Regent Physics Review

Physics Units I. Physics Skills II. Mechanics III. Energy IV. Electricity and Magnetism V. Waves VI. Modern Physics

I. Physics Skills A. Scientific Notation B. Graphing C. Significant Figures D. Units E. Prefixes F. Estimation

A. Scientific Notation Use for very large or very small numbers Write number with one digit to the left of the decimal followed by an exponent (1.5 x 105) Examples: 2.1 x 103 represents 2100 and 3.6 x 10-4 represents 0.00036

Scientific Notation Problems 1. Write 365,000,000 in scientific notation 2. Write 0.000087 in scientific notation

Answers 1.) 3.65 x 108 2.) 8.7 x 10-5

B. Graphing Use graphs to make a “picture” of scientific data “independent variable”, the one you change in your experiment is graphed on the “x” axis and listed first in a table “dependent variable”, the one changed by your experiment is graphed on the “y” axis and listed second in a table

Best fit “line” or “curve” is drawn once points are plotted Best fit “line” or “curve” is drawn once points are plotted. Does not have to go through all points. Just gives you the “trend” of the points The “slope” of the line is given as the change in the “y” value divided by the change in the “x” value

Types of Graphs 1. Direct Relationship means an increase/decrease in one variable causes an increase/decrease in the other Example below

2. Inverse(indirect) relationship means that an increase in one variable causes a decrease in the other variable and vice versa Examples

3. Constant proportion means that a change in one variable doesn’t affect the other variable Example;

4. If either variable is squared(whether the relationship is direct or indirect), the graph will curve more steeply.

C. Significant figures Uncertainty in measurements is expressed by using significant figures The more accurate or precise a measurement is, the more digits will be significant

Significant Figure Rules 1. Zeros that appear before a nonzero digit are not significant (examples: 0.002 has 1 significant figure and 0.13 has 2 significant figures) 2. Zeros that appear between nonzero digits are significant (examples: 1002 has 4 significant figures and 0.405 has 3 significant figures)

Significant figures rules(cont.) 3. zeros that appear after a nonzero digit are significant only if they are followed by a decimal point (20. has 2 sig figs) or if they appear to the right of the decimal point (35.0 has 3 sig figs)

Sig Fig problems 1. How many significant figures does 0.050900 contain? 2. How many significant figures does 4800 contain?

Answers 1. 5 sig figs 2. 2 sig figs

D. Units 1. Fundamental units are units that can’t be broken down 2. Derived units are made up of other units and then renamed 3. SI units are standardized units used by scientists worldwide

Fundamental Units Meter (m)– length, distance, displacement, height, radius, elongation or compression of a spring, amplitude, wavelength Kilogram (kg)– mass Second (s)– time, period Ampere (A)– electric current Degree (o)– angle

Derived Units Meter per second (m/s)– speed, velocity Meter per second squared (m/s2)– acceleration Newton (N)– force Kilogram times meter per second (kg.m/s)– momentum Newton times second (N.s)-- impulse

Derived Units (cont.) Joule (J)– work, all types of energy Watt (W)– power Coulomb (C)– electric charge Newton per Coulomb (N/C)– electric field strength (intensity) Volt (V)- potential difference (voltage) Electronvolt (eV)– energy (small amounts)

Derived Units (cont.) Ohm (Ω)– resistance Ohm times meter (Ω.m)– resistivity Weber (Wb)– number of magnetic field (flux) lines Tesla (T)– magnetic field (flux) density Hertz (Hz)-- frequency

E. Prefixes Adding prefixes to base units makes them smaller or larger by powers of ten The prefixes used in Regents Physics are tera, giga, mega, kilo, deci, centi, milli, micro, nano and pico

Prefix Examples A terameter is 1012 meters, so… 4 Tm would be 4 000 000 000 000 meters A gigagram is 109 grams, so… 9 Gg would be 9 000 000 000 grams A megawatt is 106 watts, so… 100 MW would be 100 000 000 watts A kilometer is 103 meters, so… 45 km would be 45 000 meters

Prefix examples (cont.) A decigram is 10-1 gram, so… 15 dg would be 1.5 grams A centiwatt is 10-2 watt, so… 2 dW would be 0.02 Watt A millisecond is 10-3 second, so… 42 ms would be 0.042 second

Prefix examples (cont.) A microvolt is 10-6 volt, so… 8 µV would be 0.000 008 volt A nanojoule is 10-9 joule, so… 530 nJ would be 0.000 000 530 joule A picometer is 10-12 meter, so… 677 pm would be 0.000 000 000 677 meter

Prefix Problems 1.) 16 terameters would be how many meters? 2.) 2500 milligrams would be how many grams? 3.) 1596 volts would be how many gigavolts? 4.) 687 amperes would be how many nanoamperes?

