Transcript for:
Introductory Physics Concepts

in this introduction video we're going to cover some basic concepts in physics we're going to talk about unit conversion kinematics vectors projectile motion forces like newton's 3 laws circular motion working energy linear momentum and also rotational motion now we won't go too deep into each of these topics but we're just gonna go over these topics briefly just a nice simple brief overview and if you want more examples on a particular topic feel free to go to my channel and look for my physics playlist you should find a lot of practice problems for each of these different topics so let's go ahead and begin let's start with unit conversions now you need to know that one kilogram is equal to a thousand grams these are some common conversions that you need to know that you're going to use frequently when you're solving problems in physics one kilometer is equivalent to a thousand meters right now it might be wise just to take notes one mile is equal to 5280 feet one inch is equal to 2.54 centimeters and there's twelve inches in a foot and there's three feet in a single yard also one meter is equivalent to 100 centimeters now let's work on a practice problem let's say if we have a mass of 470 grams and we wish to convert it to kilograms how can we do so the first thing you want to do is write the conversion factor that connects grams to kilograms we know that a thousand grams is equivalent to one kilogram so always start with what you have so we have 470 grams of mass by the way the standard unit of mass is the kilogram so in a typical physics problem if you're solving an equation or using an equation and it has mass in it make sure that you plug in the mass in kilograms not in grams in the second fraction we're going to put the conversion factor in it since we have grams on top we need to make sure the unit grams goes on the bottom so that they will cancel and kilograms has to go on top so a thousand is associated with the number of grams and one is associated with the kilograms so if you set it this way you should get it right so it's going to be 470 divided by a thousand which is equal to 0.47 kilograms so that's how you can convert grams into kilograms so i'm just going to give you common examples that you'll see throughout physics as you solve problems in the future so let's try this one how would you convert centimeters to meters so feel free to pause the video and work on this particular example so first what is the conversion factor between centimeters and meters we know that 100 centimeters is equal to a single meter so let's start with what we're given that is 4.6 centimeters and let's convert it to meters so since we have the unit centimeters on top in the next fraction we're going to put centimeters on the bottom so that it cancels which means that we need to put meters on top the number that's in front of centimeters is 100 and the number that's in front of meters is one so to convert from centimeters to meters simply divide by a hundred so this is going to be point zero four six meters whenever you divide by a hundred simply move the decimal two units to the left go ahead and try this one how can we convert 25 kilometers per hour into meters per second so you might see a problem like this whenever you're dealing with speed and if you have the units kilometers per hour and you're using an equation that involves speed or velocity make sure you convert it to meters per second 99 of the time you'll need it in this unit so what can we do to convert it so there's two parts in this problem we need to convert kilometers into meters and we need to convert hours into seconds so let's go ahead and begin so we have 25 kilometers on top and one hour in the bottom because it's two units here so we already know the conversion between kilometers and meters one kilometer we know is equal to a thousand meters so let's put meters on top kilometers on the bottom so that these two units will cancel now what about hours into seconds well it turns out that one hour is equivalent to 60 minutes so we have the unit hours on the bottom we need to put that on the top so that these will cancel and we can put minutes on the bottom so one hour is equivalent to 60 minutes and it turns out that one minute is equivalent to 60 seconds so these two cancel so the only unit that we should have left is meters over seconds or meters per second and so now we can do the math so let's take out a calculator it's going to be 25 divided by a thousand wait i did something wrong the thousands should be on top not on the bottom it's very easy to make a mistake but it's always good to catch it one is in front of kilometers and a thousand is in front of meters so it should be 25 times a thousand divided by 60 and then take that result divided by 60 again so you should have 6.94 meters per second so this is the answer let's try another problem like that convert 50 miles per hour into meters per second so go ahead and work on this example so let's begin so we're going to have 50 miles on top and per one hour on the bottom so we know how to go from hours to minutes to seconds we did so in the last example but now let's focus on mouse we need to convert miles into meters what's the best way to do that what would you suggest we could go from browse to feet to inches to centimeters to meters to kilometers but that that's going to be a long wrap we have everything to do that but if you know the conversion between miles and kilometers that will save you a lot of time and then from kilometers you can go to meters it turns out that one kilometer if i recall correctly is about 0.6214 miles and also uh one mile is 1.609 kilometers so let's go ahead and use that conversion so one kilometer equals 0.6214 miles so these units will cancel now that we have kilometers we can convert it to meters we know that there's a thousand meters for every kilometer so now we have the desired unit meters so now let's focus on hours let's convert it to minutes one hour equals 60 minutes and one minute equals 60 seconds so the unit hours cancel and also minutes cancel so we're left with just meters on top seconds on the bottom so let's go ahead and get the answer so it's going to be 50 divided by 0.6214 times 1000 divided by 60 and then divide that result by 60 as well so the answer that i got is 22.35 meters per second let me just double check that make sure i typed in everything correctly and yeah i believe that's the answer so i don't think i made a mistake here and so this should be it now you know how to convert miles per hour into meters per second now let's talk about units of length area and volume distance for example is a unit of length it's like the meter is the typical unit of length for area it's going to be square meters and in the case of volume the volume is three-dimensional so it's cubic meters so let's say if you have feet area could be described as square feet volume cubic feet or you could measure units of length such as let's say centimeters this could be square centimeters in the case of area or cubic centimeters in the case of volume so let's say if you wish to convert from one unit of area to another how can we do so how can we convert 36 square feet into let's say square yards how can we do that what would you do in order to accomplish this uh conversion so always start with what you're given 36 square feet and then simply convert feet into yards now we know that one yard is equivalent to three feet so since we have the unit feet on top we need to put it on the bottom and so yards is going to go on top and there's three feet in one yard now notice that we have the square all we need to do is simply square the conversion factor so we're not going to divide 36 by 3 we're going to divide 36 by 3 squared and 3 squared is 9. 36 divided by nine is four so the answer is four yards squared so that's how you can convert from one unit of area to another so now what about volume so based on what you've seen in the last example try this one convert 288 cubic inches into cubic feet so we know that there's 12 inches in a foot and since we have inches on the top left we need to put it in the bottom right and all we need to do is raise the conversion factor to the third power so we have 288 and we're just going to divide it by 12 three times if we divide it by 12 the first time it's going to give us 24. if we divide 24 by 12 it will equal 2 and if we divide 2 by 12 it's 1 over 6 or 0.167 cubic feet so now you know how to convert from one unit of volume into another unit now the next topic that we need to talk about is the metric system perhaps you heard of terms like mega micro nano you need to know what these prefixes mean terra is associated with 10 to the 12. let's put some space between these two mega is associated with before omega this giga which is associated with 10 to the nine and then you have mega which is ten to the six and then there's kilo ten to the three and then hecto 10 to the 2 and then deca which is 10 to the first power so you need to know what this means so one terameter is equivalent to 1 times 10 to the 12 meters on the left you have the prefix on the right you have your base unit so let's say the base unit is liters 1 gigal is equal to 1 times ten to the nine of the base unit liters let's say if we have uh units of frequency hertz one megahertz is equal to one times ten to the sixth of the base unit hertz one kilometer is one times ten to the three meters ten to the three is basically one with three zeros or as you mentioned before one kilometer is a thousand meters ten to the six is one with six zeros so a megahertz is a million hertz a gigaliter is a billion liters so you need to be able to use this table to write a conversion factor now below deco which is 10 to the 1 we have our base unit and below that you have deci which is 10 to the minus 1 and then there's centi which is 10 to the minus 2 and then milli that's 10 to the minus 3. micro is ten to the negative six nano is ten to the negative nine pico is ten to minus twelve so these are the common values there are some other ones like ato femto and some other variations but i'm just gonna stop at pico so one milliliter is equal to one times ten to the minus three liters if you multiply both sides by a thousand you'll see that a thousand milliliters is equal to one liter one nanometer is equal to one times ten to the minus nine of the base unit meters so as you can see there's always a one on the left side and on the right side it's going to be 1 times whatever the multiply is of that base unit so make sure you know how to use this so you're able to write a conversion factor so now let's work on some examples let's convert 4.3 times 10 to the minus 4 millimeters and 2 micrometers so go ahead and pause the video take a minute to work on this example when converting from one unit to another in a metric system personally i find it's helpful to convert to the base unit and then convert to your desired unit it helps if you do that we know the conversion between millimeters and meters merely represents ten to the minus three so we could say that one millimeter is equal to one times ten to the minus three meters micro is ten to the negative six so one micrometer is one times ten to the minus six meters and these are the two conversion factors that we need in order to complete this problem so let's begin let's start with four point three times ten to the minus four millimeters now let's convert millimeters into the base unit meters so we need to put millimeters on the bottom so one millimeter is one times ten to the minus three meters so these units will cancel now that we have the base unit meters let's convert it to micrometers now