10/30/13

• Bull’s Eye Lab• Our Own Hot Wheels Challenge• Projectiles and Pumpkins• Isaac Newton

11/4/2013• Turn in Bullseye Lab by

3:15• Pumpkins and Projectiles?• Hey remember that

scientific method activity?

• And so it comes to this, the mind blowing Sir Isaac Newton and his laws of motion

• Newton’s Laws

11/5/2013• Pumpkins and

Projectiles?• And so it comes t

o this, the mind blowing Sir Isaac Newton and his laws of motion

• Newton’s Laws

11/1/12–Shooting the Moon

–Normal Force• Tug of war physics, oh…okay

• Vectors in multiple directions!

10/30/2012• Bullseye Lab

• Shooting the Moon

• Newton’s Third Law

11/6/2013• Turn in Homework!• That’s some good inertia• What is this Force? (not

that force)• Newton’s 1st and 2nd Law

Demos

An object at rest tends to stay at rest and an object in motion tends to stay in motion with the same speed and in the same direction unless acted upon by an unbalanced force .

Center of gravity (mass)

What is a

Force?

First Law - InertiaLaw of Inertia – every object continues in a state of rest, or of motion in a straight line at a constant speed, unless it is compelled to change that state by forces exerted upon it.

Inertia ConceptsMass – the more mass an object has, the greater its inertia and the more force it takes to change its state of motion. Mass is the measure of the inertia of an object.

11/7/2013• Turn in Homework!• That’s some good inertia• What is this Force? (not

that force)• Newton’s 1st and 2nd Law

Demos

Representing Forces...• Forces are vectors

– Forces are drawn as arrows, the length represents the magnitude and the direction of the arrow is the direction of the force.

– forces add like vectors.

– the sum of all the forces is called the net force.

• A picture of a body with arrows drawn representing all the forces acting upon it is called a FREE BODY DIAGRAM.

Try it...

Draw a picture of your book sitting

on the desk. Identify all the

forces acting on it.

Free Body Diagrams...

Book

T (table)

W (weight)

Free Body Diagrams...What forces are acting on a

skier as she races down a hill?

Plane Free Body Diagram

Free Body Diagrams

11/14/2013• Test Make-ups• Newton’s 2nd and 3rd

Laws• Newton’s Laws Work

Newton’s Second Law:The Law of Acceleration

When an unbalanced force is applied to an object it will accelerate in the direction of the net force with an acceleration proportional to the force applied.

F = m x aForces cause

accelerations!

a = F/m

F = maMass = kg

Acceleration = m/s2

Force = kg ·m/s2

Newton (N) = kg ·m/s2

Examples• A jet thruster applies a force of 20,000N

at maximum burn. If the jet has a mass of 5,500 kg what is the acceleration of the jet? How long will it take to increase speed from 0 m/s to 80 m/s?

11/15/2013• Test Make-ups• Newton’s 2nd and 3rd

Laws• Newton’s Laws Work

Examples• What force is needed to

decelerate a 15,000 N car from 18 m/s to rest in 6 seconds?

Think About it...• A textbook rests on a table. What forces

act on the book? On the table?• True or False. When you jump the Earth

accelerates.• An 8 ton bus crashes into a 1200 lb VW

Rabbit. If the bus applies a force of 20,000 lb on the car, what is the force of the car on the bus?

Second Law of MotionNewton was the first to realize that the acceleration produced when we move something depends not only on how hard we push or pull, but also on the object’s mass. The acceleration produced by a net force on an object is directly proportional to the magnitude of the net force and is inversely proportional to the mass of the object.

The Law of Force-Counterforce

• unaccompanied forces do not exist in nature.

• ‘action-reaction’ forces are not the same as ‘balanced’ forces.

When one body exerts a force on another body the second one exerts an equal and opposite force on the first.

• Whenever a first body exerts a force F on an second body, the second body exerts a force -F on the first body. F and –F are equal in magnitude but opposite in direction.