Answers 1.) 16 000 000 000 000 meters 2.) 2.500 grams 3.) 1596 000 000 000 gigavolts 4.) 0.000 000 687 amperes

F. Estimation You can estimate an answer to a problem by rounding the known information You also should have an idea of how large common units are

Estimation (cont.) 2 cans of Progresso soup are just about the mass of 1 kilogram 1 medium apple weighs 1 newton The length of an average Physics student’s leg is 1 meter

Estimation Problems 1.) Which object weighs approximately one newton? Dime, paper clip, student, golf ball 2.) How high is an average doorknob from the floor? 101m, 100m, 101m, 10-2m

Answers 1.) golf ball 2.) 100m

II. Mechanics A. Kinematics; vectors, velocity, acceleration B. Kinematics; freefall C. Statics D. Dynamics E. 2-dimensional motion F. Uniform Circular motion G. Mass, Weight, Gravity H. Friction I. Momentum and Impulse

Kinematics; vectors, velocity, acceleration In physics, quantities can be vector or scalar VECTOR quantities have a magnitude (a number), a unit and a direction Example; 22m(south)

SCALAR quantities only have a magnitude and a unit Example; 22m

VECTOR quantities; displacement, velocity, acceleration, force, weight, momentum, impulse, electric field strength SCALAR quantities; distance, mass, time, speed, work(energy), power

Distance vs. Displacement Distance is the entire pathway an object travels Displacement is the “shortest” pathway from the beginning to the end

Distance/Displacement Problems 1.) A student walks 12m due north and then 5m due east. What is the student’s resultant displacement? Distance? 2.) A student walks 50m due north and then walks 30m due south. What is the student’s resultant displacement? Distance?

Answers 1.) 13m (NE) for displacement 17 m for distance 2.) 20m (N) for displacement 80 m for distance

Speed vs. Velocity Speed is the distance an object moves in a unit of time Velocity is the displacement of an object in a unit of time

Average Speed/Velocity Equations

Symbols

Speed/Velocity Problems 1.) A boy is coasting down a hill on a skateboard. At 1.0s he is traveling at 4.0m/s and at 4.0s he is traveling at 10.0m/s. What distance did he travel during that time period? (In all problems given in Regents Physics, assume acceleration is constant)

Answers 1.) You must first find the boy’s average speed before you are able to find the distance

Answers (cont.)

Acceleration The time rate change of velocity is acceleration (how much you speed up or slow down in a unit of time) We will only be dealing with constant (uniform) acceleration

Symbols (cont.)

Constant Acceleration Equations

Constant Acceleration Problems 1.) A car initially travels at 20m/s on a straight, horizontal road. The driver applies the brakes, causing the car to slow down at a constant rate of 2m/s2 until it comes to a stop. What was the car’s stopping distance? (Use two different methods to solve the problem)

Answers First Method vi=20m/s vf=0m/s a=2m/s2 Use vf2=vi2+2ad to find “d” d=100m

Answer (cont.) Second Method

B. Kinematics: freefall In a vacuum (empty space), objects fall freely at the same rate The rate at which objects fall is known as “g”, the acceleration due to gravity On earth, the “g” is 9.81m/s2

Solving Freefall Problems To solve freefall problems use the constant acceleration equations Assume a freely falling object has a vi=0m/s Assume a freely falling object has an a=9.81m/s2

Freefall Problems 1.) How far will an object near Earth’s surface fall in 5s? 2.) How long does it take for a rock to fall 60m? How fast will it be going when it hits the ground 3.) In a vacuum, which will hit the ground first if dropped from 10m, a ball or a feather?

Answers 1.)

Answers (cont.) 2.)