since meters is on the top left we got to put meters on the bottom right 1 times 10 to the minus 6 meters is equivalent to one micrometer so at this point we just need to work on the math we really don't need a calculator for these types of problems now we know that one over x to the negative three is the same as x to the positive three if you've taken algebra at this point whenever you move a variable from the bottom to the top or vice versa the exponent will change sign in this case from negative three to positive three so we could take this number move it to the top the negative six will change into positive six so what we have is four point three times ten to the minus four times ten to the minus three times 10 to the positive 6. we can ignore the ones because when you multiply or divide by 1 the number does not change now x squared times x cubed is x to the fifth power according to the rules of algebra when you multiply by a common base you are allowed to add the exponents two plus three is five so here we can add these exponents negative four plus negative three is negative seven negative seven plus six is negative one so the final answer is four point three times ten to the minus one micrometers so let's go ahead and try another example try this one convert 2.6 times 10 to the 11 micrometers into kilometers so let's write out the conversion factors let's convert micrometers to meters and the meters to kilometers one micrometer is one times ten to the minus six meters and one kilometer is ten to the three meters or simply a thousand meters so let's begin with the unit that we have which is uh micrometers so one micrometer is 1 times 10 to the minus 6 meters so we can cross out those two units and now let's convert meters into kilometers we know that a thousand meters or let's divide it in scientific notation one times ten to the three meters is equivalent to one kilometer so we're going to do is take this number and move it to the top so it's 2.6 times 10 to the 11 times 10 to the minus six and then the positive three will change to negative three and then let's add the exponents so 11 plus negative 6 is positive 5 positive 5 plus negative 3 is positive 2. so it's 2.6 times 10 to the 2 kilometers 10 squared 10 times 10 that's 100 so 2.6 times 100 we can also say it's 260 kilometers now let's move on to a new topic let's talk about the difference between distance and displacement now let's say if a person travels 10 miles east the displacement is positive 10. initially and then let's say he travels 18 miles west the displacement for the second part is negative eighteen and then he travels let's say three miles east what is the displacement of the person and also what is the total distance traveled now imagine if we started from a number line and he started at position zero he traveled 10 meters to the right or 10 miles to the right so he's at position 10 and then 18 miles to the left or towards the west so 10 minus 18 that means he's at negative eight and then three miles east negative eight plus three that means he would end at negative five so he started at zero and he ended at negative five which means that the displacement of the person is negative five he's five miles west from where he started that's the displacement you can simply add these three values if you add ten negative eighteen positive three you're gonna get negative five so at the end of his trip he's five miles west from where he started that's the displacement the total distance traveled you simply add the three values you add ten and three and positive 18 instead of negative 18. so if you add these three values this will give you 31. so the total distance traveled is 31 miles but the displacement is negative 5. distance is a scalar quantity it only has a magnitude distance is always positive direction is not important for distance displacement is a vector quantity meaning that not only does it have magnitude but it also has direction so let's say if a person travels 12 miles north not only you have the magnitude which is the 12 miles but you also have the direction which is north so this represents displacement now let's say if a person travels only eight miles we don't know where he's going it could be east west south if you only have the magnitude then what you have described there is the distance not the displacement displacement is distance with direction now you need to be able to calculate average speed and average velocity average speed is equal to the total distance traveled divided by the total time average velocity is equal to the displacement divided by the total time so speed like distance is a scalar quantity speed is always positive direction is not important for speed velocity is a vector quantity it can be positive or negative so if a person is traveling east the velocity is positive if they're traveling west it's negative a car travels a hundred miles north in two hours and then 40 miles south in one hour calculate the total distance traveled the displacement average speed and average velocity so let's begin let's start with the total distance to find the total distance simply add up the distance of each segment of the trip so 100 plus 40 the total distance is simply 140 miles now what about the displacement by the way feel free to pause the video and work on this example so the car traveled 100 miles north so in the positive y direction the displacement is positive 100 then he turns around travels 40 miles south with a displacement of negative 40. so we need to add these two values 100 plus negative 40 is positive 60. so the net displacement of the car is 60 miles north now what about part c the average speed the average speed is equal to the total distance traveled divided by the total time the total distance traveled is 140 miles the total time is two hours plus an hour so that's three hours total 140 divided by 3 is about 46.7 miles per hour now what about the average velocity so the average velocity is equal to the displacement which is positive 60 divided by the total time of three hours so this is going to be positive 20 miles per hour so that's it now you know how to calculate average speed and average velocity now what about average acceleration what is the equation for that and what exactly is acceleration acceleration tells you how fast the velocity changes acceleration is equal to the change in velocity that is the final velocity minus the initial velocity divided by the time now let's say if we have if there's two vehicles a sports car and a truck the sports car can get to a speed of 60 miles per hour and so can the truck speed tells you how fast an object is moving but acceleration tells you how fast the speed changes the sports car can go from 0 to 60 miles per hour in a very short time it can probably do so within five seconds the truck is going to take a longer time to go from zero to sixty it might take ten or fifteen seconds so therefore the sports car has a greater acceleration than the truck even though they can both get up to the same speed the sports car can get to 60 miles per hour a lot faster than the truck and so it has a great acceleration and acceleration is basically the rate of change of velocity tells you how fast your velocity is changing now sometimes you may want to use this equation if you wish to calculate final speed use this so this equation comes from the acceleration equation go ahead and take a minute and try this problem a car speeds up from 10 meters per second to 40 meters per second in 5 seconds what is the acceleration of the car so let's answer the first question acceleration is the change in velocity divided by the change in time the final velocity or final speed in this problem is 40 meters per second the initial is 10 and the time is 5 seconds so the change in speed is 30 meters per second per 5 seconds so 30 divided by 5 the change in speed per second is 5 meters per second squared i mean 6 meters per second squared so that is the acceleration so what this tells us is that the speed is increasing by 6 meters per second every second now what about part b what is the final speed of a truck that accelerates at a rate of 1.5 meters per second squared starting from an initial speed of 20 meters per second for 8 seconds so let's make a list of the information that we have we have the acceleration we're given the time it's eight seconds and we also have the initial speed which is uh 20 meters per second so our goal is to find the final speed we can use this formula to calculate it so the initial speed is 20 the acceleration is 1.5 and the time is 8. now 1.5 times 8 is 12 and if we add 20 to it this is going to give us 32 so the final speed is 32 meters per second now let's make sense of that last part so we have an acceleration of 1.5 meters per second squared the initial speed we said was uh 20 meters per second and the time was 8 seconds and we calculated a final speed of 32 meters per second let's make a table at t equals zero the speed was 20. one second later what will the speed be now keep in mind the acceleration is 1.5 meters per second squared that means every second the speed will increase by 1.5 meters per second so one second later it's going to be 21.5 two seconds later 23. three seconds later 24.5 four seconds later it's going to be 26. every second it's going to go up by 1.5 5 seconds later it's 27.5 6 seconds later is 29 and then 7 seconds later 30.5 8 seconds later then it's 32 which is what we have here so as you can see the acceleration tells you how much the speed changes every second make sure you understand that general concept of acceleration that's what it is acceleration like velocity is the vector acceleration can be positive or negative whenever the acceleration is positive the velocity is increasing whenever the acceleration is negative the velocity is decreasing now let's go over a few kinematic equations that you need to know whenever the object is moving with constant speed that is the acceleration is zero there's only one equation that really applies displacement is equal to velocity multiplied by time now keep in mind whenever an object moves from let's say position a to position b and if it doesn't change direction if it's moving straight then distance and displacement are the same so you can use this equation to calculate the displacement or the distance the only time distance and displacement are not the same is if the object changes direction if it moves to the right and to the left or up down whenever changes direction that's when those two will vary but for a one-dimensional kinematic problem if you're asked to find the distance or displacement you can use this but technically speaking d represents the displacement it's just that the displacement equals the distance if the object doesn't change direction now let's say if the object moves with constant acceleration in this situation here are the equations that you can use so the final velocity is equal to the initial velocity plus the acceleration multiplied by the time the final velocity squared is equal to the initial velocity squared plus two times the acceleration times the displacement the displacement is equal to the initial velocity multiplied by the time plus one half a t squared and also the displacement is equal to the average velocity multiplied by time so this is similar to the first equation that we just mentioned under constant speed d equals v t but under constant acceleration you need to use the average speed the average speed is basically the sum of the initial and the final speed divided by two so then this equation becomes d is equal to one half v initial plus v final multiplied by t so those are the four kinematic equations that you need here's