• The law of action reaction

Newton’s Third Law and the

Grasshopper

11/18/2013• Homework• Forces in Balance• Action-Reaction Lab

• Balloon force• Newton’s Laws Work

• According to legend, a horse learned newton's laws. When the horse was told to pull a cart, it refused, saying that if it pulled the cart forward, according to Newton's third law, there would be an equal force backwards: thus there would be balanced forces, and the cart would not accelerate. How would you reason with this horse?

Interaction PairsTwo forces that are in opposite

directions have equal magnitude.

You push your friend, this does not cause your friend to exert a force on you. The

forces exist together or not at all.

The conditions for a particle to be in equilibrium

• Necessary conditions for an object to settle into equilibrium (all things in balance, no change in motion):

SF = 0

Forces in Equilibrium

Dog Fight• Susan is holding her dog, its’ mass is 8.0 kg, when

Allen decides that he wants it and tries to pull it away from Susan.

• If Allen pulls horizontally on the dog with a force of 10 N and Susan pulls with a horizontal force of 11 N in the opposite direction, what is the horizontal acceleration of the dog??

• Why doesn’t the dog bite one of them?

Normal ForceThe perpendicular contact force

exerted by a surface on another object.

11/19/13

–Drag force• Tug of war physics, oh…okay

• New Lab: Balloon Cars

Drag Force• Is it true that particles in the air around an object

exert forces on it?• Yes, a huge force, but they all balance, and there is

no net effect.• What if the object is moving through the air?

– It experiences a drag force

• Drag Force: the force exerted by a fluid on an object moving through a fluid.

• There is a direct relationship between the magnitude of the drag force and the surface area of a moving object.

Terminal Velocity

The constant velocity that is reached when the drag force equals the force of gravity.

TUG OF WAR in PhysicsWhich of Newton’s laws are involved?

How do you determine the winner?

What is Friction?

• Friction is the force resisting the relative lateral (side to side) motion of solid surfaces, fluid layers, or material elements in contact.

• So far we have neglected friction, but since it is all around us, it is worth treating.

Two Main Types of Friction

• Push a book across a desk, it experiences a type of friction that acts on all moving bodies.

• KINETIC FRICTION (Fk)a force that is exerted by one surface against another when the two surfaces rub against each other because one or both of the surfaces are moving.

11/20/2013–Friction Force

• New Lab: Balloon Cars

Two Main Types of Friction

• Now try pushing a heavy couch across the floor, give it a push, and it stays where it is. Why?

• STATIC FRICTION (Fs)the force exerted on one surface by another when there is no motion between the two surfaces.

11/21/2013–Friction Force

• New Lab: Balloon Cars HW

*** At Constant Velocity: 

 Fapplied = Friction Force

 a = 0Fnet = 0 

*** When object is moving on a horizontal surface, 

the normal force equals .....

 …the weight force.

Fn = weight force 

= w = mg

On a horizontal surface:

11/22/2013–Friction Force

• New Lab: Balloon Cars HW• Friction Lab

11/25/2013–Friction Force• Friction Lab

•Remember Balloon Cars HW

11/26/2013–Friction Force

• Friction Lab Excel Work

•Remember Balloon Cars HW

12/2/2013• Balloon cars to the test• Friction Lab Excel Work

12/6/2013• Balloon car packet due!• Appling Newton’s Laws

• Universal Forces• Friction Lab Excel Work

• Two horizontal forces, 225 N and 165 N, are exerted on a canoe. If these forces are applied in the same direction, find the net horizontal force on the canoe.

• Three confused sleigh dogs are trying to pull a sled across the Alaskan snow. Alutia pulls east with a force of 35 N, Seward also pulls east but with a force of 42 N, and big Koda pulls west with a force of 53 N. What is the net force on the sled?

