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1 2012 ppt semester 1 review and tutorial 2
1 2012 ppt semester 1 review and tutorial 2
1 2012 ppt semester 1 review and tutorial 2
1 2012 ppt semester 1 review and tutorial 2
1 2012 ppt semester 1 review and tutorial 2
1 2012 ppt semester 1 review and tutorial 2
1 2012 ppt semester 1 review and tutorial 2
1 2012 ppt semester 1 review and tutorial 2
1 2012 ppt semester 1 review and tutorial 2
1 2012 ppt semester 1 review and tutorial 2
1 2012 ppt semester 1 review and tutorial 2
1 2012 ppt semester 1 review and tutorial 2
1 2012 ppt semester 1 review and tutorial 2
1 2012 ppt semester 1 review and tutorial 2
1 2012 ppt semester 1 review and tutorial 2
1 2012 ppt semester 1 review and tutorial 2
1 2012 ppt semester 1 review and tutorial 2
1 2012 ppt semester 1 review and tutorial 2
1 2012 ppt semester 1 review and tutorial 2
1 2012 ppt semester 1 review and tutorial 2
1 2012 ppt semester 1 review and tutorial 2
1 2012 ppt semester 1 review and tutorial 2
1 2012 ppt semester 1 review and tutorial 2
1 2012 ppt semester 1 review and tutorial 2
1 2012 ppt semester 1 review and tutorial 2
1 2012 ppt semester 1 review and tutorial 2
1 2012 ppt semester 1 review and tutorial 2
1 2012 ppt semester 1 review and tutorial 2
1 2012 ppt semester 1 review and tutorial 2
1 2012 ppt semester 1 review and tutorial 2
1 2012 ppt semester 1 review and tutorial 2
1 2012 ppt semester 1 review and tutorial 2
1 2012 ppt semester 1 review and tutorial 2
1 2012 ppt semester 1 review and tutorial 2
1 2012 ppt semester 1 review and tutorial 2
1 2012 ppt semester 1 review and tutorial 2
1 2012 ppt semester 1 review and tutorial 2
1 2012 ppt semester 1 review and tutorial 2
1 2012 ppt semester 1 review and tutorial 2
1 2012 ppt semester 1 review and tutorial 2
1 2012 ppt semester 1 review and tutorial 2
1 2012 ppt semester 1 review and tutorial 2
1 2012 ppt semester 1 review and tutorial 2
1 2012 ppt semester 1 review and tutorial 2
1 2012 ppt semester 1 review and tutorial 2
1 2012 ppt semester 1 review and tutorial 2
1 2012 ppt semester 1 review and tutorial 2
1 2012 ppt semester 1 review and tutorial 2
1 2012 ppt semester 1 review and tutorial 2
1 2012 ppt semester 1 review and tutorial 2
1 2012 ppt semester 1 review and tutorial 2
1 2012 ppt semester 1 review and tutorial 2
1 2012 ppt semester 1 review and tutorial 2
1 2012 ppt semester 1 review and tutorial 2
1 2012 ppt semester 1 review and tutorial 2
1 2012 ppt semester 1 review and tutorial 2
1 2012 ppt semester 1 review and tutorial 2
1 2012 ppt semester 1 review and tutorial 2
1 2012 ppt semester 1 review and tutorial 2
1 2012 ppt semester 1 review and tutorial 2
1 2012 ppt semester 1 review and tutorial 2
1 2012 ppt semester 1 review and tutorial 2
1 2012 ppt semester 1 review and tutorial 2
1 2012 ppt semester 1 review and tutorial 2
1 2012 ppt semester 1 review and tutorial 2
1 2012 ppt semester 1 review and tutorial 2
1 2012 ppt semester 1 review and tutorial 2
1 2012 ppt semester 1 review and tutorial 2
1 2012 ppt semester 1 review and tutorial 2
1 2012 ppt semester 1 review and tutorial 2
1 2012 ppt semester 1 review and tutorial 2
1 2012 ppt semester 1 review and tutorial 2
1 2012 ppt semester 1 review and tutorial 2
1 2012 ppt semester 1 review and tutorial 2
1 2012 ppt semester 1 review and tutorial 2
1 2012 ppt semester 1 review and tutorial 2
1 2012 ppt semester 1 review and tutorial 2
1 2012 ppt semester 1 review and tutorial 2
1 2012 ppt semester 1 review and tutorial 2
1 2012 ppt semester 1 review and tutorial 2
1 2012 ppt semester 1 review and tutorial 2
1 2012 ppt semester 1 review and tutorial 2
1 2012 ppt semester 1 review and tutorial 2
1 2012 ppt semester 1 review and tutorial 2
1 2012 ppt semester 1 review and tutorial 2
1 2012 ppt semester 1 review and tutorial 2
1 2012 ppt semester 1 review and tutorial 2
1 2012 ppt semester 1 review and tutorial 2
1 2012 ppt semester 1 review and tutorial 2
1 2012 ppt semester 1 review and tutorial 2
1 2012 ppt semester 1 review and tutorial 2
1 2012 ppt semester 1 review and tutorial 2
1 2012 ppt semester 1 review and tutorial 2
1 2012 ppt semester 1 review and tutorial 2
1 2012 ppt semester 1 review and tutorial 2
1 2012 ppt semester 1 review and tutorial 2
1 2012 ppt semester 1 review and tutorial 2
1 2012 ppt semester 1 review and tutorial 2
1 2012 ppt semester 1 review and tutorial 2
1 2012 ppt semester 1 review and tutorial 2
1 2012 ppt semester 1 review and tutorial 2
1 2012 ppt semester 1 review and tutorial 2
1 2012 ppt semester 1 review and tutorial 2
1 2012 ppt semester 1 review and tutorial 2
1 2012 ppt semester 1 review and tutorial 2
1 2012 ppt semester 1 review and tutorial 2
1 2012 ppt semester 1 review and tutorial 2
1 2012 ppt semester 1 review and tutorial 2
1 2012 ppt semester 1 review and tutorial 2
1 2012 ppt semester 1 review and tutorial 2
1 2012 ppt semester 1 review and tutorial 2
1 2012 ppt semester 1 review and tutorial 2
1 2012 ppt semester 1 review and tutorial 2
1 2012 ppt semester 1 review and tutorial 2
1 2012 ppt semester 1 review and tutorial 2
1 2012 ppt semester 1 review and tutorial 2
1 2012 ppt semester 1 review and tutorial 2
1 2012 ppt semester 1 review and tutorial 2
1 2012 ppt semester 1 review and tutorial 2
1 2012 ppt semester 1 review and tutorial 2
1 2012 ppt semester 1 review and tutorial 2
1 2012 ppt semester 1 review and tutorial 2
1 2012 ppt semester 1 review and tutorial 2
1 2012 ppt semester 1 review and tutorial 2
1 2012 ppt semester 1 review and tutorial 2
1 2012 ppt semester 1 review and tutorial 2
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1 2012 ppt semester 1 review and tutorial 2

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  • Welcome to the physics review for semester one. Hello everybody, this is Dr. Fiala. The tutorial you are about to begin is comprehensive. Material reviewed includes Kinematics, force, projectiles, momentum and impulse. To receive the most benefit from this video there are two important points that should be noted.
  • Point one. I have color coded the slides so you can more easily practice skills with which you have difficulty. This is a good way for you to roll up your sweatpants. Word problems are presented on red slides. Data tables can be found on olive slides. Green slides work you graph reading skills and tan slides match concepts. Solutions are presented for each problem on the following slide. I will give you an audio cue that the slide will be changing. That way if you want to stop the video before you see the solution you can do so at that time. You may choose to view the entire video or simply move to the slide color representing your weakness.