Answers (cont.) 3.) Both hit at the same time because “g” the acceleration due to gravity is constant. It doesn’t depend on mass of object because it is a ratio

Solving Anti??? Freefall Problems If you toss an object straight up, that is the opposite of freefall. So… vf is now 0m/s a is -9.8lm/s2 Because the object is slowing down not speeding up

Antifreefall Problems 1.) How fast do you have to toss a ball straight up if you want it reach a height of 20m? 2.) How long will the ball in problem #1 take to reach the 20m height?

Answers 1.)

Answers (cont.) 2.)

C. Statics The study of the effect of forces on objects at rest Force is a push or pull The unit of force is the newton(N) (a derived vector quantity)

Adding forces When adding concurrent (acting on the same object at the same time) forces follow three rules to find the resultant (the combined effect of the forces) 1.) forces at 00, add them 2.) forces at 1800, subtract them 3.) forces at 900, use Pythagorean Theorem

Force Diagrams Forces at 00 Forces at 1800 Forces at 900

Composition of Forces Problems 1.) Find the resultant of two 5.0N forces at 00? 1800? And 900?

Answers 1.) 00 5N 5N = 10N

Answers (cont.) 1.) 1800 5N 5N = 0N

Answers (cont.) 1.) 900 5N = 5N 7.1N

Resolution of forces The opposite of adding concurrent forces. Breaking a resultant force into its component forces Only need to know components(2 forces) at a 900 angle to each other

Resolving forces using Graphical Method To find the component forces of the resultant force 1.) Draw x and y axes at the tail of the resultant force 2.) Draw lines from the head of the force to each of the axes 3.) From the tail of the resultant force to where the lines intersect the axes, are the lengths of the component forces

Resolution Diagram Black arrow=resultant force Orange lines=reference lines Green arrows=component forces y x

Resolving Forces Using Algebraic Method

Equilibrium Equilibrium occurs when the net force acting on an object is zero Zero net force means that when you take into account all the forces acting on an object, they cancel each other out

Equilibrium (cont.) An object in equilibrium can either be at rest or can be moving with constant (unchanging) velocity An “equilibrant” is a force equal and opposite to the resultant force that keeps an object in equilibrium

Equilibrium Diagram Black arrows=components Blue arrow=resultant Red arrow=equilibrant = +

Problems 1.) 10N, 8N and 6N forces act concurrently on an object that is in equilibrium. What is the equilibrant of the 10N and 6N forces? Explain. 2.) A person pushes a lawnmower with a force of 300N at an angle of 600 to the ground. What are the vertical and horizontal components of the 300N force?

Answers 1.) The 8N force is the equilibrant (which is also equal to and opposite the resultant) The 3 forces keep the object in equilibrium, so the third force is always the equilibrant of the other two forces.

Answers (cont.) 2.)

C. Dynamics The study of how forces affect the motion of an object Use Newton’s Three Laws of Motion to describe Dynamics

Newton’s 1st Law of Motion Also called the law of inertia Inertia is the property of an object to resist change. Inertia is directly proportional to the object’s mass “An object will remain in equilibrium (at rest or moving with constant speed) unless acted upon by an unbalanced force”

Newton’s Second Law of Motion “When an unbalanced (net) force acts on an object, that object accelerates in the direction of the force” How much an object accelerates depends on the force exerted on it and the object’s mass (See equation)

Symbols Fnet=the net force exerted on an object (the resultant of all forces on an object) in newtons (N) m=mass in kilograms (kg) a=acceleration in m/s2

Newton’s Third Law of Motion Also called law of action-reaction “When an object exerts a force on another object, the second object exerts a force equal and opposite to the first force” Masses of each of the objects don’t affect the size of the forces (will affect the results of the forces)

Free-Body Diagrams A drawing (can be to scale) that shows all concurrent forces acting on an object Typical forces are the force of gravity, the normal force, the force of friction, the force of acceleration, the force of tension, etc.