number one number two number three and number four so these two are the same now keep in mind the displacement is the change in position it's the final position minus the initial position it can be relative to the x-axis or relative to the y-axis a car speeds up from rest at a constant rate of 3.5 meters per second squared for 12 seconds how far does the car travel during this time so let's make a list of what we have we have the acceleration and the key is the unit the units of acceleration is meters per second squared so that's 3.5 and we have the time it's 12 seconds we also have the initial speed the fact that it speeds up from rest tells us that the initial speed is zero so what equation do we need that will help us to distance what equation has a t v initial and d so the equation that we need is this one v initial is zero so this whole term is zero and then everything else we could just plug in so half of 3.5 is 1.75 12 squared is 144 so we just got to multiply these two numbers and you should get a distance of 252 meters so now what equation can we use to find the final speed how fast it's moving after 12 seconds there's two equations that we can use we're going to use both of them first let's use the equation that we've used before the initial is zero the acceleration is 3.5 and t is 12. so using this equation we're going to get 42 meters per second now we can also use this equation let's say if we didn't know the time the initial is still zero the acceleration is 3.5 and the distance is 252. so 2 times 3.5 is 7 7 times 252 is 1764. and if you take the square root of that to get the final because we have a square here this will give you the same answer 42 meters per second so as you can see you have a lot of options a lot of different equations that you can use it helps if you make a list of the variables that you have and the variables that you need to find here's another problem that we could try a bus speeds up from 15 meters per second to 35 meters per second in 8 seconds how far does it travel during this time period so let's make a list of the things that we have the initial speed is 15 meters per second the final speed is 35 meters per second and the time is eight seconds so we need to find the distance that it travels we can use this equation d equals one half v initial plus v final times t so let's add the initial and the final speed and then we'll multiply by the time so 15 plus 35 is 50 and half of eight is equal to four and four times fifty is two hundred so it's going to travel two hundred meters now what about part b what is the acceleration of the bus now if you recall the acceleration is the change in speed divided by the time so the speed changes by 20. it goes from 15 to 35 so 20 divided by 8 is equal to 2.5 meters per second squared so that's the acceleration by the way feel free to check out my channel and look for my physics playlists if you want more examples on kinematic problems now let's talk about vectors adding and subtracting vectors let's say if we have a box and if we apply a force vector of 50 newtons towards the right a force is basically a push or pull action a force can be used to do work on an object now let's say if we apply a 50 newton force towards the right and a 30 newton force towards the left what is the net force the net force is basically the difference of these two forces it's 50 minus 30 or positive 20. so there's a net force of 20 newtons directed towards the right so whenever you're whenever you wish to add two vectors if the vectors are in the same direction you can just add them directly so in this case to add these two vectors we're just going to get a larger vector of 100 newtons now if you have two vectors that are antiparallel or that are opposite to each other then to add the two vectors simply subtract them this is going to be 100 minus 40. uh this is negative 40 because it's moving to the left and so the net force is positive 60. now how can we add two vectors if they're perpendicular to each other let's say this is a third unit in force and here we have a 40 newton force what is the resultant vector or the sum of these two vectors to draw it graphically if you place the second vector if you place the tail of the second vector to the head of the first vector the resultant vector is going to be right here start it from the tail of the first vector to the head of the second and if you draw in that direction that's the resultant vector so to find the magnitude of the resultant vector you basically have to find the hypotenuse of the triangle so it's going to be the square root of 30 squared plus 40 squared if you recall this is a this is b dot c a squared plus b squared is equal to c squared so c is the square root of a squared plus b squared the square root of 30 squared plus 40 squared is 50. so the hypotenuse so the resultant vector is 50. so let's say if we have a velocity vector v you can resolve it into components this is v x the x component of v and v y so if the velocity is moving in this direction it's not only moving to the right but it's also moving up at the same time now there are four equations that you need to know and here's the angle theta to find v x from v v x is equal to v cosine theta v y is v sine theta v is the square root of v x squared plus v y squared and to find theta it's equal to the inverse tangent of v y over v x now let's explain how to get these equations if you're not sure about it perhaps you heard of the term sohcahtoa chances are if you have taken trig you've seen this at some point the sole part refers to the sine function sine theta is equal to the opposite side divided by the hypotenuse so relative to this angle the opposite side is uh v y the adjacent side is v x the hypotenuse is v so therefore sine theta is equal to v y divided by v solving for v y just multiply both sides by v and so you get the equation for v y it's uh v sine theta now you could do the same thing for cosine cosine theta is equal to the adjacent side divided by the hypotenuse so it's v x over v and if you solve for v x you'll see that v x is equal to v cosine theta the last one is tangent tangent theta is equal to notice the o and a part so it's the opposite side divided by the adjacent side and the opposite side is v y the adjacent side is v x now sometimes it might be helpful to solve for theta so if you take the inverse tangent of both sides of the equation you're going to get this expression the inverse tangent of tangent cancels so theta is simply equal to the inverse tangent or the arctan function of v y over v x sometimes you may need to find the angle theta now we talked about adding vectors when they're parallel or antiparallel and also when they're perpendicular when they're perpendicular you can use the pythagorean theorem to get the answer but now what if they're not parallel or perpendicular how can we add two vectors so let's say if we have vector a and vector b how can we add them so the first thing you want to do is resolve a and b into their components you want to find ax and you want to find a y you want you also need to find b x and b y let's say r is the resultant vector or the sum of the two vectors once you have the components add all the x components so our x is going to be the sum of all the exponents if there's three vectors a b c our x is going to be a x plus b x plus c x our y is the sum of all the y components once you have r x and ry you can find the magnitude of the resultant vector which is the square root of rx squared plus ry squared and if you want to you can find the angle of that resultant vector using this expression so this is the process that you need to take whenever you're adding two vectors so let's work on an example let's say if this is vector a and it has a magnitude of 100 and vector b has a magnitude of 200 and let's say it's 30 degrees above the horizontal so first let's calculate ax and ay so now you need to know your angles this is 0 degrees 90 180 270 and 0 is the same as 360. so the angle that corresponds to a is zero degrees because it's on the positive x-axis it's directed towards the right b is at an angle of 30 degrees so ax is a hundred cosine zero cosine zero is one so ax is a hundred a y is a hundred sine zero sine zero is zero so a y is zero because a is in the x direction there is no y component now what about bx now bx is at an angle so bx i mean b has a an x component bx and b has a y component so to find the x component it's going to be 200 cosine 30 and make sure your calculator is in degree mode by the way so 200 cosine 30 is 173.2 and to find b y it's 200 sine 30 sine 30 is a half so this is going to be a hundred so now we can find rx simply add these two values and we could find ry which is a hundred now once you have rx and ry draw a new vector so our x since x is positive we need to draw towards the right y is positive so we need to go up so r is in this direction and we're going to find the angle theta so let's calculate r first it's the square root of rx squared so that's 273.2 squared plus ry squared so the magnitude of the resultant vector is 290.9 now to find the angle theta let's use this function it's the inverse tangent of the y component divided by the x component and that's 20.1 degrees let's try one more example so let's say if we have a force factor well let's say vector a is 200 directed towards the left vector b is a hundred direct itself and vector c is uh let's say 150 directed at an angle of 60 degrees above the x-axis so go ahead and add up these three vectors find the magnitude and the direction of the resultant vector so let's calculate ax and a1 now a is horizontal which means that it doesn't have a y component so a y is zero a is at an angle of 180 degrees keep this in mind this is 090 180 270. because a is directed towards the west it's at an angle of 180. if you type in 200 sine 180 in your calculator you're gonna get zero sine 180 is zero and make sure it's in degree mode now if you type in 200 cosine 180 you should get negative 200 because it's in a negative x direction so ax has to be negative now bx is zero because b is in the y direction there's no x component b y is negative 100 because it's directed in the negative y direction now c has an x component and it has a y component because it's at an angle so let's calculate c x and c y so to find c x it's going to be 150 times cosine 60 which is a 75 and it's positive 75 because as you can see cx is in the positive x-axis cy is also positive it's going up cy is going to be 150 sine 60 which is uh 129.9 so now we can find rx and ry so let's add up all the x components negative 200 plus 75 is negative 125 and if we add the y components negative 100 plus 129.9 is positive 29.9 so now we can find the magnitude of the resultant vector so it's going to be rx squared plus ry squared so 125 squared plus 29.9 squared once you square it the negative sign will disappear and you should get a positive result you should get 16 519 before you take the square root once you take the square root of that expression r is going to be 128.5 it's very close to this answer because this value is very small when it's too small becomes insignificant so now that we have the magnitude of the resultant vector let's find the angle so x is negative so let's draw a vector towards the left that's rx our y is positive so we need to go up so therefore r is in quadrant two now let's calculate theta which is the reference angle it's inverse tangent r y divided by rx but for now ignore any negative signs just put in positive numbers so inverse tan 29.9 divided by 125 will give you an angle of 13.5 degrees so that is the reference angle or the angle inside the triangle typically you want to find the angle measured