  • Point two. You must have all of the formulas handy while going though the semester on physics review. If you have them written down in my order it would be helpful because I refer to the formulas as formula one, formula two, formula three and so on according to the order in which they appear on my card. Because all students must use the formula given out on final exam day, it is important that you are skilled at finding any variable in any formula. You can stop this video at any time and I encourage you do do so. I suggest that you complete this video review and tutorial when you can think about what we are talking about without distraction. Stay positive . You can do this!
  • Vi&f = 13.67 m/s ti = 0 s tf = 7.32 s Xi = 0 m a = 0 m/s2
    Xf=
  • So now that we have the problem KUEd we can look at the index card and choose an equation. The first equation assumes a constant velocity which we have. We know the change in time is 7.32 seconds, so we can use that formula. The second, third, and fifth formulas do not have final position in so they are of no use to us. The seventh equation assumes a velocity change and since there is no acceleration, it cannot be used here.
  • We can use the fourth, and sixth formulas to find final horizontal position with the information we are given. You should try them all to practice your math skills. The most simple equation to use is the first one. Manipulating the first equation gives us final position equals constant velocity times time.
  • Although Harley Davidson motorcycles are made in Milwaukee, Wisconsin, they still come equipped with a speedometer that registers speed in both the English and the metric systems of measurement. When I look down I see I am traveling at 49.21 km/h. How far will I travel in that same 7.32 second time period?
  • The answer lies in the ability to convert within the metric system using dimensional analysis. An hour has 60 minutes in it so in essence I am multiplying my first fraction by one therefore not changing the relative value of the 49.21 only its units. I repeated the process because I know that there are 60 seconds in each minute. The units I have now is km/s so I need to change the kilometers to meters. There are 1000 meters in a kilometer. When all the multiplication is done I find out that 49.21 km/h is equal to 13.67 m/s. That means I travel exactly the same distance, 100.07 meters. If you did not remember there are 1000 meters in a kilometer, review the pneumonic I gave you at the beginning of the year. Many kids have dropped over dead converting metrics. I have placed a one under the K for kilometers. To convert to meters on the chart I need to move three places to my right. Meters is located at 0 for over. All metric system base units, meters, liters, and grams, are located at 0. So simply place a zero in for each place moved to the right and you convert one to one thousand.
  • vi = 0 m/s ti = 0 s tf = 8 s a = 4 m/s2 Xi = 0 m
    Xf = vf
  • Looks like the beast is going to be the winner. The first three formulas and the fifth formula do not contain final position in them so they are of no use to us. We cannot use the fourth formula because we do not know final velocity right now. It does not mean we cannot currently find final velocity, but there is no need to do that work if we can find the final position without it.
  • To improve math skills it is important to recognize the symbols we use to represent physics quantities. Examples of this are ΔX to represent a change in position, and lower case ‘a’ to represent acceleration. However it is truly important to understand what each of those physics quantities mean. The first step to this understanding is to take the time to learn the definition of each physics quantity.
  • As we learn new physics quantities write them down. Study them by simply reading them after school. You will improve your math skills if you do.
    Then go over their meaning when asked to find that quantity during a quiz or exam. For example if I am asked to find the velocity of an object that travels from 1 m to 3 m in 3 seconds, it might be nice to remember that velocity is the movement of an object in a specific direction over time. The word over means divided by so in this case the definition gives you the formula of the movement of three meters divided by 3 seconds is the velocity.
  • One method of completing a data table is to find the pattern within the numbers for each object’s position. Since this is a data table of constant velocity, I know that whatever displacement occurs over one second is the same displacement that occurs for every other second thereafter. So the pattern for object number one is sixteen, thirty-two, forty-eight, and sixty-four. We can also think mathematically here. Since we have a data table of position change and time, we can apply the formula velocity equals position change divided by time. Still looking at the numbers for object number one found in column number one, we can take 48 meters and divide them by 3 seconds which is the time it took object one to travel 48 meters to find that object #1 has a constant velocity of 16 m/s. That means that the value in each row of the object #1 column increases by 16 each time.
  • Check your answers. I think you will find that you are better at this than you had originally thought you were.
  • Lets try this again only this time with constant acceleration. If we apply a constant acceleration we would expect the velocity to change over time. In fact, that is a shortened definition of acceleration. Again look for the pattern or apply a formula when you know initial velocity, final velocity, and time. If we look at object #3 we see a change of 1 m/s in 1 second. That means that during each second the object increases its velocity by 1 m/s.
  • This data table was a little harder only because there were some uneven numbers. For some of you that caused you to pause a little. Remember we are learning procedures here. To do the same thing, the same way, every time. If you stick to the procedure barring calculator error you should get the correct answer each time!
  • Lets take a look at our first graph of the review. Eventually you should be able to shut off the sound on this video and talk your way through the graph as we have learned to do in class. Okay, here we go. Horizontal position is plotted on the y-axis and is measured in meters. Time is plotted on the x-axis and is measure in seconds.
  • The y-intercept indicates an object started 3 meters to the left of the reference point. There is no x-intercept. The area (rise times run) of this graph is not useful to us because its units are meters times seconds. The slope of this graph represents the velocity of the object and is measure in m/s. As you can see there are four different slopes on this graph. Slope is equal to rise divided by run. To find the magnitude of the rise of the first slope we take our final position, which is -6 meters and subtract our initial position, which was -2 meters. The magnitude of the rise is -4 meters. To find the magnitude of the run of the first slope we take our final time, which is 2 seconds and subtract our initial time, which is 0 seconds. The magnitude of the run is 2 seconds. So the magnitude of the first slope is -4 meters divided by 2 seconds or -2 m/s. Again we are learning procedure so find the remaining three slopes using the same procedure.
  • So now we can analyze a kinematic graph for all of its useful information. Remember to keep reading slopes out loud. The first slope on this graph would be read as negative (because of its downward slope) constant (because it’s a straight line) velocity (because that is the physics quantity this slope represents). In this case we also know its magnitude is -2 m/s. Reading the remaining slopes out loud yields constant neutral which means 0 m/s, constant positive 1 m/s, followed by constant neutral velocity.
  • Try reading this position graph of a projectile on your own. Remember the six important factors when reading kinematic line graphs. Identify the quantities plotted on the x and y axis. Identify if there is an x or y- intercepts and what those intercepts represent. Identify the physics quantity of the slope by its units. You get the units of slope by dividing rise units by run units. Identify the physics quantity of the area by its units. You get the units of area by multiplying rise units by run units. Don’t worry about the area of a parabola. Remember the slope of this parabola is expressed by Yf = Yi + Vi t + ½ gt2. As you know, I often refer to this equation as the beast. Where can we get more information from this graph? What do you know about the 120 meter point? What is the time at that point? What is the velocity at this point? Good for you if you are KUEing the information as we go.
  • The 120 meter point is important for several reasons. We recognize that the 120 meter point is the maximum vertical position (ΔY) for the object. We know that time to reach the object’s maximum height is 5 seconds. With that information we can apply the beast which is the formula that describes the slope to find initial velocity. This calculation yields 48.5 m/s. We also know that at its maximum height a projectiles vertical velocity (Vf) is 0 m/s. Using the formula final velocity equals initial velocity plus gravity times time we find the initial velocity to be 48.5 m/s. So as you can see graph reading yields much information.
  • The vertical position graph of this projectile is parabolic. That means there will be more than one slope. Reading that kind of graph is not a problem because we simply read one slope at a time. With that in mind reading the first slope of the position graph out loud reveals a positive changing velocity from fast to slow.
