I. Physics Skills



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I. Physics Skills

  • I. Physics Skills

  • II. Mechanics

  • III. Energy

  • IV. Electricity and Magnetism

  • V. Waves

  • VI. Modern Physics



A. Scientific Notation

  • A. Scientific Notation

  • B. Graphing

  • C. Significant Figures

  • D. Units

  • E. Prefixes

  • F. Estimation



Use for very large or very small numbers

  • 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



1. Write 365,000,000 in scientific notation

  • 1. Write 365,000,000 in scientific notation

  • 2. Write 0.000087 in scientific notation



1.) 3.65 x 108

  • 1.) 3.65 x 108

  • 2.) 8.7 x 10-5



Use graphs to make a “picture” of scientific data

  • 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. Does not have to go through all points. Just gives you the “trend” of the points

  • 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



1. Direct Relationship means an increase/decrease in one variable causes an increase/decrease in the other

  • 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

  • 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

  • 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.

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



Uncertainty in measurements is expressed by using 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



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)

  • 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)



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)

  • 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)



1. How many significant figures does 0.050900 contain?

  • 1. How many significant figures does 0.050900 contain?

  • 2. How many significant figures does 4800 contain?



1. 5 sig figs

  • 1. 5 sig figs

  • 2. 2 sig figs



1. Fundamental units are units that can’t be broken down

  • 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



Meter (m)– length, distance, displacement, height, radius, elongation or compression of a spring, amplitude, wavelength

  • 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



Meter per second (m/s)– speed, velocity

  • 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



Joule (J)– work, all types of energy

  • 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)



Ohm (Ω)– resistance

  • 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



Adding prefixes to base units makes them smaller or larger by powers of ten

  • 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



A terameter is 1012 meters, so… 4 Tm would be 4 000 000 000 000 meters

  • 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



A decigram is 10-1 gram, so… 15 dg would be 1.5 grams

  • 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



A microvolt is 10-6 volt, so… 8 µV would be 0.000 008 volt

  • 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



1.) 16 terameters would be how many meters?

  • 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?



1.) 16 000 000 000 000 meters

  • 1.) 16 000 000 000 000 meters

  • 2.) 2.500 grams

  • 3.) 1596 000 000 000 gigavolts

  • 4.) 0.000 000 687 amperes



You can estimate an answer to a problem by rounding the known information

  • 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



2 cans of Progresso soup are just about the mass of 1 kilogram

  • 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



1.) Which object weighs approximately one newton? Dime, paper clip, student, golf ball

  • 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



1.) golf ball

  • 1.) golf ball

  • 2.) 100m



A. Kinematics; vectors, velocity, acceleration

  • 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



In physics, quantities can be vector or scalar

  • 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

  • SCALAR quantities only have a magnitude and a unit

  • Example; 22m



VECTOR quantities; displacement, velocity, acceleration, force, weight, momentum, impulse, electric field strength

  • VECTOR quantities; displacement, velocity, acceleration, force, weight, momentum, impulse, electric field strength

  • SCALAR quantities; distance, mass, time, speed, work(energy), power



Distance is the entire pathway an object travels

  • Distance is the entire pathway an object travels

  • Displacement is the “shortest” pathway from the beginning to the end



1.) A student walks 12m due north and then 5m due east. What is the student’s resultant displacement? Distance?

  • 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?



1.) 13m (NE) for displacement

  • 1.) 13m (NE) for displacement

  • 17 m for distance

  • 2.) 20m (N) for displacement 80 m for distance



Speed is the distance an object moves in a unit of time

  • Speed is the distance an object moves in a unit of time

  • Velocity is the displacement of an object in a unit of time







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)

  • 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)



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





The time rate change of velocity is acceleration (how much you speed up or slow down in a unit of time)

  • 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







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)

  • 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)



First Method

  • First Method

  • vi=20m/s

  • vf=0m/s

  • a=2m/s2

  • Use vf2=vi2+2ad to find “d”

  • d=100m



Second Method

  • Second Method



In a vacuum (empty space), objects fall freely at the same rate

  • 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



To solve freefall problems use the constant acceleration equations

  • 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



1.) How far will an object near Earth’s surface fall in 5s?

  • 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?



1.)

  • 1.)



2.)

  • 2.)