from the positive x axis so if from here to the negative x-axis is 180 then the angle measured by the purple line is 180 minus 13.5 so it's really 166.5 degrees which is located in quadrant two so that's the angle of the resultant vector measured from the positive x-axis now let's spend some time talking about projectile motion here's the first type of problem that you might see so let's say if we have a ball on top of a cliff and it's moving horizontally and eventually it's going to fall and hit the ground h represents the height of the cliff and let's say d is that actually r is the range that it travels that's the horizontal distance that the ball travels now let's say if it takes five seconds for the ball to hit the ground how can you calculate the height of the cliff and also let's say if the ball moves with an initial speed of 30 meters per second how can you find where it lands relative to the base of the cliff for this type of problem there's only two equations that you really need h is equal to one half a t squared this equation comes from this equation d equals v initial t plus one half a t squared d represents the vertical displacement which is the same as the height v initial is really v y initial because the ball is moving horizontally the 30 meters per second represents v x anytime you have an object moving horizontally there is no y component so v y is zero therefore v y initial is zero everything in this equation is in the y direction not the x direction h is the vertical height or vertical displacement and that's equal to one half times the acceleration in the y direction times t squared now this object is in free fall it's a projectile a projectile is basically an object that is under the influence of gravity only and the acceleration due to gravity is 9.8 meters per second squared you can add the negative sign if you want so this is the one equation that you'll need the second equation comes from the fact that d equals v t now the horizontal velocity v x is not affected by the vertical acceleration a y the x and y components are independent of each other so v x is constant that's why we can use this equation since the ball is moving with constant speed in the x direction the horizontal displacement dx is the range so the range is equal to vxt so these are the two equations that you'll need in this type of problem so let's go ahead and solve it so to find the height simply use this equation h is equal to one-half ac squared the acceleration is negative 9.8 t is 5. half of 9.8 is 4.9 and 5 squared is 25 so 25 times 4.9 is 122.5 now the reason why we have a negative sign is because the ball travels in a negative y direction so the displacement is negative but if you just want to find the height of the cliff make it positive so the cliff is about 122.5 meters tall now what about the range the range is equal to vxt v x is 30 t is five so the ball is going to travel 150 meters towards the right so it's going to land 150 meters from the base of the cliff and so those are the main two equations that you'll need for this type of projectile motion problem let's try another problem like the one before so a ball falls off a 200 meter cliff and lands 300 meters away from the base the cliff what was the initial speed of the ball so let's begin by drawing a picture so we have the height of the cliff it's 200 meters and we know the range it lands uh 300 meters away how can we use this information to find the speed of the ball initially we know the ball was moving in a horizontal direction so our goal is to find v x and we can use this equation to do that but we got to find t first so let's use this equation to find the time so the height of the cliff is 200 and the acceleration in the y direction is 9.8 so 200 equals half of 9.8 which is 4.9 times t squared so let's divide 200 by 4.9 and that's a 40.816 then to solve for t take the square root of both sides so 6.389 is equal to t so the ball was in the air for 6.389 seconds now we can use this equation we have the range which is 300 and we have t which is 6.389 so vx is simply the distance that it traveled in the horizontal direction 300 meters divided by the time of 6.389 so vx is equal to 46.96 meters per second so that's how you can find the speed of the ball as it uh left the cliff another type of problem that you might see in regards to projectile motion is if a ball is kicked from a level surface it's going to go up and then it's going to go back down let's call this position a b and c so if you wish to calculate the maximum height of the ball you can use this equation h is equal to v squared sine squared theta divided by 2g theta is basically the angle of the ball relative to the ground so this is the angle theta v is the velocity vector this is not v x or v y this is v so here's v here's v x and this is v y so keep in mind v x is v cosine theta v y is equal to v sine theta now to calculate the range which is the horizontal displacement you can still use the equation that we mentioned before range is equal to v x t but you can also use this equation in terms of v and theta range is equal to v squared sine 2 theta divided by g so as long as you have the initial velocity of the ball and the angle theta you can find the maximum height of the ball and the range that it travels now what about the time how can we calculate the time it takes to go from position a to position b what equation can we use the time it takes to go from a to b is simply equal to v sine theta divided by g now if you want to find the total time it takes to go from a to c you simply need to multiply by two because the left side is exactly the same as the right side the graph is symmetrical at point b so this is going to be two v sine theta divided by g now sometimes you may need to use this equation this equation really comes from that equation at the top we need to realize that v y is equal to zero so let's talk about how to derive the equation on the bottom so we're dealing with the y direction so we can rewrite it as v y final is equal to v y initial plus a y t now going from a to b the final position point b the vertical velocity is zero and v y initial is simply v sine theta or v initial sine theta acceleration in the y direction is the same as g which is 9.8 so moving this to the other side we have negative v sine theta equals gt so we could divide by g so we get negative v sine theta divided by g which is equal to t and keep in mind g is really negative 9.8 so the two negative signs will cancel and ultimately time should always be positive so you could just get rid of the negative sign and that's how you can get this equation now let's talk about how to derive the other equations so we said that range is equal to v x t and v x we know it's v cosine theta and t the time it takes to go from position a to position c it's two times v sine theta divided by g so v x is v cosine theta and let's replace t with this expression let's get rid of this and just simply move this over here so v times v is equal to v squared and then we have 2 sine theta cosine theta if you have taken trig at this point you know that this is a double angle formula which you can also look it up on google sine 2 theta is equal to 2 sine theta cosine theta so therefore we have this equation so the range is equal to v squared sine 2 theta divided by g so now you know how to derive that equation now let's see if we can come up with an equation that describes the height so let's see if we can derive this equation so let's draw a picture so let's travel from position a to position b so the height is basically the displacement so what equation will help us to calculate the displacement what would you say here's one equation that we can use here's another equation v final squared is equal to v initial squared plus 2ad we can replace d with h but which equation should we use now we don't have time in this equation t is not there so we probably shouldn't use this one so let's try using this equation so we know that v y final is zero at point b so this disappears so we have zero is equal to v initial squared plus two a d and a is the same as g and d is basically the height that's the vertical displacement so let's solve for g so we have negative v initial squared is equal to 2gh so let's divide by 2g oh by the way there's supposed to be a 2 here i forgot about that so this is going to be so we have this equation now keep in mind v y initial this is supposed to be a y is equal to v sine theta so this is really negative v sine theta squared over 2g which is the same as if you distribute the 2 you're going to get v squared sine squared over 2g now you can ignore the negative sign if you want the height to be positive but that's how you can derive this equation now let's go over the same trajectory that we had in the last example so let's say the ball goes up and then it goes back down now the acceleration due to gravity is always 9.8 but to keep things simple we're going to round it and use a value of 10. so we're going to say g is 10 meters per second squared just for this particular example now let's say that the initial vertical speed in the y direction is 30. so that's v y and let's say v x is five meters per second what is v x and v y one second later now the velocity vector is directed in the upward direction that's v y and the acceleration is opposite to it whenever velocity and acceleration are in the opposite direction the velocity will decrease the acceleration tells you how much the velocity changes every second so every second is going to decrease by 10. so one second later v y is going to be 10 less than what it was so it's now 20. vx is constant and you need to understand that for any projectile motion problem where it's only where the object is only affected by gravity v x is constant and v y changes by whatever the amount of g is in a typical problem it changes 9.8 meters per second every second for this example we're going to round it to 10. so one second later v y is going to be 10. v x is going to be five still at the top v y is equal to zero v x is still five but i'm not gonna write it anymore because i'm running out of space now one second after that v y is going to be negative 10 another second later v y is now negative 20 and when it reaches back to the ground just before it hits the ground v y is going to be negative 30. so as you can see the velocity of the ball is the same at the same height so these two are at the same height my drawn is not perfect whenever the height is the same the velocity of the ball will be the same well not the velocity i should say the speed because here the velocity is negative but here it's positive but the magnitudes are the same now notice that when the ball is going upward that is in the first half of the graph the velocity is decreasing anytime the acceleration is negative the velocity will always be decreasing even when it's going down as you can see the velocity is still decreasing because the acceleration is negative so on the left side and the right side is decreasing now what about the speed on the left side is it speeding up or slowing down whenever the ball is going up it's going to be slowing down the speed is decreasing now what about when it's going down it turns up that i meant to say it turns out that it's a speeding up and when it's going down keep in mind velocity can be positive or negative but speed is always positive so here the speed is positive 10 and then 20 and then 30 notice that the speed is increasing as it goes down so it's speeding up so while it's going up actually before we talk about that here's something that you need to know whenever velocity and the acceleration have the same signs that is if they're both positive or