  • Since this is positive changing velocity the slope of the velocity graph will have to be in the 1st quadrant and will be sloping towards the x-axis. I know that because nothing is slower than 0 m/s and 0 m/s is found on the x –axis. The object is slowing down while traveling to the right. This leaves us with two possibilities. Lets see if reading the second slope on the position graph sheds any light on which of these two choices is correct. The second slope is read as negative changing velocity from slow to fast.
  • This means the correct choice can only be the graph on the bottom right. Now I can read the slope made by those two slopes put together. Here I indicated it by a yellow arrow. I would read this out loud as constant because it is a straight line, negative because it has a downward slope, acceleration because that is the quantity that is measured by the slope of a velocity versus time graph.
  • The only graph that has a slope shape and orientation that reads as constant negative acceleration is the center graph.
  • The negative acceleration is due to gravity, which works against the velocity that is taking the projectile away from the Earth until there is no positive velocity remaining. Since the only force acting on a projectile is due to gravity, the projectile immediately begins to return to Earth with increasing velocity. Since the object is falling we agreed to refer to the velocity as negative. When velocity changes from positive to negative, in other words crosses the x-axis, the object has changed directions. Whenever an object changes directions the position graph is parabolic. Whenever the position graph is parabolic there is acceleration present. Remember that acceleration causes the shape of the position graph’s slope to be curved.
  • Vi = 0 m/s Xi = 0 m Xf = 1530 m a = 2.25 m/s2 ti = 0 s
    Vf= tf =
  • You cannot use formula one, two, four, five, and six because you do not know the final time, You cannot use formula three because you do not know average velocity. That leaves the last and most ignored of the kinematic formulas.
  • Vi = 0 m/s Xi = 0 m Xf = 1530 m a = 2.25 m/s2 ti = 0 s
    tf= Vf =
  • Here you again have a bunch of ways to solve the problem. Once again I encourage you to practice each of these formulas. The more comfortable you become with using the formulas the easier physics becomes as well. The first formula is for constant velocity which could be provided by using the third formula first but there is no reason to make this a two step problem!
  • Please pause if you want to practice your graph reading before I begin mine. You may want to pause after each segment. Okay, here goes…Horizontal velocity is plotted on the y-axis and is measured in meters per second. Time is plotted on the x-axis and is measure in seconds. The y-intercept indicates an object had an initial velocity of 15 m/s. There is no x-intercept. The units for the area (rise times run) of this graph is meters, indicating the change in position of the object. The area is a rectangle so we simply multiply base times height to find it. 15 m/s times 2 seconds is 30m. Thirty meters is the objects change of position in 2 seconds. The slope of this graph represents the acceleration of the object and is measure in m/s2. As you can see there are four different slopes and therefore four different areas on this graph. Slope is equal to rise divided by run. To find the magnitude of the rise of the first slope we take our final velocity, which is 15 m/s and subtract our initial velocity, which was 15 m/s. The magnitude of the rise is 0 m/s. To find the magnitude of the run of the first slope we take our final time, which is 2 seconds and subtract our initial time, which is 0 second. The magnitude of the run is 2 seconds. So the magnitude of the first slope is 0 m/s divided by 2 seconds or 0 m/s2. Again we are learning procedure so find the remaining three slopes using the same procedure.
  • Check your answers to see how successful you were at graph reading.
  • Notice first of all that we are reading a velocity graph and predicting what the position graph will look like. So look at the velocity graph on top of the page and match one of the position graphs to it. For the first two seconds the object has constant positive velocity. This does not rule out any position graphs since they are all in the first quadrant at the two second mark. From two seconds to five second the object is still moving with positive velocity. The value at the five second mark is still positive. Now if we look back at the position graph choices we can see that the graph on the upper right and lower right cannot be correct. Both of those show a negative slop between two and five seconds indicating movement to the left. The next slope of the velocity graph shows no change in the positive velocity so the position graph on the lower left can be ruled out.
  • The corresponding position graph would be read as constant positive velocity followed by positive changing velocity from fast to slow, followed by positive constant velocity followed by positive changing velocity from slow to fast.
  • Vector arrows can be used to represent any of the vectors we have studied so far, displacement, velocity, acceleration, force, and momentum. When we use the vectors arrows as a motion map we are showing the movement of an object. The arrow heads indicate direction of motion and the length of the arrows indicate the relative amount of motion. So this motion map illustrates an object moving to the left first slowing down then speeding back up. We could also say the object went from fast to slow to fast. Lets first look at all of the position graphs that fit this motion.
  • The object is moving to the left. So I circled the only graphs that show movement to the left if it is read as a position graph. Now consider that the object is changing velocity according to the changing length of the motion map arrows. Fast to slow to fast. That means the slope of the position graph has be a curved line. A straight line would mean constant velocity but having acceleration automatically means changing velocity. Fast to slow to fast. The slope of an object moving at the speed of light would be vertical. So something moving fast would have a more vertical slope. That means something moving slow would have a more horizontal slope.
  • Reading this position graph is broken down into two slopes circled in red. The first slope is read as negative changing velocity from fast to slow. Since this slope is negative the slope of the next kinematic graph must be in the 4th quadrant. I have circled those two for you in blue. Again let’s see if reading the second slope, also circled in red, of the position graph gives us an insight into predicting the velocity graph.
  • Reading the second slope of the position graph illustrates an object that continues to move left but is picking up velocity. In other words, moving from slow to fast. I have place yellow arrows to indicate the direction of the slope of the velocity graph. This gives the direction of the acceleration. Can you now pick out the acceleration graph?
  • There must be acceleration for an object to speed up or slow down or change direction. Since the direction of the object did not change, it traveled to the left, a negative movement, the only acceleration that could slow it down was positive. Remember that opposite signs on velocity and acceleration cause an object to slow down. Then since the object still had negative velocity, in other words moving to the left, the only acceleration that can speed the object back up is negative. Like signs on the velocity vector and the acceleration vector cause an object to speed up.
  • This motion map illustrates an object that moved to the left slowed down, stopped, changed directions and sped back up. My knowledge of kinematic graphing tells me that the slope of the first graph will be negative because the object is traveling left, then positive because the object is traveling right and will be curved twice because it changed velocity twice. Be careful here because two curves can sometimes appear as one big curve.
  • Now lets see if the vector arrows tell us anything about the orientation and shape of the velocity versus time slope . For purposes of this slide only I use the color blue for left or negative and the color red for right or positive. So I am looking to match a pattern of left or negative getting slower then becoming right or positive and getting faster.
  • Now lets see if the vector arrows tell us anything about the orientation and shape of the velocity versus time slope . For purposes of this slide only I use the color blue for left or negative and the color red for right or positive. So I am looking to match a pattern of left or negative getting slower then becoming right or positive and getting faster. The graph located in the top row center column fits the description.
  • Now lets take a look at the orientation and shape of the slope of the velocity versus time graph. I can see we have constant positive acceleration.
  • Read all of the slopes out loud for practice. The position graph shows negative changing velocity from fast to slow followed by positive changing velocity from slow to fast. The velocity graph shows positive constant acceleration. The acceleration graph shows constant or neutral jerk. Jerk is a quantity that you will not have to calculate. Read the area of the velocity and acceleration graphs. The area of the velocity graph shows decreasing displacement to the left followed by increasing displacement to the right. The area of the acceleration graph shows constantly increasing positive velocity.
  • Now we have an object that is starting from rest and speeding up to the right. Displacement is on the y-axis of the position graph so moving to the right is up on the graph. We would expect to see a curved line since the object is speeding up. On a position graph a horizontal slope means slow and a vertical slope means fast. So we are looking for a slope going from horizontal to vertical.