• Abe an Anne simultaneously grab a 0.75 kg piece of rope and begin tugging on it in opposite directions. If Anne pulls with a force of 16 N and the rope accelerates away from her at 1.25 m/s2, with what force is Abe pulling?

• Ray, with a mass of 85 kg, is standing by the boards at the side of an ice-skating rink. He pushes off the boards with a force of 9 N. What is his resulting acceleration?

Master thief Manny is running across the roof top of International Jewelry, jumps off the roof and grabs on to a mass-less rope hanging from the escape helicopter. In the process of running Manny’s left shoe becomes untied. Self appointed superhero Captain Hammer is chasing Manny and jumps off the roof and grabs onto the thief’s shoe lace. Manny has a mass of 65 kg, Captain Hammer has a mass of 115 kg, what is the tension on Manny’s rope and the shoe lace?

Which fundamental interaction is responsible for:Friction? Planetary orbits?Nuclear bonding?

What does the slope represent?• In this case the slope represents

the coefficient of kinetic friction.• We use it to find Kinetic Friction

Force (Fk), as follows: Highly Polished TableRough Table

Sand

pape

r

Kinetic Frictional Force vs. Normal Force

• Think of the ropes cut in two halves.• The left hand is not moving, so the net force is 0.• Thus, F A on rope = Fright on left = 500 N.

• Similarly, F B on rope = Fleft on right = 500 N.

• But the two tensions Fright on left and Fleft on right are an interaction pair, so they are equal and opposite.

• So the tension on the rope equals the force each team pulls with, 500 N.

How much force is needed to keep a 20-N stone from falling?

F = 20 N

3. What applied force accelerates a 20-kg stone straight up at 9.8 m/s2?

9.8 m/s2

Fnet = Fup + Fdown

Fup = Fnet - Fdown

Fup

Fdown

Fup = manet - madown

Fup = m(anet - adown)

Fnet

Fup = 20kg[9.8m/s2-(-9.8m/s2)]

Fup = 20kg[19.6 m/s2]

Fup = 392 N9.8 m/s2

4 m/s2

W = 9800 N

4. A rocket weighs 9800 N a) What is its mass?

W = mgm = 9800n/9.8m/s2

m = 1000 kg

m= W/g

4 m/s2

W = 9800 N

4. A rocket weighs 9800 N b) What force gives it a vertical acceleration of 4 m/s2?

Fnet = Fup + Fdown

Fup = Fnet - Fdown

Fup = manet - Fdown

4 m/s2

Fup = (1000 kg·4 m/s2) - (-9800 N)

Fup = 4000 N + 9800 N

Fup = 13800 N

Shooting the Moon

Mad Hatter Harry is sick of being watched by the man in the moon every night. So he sets out on a mission to rid him self of the moon once and for all. He is trying to build a cannon that can shoot the peeping tom in the sky. If the combustion process of the cannon’s “fuel” takes 0.2 seconds, the cannon ball is 20 kg, and escape velocity is 11,201 m/s. What magnitude of force must he impart on the cannon shell?

Elevator Problem

Your mass is 75kg, and you are standing on a bathroom scale in an elevator. Starting from rest, the elevator accelerates upward at 2 m/s2 for 2 seconds and then continues at a constant speed. Is the scale reading during the acceleration greater than, equal to, or less than the scale reading when the elevator is at rest?

11/5/12–Bullseye reward?–Old Homework

• Tug of war physics, oh…okay• Chapter 5 application of Newton’s

Three Laws• New Lab: Circular motion

Tension:–A specific name for the force exerted

by a string or rope.

Forces on Ropes and Strings

Goals for Chapter 5

• To study conditions that establish equilibrium.• To study applications of Newton’s Laws as they

apply when the net force is not zero.• To consider contact forces and the effects of

friction.• To study elastic forces (such as spring force).• To consider forces as they subdivide in nature

(strong, electromagnetic, weak, and gravitational).

Tension:–A specific name for the force exerted

by a string or rope.