Free-body Forces Fg is the force of gravity or weight of an object (always straight down) FN is the “normal” force (the force of a surface pushing up against an object) Ff is the force of friction which is always opposite the motion Fa is the force of acceleration caused by a push or pull

Free-Body Diagrams If object is moving with constant speed to the right…. Black arrow=Ff Green arrow=Fg Yellow arrow=FN Blue arrow=Fa

Free-Body Diagrams on a Slope When an object is at rest or moving with constant speed on a slope, some things about the forces change and some don’t 1.) Fg is still straight down 2.) Ff is still opposite motion 3.) FN is no longer equal and opposite to Fg 4.) Ff is still opposite motion More……

Free-Body Diagrams on a Slope Fa=Ff=Ax=Acosθ FN=Ay=Asinθ Fg=mg (still straight down) On a horizontal surface, force of gravity and normal force are equal and opposite On a slope, the normal force is equal and opposite to the “y” component of the force of gravity

Free-Body Diagram on a Slope Green arrow=FN Red arrow=Fg Black arrows=Fa and Ff Orange dashes=Ay Purple dashes=Ax

Dynamics Problems 1.) Which has more inertia a 0.75kg pile of feathers or a 0.50kg pile of lead marbles? 2.) An unbalanced force of 10.0N acts on a 20.0kg mass for 5.0s. What is the acceleration of the mass?

Answers 1.) The 0.75kg pile of feathers has more inertia because it has more mass. Inertia is dependent on the “mass” of the object

Answers (cont.) 2.)

More Problems 3.) A 10N book rests on a horizontal tabletop. What is the force of the tabletop on the book? 4.) How much force would it take to accelerate a 2.0kg object 5m/s2? How much would that same force accelerate a 1.0kg object?

Answers 1.) The force of the tabletop on the book is also 10N (action/reaction)

Answers (cont.) 2.)

E. 2-Dimensional Motion To describe an object moving 2-dimensionally, the motion must be separated into a “horizontal” component and a “vertical” component (neither has an effect on the other) Assume the motion occurs in a “perfect physics world”; a vacuum with no friction

Types of 2-D Motion 1.) Projectiles fired horizontally an example would be a baseball tossed straight horizontally away from you

Projectile Fired Horizontally Use the table below to solve these type of 2-D problems Quantities Horizontal Vertical vi Same as vf 0m/s vf Same as vi a 0m/s2 9.81m/s2 d t Same as vertical time Same as horizontal time

Types of 2-D Motion 2.) Projectiles fired at an angle an example would be a soccer ball lofted into the air and then falling back onto the ground

Projectile Fired at an Angle Use the table below to solve these type of 2-D problems Ax=Acosθ and Ay=Asinθ Quantities Horizontal Vertical vi Ax Ay vf 0m/s a 0m/s2 - 9.81m/s2 d t twice vertical time

2-D Motion Problems 1.) A girl tossed a ball horizontally with a speed of 10.0m/s from a bridge 7.0m above a river. How long did the ball take to hit the river? How far from the bottom of the bridge did the ball hit the river?

Answers 1.) In this problem you are asked to find time and horizontal distance (see table on the next page)

Answers (cont.) Quantities Horizontal Vertical vi 10.0m/s 0.0m/s vf Don’t need a 9.81m/s2 d ? 7.0m t

Answers (cont.) Use

More 2-D Motion Problems 2.) A soccer ball is kicked at an angle of 600 from the ground with an angular velocity of 10.0m/s. How high does the soccer ball go? How far away from where it was kicked does it land? How long does its flight take?

Answers 2.) In this problem you are asked to find vertical distance, horizontal distance and horizontal time. Finding vertical time is usually the best way to start. (See table on next page)

Answers (cont.) Quantities Horizontal Vertical vi Ax=5.0m/s Ay=8.7m/s vf 0.0m/s a - 9.81m/s2 d ? t Need to find

Answers (cont.) Find vertical “t” first using vf=vi+at with…. vf=0.0m/s vi=8.7m/s a=-9.81m/s2 So…vertical “t”=0.89s and horizontal “t” is twice that and equals 1.77s

Answers (cont.) Find horizontal “d” using

Answers (cont.) Find vertical “d” by using

F. Uniform Circular Motion When an object moves with constant speed in a circular path The force (centripetal) will be constant towards the center Acceleration (centripetal) will only be a direction change towards the center Velocity will be tangent to the circle in the direction of movement

Uniform Circular Motion Symbols Fc=centripetal force, (N) v=constant velocity (m/s) ac=centripetal acceleration (m/s2) r=radius of the circular pathway (m) m=mass of the object in motion (kg)

Uniform Circular Motion Diagram Fc a v

Uniform Circular Motion Equations

Uniform Circular Motion Problems 1.) A 5kg cart travels in a circle of radius 2m with a constant velocity of 4m/s. What is the centripetal force exerted on the cart that keeps it on its circular pathway?