  • Read the slope of the position graph out loud to determine the orientation and shape of the velocity graph. Again look at the vector arrows to help you with your verbiage. Positive movement from slow to fast. Positive means it will be found in the 1st quadrant. Of the graphs above identify the ones that are found entirely in the 1st quadrant.
  • Here are the graphs that have slopes starting in the 1st quadrant. Two of them marked in orange can be eliminated because they are curved. In this physics class the velocity graph will never be curved. One of the graphs can be eliminated because it has a horizontal slope indicating no change in velocity. This graph is circled in green.
  • So now we know what the velocity graph will look like. I have placed an arrow in the direction of the slope. Read it out loud and pick an acceleration versus time graph that would match.
  • The motion map indicates that an object starts from rest and speeds up to the right. The position graph is read as positive changing velocity from slow to fast. The velocity graph is read as constant positive acceleration. The acceleration graph is read as neutral or no jerk.
  • The quantity plotted on the y-axis is acceleration measured in m/s2. The quantity plotted on the x-axis is time measured in seconds. He y-intercept indicates the object had an initial acceleration of 6 m/s2. There is no x-intercept. The slope of an acceleration versus time graph is jerk measure in m/s3. The area between the slope and the x-axis is the change in velocity measured in m/s. So we need to calculate the slopes and the areas give magnitudes to each part of the graph. The first slope has a rise of 0 m/s and a run of 3 seconds. The magnitude of the first slope is 0 m/s divided by 3 seconds which equals 2 m/s3. The first area has a rise of 6 m/s and a run of 3 seconds. The magnitude of the first area is 6 m/s times 3 seconds which equals 18 meters.
  • Now continue with the procedure you have just confirmed concerning slope and areas to complete the remaining two slopes and two areas.
  • An acceleration graph is important for giving acceleration, time, jerk and the change in velocity.
  • Remember that the only math we ever perform with vectors is addition. Sometimes the addition is as simple as adding two positive numbers. Sometimes it can become increasingly complex such as adding negative numbers, of numbers at right angles. Here identify the vector magnitude by its value and the direction of the vector by the arrow. The first one is adding two positive numbers. The second problem is solved by adding a positive number with a negative number. The final problem is solved by using the pythagoreum theorem.
  • Positive integer plus a positive integer. Positive integer plus a negative integer. A2 + B2 = C2
  • W = 153.08 N g = -9.8 m/s2
    m =
  • Remember that weight is a force so we can use the first or third formulas. But since we dealing with weight here I would go with the simplest formula possible.
  • m = 51.62 kg a = 0 m/s2 g = -9.8 m/s2 Fa = 6 N

    Ff =
    Here you must recognize that you are dealing with more than one force in the horizontal. Seek out a formula that contains the sum of all forces in the horizontal.
  • Combine the second and fourth formulas to calculate the force of friction. The sum of the horizontal forces is what is happening to the object. This object is not accelerating so the max side of the equation is equal to zero. That means that the applied force and the force of friction have to be equal in magnitude but opposite in direction.
  • Here is the motion map of the object moving at constant velocity to the right. Predict what the force diagram would look like.
  • For an object to move at constant velocity, remember that 0m/s is a constant velocity, the sum of all forces in the x and y must equal zero. Now predict what the position, velocity and acceleration graphs would look like.
  • The motion map depicts an object moving to the right at constant velocity. The position graph is read as constant positive velocity. The velocity graph is read as neutral acceleration. The acceleration graph is read as neutral jerk.
  • m = 1315 kg a = 3.23 m/s2
    F =
  • Again this is a problem with only one force in it. Go with the simplest formula that you can.
  • m = 1315 kg F = 4247.45 N a = 3.00 m/s
    Ff =
  • Notice now that we have two forces in the horizontal for which to account. Lets use the sum of all forces in the x formula. Remember that the sum of all forces in the x is what is actually happening to the car so use the 3 m/s2 and the mass of the car and occupants for that side of the equation.
  • The motion map of my car and occupants shows that we are speeding up to the right. We must have an unbalanced force to the right for this to occur. Since the car is not rising up off the road nor burying itself into the road, the sum of all forces on the y-axis must be zero.
  • The motion map of my car and occupants shows that we are speeding up to the right. We must have an unbalanced force to the right for this to occur. Since the car is not rising up off the road nor burying itself into the road, the sum of all forces on the y-axis must be zero.
  • A read of the position graph slope reveals the car has positive changing velocity from slow to fast. The velocity graph slope is read as constant positive acceleration. The slope of the acceleration versus time graph is read as positive constant jerk.
  • m = .448 kg g = -9.8 m/s2
    Fs =
  • The force diagram of a projectile is easy to understand. The only force acting on the projectile is gravity. The sum of all forces is equal to what is happening to the apple. The .448 kg apple is falling at -9.8 m/s2. Notice that the force of gravity in this case will also be the product of the .448 kg apple is falling at -9.8 m/s2. Since the force of support acting on an object is equal to its weight and the force of support is zero, the apple is “weightless” during freefall.
  • Since the only force acting on this object is gravity, the object is said to be a projectile in free fall. The motion map should have the object starting from rest and gaining velocity on the way down. The position graph slope should show an object falling and being displaced at an increasing rate. The velocity graph slope should be in the 4th quadrant because the velocity is negative. The acceleration graph slope should indicate that the acceleration was due to gravity.
  • Keep practicing your graph reading out loud.
  • m = 44.10 kg a = 1.1 m/s2
    F =
  • Since the problem stated that the surface was frictionless, we really do not need to use a formula with more than one force in it. However, you can combine formula two and four together and plug in zero for your force of friction.
  • Remember that you now know the acceleration generated by the gizmo ever time it fires. Use this information to help complete the data table.
  • We ended up using Newton’s 2nd law of motion to determine the masses of the students.
  • m1 = 42 kg m2 = 43.25 kg m3 = 44 kg m4 = 44.23 kg m5 = 44.77 kg m6 = 45.01 kg m2 = 45.45 kg g = -9.8 m/s2
    Fs Fg
  • We needed to add the masses of everyone on the elevator to solve this problem. Since elevators travel in the vertical, we will use the sum of all forces in the y. Since the elevator is not accelerating, the force of support and the force of gravity will be equal in magnitude but opposite in direction.
  • m = 308.71 kg Fg = -3025.36 N g = -9.8 m/s2 a = .75 m/s2
    Fs=
    Experience tells us that if we are speeding up on the way up in an elevator we feel pressure in the bottom of our feet.
  • This indicates that the force of support is larger than the force of gravity at that time. We will again use the formula for the sum of all forces in the y. This time the elevator is accelerating so the force of support and the force of gravity cannot be equal in magnitude because there has to be an unbalanced force to cause this acceleration. The sum of all forces in the y is what is actually happening to the elevator. The elevator is accelerating at .75 m/s2 . Remember that the force of gravity cannot change since we stay on planet Earth and no one falls out of the elevator. So the only force that changes in an elevator is the force of support.
  • The force diagram shows an unbalanced force upward. . The force of gravity is less than the force of support. This diagram could explain an elevator slowing down on the way down or speeding up on the way up. In this case the motion map should have the object speeding up on the way up. The position graph slope should show an object traveling upward and being displaced at an increasing rate. The velocity graph slope should be in the 1st quadrant because the velocity is positive. The acceleration graph slope should indicate that the acceleration was positive because positive velocity and positive acceleration cause an object to sped up.
  • Again, keep practicing your graph reading out loud to improve your ability.