Forces on Ropes and Strings

11/6/2012• Old Homework• Applying Newton’s Three

Laws• Circular motion lab

Two dimensional equilibrium – Example 5.2• Both x and y forces must be considered separately.• Follow worked example 5.2 on page 130.

Two dimensional equilibrium – Example 5.2

Problem 4 on page 152

11/7/2012• Watch Inclined Plane Force Components on

YouTube from KhanAcademy.org• Practice Problem in back• Forces not in equilibrium

Inclined Plane Force Components

Problem in the back of the room

Problem 6

An example involving two systems – Example 5.4

•See the worked example on page 132 and 133.•This example brings nearly every topic we have covered so far in the course.

11/8/2012• Newton’s Laws in dynamic situations• Sample problems• Contact Forces• Circular motion lab

Forces in Dynamics

• An object is no longer in equilibrium due to forces acting on it.

• Same as the things we have discussed before, now we just apply Newton’s Laws.

• Can you think of any examples of an object experiencing unbalanced forces?

Superman Problem• A train is approaching a washed out railroad

bridge and it is accelerating uncontrollably at 2 m/s2. Superman arrives just in time and begins pushing on the front of the train when there is only 500 m of track left. Assuming Superman causes the 750,000 kg train to uniformly decelerate from 30 m/s to 0 m/s just in time to keep from going off the end of the track, what total force did Superman have to apply to the train in order to stop it?

11/9/2012• Projectile motion problem solving.• Newton’s Laws in dynamic situations• Sample problems• Circular motion lab

Air cannon problem• How fast is the air cannon shooting?

– Find horizontal and vertical components, then velocity.• What kind of information can we get from experimentation?

Problem 19 on 153

Problem 22 conceptually in back.

11/13/2012• Test on Chap 3 and 4• Circular motion lab

• Nate is driving along a cliff side road when a wayward moose crosses his path. Nate slams on the brakes and swerves into the guard rail. He gets out to inspect the damage and sees the front right side of his car is completely wrecked. In frustration he throws his keys horizontally at 8 m/s off a 64 meter high cliff. How far from the base of the cliff should Nate look for his keys?

• An outfielder is throwing a baseball to the third baseman. The ball is released from shoulder height with an initial velocity of 29.4 m/s at an initial angle of 30° with respect to the ground. If the ball flies through the air for 3 seconds before being caught by the third baseman at an equal shoulder height, what was the maximum height of the ball above the outfielders shoulder height as it flew through the air?

• The mass of the space shuttle is approximately 2.0 x 106 kg. During lift-off, the net force on the shuttle is 1.0 x 107 N directed upward. What is the velocity of the shuttle 10 s after lift-off?

• A 50-kg woman wearing a seat belt is traveling in a car that is moving with a velocity of 10 m/s. In an emergency, the car is brought to a stop in 0.5 s. What force does the seat belt exert on the woman so that she remains in her seat?

• What is the mass of an object weighing 196 N?

• A 15-kg mass weighs 60.0 N on Planet X. The mass is allowed to fall freely from rest near the surface of the planet. What will be the velocity of the mass after falling for 6.0 seconds?

11/15/2012• Test• Lab due Monday• Homework Problem• Friction and motion• Two main types of friction• Friction Lab

• Number 12 on page 153.• In a rescue, the 73 kg police officer is suspended by

two cables as shown below. a) Sketch a free body diagram of the officer.b) Find the tension in each cable.

Pull a block of known mass along a table at a constant velocity, stack more blocks on top to increase the normal force and note the effect. If you do all this, collect your data and then try it with different surfaces coming into contact with the table, you can make a graph like this one.

Normal Force

Kine

tic fr

ictio

nal f

orce

Highly Polished Table

Rough Table

Sand

pape

r

Kinetic Frictional Force vs. Normal Force

The slope of each line is related to the magnitude of the resulting frictional force.The steeper the slope, the greater the force needed to pull the object across the surface.

