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Time (s) 0 1 2 3 4

Distance (m) 0 25 50 75 100

Velocity (m/s) 25 25 25 25 25

Acceleration (m/s2) 0 0 0 0 0

Example: Two friends bicycle 3.0

kilometers north and then turn to bike

4.0 kilometers east in 25 minutes.

b) What is their average velocity?

displacement/time

5.0 km/ 25 min =

0.20 km/min x 60 min/hr =

12 km/hr

Angle: 53o west of north

a) What is their average speed?

distance/time

7.0 km / 25 min =

0.28 km/min x 60 min/hr =

16.8 = 17 km/hr

Mechanics Kinematics – The Study of Motion

average vs. instantaneous: over a period of time vs. at one instant

Symbols: s = distance or displacement

u = initial speed or velocity

v = final speed or velocity

a = average acceleration

or 2

also 2

so

u vs vt s t

s u vv v

t

+= =

+= =

v v ua

t t

v u at

∆ −= =∆

= +

2

2 2

1

2

2

s ut t

v u as

= +

= +

Equations:

Condition for applying

equations for uniformly

accelerated motion:

must be constant, smooth acceleration

equations use average acceleration = instantaneous if acceleration is constant

Constant Velocity

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Time (s) 0 1 2 3 Distance (m) 0 3 10 23

Velocity (m/s) 0 5 10 15

Acceleration (m/s2) 5 5 5 5

Constant Acceleration

b) What does the slope of a velocity-time graph represent? Instantaneous acceleration - derivative

c) What does the area under a velocity time graph represent? displacement - integral

a) What does the slope of a position-time graph represent? Instantaneous velocity - derivative

Dropping

1. A stone is dropped from rest from the top of a tall building. After 3.00 s of free-fall, what is the displacement of the stone?

What is its velocity?

How would these graphs change in the presence of air resistance?

Terminal velocity: no acceleration – constant velocity – Fg = Fair

discuss change of frame of reference – downward is positive

s = ut + ½ at2

s = ½ at2 = -45 m

v = u + at

v = -30 m/s

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Throwing Up

A ball is thrown straight up in the air (shown

here stretched out for clarity.) Sketch velocity

and acceleration vectors at each instant.

A football game customarily begins with a coin toss to determine who

kicks off. The referee tosses the coin up with an initial speed of 6.00

m/s. In the absence of air resistance, how high does the coin go above

its point of release? How long is it in the air?

v = u + at

0 = 6.00 + (-10)t

t = 0.6 s x 2 = 1.2 s

v2 = u

2 + 2 as

02 = (6)

2 + 2(-10)s

s = 1.8 m

discuss change of frame of reference – downward is positive

Projectile Motion: resultant of two independent components of motion

1. Vertical: constant acceleration (in absence of air resistance)

2. Horizontal: constant velocity – no horizontal acceleration

Horizontal Projectile A ball is shot horizontally off a cliff that is 100. m high at a speed of 25 m/s. How long

does it take to hit the ground? How far away from the base of the cliff does it land?

y-direction:

s = ut + ½ at2

-100 = 0 + ½(-10)t2

t = 4.5 s

x-direction:

s = ut + ½ at2

s = 25 (4.5) + 0

s = 113 m

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Angled Projectile

1. Break initial velocity into horiz and vert components

2. maximum height occurs after ½ time

3. maximum range occurs after fulltime and at 450

4. air resistance: not as high nor as far – show on diagram

A football was kicked with a speed of 25 m/s at an angle of 30.0o to the horizontal. Determine how high it went and where it landed.

y: sin 300 = vi/25 m/s

vi = 12.5 m/s

Components

x: cos 300 = vi/25 m/s

vi = 21.7 m/s

time:

v = u + at

0 = 12.5 m/s + (-10)(t)

t = 1.25 s

total time = 2(1.25) = 2.5

Height:

s = ut + ½ at2

s = (12.5 m/s)(1.25 s) + ½ (-10 m/s2)(1.25 s)

2

s = 7.8 m

Range:

s = ut + ½ at2

s = (21.7 m/s)(2.5 s) + 0

s = 54 m

Newton’s Laws of Motion

1. An object at rest remains at rest and an object in motion remains in motion at a constant speed in a straight

line (constant velocity) unless acted on by unbalanced forces. (An object continues in uniform motion in a

straight line or at rest unless a resultant (net) external force acts on it.)

2. When unbalanced forces act on an object, the object will accelerate in the direction of the resultant (net)

force. The acceleration will be directly proportional to the net force and inversely proportional to the object’s

mass. (The resultant force on an object is equal to the rate of change of momentum of the object.)

3. For every action on one object, there is an equal and opposite reaction on another object. (When two

bodies A and B interact, the force that A exerts on B is equal and opposite to the force that B exerts on A.)

Statics and Dynamics – The Study of Forces

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Newton’s Second Law For constant mass

F = m (∆v / ∆t)

F = m a

F = ∆p / ∆t

F = ∆(mv) / ∆t

2nd Law or 3

rd Law?