if they're both negative the object is speeding up whenever velocity and acceleration have opposite signs let's say this is positive and that's negative or if this is negative and that's positive the object is slowing down so now let's analyze it while the ball was traveling upward the velocity was positive the velocity was decreasing but it was still positive it was going from 30 to 20 to 10 and the acceleration was negative in free fall the acceleration is always negative is always directed towards the ground so because these two have opposite signs the ball was slowing down the speed was decreasing now on the right side when the ball was going down the velocity was negative it was still decreasing but it was negative the acceleration remained negative and because these two have the same signs the ball was speeding up so anytime velocity and acceleration have the same sign the object is speeding up but whenever it has opposite signs it's slowing down now keep in mind the examples that we went over in projectile motion we're simply scratching the surface there's a lot more problems that you'll need to know how to solve in a typical physics course so just check out my channel and just look for my physics playlist and then you can find a video on projectile motion with a lot more examples so be sure to check that but now let's move on into forces you need to be familiar with newton's three laws his first law states that an object in motion will continue in motion unless acted on by net force and an object at rest will remain at rest unless acted on by a force so if you have a box it's just going to sit there unless you apply a force on that box is not going to move now imagine if you have an object that is already moving on a frictionless surface if there's no forces acting on that object it will continue to move forever now in real life there's friction so due to friction friction opposes motion it's a force that slows things down so if you were to roll a ball on the table it's gonna move for a significant amount of time but eventually it's going to come to a stop the less friction you have the longer it's going to keep rolling if you roll a ball on a rough surface it's going to come to a stop quickly let's say if you try to roll a ball on carpet it's not going to roll very far but now if you roll it on a smooth surface it's going to roll for a longer distance likewise let's say if you hit a puck on ice let's say on top of a lake that's frozen there's not much friction between the puck and the ice so it's going to slide for a very very very long time until it comes to a stop now there's no friction it can move forever so for example an object in space there's basically almost no air in space so if you have an object that's moving in space it will continue to move forever if there's no other forces acting on it now what about newton's second law his second law is basically an equation f is equal to ma the net force is equal to the mass multiplied by the acceleration now if the force is constant if you increase the mass the acceleration will decrease these two are inversely related when the force is constant if you increase the acceleration assuming the mass is constant the force will increase and if you increase the mass under constant acceleration the force will increase so force is directly related to mass and acceleration but when the force is constant the mass and acceleration are inversely related now this another law newton's third law which states that for every action force there is an equal and opposite reaction force basically f of a is equal to negative fb so let's say if you have a smaller person and a larger person and let's say they push each other let's say this person applies a force of 100 newtons what force will the other person apply to the smaller person according to newton's third law the forces are the same they're equal and opposite in direction now let's say if we have an astronaut in space and he throws a ball away from himself so let's say he throws the ball towards the right now while in contact let's say he applies a force of 200 newtons on the ball what force does the ball exert on him during contact according to newton's third law the forces have to be the same they have to be equal but opposite in direction so in space if he throws the ball forward he's going to feel a force that's going to propel him backward now let's say that the mass of the astronaut with his suit is 100 kilograms and let's say the mass of the ball is only let's say two kilograms what is the acceleration on the person and on the ball now if you recall f is equal to ma so the acceleration if we solve for it it's the force divided by the mass so we have a force of 200 and the mass of two so 200 divided by two is a hundred so the acceleration on a ball is a hundred now what about the acceleration on the person if we take the force and divided by the mass 200 divided by 100 is two so even though the same force is being exerted on the astronaut and the ball the acceleration is not the same because the ball has less mass and therefore less inertia it's going to move further away the ball is going to have a greater velocity due to the large acceleration that was exerted upon it so the ball is going to fly further than the person because the person has more mass the acceleration is less so remember we said that whenever the force is the same the person or the object that has more mass will have a smaller acceleration these two are inversely related so the person he's going to move backwards but not as much as the ball as it moves forward so make sure you understand this concept now let's talk about friction we know that friction opposes motion and there's two types of frictions that you need to be familiar with imagine if you have a large box on a rough carpet and you try to push the box initially as you apply a force the box won't move it's going to sit there and as you increase the force eventually the box will begin to move against the rough carpet now once you get it going it's easy to keep it going why is that why is it easier to keep the box moving than to get it started to start moving it you've seen this effect but have you ever wondered why now there's two types of frictions that you need to be familiar with static friction which occurs when the surfaces between the the box and the carpet are not sliding past each other so when the box is not moving you have static friction when a person applies a force keep in mind if the person applies a force and the box doesn't move that means that there's another force that opposes the motion that prevents it from moving and that is static friction now once you get it to move static friction no longer applies another force is present and that's kinetic friction and the reason why it's easier to keep a box moving on a rough carpet as opposed to getting it started is because static friction is usually greater than kinetic fiction i haven't seen a case where it's not greater so for the most part it's greater than kinetic friction to calculate static friction it's always less than or equal to the coefficient of static friction times the normal force the kinetic friction is equal to the coefficient of kinetic friction times the normal force we'll talk about the normal force later but now let's say that static friction can be less than or equal to 100 newtons and let's say that kinetic friction is equal to 50. now let's say f is the applied force lowercase f represents uh static friction let's make a table so you can understand this concept so f is the applied force f s static friction f of k kinetic friction and let's say f net the net force so if we apply a force of zero what are the values for f of s f k and then that force if you don't apply any force there's not going to be any friction and so the net force is zero now what if we apply a force of 20 what are the values will the box begin to move if we apply a force of 20 it turns out that it will not move until we exceed the maximum static frictional force it's not going to move so what is f of s then well let me tell you what f of s is not f of s is not a hundred now if you chose this value let's think about what this means if the static frictional force was 100 and if the applied force was 20 that means that there would be a net force of 80. that means that static friction would actually push the box towards the left as you try to push it towards the right and that's not going to happen so static friction is going to match the applied force until you exceed 100 so it's going to be 20. so if you push the box with a force of 20 static friction pushes back on you with a force of 20. so the net force is 0 it doesn't move if you apply a force of 50 f of s is going to be 50 the box is not going to move if you apply a force of 100 static friction will still be 100 and it's not going to move now what happens if you exceed the static frictional limit let's say if you go past 100 what's going to happen if you apply a force of 120 the box will begin to slide against the carpet and once it slides static friction is no longer present so it's zero now kinetic friction takes over and it's always 50. so now the net force is a difference of 70. so if you increase it to let's say 200 f of s is still zero the box is sliding f k will remain 50. notice that we don't have an inequality sign it's always equal to 50 and so the net force is 150. so static friction is the force that prevents the object from moving and until you exceed the maximum static frictional force it won't slide now once you exceed it the box begins to slide and then kinetic friction applies now let's say if you have a 10 kilogram block what is the weight of the block what's the difference between mass and weight mass represents the quantity of matter whereas weight represents the force of gravity acting on that object for example the weight of an object on the earth as opposed to the moon are not the same it's different now the mass is the same the 10 kilogram object is going to have the exact same quantity of matter on the earth as on the moon however the weight will be different weight is equal to mg it's the mass times the gravitational acceleration now even though the mass is the same the gravitational acceleration on the earth is different from that on the moon on the earth g is about 9.8 on the moon it's 1.6 so the weight of the object on the earth is 10 times 9.8 or it's 98 newtons weight is a force so it's measured in newtons on the moon it's going to be 10 times 1.6 which is about 16 newtons so on the moon you will feel a lot lighter perhaps you've seen astronauts in the moon they could jump pretty high it's because the force of gravity is very weak on the moon compared to the earth so make sure you understand the difference between the weight force and mass now let's say if we have a a 10 kilogram block on a table now we know that there's a downward wave force that the block exerts on a table and that wave force we set it to 98 newtons now if the block is remaining at rest that means that the net force has to be zero which implies there's another force that keeps the object from falling down what is that force called the table or the surface exerts a force of 98 newtons on a block these forces have to be the same because the block is not accelerating upward or downward it's at rest whenever an object is at rest and if it's remaining at rest then that force has to be zero so these two forces have to balance each other so the force exerted by the table on the block is known as the normal force the normal force is always directed perpendicular to the surface now how can we increase or decrease the value of the normal force without changing the