Answers 1.)

G. Mass, Weight and Gravity Mass is the amount of matter in an object Weight is the force of gravity pulling down on an object Gravity is a force of attraction between objects

Mass Mass is measured in kilograms (kg) Mass doesn’t change with location (for example, if you travel to the moon your mass doesn’t change)

Weight Weight is measured in newtons (N) Weight “does” change with location because it is dependent on the pull of gravity Weight is equal to mass times the acceleration due to gravity

Weight/Force of Gravity Equations

Gravitational Field Strength g=acceleration due to gravity but it is also equal to “gravitational field strength” The units of acceleration due to gravity are m/s2 The units of gravitational field strength are N/kg Both quantities are found from the equation:

Mass, Weight, Gravity Problems 1.) If the distance between two masses is doubled, what happens to the gravitational force between them? 2.) If the distance between two objects is halved and the mass of one of the objects is doubled, what happens to the gravitational force between them?

Answers 1.) Distance has an inverse squared relationship with the force of gravity. So…since r is multiplied by 2 in the problem, square 2……so….22=4, then take the inverse of that square which equals ¼….so….the answer is “¼ the original Fg”

Answers (cont.) 2.) Mass has a direct relationship with Fg and distance has an inverse squared relationship with Fg. First…since m is doubled so is Fg and since r is halved, square ½ , which is ¼ and then take the inverse which is 4. Then…combine 2x4=8 So…answer is “8 times Fg”

More Problems 3.) Determine the force of gravity between a 2kg and a 3kg object that are 5m apart. 4.) An object with a mass of 10kg has a weight of 4N on Planet X. What is the acceleration due to gravity on Planet X? What is the gravitational field strength on Planet X?

Answers 3.) 4.)

H. Friction The force that opposes motion measured in newtons (N) Always opposite direction of motion “Static Friction” is the force that opposes the “start of motion” “Kinetic Friction” is the force of friction between objects in contact that are in motion

Coefficient of Friction The ratio of the force of friction to the normal force (no unit, since newtons cancel out) Equation Ff= μFN μ=coefficient of friction Ff=force of friction FN=normal force

Coefficient of Friction The smaller the coefficient, the easier the surfaces slide over one another The larger the coefficient, the harder it is to slide the surfaces over one another Use the table in the reference tables

Coefficient of Friction Problems 1.) A horizontal force is used to pull a 2.0kg cart at constant speed of 5.0m/s across a tabletop. The force of friction between the cart and the tabletop is 10N. What is the coefficient of friction between the cart and the tabletop? Is the friction force kinetic or static? Why?

Answers 1.) The friction force is kinetic because the cart is moving over the tabletop

I. Momentum and Impulse Momentum is a vector quantity that is the product of mass and velocity (unit is kg.m/s) Impulse is the product of the force applied to an object and time (unit is N.s)

Momentum and Impulse Symbols p=momentum Δp=change in momentum= (usually) m(vf-vi) J=impulse

Momentum and Impulse Equations p=mv J=Ft J=Δp pbefore=pafter

Momentum and Impulse Problems 1.) A 5.0kg cart at rest experiences a 10N.s (E) impulse. What is the cart’s velocity after the impulse? 2.) A 1.0kg cart at rest is hit by a 0.2kg cart moving to the right at 10.0m/s. The collision causes the 1.0kg cart to move to the right at 3.0m/s. What is the velocity of the 0.2kg cart after the collision?

Answer 1.) Use J=Δp so…… J=10N.s(E)=Δp=10kg.m/s(E) and since original p was 0kg.m/s and Δp=10kg.m/s(E), new p=10kg.m/s(E) then use….. p=mv so…….. 10kg.m/s(E)=5.0kg x v so…. v=2m/s(E)

Answers (cont.) 2.) Use pbefore=pafter Pbefore=0kg.m/s + 2kg.m/s(right) =2kg.m/s(right) Pafter=2kg.m/s(right)=3kg.m/s + P(0.2kg cart) so….p of 0.2kg cart must be -1kg.m/s or 1kg.m/s(left) more…..