  • m = 308.71 kg Fg = -3025.36 N g = -9.8 m/s2 a = .-50 m/s2
    Fs=
    Recall once again your experience about slowing down on the way up in an elevator. Should the force of support be larger or smaller than the force of gravity? I think I feel less pressure on the bottom of my feet at this time.
  • We will again use the formula for the sum of all forces in the y. Again the elevator is accelerating so the force of support and the force of gravity cannot be equal in magnitude because there has to be an unbalanced force to cause this acceleration. The force diagram shows us that the force of support is smaller than the force of gravity.
  • This force diagram shows an unbalanced force in the vertical. The force of support is less than the force of gravity. This diagram could be from an elevator slowing down on the way up or speeding up on the way down. In this case the motion map should have the object slowing down on the way up. Since we need to pick graphs for the entire trip we will need the position graph slope to show an object traveling upward and being displaced at an increasing rate and then at a decreasing rate. The velocity graph slope should be in the 1st quadrant because the velocity is positive. The acceleration graph slope should indicate that the acceleration was positive because positive velocity and positive acceleration cause an object to sped up.
  • Keep practicing your graph reading out loud.
  • mt = 2000 kg mb = .0002 kg g = -9.8 m/s2 F = -50 N
    at =
    ab =
  • Both the truck and the bug experience the same amount of force at the same time. Remember we have spoken in the past about the effect of g-forces on the body. You may recall that around 4-5 g’s most people will black out. To calculate the g-force on the body simply divide the acceleration experienced by the object by 9.8. So the bug has over 25,000 g-forces acting on it! The outside of the bug stopped moving when it hit the truck, however, the insides of the bug kept accelerating until they hit the truck.
  • T1 = 375.4 N g = -9.8 m/s2 Θ = 7.5°
    m=
  • The weight of an object in equilibrium must equal the total force of support. Notice that the two cables are hung at the same angle. How do we determine the angle? You can see that there is 165° between the two cables. In a straight line there is 180°. If you subtract 165° from 180° it means there is 15° left over to be divided up for the cable angles. Each side therefore is hung at 7.5°. Force of support is in the vertical so if you measure from positive X to the vector you use the sine of 7.5°. Remember there are two cables so you need to double your answer to account for the opposite cable’s support.
  • Vi = 30 m/s Vi = 5 m/s a2 + b2 = c2
    Vf=
  • Here we are given two velocity vectors that are directly on the y and x-axis. When we want to create a resultant vector out of the components we use the pythagoreum theorem. In this case you can see that I am being pushed southeast by the wind.
  • Remember that the formulas for finding angles will give you the angles on either side of the vector inside the quadrant in which the vector is located. Once you have made your calculation you need to assign that angle to the correct location within the quadrant.
  • We have two choices to find the angle. One will find theta and one will find phi. You can see that the 9.46° is located between the 270° y-axis and the resultant vector. So you can add 9.46° to 270° to find the angle as measured from positive x. You can also see that the 80.54° angle is located between the 360° x-axis and the resultant vector. You can subtract 80.54° from 360° to find the angle as measured from positive x.
  • Yf = -15 m Yi = 0 m m = 15 kg g = -9.8 m/s2 Vi = 0 m/s Vf = 31.41 m/s
    ti = 0 s ti = 4 s
    a =
    F =
  • First we need to find the acceleration of the box to find the force acting on the box. We can use the change in velocity over the change in time formula to find this acceleration. We can now use Newton’s 2nd law to find the applied force on the box. Since this is a horizontal problem and there is more than one force acting on the box we will use formulas two and four to get the sum of all forces in the horizontal.
  • The motion map of the box shows that it is speeding up to the right. We must have an unbalanced force to the right for this to occur. Since the box is not rising up off the ground nor burying itself into the ground, the sum of all forces on the y-axis must be zero.
  • A read of the position graph slope reveals the car has positive changing velocity from slow to fast. The velocity graph slope is read as constant positive acceleration. The slope of the acceleration versus time graph is read as positive constant jerk. Now I want you to complete these graphs by adding the magnitudes of forces, displacement, velocity and acceleration.
  • The force of gravity acting on an object is found by multiplying mass times the acceleration due to gravity. The force of support is equal and opposite since the box is not accelerating in the vertical. The slope of the position graph can be described by the beast formula in the horizontal. Velocity is found by using Vf = Vi + at. We have already found the acceleration in a previous slide.
  • Yf = -15 m Yi = 0 m m = 15 kg g = -9.8 m/s2 Vi = 0 m/s ti = 0 s
    tf =
    Vf =
  • This package is now in free fall. It had no initial velocity in the vertical. What does the motion map look like?
  • In the vertical the motion map should have the object starting from rest and gaining velocity on the way down. In the horizontal the motion map should have the object having constant horizontal velocity for the entire trip. The position graph slope should show an object falling and being displaced at an increasing rate. The velocity graph slope should be in the 4th quadrant because the velocity is negative. The acceleration graph slope should indicate that the acceleration was due to gravity.
  • Here are the correct answers. Keep practicing your graph reading out loud.
  • Lets try our hand at data table reading again. Each of the columns has a mathematical equation with which it is associated. Check your packet on projectile motion to reaffirm these equations.
  • Vertical position is found by the free fall equation, Yf = ½ gt2., since there is not initial velocity in the vertical. Horizontal position is found by the range equation Xf = Vit. Remember that horizontal velocity cannot change. Vertical velocity is found by using Vf = Vi + gt.
  • V = 27 m/s Θ= 27° g = -9.8 m/s2
    Viy =
    Vix =
  • The same concept that we have used in force is also true for velocity because they are both vector quantities. We resolve a vector into its horizontal and vertical components using trigonometry. If the angle is measured from positive x then we use cosine for the horizontal component and sine for the vertical component.
  • g = -9.8 m/s2 Θ= 27° Viy = 12.26 m/s Viy = 24.06 m/s ti = 0 s Yi = 0 m Yf = 0 m
    tf =
  • Using the beast in the vertical will give us the overall time that the baseball will spend in the air. Remember to set your launch and land positions at 0 meters since the baseball will be launch and land at the same elevation. You will need to factor out time in the equation. If you worried about doing that, you can find the time to reach maximum height and then multiply by two. Remember that you can only do that if the projectile is launched and lands at the same elevation. At maximum height the baseball will have a vertical velocity of 0 m/s and you would use formula two on this slide.
  • g = -9.8 m/s2 Θ= 27° Viy = 12.26 m/s Viy = 24.06 m/s ti = 0 s Yi = 0 mYf = 0 m
    Xi = 0 m tf = 2.5 s
    Xf =
  • Range is the product of horizontal velocity and time in the air.
  • g = -9.8 m/s2 Θ= 27° Viy = 12.26 m/s Vix = 24.06 m/s ti = 0 s Yi = 0 mYf = 0 m
    Xi = 0 m tf = 2.5 s Xf = 60.15 m
    Δy =
  • Since we are looking for maximum height and we have an initial vertical velocity we have three formula choices. We can average the velocity and multiply by half the time in the air, use the beast in the vertical using half the time in the air or formula four on the card where you do not need to worry about time.
  • Vertical velocity is plotted on the y-axis and is measured in meters per second. Time is plotted on the x-axis and is measure in seconds. The y-intercept indicates the projectile had an initial vertical velocity of 30 m/s. The x-intercept indicates that around 3 seconds the projectile had no vertical velocity and changed direction. We can find the exact seconds mathematically. The slope of this graph indicates constant negative acceleration. The area between the slope and the x-axis is the change in position of the projectile.
  • The area between the slope and x-axis describes the same change of position as the position graph would. The beast describes the slope of the position graph and should result in the same calculation as the area of the triangle. We can get the magnitude of the x-intercept by using Vf = Vi + gt.