Maximum Static Friction Force is related to the normal force in a similar way.

• Static Friction Force (Fs)is the force that responds to a force trying to cause a stationary object to move.

• If no force is acting on an object, Fs is zero.• If a force begins to act on an object, then Fs will

increase to a maximum value before it is overcome and the object will begin moving.

• We calculate Fs using the coefficient of static friction as follows:

11/16/2012

• Test Grades• Lab due Monday• Friction Problems• Friction Lab

• A girl exerts a 36 N horizontal force as she pulls a 52 N sled across a cement sidewalk. What is the coefficient of kinetic friction between the sidewalk and the metal sled runners? Ignore air resistance.

Girl and her sled

Logan on the icy hillLogan is at the top of an ice covered hill and he is wearing a

slippery snow suit (very little friction between the two). Assuming he weights 15 kg and the hill is at a 20 degree angle to the horizontal, what is his acceleration down the hill?

Now add frictionA 62 kg person on skis is going down a hill sloped at 37. The coefficient of kinetic friction between the skis and the snow is 0.15. How fast is the skier going 5 s after starting?

11/20/2012• Homework ?’s• Practice with Friction• Apophis• Friction Lab

Problem # 38 on page 155• Tires on a road, rolling friction:

11/26/12• New Homework• Test Make-up Question and review

session (Tues. and Wed. after school)• Elastic Forces (Phet Demo)• Friction Lab

Hooke's lawis the relationship between the force

exerted on the mass attached to a spring and its position x.

Consider a object with mass m, that is on a frictionless surface and is attached to a spring with spring constant k. The force the spring exerts on the mass depends on how much the spring is stretched or compressed, and so this force is a function of the mass's position.

Fs = kx

Hooke's Law

0

100

200

300

400

500

600

0 5 10 15 20 25

X - displacement (cm)

F - w

eigh

t (g)

k = F/ x Fs = kx

Hooke's Law

0

100

200

300

400

500

600

0 5 10 15 20 25

X - displacement (cm)

F - w

eigh

t (g)

Ws = ½ Fsx A = ½bh

Problem 57 on page 156

11/27/12• New Homework• Test Make-up Question and review

session (Tues. and Wed. after school)• Friction Lab

Problem 77 on page 158

Goals for Chapter 6• To understand the dynamics of circular

motion.• To study the unique application of circular

motion as it applies to Newton’s Law of Gravitation.

• To examine the idea of weight and relate it to mass and Newton’s Law of Gravitation.

• To study the motion of objects in orbit as a special application of Newton’s Law of Gravitation.

• A review of the relationship between v and ac.

• The velocity changes direction, not magnitude.

ac = v2/r

The force needed to keep an object moving in a circular path is called the centripetal force. It is the force that produces the acceleration and is always directed toward the center.

Fc= mv2/r

In terms of the force of gravity and centripetal force what is happening

in the cup?

What centripetal force is needed to keep a 4-kg mass moving at a constant speed of 3 m/s in a circle having a radius of 8 m? Fc = mv2/r

Fc = (4 kg)(3 m/s)2/8 m

Fc = 4.5 kg-m/s2

Fc = 4.5 N

Practice

• Notice how v becomes linear when Fc vanishes.

p. 166 Rounding the curve

11/29/12

• Homework Questions?• Circular Motion• Cavendish Measures Gravity• Force of gravity between two objects• Objects in orbit• What about the guy that said there is no gravity?

p. 167 Rounding a banked curve

A frictionless rollercoaster does a vertical loop with a radius of 6.0m. What is the minimum speed that the roller coaster must have at the top of the loop so that it stays in touch with the rail?

mv2/r = mg

Fc = 0 + Fg

g = v2/r

v = 7.7 m/s

Fnet = FN + Fgv2 = gr

v2 = 9.8 m/s2 x 6 m

Practice

A

B

C

Draw free-body diagrams at each location on the roller coaster.