Answers (cont.) So if p after collision of 0.2kg cart is 1kg.m/s(left) and p=mv 1kg.m/s(left)=0.2kg x v And v=5m/s(left)

III. Energy A. Work and Power B. Potential and Kinetic Energy C. Conservation of Energy D. Energy of a Spring

A. Work and Power Work is using energy to move an object a distance Work is scalar The unit of work is the Joule (J) Work and energy are manifestations of the same thing, that is why they have the same unit of Joules

Work and Power (cont.) Power is the rate at which work is done so there is a “time” factor in power but not in work Power and time are inversely proportional; the less time it takes to do work the more power is developed The unit of power is the watt (W) Power is scalar

Work and Power Symbols W=work in Joules (J) F=force in newtons (N) d=distance in meters (m) ΔET=change in total energy in Joules (J) P=power in watts (W) t=time in seconds (s)

Work and Power Equations W=Fd=ΔET ***When work is done vertically, “F” can be the weight of the object Fg=mg

Work and Power Problems 1.) A 2.5kg object is moved 2.0m in 2.0s after receiving a horizontal push of 10.0N. How much work is done on the object? How much power is developed? How much would the object’s total energy change? 2.) A horizontal 40.0N force causes a box to move at a constant rate of 5.0m/s. How much power is developed? How much work is done against friction in 4.0s?

Answers 1.) to find work use W=Fd So…W=10.0N x 2.0m=20.0J To find power use P=W/t So…P=20.0J/2.0s=10.0W To find total energy change it’s the same as work done so…… ΔET=W=20.0J

Answers (cont.) 2.) To find power use So… P=40.0N x 5.0m/s=200W To find work use P=W/t so…200W=W/4.0s So….W=800J

More problems 3.) A 2.0kg object is raised vertically 0.25m. What is the work done raising it? 4.) A lift hoists a 5000N object vertically, 5.0 meters in the air. How much work was done lifting it?

Answers 3.) to find work use W=Fd with F equal to the weight of the object So…..W=mg x d So...W=2.0kgx9.81m/s2x0.25m So…W=4.9J

Answers (cont.) 4.) to find work use W=Fd Even though it is vertical motion, you don’t have to multiply by “g” because weight is already given in newtons So…W=Fd=5000N x 5.0m And W=25000J

B. Potential and Kinetic Energy Gravitational Potential Energy is energy of position above the earth Elastic Potential Energy is energy due to compression or elongation of a spring Kinetic Energy is energy due to motion The unit for all types of energy is the same as for work the Joule (J). All energy is scalar

Gravitational Potential Energy Symbols and Equation ΔPE=change in gravitational potential energy in Joules (J) m=mass in kilograms (kg) g=acceleration due to gravity in (m/s2) Δh=change in height in meters (m) Equation ΔPE=mgΔh ***Gravitational PE only changes if there is a change in vertical position

Gravitational PE Problems 1.) How much potential energy is gained by a 5.2kg object lifted from the floor to the top of a 0.80m high table? 2.) How much work is done in the example above?

Answers 1.) Use ΔPE=mgΔh to find potential energy gained so ΔPE=5.2kgx9.81m/s2x0.80m So…ΔPE=40.81J 2.) W=ΔET so..W is also 40.81J

Kinetic Energy Symbols and Equation KE=kinetic energy in Joules (J) m=mass in kilograms (kg) v=velocity or speed in (m/s)

Kinetic Energy Problems 1.) If the speed of a car is doubled, what happens to its kinetic energy? 2.) A 6.0kg cart possesses 75J of kinetic energy. How fast is it going?

Answers 1.) Using KE=1/2mv2 if v is doubled, because v if squared KE will be quadrupled. 2.) Use KE=1/2mv2 so….. 75J=1/2 x 6.0kg x v And…..v=5m/s

C. Conservation of Energy In a closed system the total amount of energy is conserved Total energy includes potential energy, kinetic energy and internal energy Energy within a system can be transferred among different types of energy but it can’t be destroyed

Conservation of Energy in a Perfect Physics World In a perfect physics world since there is no friction there will be no change in internal energy so you don’t have to take that into account In a perfect physics world energy will transfer between PE and KE

In the “Real” World In the real world there is friction so the internal energy of an object will be affected by the friction (such as air resistance)

Conservation of Energy Symbols ET=total energy of a system PE=potential energy KE=kinetic energy Q=internal energy ***all units are Joules (J)

Conservation of Energy Equations In a real world situation, ET=KE+PE+Q because friction exists and may cause an increase in the internal energy of an object In a “perfect physics world” ET=KE+PE with KE+PE equal to the total “mechanical energy of the system object

Conservation of Energy Examples (perfect physics world) position #1 position #2 position #3 more..