mass of the object what would you say now let's say g is 10 just to keep the math simple and let's say we have a 20 kilogram block what is the weight force now the weight force is going to be 20 times 10 which is about 200 newtons the normal force at this point is also 200 newtons but what's going to happen if we apply a downward force of 80 newtons on a block will that increase or decrease the normal force so before we applied the downward force of 80 the ground had to exert a normal force of 200 newtons to keep the block at rest now that we've applied a downward force of 80 the ground has to do more work not only does it have to support the weight of the block but it must supply a force to counteract the downward 80 newton force so the normal force exerted by the surface is now 280. so that's how you can increase the normal force exerted by the ground on a block is by applying a downward force on a block now what's going to happen if we try to lift up the block let's say with a rope so let's say we apply a force of 150 newtons to lift up the block now the block is not going to go up because unless we apply a force that's greater than the weight force of the block it's going to remain on the surface by the way the tension force is a force that acts through a rope or a cord that's all it is so if the tension force is 150 and the weight force is 200 what's the normal force the normal force is going to be less it's only going to be 50 newtons the block exerts a downward force of 200 newtons but you're lifting it with 150 so there's a net downward force of 50 and the surface is going to apply a force that's going to counteract that net downward force now what if we have an incline what is the normal force acting on a block when it's on an incline the normal force is always perpendicular to the surface so it's directed in this direction on an incline the normal force is equal to mg cosine theta where theta is the angle of the incline relative to the horizontal on a flat surface if we don't apply a downward force and if we don't lift up the box with an upward tension force the normal force is simply equal to the weight force and we know that the weight force is mg so the normal force is equal to mg on a flat surface with no other forces acting in the y direction except these two but on an incline it's mg cosine theta now there is a component of the force of gravity that accelerates the block downward and i'm going to call it fg this value is equal to mg but times sine theta so gravity is going to accelerate it downward and if friction is present if the block is moving you can use f of k f k is mu k times the normal force and you know how to find the normal force using the equations that you've learned so far how can we answer these questions go ahead and try it feel free to pause the video so we have an incline and then we have a surface containing friction so the box it slides down the incline and the distance of the incline is 10 meters and we need to find the acceleration of the box as it slides down the incline and the angle of the incline is 30 degrees so how can we find the acceleration of the box now there's no friction on this surface so the only force that accelerates it downward is the force fg and we know that fg is equal to mg sine theta and since it's the only force actin on the box in that direction that means it's equal to the net force the net force is always equal to m a so m a is equal to mg sine theta we can cancel m so the acceleration down the incline is simply g sine theta g is 9.8 and theta is 30. sine 30 is a half half of 9.8 is 4.9 so that's the acceleration of the block down the incline it's 4.9 meters per second squared now what about part b what is the final speed of the box as it reaches the bottom of the incline so let's think about what we know it accelerates from rest meaning that the initial speed is zero we're looking for the final speed we have the acceleration and we have the distance what equation has v final venus show a and d so if you go back to your list of kinematics equation you notice that v final squared is equal to v initial squared plus 2ad so v initial is zero a is positive 4.9 and d is 10. so 2 times 4.9 is 9.8 times 10 that's 98 and the square root of 98 is 9.9 so that is the speed of the box as it reaches the bottom of the incline now what about part c so once the box reaches the bottom of the incline once it's at this position it's going to have a speed of 9.9 meters per second now at this point it encounters a rough surface so this friction and we have the coefficient of kinetic friction it's point twenty how can we use this information to calculate how far it will travel before it comes to rest now during the second part of the problem the speed of 9.9 is now the initial speed when the box comes to rest the final speed is zero and our goal is to find d so if we can find the acceleration we can find d using the same kinematic equation that we just used so let's calculate friction once it reaches the bottom of the incline there's no fg there's no gravitational force that accelerates it downward on the bottom of the incline the only force acting on the box is friction there's no force that's accelerating it friction is going to slow it down bringing it to rest so therefore we could say the net force is equal to fk net force according to newton's second law is always equal to ma and we know that the frictional force on a horizontal surface with no other vertical forces applied to it is equal to mu k times the normal force and the normal force on a horizontal surface is simply mg so we can cancel m so the acceleration on this surface is simply mu k times g or 0.20 times 9.8 0.20 times 9.8 is about 1.96 so that's the acceleration and it should be negative 1.96 because it's slowing the object down it's bringing it to rest the object is decelerating and not accelerating so now with the acceleration we can now use this equation so we know v final is zero v initial is 9.9 a is negative 1.96 and d we're looking for it so we need to move this term to the left side so let's make some space so this point is just math 9.9 squared is about 98. on the left side is going to be negative 98 2 times 1.96 is 3.92 so negative 98 divided by negative 3.92 is equal to 25. so that's how far the box is going to travel once it encounters the horizontal surface or friction so don't forget the old kinematic equations it's going to be needed throughout your course of physics now let's talk about energy there's three forms of energy that you're going to encounter in physics kinetic energy potential energy and mechanical energy anytime an object is moving if it has some sort of speed or velocity component it has kinetic energy kinetic energy is energy in motion it's equal to one half times the mass times the velocity squared so here's a question for you if you double the mass of the object what effect will it have on the kinetic energy if you see a question like this all you got to do is plug in the number for m and everything else replace it with one so the one half just put a one for it v plug in one if you double the mass it's two to the first power the kinetic energy will double if you double the speed plug in two into v everything else just put a one two squared is four so if you double the speed the kinetic energy will quadruple if you triple the speed 3 squared is 9 the kinetic energy will increase by a factor of 9. now anytime an object has the ability to fall it has gravitational potential energy and the greater the height difference between the object and the surface the more potential energy it has gravitational potential energy is equal to mgh is proportional to the mass the gravitational acceleration and the height so potential energy is a form of stored energy whereas kinetic energy is energy in motion the mechanical energy is the sum of the kinetic and the potential energy whenever you have an object that is in free fall and that is only under the influence of let's say gravity gravity is a conservative force gravity can slow down the object when it's going up and it can increase the speed of the object when it's going down so gravity is a conservative force in the sense that it doesn't change the total mechanical energy of the ball as it travels when the ball goes up it's slowing down which means that the kinetic energy is decreasing however because the height is increasing because it's moving away from the surface the potential energy increases such that the mechanical energy is constant as the ball falls down it speeds up and so the kinetic energy increases but the height is decreasing so the potential energy is decreasing so as it falls down potential energy is converted into kinetic energy such that the mechanical energy is constant so gravity is a conservative force it keeps the mechanical energy constant now conservative force doesn't really depend on the path taken so regardless if the ball goes like this or like that gravity will still have the same effect friction and let's say an applied force those are non-conservative forces friction will always oppose motion so friction always act in such a way to decrease the total mechanical energy of a system an applied force can be used to increase the mechanical energy or decrease it if the applied force is in the same direction as the motion of the object the mechanical energy will go up if the applied force is opposite to the motion of the object the object slows down and the mechanical energy will decrease so whenever the force and the velocity are in the same direction the kinetic energy will increase the object will speed up and whenever force and velocity if they're in opposite directions the kinetic energy will decrease the object will slow down to transfer energy into or out of a system there's two ways you can do it heat and work we're not going to talk about heat that's for like thermodynamics but let's talk about work whenever you apply a force on an object you can increase or decrease the kinetic energy of the object you can also increase its potential energy if the force is used to increase the height of the object if you're lifting it above ground level work is equal to the force multiplied by the displacement so if the displacement vector and the force vector are in the same direction work is positive the kinetic energy will increase if the displacement vector and the force vector are opposite to each other the work performed on the object is negative the kinetic energy will decrease so keep that in mind not only is work equal to force times displacement but work is equal to the change in kinetic energy that is is equal to the final kinetic energy minus the initial kinetic energy and it's also equal to the negative change in potential energy this is the work energy theorem now let's say if we have a box on a surface and we have a rope and we're pulling the rope with a force and let's say that force is at an angle and the object is moving this way so this force has an x component and it has a y component so the x component is called f of x and the y component f y if the displacement vector and if the force vector are perpendicular to each other no work is done the only time maximum work can be done is if the displacement vector and the force vector are parallel to each other in such a case work is equal to fd so in this problem we have a force that is not parallel to the displacement vector f is not parallel to d however f x is parallel to d so therefore the work is equal to f x times d and we know f of x is f cosine theta so therefore relative to this force the work is going to be f d cosine beta now let's see if we can come up with a problem that connects work and kinetic energy so take a