Net force on ball: not zero

so it accelerates – not in

equilibrium

Fnet = Fg = mg

Action-Reaction pairs:

Earth pulls ball down

Ball pulls Earth up

FEB = -FBE

mA = -Ma

Fg

Net force on block: zero

at rest – in equilibrium

Fnet = FN - Fg = 0

Action-Reaction pairs:

Earth pulls block down

Block pulls Earth up

block pushes down on table

table pushes up on block

Translational equilibrium: net force acting on object is zero – no acceleration

1. Find the resistive force F caused by the drag of the water on the boat moving at a constant velocity in the diagram shown.

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2. Find the tension in each cable supporting the 600 N cat burglar pictured.

600 997 796

3. A 20.0-kg floodlight in a park is supported at the end of a horizontal beam of negligible mass that is hinged to a pole, as shown. A cable at

an angle of 30.0° with the beam helps to support the light. Find (a) the tension in the cable and (b) the horizontal and vertical forces

exerted on the beam by the pole.

H

V196 N

30.0°

T

d

a) 400N

b) Rx = 346 Ry = 0

4. How would your answers change if the mass of

the beam shown above was not negligible?

T increase

show how reaction force is oriented

FW

Fg1

Fg2

FN

Ff

5. Indicate the direction of the reaction force from the floor and

the reaction force from the wall for the situation shown below.

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Weight, Mass, and the Normal Force

Mass 1) the amount of matter in an object

2) the property of an object that determines its resistance to a

change in its motion (a measure of the amount of inertia of

an object)

Property:

Remains constant

Symbol : m Units : kg

Weight:

the force of gravity acting on an object

Property:

Varies from place to place Symbol : Fg or W

Units : N

Σ F = 0

FN – Fg = 0

FN – 700 = 0

FN = 700 N

Σ F = ma

FN – Fg = m a

1000 – 700 = 70 a

a = +4.3 m/s2

Σ F = m a

FN – Fg = m a

400 – 700 = 70 a

a = - 4.3 m/s2

Σ F = m a

– Fg = m a

– 700 = 70 a

a = -10 m/s2

Elevators: In each case, the scale will read . . . the normal or reaction force, not the weight

Calculate

the

acceleration

of the man

in each

case.

Inclined Plane – Assume the box shown is in equilibrium and draw the . . .

Head-to-tail vector diagram Concurrent vector diagram

θ

FN

Ff

Fg

Free-body diagram

Concurrent vector diagram with

perpendicular components

θ

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200

4.5 kg

Calculate the force of friction acting on this box if it accelerates down the

incline at a rate of 0.67 m/s2.

Fnet = ma = 4.5 (0.67) = 3.0 N

Fg|| = mg sin θ = 15 N

Ff = 12 N

Formulas:

Uniform Circular Motion constant speed and constant radius

V bar = distance / time

V bar = circumference / period

V bar = 2πr / T

V = 2πr / T

Period time take for one complete cycle symbol: T units: s

Uniform Circular Motion

1. The direction of the object’s instantaneous velocity is always tangent to the circle in

the direction of motion.

2. Since the direction of the object’s motion is always changing, its velocity is always

changing therefore the object is always accelerating and is never in equilibrium.

3. Direction of net force – towards the center - centripetal

V bar = 2πr / T

ac = v2 / r

F = m a

Fc = m ac

Fc = mv2 / r

ac = (2πr / T)2 / r

ac = (4π2r

2 / T

2) / r

ac = 4 π2 r / T

2

The phrase “centripetal force” does not denote a new and separate force created by nature. The phrase merely labels the net force pointing

toward the center of the circular path, and this net force is the vector sum of all the force components that point along the radial direction.

1. The model airplane shown has a mass of 0.90 kg and moves at a constant

speed on a circle that is parallel to the ground. Find the tension T in the

guideline (length = 17 m) for a speed of 19 m/s.

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2. At amusement parks, there is a popular ride where the floor of a rotating cylindrical room falls away, leaving the backs of

the riders “plastered” against the wall. For a particular ride with a radius of 8.0 m and a top speed of 21 m/s, calculate the

reaction force and the friction force from the wall acting on a 60. kg rider. Which of these is the centripetal force?

3. A 2100-kg demolition ball swings at the end of a 15-m cable on the arc of a vertical circle. At the lowest point of the

swing, the ball is moving at a speed of 7.6 m/s. Determine the tension in the cable.

Work, Power and Efficiency

Work: product of force and displacement in the direction of the force

Power:

1) the rate at which work is done

2) the rate at which energy is transferred

Type: Scalar but can be positive or

negative

Formula: W = (F cos θ) s

W = F s cos θ

θ is angle between F and s

Units:

N m or Joules (J)

Formula: P = W/ t

= E / t

= Q/t

Units:

J / s

= Watts (W)

Alternate Formula: P = W / t

= (F cos θ d) / t

= F v cos θ

Type: Scalar

Efficiency:

1) ratio of useful work done by a system to the total work done by the system

2) ratio of useful energy output of a system to the total energy input to the system

3) ratio of useful power output of a system to total power input to the system

Formula:

e = useful out/

total in

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2. a) How much work is done dragging the 5.00 kg box to the top of the hill shown if the hill exerts an

average friction force of 5.0 N?

b) Compare your answer to the amount of work done lifting the box straight up to the top of the hill.

c) Calculate the power expended if the box is dragged to the top in 15 seconds.

d) Calculate the efficiency of dragging the box rather than lifting the box.