Conservation of Energy Examples (cont.) Position #1 Position #2 Position #3

Conservation of Energy Examples (cont.) Position #1 Position #2 Position #3

Conservation of Energy (perfect physics world) Position #1 the ball/bob has not starting falling yet so the total energy is all in gravitational potential energy Position #2 the ball/bob is halfway down, so total energy is split evenly between PE and KE Position #3 the ball/bob is at the end of its fall so total energy is all in KE

Conservation of Energy Problems 1.) A 2.0kg block starts at rest and then slides along a frictionless track. What is the speed of the block at point B? A 7.0m B

Answer Since there is no friction, Q does not need to be included So…use ET=PE+KE At position B, the total energy is entirely KE Since you cannot find KE directly, instead find PE at the beginning of the slide and that will be equal to KE at the end of the slide more…..

Answer (cont.) PE (at position A) =mgΔh=2.0kgx9.81m/s2x7.0m =137.3J KE (at position A) =0J because there is no speed So ET (at position A)=137.3J At position B there is no height so the PE is 0J More….

Regents Physics Answer (cont.) At position B the total energy still has to be 137.3J because energy is conserved and because there is no friction no energy was “lost” along the slide So….ET(position B)=137.3J=0J+KE So…KE also equals 137.3J at position B More…

Answer (cont.) Use KE=1/2mv2 So…KE=137.3J=1/2x2.0kgxv2 So v (at position B)=11.7m/s

More Conservation of Energy Problems position #1 position #2 position #3 1.) From what height must you drop the 0.5kg ball so that the it will be traveling at 25m/s at position #3(bottom of the fall)? 2.)How fast will it be traveling at position #2 (halfway down)? *Assume no friction

Answers 1.) At position #3, total energy will be all in KE because there is no height and no friction So…use ET=KE=1/2mv2 KE=1/2 x 0.5kg x (25m/s)2 So…KE=156.25J=PE (at position#1) So…ΔPE=156.25J=mgΔh And Δh=31.86m

Answers (cont.) 2.) Since position #2 is half way down total energy will be half in PE and half in KE So…KE at position #2 will be half that at position #3 So…KE at position #2 is 78.125J Then use KE (at #2)=78.125J =1/2 x 0.5kg x v2 v=17.68m/s at position #2

D. Energy of a Spring Energy stored in a spring is called “elastic potential energy” Energy is stored in a spring when the spring is stretched or compressed The work done to compress or stretch a spring becomes its elastic potential energy

Spring Symbols Fs=force applied to stretch or compress the spring in newtons (N) k=spring constant in (N/m) ***specific for each type of spring x=the change in length in the spring from the equilibrium position in meters (m)

Spring Equations Fs=kx PEs=1/2kx2

Spring Diagrams x

Spring Problems 1.) What is the potential energy stored in a spring that stretches 0.25m from equilibrium when a 2kg mass is hung from it? 2.) 100J of energy are stored when a spring is compressed 0.1m from equilibrium. What force was needed to compress the spring?

Answers 1.) Using PEs=1/2kx2 you know “x” but not “k” You can find “k” using Fs=kx With Fs equal to the weight of the hanging mass So… Fs=Fg=mg=2kgx9.81m/s2 Fs=19.62N=kx=k x 0.25m k=78.48N/m More…

Answers (cont.) Now use PEs=1/2kx2 PEs=1/2 x 78.48N/m x (0.25m)2 So PEs=2.45J

Answers (cont.) 2.) To find the force will use Fs=kx, but since you only know “x” you must find “k” also Use PEs=1/2kx2 to find “k” PEs=100J=1/2k(0.1m)2 k=20 000N/m use Fs=kx=20 000N/m x 0.1m Fs=2000N

Examples of Forms of Energy 1.) Thermal Energy is heat energy which is the KE possessed by the particles making up an object 2.) Internal Energy is the total PE and KE of the particles making up an object 3.) Nuclear Energy is the energy released by nuclear fission or fusion 4.) Electromagnetic Energy is the energy associated with electric and magnetic fields