

If a base ball and an big spaceship were sent to the same distance r from the center of Earth, both will orbit Earth with same speed and have the same acceleration.

The speed and acceleration of satellite

2r

MGa central

Check your understanding

• A satellite is orbiting the earth. Which of the following variables will affect the speed of the satellite?

a. mass of the satellite

b. height above the earth's surface

c. mass of the earth

Weightlessness in Orbit• Astronauts who are orbiting the Earth often experience

sensations of weightlessness. These sensations experienced by orbiting astronauts are the same sensations experienced by anyone who has been temporarily suspended above the seat on an amusement park ride.

• You can only feel the force of gravity pulling upon your body through contact force, such as standing on a floor, sitting in a chair, or lying on a bed. Without the contact force (the normal force), there is no means of feeling the non-contact force (the force of gravity).

• When you are falling on an amusement park ride, there is still gravity act on you even thought you can not feel it.

Test your preconceived notions about weightlessness:

• Astronauts on the orbiting space station are weightless because...

a. there is no gravity in space and they do not weigh anything.

b. space is a vacuum and there is no gravity in a vacuum.

c. space is a vacuum and there is no air resistance in a vacuum.

d. the astronauts are far from Earth's surface at a location where gravitation has a minimal affect.

Scale Readings and Weight

If you want to know your weight, you may stand on a scale to find it out. However, the scale does not always measure your weight. When you stand on a scale in an elevator going up and down or when you standing on a scale on an inclined plane, the scale reading is different than when you are at rest or you are traveling at constant speed.

Fnet = m*a

Fnet = 0 N

Fnet = m*a

Fnet = 400 N, up

Fnet = m*a

Fnet = 400 N, down

Fnet = m*a

Fnet = 784 N, down

Fnorm equals

Fgrav

Fnorm = 784 N

Fnorm > Fgrav by

400 N Fnorm = 1184 N

Fnorm < Fgrav by

400 N Fnorm = 384 N

Fnorm < Fgrav by

784 N Fnorm = 0 N

What does the scale measure?

• Otis stands on a bathroom scale and reads the scale while ascending and descending the John Hancock building. Otis' mass is 80 kg. Use a free-body diagram and Newton's second law of motion to solve the following problems.

a. What is the scale reading when Otis accelerates upward at 0.40 m/s2?

b. What is the scale reading when Otis is traveling upward at a constant velocity of at 2.0 m/s?

c. As Otis approaches the top of the building, the elevator slows down at a rate of 0.40 m/s2. Be cautious of the direction of the acceleration. What does the scale read?

d. Otis stops at the top floor and then accelerates downward at a rate of 0.40 m/s2. What does the scale read?

e. As Otis approaches the ground floor, the elevator slows down (an upward acceleration) at a rate of 0.40 m/s2. Be cautious of the direction of the acceleration. What does the scale read?

f. Use the results of your calculations above to explain why Otis feels less weighty when accelerating downward on the elevator and why he feels heavy when accelerating upward on the elevator.

 

FN = 816 N

FN = 784 N

FN = 752 N

FN = 752 N

FN = 816 N

12/11 do now• Packet page 76 #154-157

• Homework – castle learning

• Review sheets

• test is on castle learning

• Food drive – 10 items for 10 points – add points to any assignment, your choice – last day to bring food is Tuesday, 12/16/14

• 12/19 – physics shirt day

12/12 do now• #179-180

• All castle learning assigned for this chapter are opened for review – close tonight

• Class work - Review sheets,

• Take home quiz – due Monday

• test is on castle learning

• Project is due Thursday

• Food drive – 10 items for 10 points – add points to any assignment, your choice – last day to bring food is Tuesday, 12/16/14

• 12/19 – physics shirt day

Do now• The path of a stunt car driven horizontally off a cliff is

represented in the diagram below. After leaving the cliff, the car falls freely to point A in 0.50 second and to point B in 1.00 second. [Neglect friction.]

1. Determine the magnitude of the horizontal component of the velocity of the car at point B.

2. Determine the magnitude of the vertical velocity of the car at point A.

3. Calculate the magnitude of the vertical displacement, dy, of the car from point A to point B.

• A skier on waxed skis is pulled at constant speed across level snow by a horizontal force of 39 newtons. Calculate the normal force exerted on the skier. Round your answer to the nearest whole number.

• A 10.-kilogram rubber block is pulled horizontally at constant velocity across a sheet of ice. Calculate the magnitude of the force of friction acting on the block. Round your answer to the nearest tenth of a newton.

• A student and the waxed skis he is wearing have a combined weight of 850 newtons. The skier travels down a snow-covered hill and then glides to the east across a snow-covered, horizontal surface. Calculate the magnitude of the force of friction acting on the skis as the skier glides across the snow-covered, horizontal surface.

• An ice skater applies a horizontal force to a 20.-kilogram block on frictionless, level ice, causing the block to accelerate uniformly at 1.4 meters per second2 to the right. After the skater stops pushing the block, it slides onto a region of ice that is covered with a thin layer of sand. The coefficient of kinetic friction between the block and the sand-covered ice is 0.28. Calculate the magnitude of the force of friction acting on the block as it slides over the sand-covered ice.

• 747 jet, traveling at a velocity of 70. meters per second north, touches down on a runway. The jet slows to rest at the rate of 2.0 meters per second2. Calculate the total distance the jet travels on the runway as it is brought to rest.

• A 60-kilogram skydiver is falling at a constant speed near the surface of Earth. What is the magnitude of the force of air friction acting on the skydiver?