  • Use one vector arrow to represent the direction of the force at the beginning and at the top of the projectiles motion. Use one vector arrow to represent the direction of the at the beginning and at the top of the projectiles motion. Do the same thing for velocity and acceleration.
  • We can see that the force is caused by gravity and faces the center of the Earth the entire trip. Since the acceleration is caused by this force, the acceleration vector is in the same direction as the force vector. Position is determined by the resultant velocity so those arrows will mirror each other as well.
  • m = .448 kg Vi = 30 m/s g = -9.8 m/s2
    p =
  • Momentum is the product of mass times velocity.
  • m = .448 kg Vi = 30 m/s g = -9.8 m/s2 ti = 0 s tf = 3.06 s Δp = -13.44 kgm/s

    F =
  • That should make sense because gravity is what changes the apples momentum from positive to zero and them to negative. At zero velocity the apple has no momentum abut is still under the influence of gravity. The acceleration due to gravity is -9.8 m/s2 and the mass of the apple is .448 kg. So when the apple is a projectile it has a force of gravity of -4.9 N acting on it at all times!
  • m1 = .448 kg m2 = .577 kg g = -9.8 m/s2 Vi1 = 10.4 m/s Vi2 = 0 m/s
    p =
    Vf =
  • To find the final velocity of the two apples stuck together we need to know the momentum of the system before the collision. Since the second apple was at rest it contributed no momentum to the system. The momentum of the first apple ends up being 4.66 kgm/s2. Since momentum is a conserved quantity and the momentum of the system initially was 4.66 kgm/s2 it has to remain 4.66 kgm/s2. Now you need to combine the masses to find the new velocity because the apples are stuck together and have to be moving at the same rate.
  • m1 = .448 kg m2 = .577 kg g = -9.8 m/s2 Vi1 = 4.55 m/s Vi2 = 0 m/s p = 4.66 kgm/s
    F =
  • Here it is important to remember that impulse equals the change in momentum. If we set these two formulas equal to each other we can find force because we know the time of contact with the wall.
  • This completes this comprehensive review for the semester one final exam. Thank you for taking the time to review with me. If you have studied hard you deserve to do well. Good luck and have a nice day.
  • Transcript

    • 1. The slides used in this video are color coded. If you are experiencing difficulty with one aspect of your understanding than another you might find this coding… useful! Slides with Slides with tan red backgrounds backgrounds involve involve word matching Slides with green problems. concepts. backgrounds Slides with olive involve backgrounds graphing. involve reading data tables.
    • 2. Dr. Fiala is traveling on his Harley at a constant 13.67 m/s. What is the distance traveled by Doc in 7.32 seconds?
    • 3. SOLUTION: Find the distance traveled. Vi&f = 13.67 m/s Xf ti = 0 s tf = 7.32 s Xi = 0 m Xf = m a = 0 m/s2
    • 4. SOLUTION: Find the distance traveled. Vi&f = 13.67 m/s Xf ti = 0 s tf = 7.32 s Xi = 0 m Xf= 100.07 m a = 0 m/s2
    • 5. Dr. Fiala notices he is now traveling at a constant 49.21 km/h. What is the distance in meters traveled by Doc in 7.32 seconds?
    • 6. SOLUTION: Dimensional analysis. M K H D 1 x x X 0 d c x X X 49.21 km/h = 13.67 m/s So the Xf remains 100.07 m m
    • 7. Dr. Fiala jumps in his unstarted car. He accelerates at a rate of 4 m/s2 for 8 seconds. How far did Doc travel?
    • 8. Doc’s final position. vi = 0 m/s ti = 0 s tf = 8 s a = 4 m/s2 Xi = 0 m vf = Xf = Xf = 128 m
    • 9. • Displacement Velocity Acceleration Inertia Force Momentum • • • • • The change in the rate or direction of motion. The resistance to a change in an object’s current state of motion. A change in position. A push or a pull that tends to accelerate an object. The movement of an object in a specific direction over time. The product of mass times velocity.
    • 10. Displacement is a change in position. Velocity is the movement of an object in a specific direction over time. Acceleration is the change in the rate or direction of motion of an object. Inertia is the resistance to a change in an object’s current state of motion. Force is a push or a pull that tends to accelerate an object. Momentum is the product of mass times velocity.
    • 11. Time (s) 1 Object #1 Object #2 Object #3 Object #4 Position (m) Position (m) Position (m) Position (m) 16 4 2 3 4 4 48 24 16 32 6
    • 12. Time (s) Object #1 Object #2 Object #3 Object #4 Position (m) Position (m) Position (m) Position (m) 1 16 4 8 2 2 32 8 16 4 3 48 12 24 6 4 64 16 32 8
    • 13. Time (s) 1 Object #1 Object #2 Object #3 Object #4 Velocity (m/s) Velocity (m/s) Velocity (m/s) Velocity (m/s) 16 8 9 2 3 4 13 8 10 2 4
    • 14. Time (s) Object #1 Object #2 Object #3 Object #4 Velocity (m/s) Velocity (m/s) Velocity (m/s) Velocity (m/s) 1 16 4 8 2 2 14.5 6 9 4 3 13 8 10 6 4 11.5 10 11 8
    • 15. #1 #3 #2 #4
    • 16. 2m /s m/s 1 0 m/s 0 m/s
    • 17. Vy = 0 m/s Yf = Yi + Vi t + ½ gt2 Vi = 48.5 m/s ty = 5 s Vf2 = Vi2 + 2g Δy Vi = 48.5 m/s
    • 18. Graph Options Position Graph Can you predict the slope shape and orientation of both the velocity and acceleration graphs?
    • 19. Graph Options Position Graph Can you predict the slope shape and orientation of both the velocity and acceleration graphs? V = 0 m/s
    • 20. Graph Options Position Graph Can you predict the slope shape and orientation of both the velocity and acceleration graphs? V = 0 m/s
    • 21. Graph Options Position Graph Can you predict the slope shape and orientation of both the velocity and acceleration graphs?
    • 22. Position Graph Vy = 0 m/s Velocity Graph Acceleration Graph +V Vy = 0 m/s -V g = -9.8 m/s2
    • 23. If Dr. Fiala starts from a full stop and accelerates at 2.25 m/s2, how incredibly fast will he be traveling when he has traveled 1530 meters?
    • 24. SOLUTION: Calculate final velocity without knowing time. Vi = 0 m/s Xi = 0 m Xf = 1530 m a = 2.25 m/s2 ti = 0 s Vf tf Vf= 82.98 m/s
    • 25. If Dr. Fiala starts from a full stop and accelerates at 2.25 m/s2, how long will it take him to drive 1530 meters?
    • 26. SOLUTION: Solve for time. Vi = 0 m/s Xi = 0 m Xf = 1530 m a = 2.25 m/s2 ti = 0 s Vf = 82.98 m/s tf Vf t = 36.88 s
    • 27. 0 m/s2 2.3 3 m 2 /s 2 .8 m/s 75 0 m/s2 30 m 33.75 m 15 m 92 m
    • 28. Velocity (m/s) 16 14 12 10 8 6 4 2 0 0 2 4 6 8 10 12 14 16 Time (s)
    • 29. Graph Options Motion Map Can you predict the slope shape and orientation of the position, velocity, and acceleration graphs?
    • 30. Graph Options Motion Map Can you predict the slope shape and orientation of the position, velocity, and acceleration graphs?
    • 31. Graph Options Motion Map Can you predict the slope shape and orientation of the position, velocity, and acceleration graphs?