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

  • 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



If you toss an object straight up, that is the opposite of freefall.

  • 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



1.) How fast do you have to toss a ball straight up if you want it reach a height of 20m?

  • 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?



1.)

  • 1.)



2.)

  • 2.)



The study of the effect of forces on objects at rest

  • 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)



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)

  • 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



Forces at 00

  • Forces at 00

  • Forces at 1800

  • Forces at 900



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



1.) 00

  • 1.) 00

  • 5N 5N = 10N



1.) 1800

  • 1.) 1800

  • 5N 5N = 0N



1.) 900

  • 1.) 900

  • 5N =

  • 5N 7.1N



The opposite of adding concurrent 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



To find the component forces of the resultant force

  • 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


Black arrow=resultant force

  • Black arrow=resultant force

  • Orange lines=reference lines

  • Green arrows=component forces

  • y

  • x





Equilibrium occurs when the net force acting on an object is zero

  • 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



An object in equilibrium can either be at rest or can be moving with constant (unchanging) velocity

  • 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



Black arrows=components

  • Black arrows=components

  • Blue arrow=resultant

  • Red arrow=equilibrant

  • =

  • +



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.

  • 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?



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.

  • 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.



2.)

  • 2.)



The study of how forces affect the motion of an object

  • The study of how forces affect the motion of an object

  • Use Newton’s Three Laws of Motion to describe Dynamics



Also called the law of inertia

  • 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”



“When an unbalanced (net) force acts on an object, that object accelerates in the direction of the force”

  • “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)



Fnet=the net force exerted on an object (the resultant of all forces on an object) in newtons (N)

  • 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



Also called law of action-reaction

  • 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)



A drawing (can be to scale) that shows all concurrent forces acting on an object

  • 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.



Fg is the force of gravity or weight of an object (always straight down)

  • 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



If object is moving with constant speed to the right….

  • If object is moving with constant speed to the right….

  • Black arrow=Ff

  • Green arrow=Fg

  • Yellow arrow=FN

  • Blue arrow=Fa



When an object is at rest or moving with constant speed on a slope, some things about the forces change and some don’t

  • 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……



Fa=Ff=Ax=Acosθ

  • 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



Green arrow=FN

  • Green arrow=FN

  • Red arrow=Fg

  • Black arrows=Fa and Ff

  • Orange dashes=Ay

  • Purple dashes=Ax



1.) Which has more inertia a 0.75kg pile of feathers or a 0.50kg pile of lead marbles?

  • 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?



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

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



2.)

  • 2.)



3.) A 10N book rests on a horizontal tabletop. What is the force of the tabletop on the book?

  • 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?



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



2.)

  • 2.)



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)

  • 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



1.) Projectiles fired horizontally

  • 1.) Projectiles fired horizontally

  • an example would be a baseball tossed straight horizontally away from you



Use the table below to solve these type of 2-D problems

  • Use the table below to solve these type of 2-D problems



2.) Projectiles fired at an angle

  • 2.) Projectiles fired at an angle

  • an example would be a soccer ball lofted into the air and then falling back onto the ground



Use the table below to solve these type of 2-D problems

  • Use the table below to solve these type of 2-D problems

  • Ax=Acosθ and Ay=Asinθ



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?

  • 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?



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





Use

  • Use



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?

  • 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?



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)

  • 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)





Find vertical “t” first using

  • 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



Find horizontal “d” using

  • Find horizontal “d” using



Find vertical “d” by using

  • Find vertical “d” by using



When an object moves with constant speed in a circular path

  • 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



Fc=centripetal force, (N)

  • 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)







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?

  • 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?



1.)

  • 1.)



Mass is the amount of matter in an object

  • 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 is measured in kilograms (kg)

  • 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 is measured in newtons (N)

  • 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





g=acceleration due to gravity but it is also equal to “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:



1.) If the distance between two masses is doubled, what happens to the gravitational force between them?

  • 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?



1.) Distance has an inverse squared relationship with the force of gravity.

  • 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”



2.) Mass has a direct relationship with Fg and distance has an inverse squared relationship with Fg.

  • 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”



3.) Determine the force of gravity between a 2kg and a 3kg object that are 5m apart.

  • 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?



3.)

  • 3.)

  • 4.)