What is the net force on the rider at point A?

What is Fnet at point A also called?What is FN at point A acting on the rider(apparent weight)?

What velocity is needed at point A to produce an FN on the rider of 0?

Fnet = -Fg + (FN)

Fnet = Fcentripetal

FN = Fg - Fc

0 = Fg - Fc Fg = Fc

mg = mv2/rv = (gr)½

A

What is the net force on the rider at point B?

What is Fnet at point B also called?What is FN at point B acting on the rider(apparent weight)?

What is the “g-force” at point B?

Fnet = FN + (-Fg)

Fnet = Fcentripetal

FN = Fc + Fg

G-force = FN / Fg

B

What is the net force on the rider at point C?

What is Fnet at point C also called?What is FN at point C acting on the rider?

What does a negative FN mean?

Fnet = FN + Fg

Fnet = Fcentripetal

FN = Fc - Fg

- FN = “I’m falling”

C

Gravitation

Newton and

Gravity

Law of Universal Gravitation

11/30/12

• New Homework • Force of gravity between two objects• Objects in orbit• Does Gravity exist?• Gravitation Lab

M1 M2

R

Fg =GM1 M2

R2

Fg = mg

m1 m2

R

Fg =

Fg = m2gR2

Gm1 m2 m2g =Gm1 m2

R2

g = Gm1

R2

Me m2

R

g = G Me

R2

Me m2

R

g =6.7 x 10-11N•m2/kg2(6.0 x 1024 kg)

(6.4 x 106 m)2

Me m2

R

g = 9.8 m/s2

50 kg 6 kg

2 .0 m

Fg =GM1 M2

R2

50 kg 6 kg

2 .0 m

Fg =6.7 x 10-11N•m2/kg2(50 kg)(6 kg)

(2 m)2

50 kg 6 kg

2 .0 m

Fg = 5.0 x 10-8N

It looks like a shooting starThe International space station is orbiting above

the Earth at approximately 350 km. If it has a mass of 450000 kg, what is the force of gravity between it and the Earth? SEE IT?

It looks like a shooting starGiven your answer to the following question,

how fast must the ISS be moving to stay in orbit?

m1 m2

R

Fg =

Fg = m2gR2

Gm1 m2 m2g =Gm1 m2

R2

g = Gm1

R2

M1 M2

R

Fg =GM1 M2

R2

100 kg 50 kg

1.0 m

M1 M2

R

100 kg 50 kg

1.0 m

Fg =6.7 x 10-11N•m2/kg2(50 kg)(100 kg)

(1.0 m)2

Fg = 3.35 x 10-7 N

M1 M2

R

100 kg 50 kg

1.0 m

Fg = 3.35 x 10-7 N = 50 kg x g

g = 6.7 x 10-9 m/s2

M1 M2

R

100 kg 50 kg

1.0 m

Fg = 3.35 x 10-7 N = 100 kg x g

g = 3.35 x 10-9 m/s2

A radioactive cesium nucleus emits a beta particle of mass 9.1 x 10-31 kg and transmutes (changes) into a barium nucleus that has a mass of 2.2 x 10-25 kg. What is the gravitational force of attraction between the barium nucleus and the beta particle when they are 2.0 x 10-8 m apart? Based on your answer, is the force of gravity important in holding subatomic particles together? Explain.

Concept Problem #5

Concept Problem #5

3.3 x 10-50 N

12/15/11

• Lunar Mystery question• Homework• Pass back Old Homework• Universal Gravitation practice• What about the guy that said there

is no gravity?

Jupiter and EarthJupiter is 5.2 times farther from the Sun than Earth. Find Jupiter’s orbital period in Earth yrs.

g =GMR2

Orbiting the Earth

• To maintain a constant distance around the Earth, a satellite must maintain a certain speed.

• If it did not it would fall into the atmosphere.