    • 32. Graph Options Motion Map Can you predict the slope shape and orientation of the position, velocity, and acceleration graphs?
    • 33. Position Graph Motion Map Velocity Graph Acceleration Graph
    • 34. Graph Options Motion Map Can you predict the slope shape and orientation of the position, velocity, and acceleration graphs?
    • 35. Graph Options Motion Map Can you predict the slope shape and orientation of the position, velocity, and acceleration graphs?
    • 36. Graph Options Motion Map Velocity Graph Position Graph Can you predict the slope shape and orientation of the position, velocity, and acceleration graphs?
    • 37. Graph Options Motion Map Velocity Graph Position Graph Can you predict the slope shape and orientation of the position, velocity, and acceleration graphs?
    • 38. Position Graph Motion Map Velocity Graph Acceleration Graph
    • 39. Graph Options Motion Map Can you predict the slope shape and orientation of the position, velocity, and acceleration graphs?
    • 40. Graph Options Motion Map Can you predict the slope shape and orientation of the position, velocity, and acceleration graphs? Position Graph
    • 41. Graph Options Motion Map Can you predict the slope shape and orientation of the position, velocity, and acceleration graphs? Position Graph
    • 42. Graph Options Motion Map Velocity Graph Can you predict the slope shape and orientation of the position, velocity, and acceleration graphs? Position Graph
    • 43. Position Graph Motion Map Velocity Graph Acceleration Graph
    • 44. 0 m/s3 18 m/s
    • 45. 0 m/s3 -1 18 m/s m/ s3 17.5 m/s 0 m/s3 2 m/s
    • 46. 2.25 + 3.25 = 2.25 + 3.25 = 2.25 + 3.25 =
    • 47. SOLUTION: Adding vectors. + 2.25 + 3.25 = 5.5 + 2.25 + 3.25 = 1.00 + 2.25 + 3.25 = 3.95 + - + + - +
    • 48. ? kg 153 N Determine the mass of a 153.08 N object.
    • 49. SOLUTION: Calculate mass. W = 153.08 N g = -9.8 m/s2 m m = 15.62 kg
    • 50. Determine the force of friction on a 15.62 kg object traveling at a constant horizontal velocity of 3.62 m/s while experiencing an applied force of 6 N.
    • 51. SOLUTION: Calculate force of friction. m = 51.62 kg a = 0 m/s2 g = -9.8 m/s2 Fa = 6 N Ff Ff = -6 N
    • 52. Force Diagrams Motion Map Match the force diagram to the motion map. Can you also predict the slope shape and orientation of the position, velocity, and acceleration graphs?
    • 53. Force Diagrams Motion Map Match the force diagram to the motion map. Can you also predict the slope shape and orientation of the position, velocity, and acceleration graphs?
    • 54. Force Diagram Motion Map Position Graph Velocity Graph Acceleration Graph
    • 55. Determine the force needed to accelerate Dr. Fiala’s car and its occupants at a rate of 3.23 m/s2 if the total mass of car and occupants is 1315 kg and there is no friction force.
    • 56. SOLUTION: Find applied force. m = 1315 kg a = 3.23 m/s2 F F = 4247.45 N (kg)(m/s ) 2
    • 57. This time, when we apply that 4247.45 N force to Dr. Fiala’s car and its occupants, the resulting acceleration is actually lower. It registers at a rate of only 3.00 m/s2. What is the magnitude for the force of friction causing the acceleration to be decreased?
    • 58. SOLUTION: Find applied force. m = 1315 kg F = 4247.45 N a = 3.00 m/s2 Ff Ff = - 302.45 N
    • 59. Force Diagrams Motion Map Match the force diagram to the motion map. Can you also predict the slope shape and orientation of the position, velocity, and acceleration graphs?
    • 60. Force Diagrams Motion Map Match the force diagram to the motion map. Can you also predict the slope shape and orientation of the position, velocity, and acceleration graphs?
    • 61. Force Diagram Motion Map Position Graph Velocity Graph Acceleration Graph
    • 62. When an object is freefalling it is weightless. Prove mathematically that a .448 kg apple is weightless during its freefall from a tree. Draw a force diagram of the apple during its fall from the tree.
    • 63. SOLUTION: Find force of support. m = .448 kg g = -9.8 m/s2 Fs Fs = 0 N Fg = -4.39 N
    • 64. Motion Map Force Diagram Position (ΔY) Graph Velocity (Vy) Graph Can you predict the motion map, and kinematic graphs for this freefalling object? Acceleration Graph
    • 65. Motion Map Force Diagram Position (ΔY) Graph Velocity (Vy) Graph Acceleration Graph
    • 66. Assuming a perfectly frictionless surface, ideal for launching students in a game of faculty bowling, Dr. Fiala uses a brand new gizmo that automatically applies a force that results in an acceleration of 1.1 m/s2. Experimentation resulted in a student with a mass of 44.10 kg, accelerating at 1.1 m/s2. Find the force generated by the gizmo for that student.
    • 67. SOLUTION: Find force in the horizontal. m = 44.10 kg a = 1.1 m/s2 F F = 48.51 N
    • 68. Mass (kg) Force (N) 0 0 42 46.2 47.56 44 48.4 48.65 49.25 49.51 45.45 50
    • 69. Mass (kg) Force (N) 0 0 42 46.2 43.25 47.56 44 48.4 44.23 48.65 44.77 49.25 45.01 49.51 45.45 50
    • 70. All of the students from the previous problem (combined mass) step into an elevator at the same time. Draw a force diagram of this situation including the magnitude of Fg and Fs.
    • 71. SOLUTION: Find force of gravity and force of support. m1 = 42 kg Fg m2 = 43.25 kg Fs m3 = 44 kg m4 = 44.23 kg m5 = 44.77 kg m6 = 45.01 kg m2 = 45.45 kg Fg= -3025.36 N g = -9.8 m/s2 Fs= 3025.36 N
    • 72. This same elevator accelerates at a rate of .75 m/s2 towards the second floor. Draw a force diagram of this situation including the magnitude of Fg and Fs.
    • 73. SOLUTION: Find force of support. m = 308.71 kg Fg = -3025.36 N g = -9.8 m/s2 a = .75 m/s2 Fs Fs = 3256.89 N Fs= N Fg = 3025.36 N 3256.89
    • 74. Motion Map Force Diagram Position Graph Velocity (Vy) Graph Can you predict the motion map, and kinematic graphs for this elevator? Acceleration Graph
    • 75. Motion Map Force Diagram Position Graph Velocity (Vy) Graph Acceleration Graph
    • 76. This same elevator accelerates 2 at a rate of .50 m/s as it begins its stop for the second floor. Draw a Force diagram of this situation including the magnitude of Fg and Fs.
    • 77. SOLUTION: Find force of support. m = 308.71 kg Fg = -3025.36 N g = -9.8 m/s2 a = .-50 m/s2 N Fs = 2871.01 N Fg = 3025.36 N Fs Fs= 2871.01
    • 78. Motion Map Force Diagram Position Graph Velocity (Vy) Graph Can you predict the motion map, and kinematic graphs for the ENTIRE TRIP? Acceleration Graph
    • 79. Motion Map Force Diagram Position Graph Velocity (Vy) Graph Acceleration Graph
    • 80. According to Newton’s 3rd law, an action force causes an equal on opposite reaction force. It is no wonder a truck windshield squashes a bug and not vice versa. A 2000 kg truck and a .0002 kg bug hit with a 50 N force. Take a closer look at why the truck wins the collision by calculating the acceleration exerienced by the bug and by the truck.