The force that opposes motion measured in newtons (N)

  • 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



The ratio of the force of friction to the normal force (no unit, since newtons cancel out)

  • 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



The smaller the coefficient, the easier the surfaces slide over one another

  • 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



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?

  • 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?



1.)

  • 1.)

  • The friction force is kinetic because the cart is moving over the tabletop



Momentum is a vector quantity that is the product of mass and velocity (unit is kg.m/s)

  • 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)



p=momentum

  • p=momentum

  • Δp=change in momentum= (usually) m(vf-vi)

  • J=impulse



p=mv

  • p=mv

  • J=Ft

  • J=Δp

  • pbefore=pafter



1.) A 5.0kg cart at rest experiences a 10N.s (E) impulse. What is the cart’s velocity after the impulse?

  • 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?



1.) Use J=Δp so……

  • 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)



2.) Use pbefore=pafter

  • 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…..



So if p after collision of 0.2kg cart is 1kg.m/s(left) and

  • 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)



A. Work and Power

  • A. Work and Power

  • B. Potential and Kinetic Energy

  • C. Conservation of Energy

  • D. Energy of a Spring



Work is using energy to move an object a distance

  • 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



Power is the rate at which work is done so there is a “time” factor in power but not in work

  • 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



W=work in Joules (J)

  • 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)



W=Fd=ΔET

  • W=Fd=ΔET

  • ***When work is done vertically, “F” can be the weight of the object Fg=mg



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?

  • 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?



1.) to find work use W=Fd

  • 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



2.) To find power use

  • 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



3.) A 2.0kg object is raised vertically 0.25m. What is the work done raising it?

  • 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?



3.) to find work use W=Fd with F equal to the weight of the object

  • 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



4.) to find work use W=Fd

  • 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



Gravitational Potential Energy is energy of position above the earth

  • 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



ΔPE=change in gravitational potential energy in Joules (J)

  • 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



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?

  • 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?



1.) Use ΔPE=mgΔh to find potential energy gained so ΔPE=5.2kgx9.81m/s2x0.80m

  • 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



KE=kinetic energy in Joules (J)

  • KE=kinetic energy in Joules (J)

  • m=mass in kilograms (kg)

  • v=velocity or speed in (m/s)



1.) If the speed of a car is doubled, what happens to its kinetic energy?

  • 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?



1.) Using KE=1/2mv2 if v is doubled, because v if squared KE will be quadrupled.

  • 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



In a closed system the total amount of energy is conserved

  • 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



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 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 there is friction so the internal energy of an object will be affected by the friction (such as air resistance)

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



ET=total energy of a system

  • ET=total energy of a system

  • PE=potential energy

  • KE=kinetic energy

  • Q=internal energy

  • ***all units are Joules (J)



  • 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



  • position #1

  • position #2

  • position #3 more..





Position #1

  • Position #1

  • Position #2

  • Position #3



Position #1 the ball/bob has not starting falling yet so the total energy is all in gravitational potential energy

  • 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



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?

  • 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



Since there is no friction, Q does not need to be included

  • 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…..



PE (at position A) =mgΔh=2.0kgx9.81m/s2x7.0m =137.3J

  • 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….



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

  • 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…



Use KE=1/2mv2

  • Use KE=1/2mv2

  • So…KE=137.3J=1/2x2.0kgxv2

  • So v (at position B)=11.7m/s



  • position #1

  • position #2

  • position #3



1.) At position #3, total energy will be all in KE because there is no height and no friction

  • 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



2.) Since position #2 is half way down total energy will be half in PE and half in KE

  • 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



Energy stored in a spring is called “elastic potential energy”

  • 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



Fs=force applied to stretch or compress the spring in newtons (N)

  • 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)



Fs=kx

  • Fs=kx

  • PEs=1/2kx2





1.) What is the potential energy stored in a spring that stretches 0.25m from equilibrium when a 2kg mass is hung from it?

  • 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?



1.) Using PEs=1/2kx2 you know “x” but not “k”

  • 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…



Now use PEs=1/2kx2

  • Now use PEs=1/2kx2

  • PEs=1/2 x 78.48N/m x (0.25m)2

  • So PEs=2.45J



2.) To find the force will use Fs=kx, but since you only know “x” you must find “k” also

  • 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



1.) Thermal Energy is heat energy which is the KE possessed by the particles making up an object

  • 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



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