• We can determine the speed with which something orbits the Earth by the radius of its orbit.

Orbiting the Earth

• We can also use the radius of a satellites orbit to determine the period of its orbit.

• A satellite is in orbit around a small planet. The orbital radius is 6.7 X 104 km and its speed is 2.0 X 105 m/s. What is the mass around which the satellite orbits?

Gravity is all aroundA moon in orbit around a planet, like ours, experiences a gravitational force not only from the planet, but also from the Sun. The illustration below shows a moon during a solar eclipse, when the planet, the moon and the Sun are aligned. The moon has a mass of 3.9x1021 kg, the planet is 2.4x1026 kg and the Sun is 2.0x1030 kg. The distance from the moon to the center of the planet is 6.0x108 m. The moon to the Sun is 1.5x1011 m. What is the ratio of the gravitational force on the moon due to the planet compared to the gravitational force on the moon due to the Sun?

12/16/11

• Lunar Mystery question• Universal Gravitation practice• Newton/Einstein Gravity• What about the guy that said there

is no gravity?

R

(x = distance stretched = d)

Ws = Fs x d

Fs = kx

Ws = ½ Fsx

Ws = PEs = ½ kx2

What does Frictional Force depend on?

Plays a role Does not Play a role

What does Frictional Force depend on?

Plays a role Does not Play a role

11/16/11• Homework due tomorrow • Friction sample problems• Equilibrium and the Equilibrant • Friction on an inclined plane• Friction Lab

Find the equilibrant

More Feng shui Problem

• You need to move a 105 kg sofa to a different location in the room. It takes 102 N to start it moving. What is the coefficient of static friction between the sofa and the carpet?

Simple Push Problem

You push a box across a wooden floor at a constant speed of 1 m/s. The coefficient of kinetic friction between the box and floor is 0.2. How much force do you exert on the box?

Simple Push Problem continued

If you double the force you exerted on the box in the previous problem, what is the resulting acceleration of the box?

When on a ramp, gravity must be broken into the effect it has on the angle of the ramp.

Gravity is diluted by the angle of the ramp.

11/17/11• Homework• Nanotechnology Applications,

things to ask about tomorrow.• Friction on an inclined plane• Friction Lab

Apply motion to an inclined plane

• A crate that weighs 562 N is resting on a plane that is inclined 30 above the horizontal. Find the components of weight force that are parallel and perpendicular to the plane.

Logan on a slideLogan, who has a mass of 15 kg, starts down a slide that is

inclined at 45⁰ with the horizontal. If the coefficient of kinetic friction between the slide and Logan is 0.25, what is his acceleration?

11/22/11• http://www.lifesaverusaonline.com/• Friction on an inclined plane• Work:

–Friction Lab–Homework –Case Study related assignment

Pulling up a slope• A 100N block is pulled up a ramp that is at a

45⁰ angle. The coefficient of friction is 0.3. What force is needed to move it up the ramp at a constant speed?

10/23/11• Robots: just sayin’• Work:

–Friction Lab–Homework –Case Study related assignment

Enjoy your time with Family and friends!

Accelerating Block

• The coefficient of kinetic friction between the block and the ramp is (0.20). The pulley is frictionless.

• What is the acceleration of the system?

11/1/10• Hand in Homework• Test on vectors and friction• Go over Homework• Pushing up a slope against friction and gravity• Practice Problems • Friction worksheet• Projectile Motion

When on a ramp, gravity must be broken into the effect it has on the angle of the ramp.

Gravity is diluted by the angle of the ramp.

Vert

ical

Dis

plac

emen

t (m

)

Horizontal Displacement (m)

A stone is thrown at a speed of 10 m/s from the top of a 100 meter high cliff.a) How long does it take the stone to reach the bottom of the cliff?b) How far from the base of the cliff does the stone hit the ground?c) What are horizontal and vertical components of the stones velocity just before it

hits the ground?