1. A 45.0-N force is applied to pull a luggage carrier an angle θ = 50° for a distance of 75.0 m at a constant speed.

Find the work done by the applied force.

W = F s cos θ

WA = (45.0 N)(75 m) cos 50o = 2170 J

Determining Work Done Graphically

1. Work done by a constant force 2. Work done by a constantly varying force

Work = area under curve

W = b h

W = f s

Work = area under curve

W = ½ b h

W = ½ F s

W = (avg force) (displacement)

ex. stretching a spring

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Energy

Types of Energy

1. 2. 3. 4. 5.

1. Kinetic energy (energy of motion)

2. Gravitational Potential energy (energy of position)

3. Elastic potential energy

4. Internal energy (thermal energy)

5. Chemical Potential energy (stored in chemical bonds) Electrical energy Light energy

Formulas:

1. EK = ½ mv2

2. EP = mgh

3. Eelas = ½ kx2

4. Q = mc∆t Q = mL

5. Ee = Pt = VIt = I2Rt = V

2/R t

Conservation of Energy Principle

In an isolated system, the total amount of energy remains constant.

1. A motorcyclist is trying to leap across a canyon by driving horizontally off the

cliff at a speed of 38.0 m/s. Ignoring air resistance, find the speed with which the

cycle strikes the ground on the other side.

300

20 meters

160 N 2. What is the speed of the box at the bottom of the incline if an average frictional force

of 15 N acts on it as it slides?

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Linear Momentum and Impulse

Linear Momentum: the product of an object’s mass and velocity Formula:

p = mv

Units:

kg m/s

A 1500. kg car is traveling east at a

speed of 25.0 m/s.

Compare its inertia, momentum, and

kinetic energy.

inertia (inertial mass)

m = 1500 kg

scalar

momentum

p = (1500) (25)

= 3.75 x 104 kg m/s, east

vector

kinetic energy

Ek = ½ (1500)(25)2

= 468750 J = 4.69 x 105 J

scalar

Alternate formula for

kinetic energy:

How does the momentum of an object change?

A net external force acts for a finite amount of time

Impulse:

(the change in momentum of a system)

the product of the average force and the time interval over which the force acts

Derivation of Impulse Formula

( )

(if mass is constant)

pF

t

p F t

mv F t

m v F t

J p F t m v

∆=∆

∆ = ∆

∆ = ∆

∆ = ∆

= ∆ = ∆ = ∆

Units:

Ns or

kg m/s

Type:

vector

Usually . . . Force is not instantaneous and is not constant

If force is linear:

J = ½ Fmax ∆t

Graphically, the impulse is

J = Favg ∆t = area

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0.50 ( 4.0 ( 3.0))

7.0 /

p m v

p kg

p kgm s

∆ = ∆

∆ = + − −

∆ =

Bouncing and Impulse

A 0.50 kg basketball hits the floor at a speed of 4.0 m/s and rebounds at 3.0 m/s. Calculate the impulse applied to it by the floor.

( )

( )

f i

f i

p m v

p m v v

p m v v

∆ = ∆

∆ = −

∆ = +

v v

v v v

v

In general: Calculation:

Force vs. time graph for bounce Velocity vs. time graph for bounce

Vel

oci

ty

Time

Fo

rce

Time

The Principle of Conservation

of Linear Momentum:

The total momentum of an isolated system remains constant.

pbefore = pafter

Types of Interactions

1. Bouncy

2. Sticky

3. Explosion

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1. A freight train is being assembled in a switching yard, and the

figure below shows two boxcars. Car 1 has a mass of m1 =

65×103 kg and moves at a velocity of v01 = +0.80 m/s. Car 2, with

a mass of m2 = 92×103 kg and a velocity of v02 = +1.3 m/s,

overtakes car 1 and couples to it. Neglecting friction, find the

common velocity vf of the cars after they become coupled.

2. Is this collision elastic or inelastic? Justify your answer.

Elastic collision: a collision in

which the total kinetic energy is

conserved

Inelastic collision: a collision in

which the total kinetic energy is not

conserved

Where does some of the mechanical energy go in an inelastic collision? energy of deformation, internal energy, sound energy

Collisions

3. A ballistic pendulum is sometimes used in laboratories to

measure the speed of a projectile, such as a bullet. A ballistic

pendulum consists of a block of wood (mass m2 = 2.50 kg)

suspended by a wire of negligible mass. A bullet (mass m1 =

0.0100 kg) is fired with a speed v01. Just after the bullet

collides with it, the block (with the bullet in it) has a speed vf

and then swings to a maximum height of 0.650 m above the

initial position (see part b of the drawing). Find the speed v01

of the bullet, assuming that air resistance is negligible.