minute and try this problem so we have a box and we're applying a force of 100 newtons and we're going to do so over a distance of 10 meters so what is the work done on the box so work is equal to force times distance that's 100 newtons times 10 meters so the work done on the box is equal to 1000 joules now what about part b what is the acceleration of the box now nothing was said about friction so we're going to assume that there's no friction on this surface so the only force is the applied force which means that it's also equal to the net force and the net force is equal to ma so we have a net applied force of 100 newtons a mass of 5 kilograms so 100 divided by 5 is 20 which is the acceleration now what is the final speed of the box we can go back to our kinematic equation v final squared is equal to v initial squared plus 2ad we're going to assume that the box started from rest since no initial speed was given the acceleration is 20 d is 10 so 2 times 20 is 40 times 10 that's 400 and the square root of 400 is 20. so the final speed is 20 meters per second now let's calculate the final kinetic energy of the box the initial kinetic energy is zero because the box started from rest but to find the final kinetic energy it's going to be one-half mv squared using the final speed so the mass is 5 and the speed is 20. so 20 squared we know it's 400 400 times five is two thousand and half of two thousand is a thousand so as we can see the work is equal to the change in kinetic energy the kinetic energy increased from zero to a thousand because the work is equal to a thousand if the initial kinetic energy was a hundred the final kinetic energy should be eleven hundred the difference between a two will remain a thousand a box slides down a 20 meter tall frictionless incline starting from rest what is the speed of the box when it reaches the bottom of the incline so here's the incline and here's the box so the height of the incline is 20 meters not the length of the incline but the height and our goal is to find the speed of the box once it reaches the bottom of the incline what we can do is solve it using conservation of energy so initially the only form of energy that we had was potential energy the object was moving and at the bottom it no longer has the ability to fall so it doesn't have any gravitational potential energy it only has kinetic energy at this point the gravitational potential energy is equal to mgh the kinetic energy is one-half mv squared so notice that we don't need the mass of the object if we multiply both sides by two we're going to get the equation 2gh is equal to v squared so if we take the square root of both sides the final speed is going to be the square root of 2 times gh so let's go ahead and solve it so that's the square root of 2 times 9.8 times the height of 20. so 2 times 9.8 times 20 is 392 and the square root of 392 is 19.8 meters per second so that's the final speed of the box as it slides down the incline or once it reaches the bottom of the incline now let's spend some time talking about circular motion but i'm going to focus mostly on the concepts and equations instead of word problems now as you mentioned before whenever the force and the velocity vector are moving in the same direction the object is speeding up and when the force and velocity vector are in opposite directions the object is slowing down now what's going to happen if the force and the velocity vector are perpendicular let's say if the force is directed north and the velocity is directed e so what's going to happen in this case the object is going to turn it's going to turn in the direction of the force now that the velocity is directed north what's going to happen if the force is directed west well it's going to turn towards the west and if the force is directed towards the south the object will turn towards the south so if the force is always directed towards the center the object will remain in a circle so that's we have circular motion this force the center seeking force is known as the centripetal force the centripetal force is not a force itself it's provided by or caused by other forces the centripetal acceleration is equal to v squared divided by r and since f is equal to m a according to newton's second law the centripetal force is mv squared divided by r now whenever an object moves around a circle you can also calculate the velocity this way we know that d is equal to v t for an object moving with constant speed so v is d over t the distance around a circle is known as the circumference which is two pi r and the time it takes to complete one revolution around the circle is known as the period represented by capital t the frequency which represents the number of cycles traveled per second is one divided by the period one revolution is equal to two pi radians which is a form of angle measure and that's also 360 degrees now here's a question for you consider the earth and the moon orbiting the earth the moon behaves as a satellite around the earth now what keeps the moon in orbit in the earth what would you say chances are you'll say gravity gravity is a force that brings matter together so the gravitational force between the earth and the moon pulls them toward each other now if that's the case what prevents the moon from crashing into the earth since it's being pulled by the earth now the moon has inertia and it has a tendency to continue moving towards outer space right now it's like just flying towards outer space however the earth pulls it toward itself and as we recall whenever you have a force vector and a velocity vector that are perpendicular to each other the object is not going to move directly in the direction of the force neither is it going to move in the direction of v instead it turns it's going to move this way and at this point the moon is going to have a velocity vector that is directed towards the west it's going to feel a gravitational force that pulls it south towards the earth and as a result it's going to continue to turn so these two things imbalances each other the tangential velocity of the moon and the gravitational force between the moon and the earth and that keeps the moon in orbit so in this case gravity provides the centripetal force for planetary motion the force of gravity between two objects is g m1 m2 over r squared so in our example we have the mass of the earth the mass of the moon divided by r squared r is the distance between the center of the earth and the center of the moon it's not the distance between the surface of the earth and the surface of the moon now how can we find the speed of the moon or of any satellite orbit in the earth to do that we need to set the gravitational force equal to the centripetal force the gravitational force between the earth and the moon is gm times m over r squared now we know the centripetal force is m v squared over r but which m is this is it the mass of the moon or the earth now which one is moving in a circle in this picture the moon orbits the earth the earth does not orbit the moon so this is the mass of the moon and v represents the speed of the moon or the speed of the satellite orbiting the earth so we can cancel this particular m and we can cancel an r so we're going to have this equation so we have left over is g m divided by r is equal to v squared so the speed of a satellite orbiting any planet is the gravitational constant times the mass of the center planet divided by the orbital radius that is the distance between the center of the planet and the center of the satellite so that's how you can calculate the speed of a satellite now what about the gravitational acceleration of a planet we know that the weight force is equal to mg the weight force is the same as the gravitational force by the way and the gravitational force is g m1 m2 divided by r squared so if we separate one of the masses we have m times g times m over r squared so these two are the same the weight force is equivalent to the gravitational force between two objects these two masses are the same therefore the gravitational acceleration is equal to this quantity so the gravitational acceleration of the earth is g times m over r squared and let's go ahead and calculate that answer we know it's 9.8 if you do some research you'll see that the gravitational constant is 6.67 times 10 to the minus 11. you can look it up on google the mass of the earth is 5.98 times 10 to the 24. and the radius of the earth if i recall it's 6.38 times 10 to the 6 meters so let's go ahead and plug this data into the equation and you should get 9.799 which is approximately 9.8 meters per second squared so this is the gravitational acceleration of the earth at the surface of the earth so g depends on the mass of the planet and the radius of the planet if you increase the mass of the planet the gravitational acceleration will increase if you move away from planet earth if you increase the r value g will decrease so as you move further away from a planet the gravitational acceleration will decrease so g is not always 9.8 it's 9.8 on the surface of the earth but if you move far into outer space like very far away from earth then g will decrease in value now let's say if we have a car that is making a circular turn which tells us that there must be some sort of centripetal force that causes it to turn in a circle and the centripetal force always points towards the center of the circle so what provides the centripetal force we know it's not gravity so what is it in this case whenever a car makes a turn you need to know that friction provides the centripetal force now i'm going to leave it up to you to research it and as to how that works but make sure you just know that and it's really static friction that creates it static friction is equal to mu s times the normal force that's the maximum static frictional value and the centripetal force is mv squared over r now if we don't have a bank curve if the road is leveled even though the car is making a turn the normal force is equal to mg so in a problem like this where you have a car making a turn sometimes they may ask you what is the maximum speed that the car can safely make the turn set these two equal to each other sometimes they won't give you the mass because as you can see m cancels so the equation that you need is this the coefficient of static friction times the gravitational acceleration is equal to the max speed divided by r so solving for v will give you the maximum velocity or speed that the car can have to safely make a turn without skidding if it goes too fast the car may not be able to turn instead it can skid off the road which is not good now what if we have a rope and a ball attached to the rope and it's spinning at a very high speed what provides the centripetal force that keeps it moving in a circle whenever you're dealing with a rope tension is the force that acts on a rope so tension provides the centripetal force the tension force for a ball moving in a horizontal circle is approximately equal to mv squared of r now what about a vertical circle instead of a horizontal circle what's the tangent force at points a b and c is the tension greater at a b or c at a the tension is going to be at its maximum because not only do you have to support the weight of the ball but the rope must also lift the ball to its new position so going from a to b the ball has to lift it up i mean the rope has to lift up the ball at point b and d the tension force is approximately equal to the centripetal force at point a the tension force is at a maximum it's equal to the sum of the centripetal force and the weight force at point c is the lightest the tension force is the difference between the centripetal force and the wave