    • 81. SOLUTION: Why the bug doesn’t survive. mt = 2000 kg at mb = .0002 kg ab g = -9.8 m/s2 F = -50 N at = -.025 m/s2 ab = -250,000 m/s2
    • 82. .40 N 375 These cables will snap if the mass of the trafffic light exceeds 10.1 kg. Does the traffic light exceed 10.1 kg?
    • 83. SOLUTION: The cable does not break. T1 = 375.4 N g = -9.8 m/s2 Θ = 7.5° m T1y m= 10 kg
    • 84. Dr. Fiala attempts to walk due east at 5 m/s at the same time as a 30 m/s cold, winter wind is blowing due south. What is the magnitude of Dr. Fiala’s velocity.
    • 85. SOLUTION: Resultant velocity magnitude. Vi = 30 m/s Vi = 5 m/s Vx = 5 m/s Vy = 30 m/s Vf Vf= 30.41 m/s 2 2 a +b =c 2
    • 86. Vx = 5 m/s V =3 0.4 1m /s Vy = 30 m/s If Dr. Fiala continues his velocity and the wind continues to blow steadily, at what angle, as measured from positive “X”, is Dr. Fiala’s velocity.
    • 87. SOLUTION: Resultant velocity angle measured from positive x. tan Θ = x y Θ = 9.46° Vx = 5 m/s Vy = 30 m/s tan Φ = y x Φ = 80.54° Θ (from +x) = 279.46°
    • 88. Because of this wind, a 15 kg package is blown from Dr. Fiala’s arms and onto the ground. The 15 kg package reaches a velocity of 30.41 m/s in a time of 4 seconds. Find the force acting on the box horizontally if there is no friction.
    • 89. SOLUTION: Find applied force. Yf = -15 m a Yi = 0 m F m = 15 kg g = -9.8 m/s2 Vi = 0 m/s Vf = 31.41 m/s ti = 0 s a = 7.60 m/s2 ti = 4 s F = 114 N
    • 90. Force Diagrams Motion Map Match the force diagram to the motion map. Can you also predict the slope shape and orientation of the position, velocity, and acceleration graphs?
    • 91. Force Diagram Motion Map Position Graph Velocity Graph Acceleration Graph
    • 92. Force Ff = 0 N Diagram Motion Map Fs = 147 N Fa = 117.79 N Fg = 147 N Position Graph Velocity Graph 31.4 m/s 62.8 m 4s Acceleration Graph 7.85 m/s2 4s 4s
    • 93. If the package is blow horizontally at 30.41 m/s off a ledge onto a parking lot that is 15 meters below how much time will it spend in the air before striking the ground? What does the motion map look like?
    • 94. SOLUTION: Find time package spends in the air. Yf = -15 m Yi = 0 m m = 15 kg g = -9.8 m/s2 Vi = 0 m/s ti = 0 m/s Vf tf tf = 1.75 s
    • 95. Force Diagram Motion Map Position (ΔY) Graph Velocity (Vy) Graph Can you predict what the force diagram, and vertical kinematic graphs for this freefalling object? Acceleration Graph
    • 96. Force Diagram Motion Map Position (ΔY) Graph Velocity (Vy) Graph Acceleration Graph
    • 97. Time (s) Vertical Position (m) Horizontal Position (m) Vertical Velocity Horizontal Velocity (m/s) (m/s) 0.11 0.27 31.41 -0.36 0.63 -5.61 -4.80 -1.94 1.07 -2.65 15.39 0.49 8.48 1.22 -6.17 33.61 38.32 1.35 -8.93 1.46 -10.44 1.75 - -11.96 -13.23 45.86
    • 98. Time (s) Vertical Position (m) Horizontal Position (m) Vertical Velocity Horizontal Velocity (m/s) (m/s) 0.11 -0.06 3.46 -1.08 31.41 0.27 -0.36 8.48 -2.65 31.41 0.49 -1.18 15.39 -4.80 31.41 0.63 -1.94 19.79 -6.17 31.41 1.07 -5.61 33.61 -10.49 31.41 1.22 -7.29 38.32 -11.96 31.41 1.35 -8.93 42.40 -13.23 31.41 1.46 -10.44 45.86 -14.31 31.41 1.75 -15.01 54.97 -17.15 31.41
    • 99. Dr. Fiala throws a baseball in the air with an initial velocity of 27 m/s at an angle of 27° to the horizon. Create a velocity vector diagram and show, by parallelogram method, the “X” and “Y” components of the baseball’s velocity.
    • 100. SOLUTION: Resolve velocity vector into “x” and “y” components just like force or any other vector. V = 27 m/s Θ= 27° g = -9.8 m/s2 m/s 27 27° Viy Vix Viy = 12.26 m/s Vix = 24.06 m/s Vx = V Vy = V
    • 101. How much time will it take for the baseball to reach the same height from which it was thrown?
    • 102. SOLUTION: Find time in the air. g = -9.8 m/s2 Θ= 27° Viy = 12.26 m/s Viy = 24.06 m/s ti = 0 s Yi = 0 m Yf = 0 m tf tf = 2.5 s
    • 103. How far will the baseball travel in 2.5 seconds?
    • 104. SOLUTION: Find range. g = -9.8 m/s2 Xf Θ= 27° Viy = 12.26 m/s Viy = 24.06 m/s ti = 0 s Yi = 0 m Yf = 0 m Xi = 0 m tf = 2.5 s Xf = 60.15 m
    • 105. What is the maximum height the baseball attained during its flight?
    • 106. SOLUTION: Find Δy. g = -9.8 m/s2 Δy Θ= 27° Viy = 12.26 m/s Vix = 24.06 m/s ti = 0 s Yi = 0 m Yf = 0 m Xi = 0 m Δy = 7.67 m tf = 2.5 s
    • 107. Vy = 0 m/s a= 9.8 m Vf = Vi + 2g Δt tf = 3.06 s - /s 2 Yf = Yi + Vi t + ½ at2 Yf = 45.9 m AREA = ½ Base x Height
    • 108. Force Vector Arrows for this Projectile Position  Velocity Using these vector arrows can you predict what the position, force, velocity and acceleration vector arrows would look like for this projectile at the start and at the top? Acceleration
    • 109. Force Position Velocity Acceleration        
    • 110. If it was a .448 kg apple that was thrown into the air at 30 m/s what was the apple’s intial momentum?
    • 111. SOLUTION: Find momentum of apple. m = .448 kg Vi = 30 m/s g = -9.8 m/s2 p p = 13.44 kgm/s
    • 112. What constant force is needed to get a change in the apple’s momentum from 13.44 kgm/s to 0 In 3.06 seconds?
    • 113. SOLUTION: Find force necessary to change momentum. m = .448 kg F Vi = 30 m/s g = -9.8 m/s2 ti = 0 s tf = 3.06 s Δp = -13.44 kgm/s F = 4.39 N
    • 114. After falling to the ground the .448 kg apple rolled at a constant 10.4 m/s where collided with a stationary .577 kg apple. If the two apples stuck together, at what velocity would they roll?
    • 115. SOLUTION: Find the velocity of two apples stuck together. m1 = .448 kg m2 = .577 kg g = -9.8 m/s2 Vi1 = 10.4 m/s Vi2 = 0 m/s Vf p p = 4.66 kgm/s2 Vf = 4.55 m/s
    • 116. Determine the force applied if the rolling apples strike a wall and a come to a stop in .311 seconds.
    • 117. SOLUTION: Find force needed to stop apples. m1 = .448 kg m2 = .577 kg ti = 0 s tf = .311 s g = -9.8 m/s2 Vi1 = 4.55 m/s Vi2 = 0 m/s p = 4.66 kgm/s F F = 14.98 N

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