force now what about if we have a horizontal circle but if the ball is not moving very fast you can try this if you take a ball attached to a rope if you spin it slowly and horizontally you'll see that it's going to be at an angle let me draw this better if you spin it fast it's going to be it's going to appear almost horizontal but if you spin it slowly the ball will be at an angle and let's say this is the angle theta so t is the tension force t y is the y component of the tension force and this is the x component so technically speaking t x is equal to the centripetal force tx provides the centripetal force because you can see it's pointing towards the center t y supports the weight of the object otherwise the object will fall so t y equals mg now this equation becomes important if the object is not moving fast enough if it's moving fast enough tx is going to be significantly larger than t y such that t y is insignificant and t x will be approximately t so if v is large t x is approximately equal to t and that's why we could say t is approximately m v squared over r only when it's moving fast if it's moving slow then you have to find tx and ty separately so if you just want to find a centripetal force use this equation now keep in mind if you need to calculate t use this t is t x squared plus t y squared so if you want to find a tension force use the speed to calculate t x use the weight of the object to find t y then once you have t x and t y you can plug it into this equation to find t and if you need to find the angle use this equation inverse tangent t y divided by t x now if you want more practice problems on circular motion just search my video on it and you can find it on youtube now let's move on to momentum momentum is basically mass in motion an object that has mass and an object that is moving has momentum momentum is represented by lowercase p is the product of mass and velocity so like velocity momentum is a vector if an object is moving towards the right the momentum is positive if it's moving towards the left the momentum is negative the next equation that you need to know is the impulse momentum theorem impulse i'm going to use the symbol i for impulse impulse is equal to force multiplied by time so it's newtons times seconds momentum we said was p is equal to mv so the units for momentum is kilograms times meters per second it turns out that the impulse is equal to the change in momentum so you have this equation so let's say if we have a wall and there's a two kilogram ball it's moving at a speed of 30 meters per second it strikes the wall and it bounces back with an equal speed of 30 meters per second so and let's say the contact time that is the time where the ball and the wall is in contact let's say it's 0.1 seconds what is the momentum of the ball before it strike the wall so momentum is mass times velocity 2 times 30 so it's positive 60. now what's the momentum after it collides with the wall notice that the velocity is negative 30 because it's moving to the left so the momentum is 2 times negative 30 or negative 60. now how can we calculate the average force exerted by the wall on the ball the wall exerted a force towards the left because the object changed direction and went from moving from the right to the left so we can use this equation to figure out the answer the impulse momentum theorem so the change in time is 0.1 seconds the mass is 2 and the change in velocity the final velocity is negative 30 the initial velocity is 30. so final minus initial negative 30 minus the initial of positive 30. so that's the change of negative 60. so it's 2 times negative 60 which is a change of 120 and as you can see the momentum changed by negative 120. it went from positive 60 to negative 16. so f times 0.1 is equal to the change momentum of negative 120. negative 120 divided by 0.1 is equal to negative 1200 newtons so the force is negative because it's directed towards the left and so that's how you can find the average force exerted by an object now that equation comes from newton's second law f is equal to m a and if recall the acceleration is the final velocity divided by initial velocity over the time which is really the change in velocity divided by the change in time so if you replace a with delta v divided by delta t all you need to do is multiply both sides by delta t so that these two will cancel and now we have the impulse momentum theorem so the force multiplied by the time is equal to the change in momentum so now let's think about that so f multiplied by delta t is equal to the change in p so if we divide both sides by delta t we can understand the definition of force so force can also be defined as the rate of change of the momentum so it's the change in momentum divided by the change in time now let's say that we have two objects object one and object two and they collide what is the effects on the total momentum of the system before the collision and after the collision what would you say now there's two types of collisions an inelastic collision and an elastic collision for an inelastic collision kinetic energy is not conserved however momentum is conserved for any collision so the total momentum before is equal to the total momentum after the system that is the momentum of object one plus the momentum of object two is equal to the momentum of object one after the collision so that's v1 final plus the momentum of object 2 after the collision so this little apostrophe here tells you that it represents the final speed after the collision now for elastic collisions momentum is still conserved so this first equation still applies for elastic collisions so you can use it for any collision however you can also use another equation for these type of problems you may have to solve it using a system of equations here's the second equation v1 plus v1 final is equal to v2 plus v2 final this equation comes from the conservation of kinetic energy and you can only use it during elastic collisions so for an elastic collision you can use both of these two equations but for an inelastic collision you can only use the equation below so here's a problem let's see if there's a box at rest and it's a 10 kilogram box and a bullet is fired at the box now the bullet has a of 0.01 kilograms but it's moving at a very high speed let's say 500 meters per second once the bullet strikes the box the bullet is embedded in the box and they move together typically when two objects collide and if they stick together it's usually an inelastic collision if they bounce off most of the time it represents an elastic collision what is the final speed of the bullet block system after the collision so we can use this equation m1v1 plus m2 v2 is equal to m1v1 prime plus m2 v2 prime however since the bullet and the block move at the same speed because they move together as one unit these two are the same so we can just write m one v final plus m two v final or if we factor out v final we could say it's just m one plus m two times v final so let's go ahead and calculate the final speed so the mass of the bullet is 0.01 the speed is 500 the block is initially at rest so it has no momentum the total mass is 0.01 plus 10 or 10.01 so the final speed is going to be 0.01 times 500 which is 5 divided by 10.01 so the final speed of the bullet and the block is going to be point 5 meters per second it's really 0.4995 but i'm going to round it to 0.5 now i'm briefly going to review rotational motion i won't go too deep into it just some equations and some simple concepts imagine this is the top view of a door if we apply a force on its door the door is going to rotate and this action created by the force is known as a torque the torque is the force multiplied by the lever arm the lever arm is the distance between where you apply the force and the axis of rotation so torque is equal to f times l now if there's an angle it's also equal to f times l times sine theta now there are some other equations to know what is the difference between linear speed and rotational speed linear speed tells you how fast an object is moving forward rotational speed tells you how fast an object is spinning or rotated rotational speed has the symbol omega you can use rotational speed or velocity but typically just rotational speed now all of the kinematic equations that you've learned before for translational motion that's like linear motion is similar to rotation of motion for example i'm going to make two columns linear and rotational at constant speed we know that d is equal to vt for rotation of motion distance is represented by theta theta is the angular displacement this is linear displacement so theta is equal to omega times t v is the linear speed omega is the rotational speed also known as the angular speed now we know that v final equals v initial plus a t whenever you have constant acceleration omega final is equal to omega initial plus alpha t a is the linear or tangential acceleration alpha is the angular acceleration and then there's this equation v final squared is equal to v initial squared plus 2ad and here you have another similar equation so you need to know the counterparts in each of these equations and there's also this one d is equal to one half the initial plus v final times t and here omega is one half i mean i mean theta excuse me is one half omega initial plus omega final times t now there are some more equations also this one d is equal to v initial t plus one half a t squared and so the angular displacement is going to be omega initial plus t plus one half alpha t squared now we know that uh work is equal to force times displacement rotational work is equal to torque times angular displacement linear momentum is mass times velocity angular momentum is inertia times angular velocity or angular speed now just as alpha is the rotational equivalent of a you have this equation linear acceleration is alpha times r so alpha is the rotational equivalent of a omega is the rotational equivalent of v so v is equal to w times r and it turns out that theta is the rotational equivalent of d displacement is equal to angular displacement times r those are some other equations to know but interestingly l is equal to m v times r if you like rearrange the equations now you might be wondering what is i i represents inertia and inertia is different depending on the types of shapes that you have it changes for instance for a disc the inertia is one half m r squared where r is the radius of the disk and m is the mass for a sphere the inertia is i believe two over five m r squared so the inertia doesn't just depend on the mass only but it depends on how the mass is distributed and there are some more equations that you need to know for inertia but those are the two common ones for a disc and for a sphere so these are the main equations that you'll see in rotational motion and there's also our energy kinetic energy is one half mv squared rotational kinetic energy is one half inertia times omega squared as you can see these two are similar so inertia you can view it as the rotational equivalent of mass just as omega is the rotational equivalent of linear speed so i'm going to stop here for today just want to give you an intro into rotation of motion and so that's it for this video so keep in mind if you need to find more physics topics or examples and practice problems just check out my channel you'll find my physics playlist on it so if you really appreciate this video feel free to comment below like it as well feel free to subscribe to my channel share with your friends or if you like you can also donate at my channel page so thanks again for watching and have a great day and i wish you all on any exams that you have to take