Newton's Second Law Blog Reflection

In this unit I learned how to apply particular equations into certain aspects of gravitational forces. When we did our first lab we learned Newton's Second Law which is a=f/m. Acceleration is directly proportional to force and acceleration is indirectly proportional to 1/mass. This means that when acceleration increases, force also increases. When acceleration decreases, mass increases. This was useful in our lab when we would add mass to the cart. When we added mass to the cart, mass increased but we noticed that acceleration decreased. When the force increased, the acceleration of the cart also increased.

In this lab we also looked at the equation of a line. The equation of a line is y= mx+b. When we are asked to translate the equation, y is always the thing placed on the y-axis. m is always the constant. *(THE M IN Y=MX+B IS NOT THE SAME AS M = MASS)* b is always too small in this situation to really pay attention to so don't really worry about b. x is whatever thing is left over. Example: Force is kept constant. Acceleration is on the y-axis and mass is increasing. y=mx+b --> acceleration=force(mass)

When we did our study on skydiving with air resistance, newton's second law was extremely important here. Remember, a=f/m. When looking at the fall of a skydiver, -f-weight is constant. f-weight is the weight of the skydiver. In order to discover our f-net which is air resistance, we must subtract f-weight by fair. When first jumping out of the plane, there is no air resistance so the f-air is 0N. So, if our f-weight is 100 N, our f-net would be 100 N because 100 - 100 = 0 N. When the skydiver gets further down, the f-net decreases and the f-air increases. When the f-net reaches 0 N and the f-weight and f-air are equal, The diver is in terminal velocity. After the diver hits terminal velocity for the first time, he opens the parachute. When the parachute is opened, the air resistance increases causing the f-net to decrease in the opposite direction. When the diver hits second velocity for the second time, he is nearly to the end of his path. The first terminal velocity the man hits, he is going a lot faster than when he hits 2nd terminal velocity. This is because when the f-air increases, the man is being pulled in the opposite direction, balancing out the mans speed, causing him to slow down. Like Newton's second law states, acceleration is directly proportional to force. When f-net decreases, so does the acceleration. Remember, the whole point of a parachute is for the skydiver to SLOW DOWN!


When studying free fall, we learned that gravity = acceleration which equals 10 m/s^2. THIS IS ALWAYS CONSTANT NO MATTER WHERE THE BALL IS!!!!! When dropping an object straight down, we always start with 0 seconds and whichever velocity the object starts with. For example, if the ball was falling at 30 m/s, we would start with 0s and 30 m/s. For each second the object falls, the velocity decreases by 10 m/s. So, after 2s the velocity would be 20 m/s and after 3s the velocity would be 10 m/s and after 4s the velocity would be 0 m/s meaning the object is at the end of its path. When asked how long the ball was in the air, we just look at the total seconds, 4s. When asked what the acceleration of the object was at 3 seconds and 1 second, remember that acceleration/gravity is constant. At both of these times the acceleration is 10 m/s^2 simply because the acceleration is the same no matter where the object is in free fall. If asked to find the distance of the ball at 3 seconds, we would use the equation d= 1/2gt^2. We only use this equation from the top of an objects path. We first use this equation to solve for the distance of how high the ball went. When we do this, we will subtract by the equation for 1 second because 4-3 =1 and we are solving for how high the ball is at 3 seconds. So, d=1/2gt^2. d=1/2(10)(4)^2. d=1/2(10)(16). d=1/2(160) d=80m.
80m = distance the ball is 4 seconds off of the ground. Now we will do the equation for 3 seconds off of the ground. Remember to use 1 for t because we are only subtracting the full distance by 1 second.
So, d=1/2gt^2. d=1/2(10)(1)^2. d=1/2(10)(1). d=1/2(10). d=5m. So, 80-5= 75 m high off the ground at 3 seconds. *Side Note* The vertical distance is the only factor that affects how long an object is in the air.


While studying throwing an object straight up into the sire using free fall, I found that basically the same rules applied as when we dropped a ball straight down. When we are throwing a ball straight up, we start at 0s and begin at the starting velocity. If we were throwing a ball up at 30 m/s we would start at 0s and 30 m/s. The 2nd second, the velocity would decrease by 10 m/s so we would be at 20 m/s. The third second the ball would be at 10 m/s and the 4th second is at 0 m/s. The ball is always at the top of its path at 0 m/s. Because the ball has to come back down, we are still increasing our time but now we are increasing our velocity by 10 m/s. So, at 5 seconds as the ball is going back down, the velocity is 10 m/s. At 6 seconds the velocity is at 20 m/s. And at 7 seconds, the velocity is at 30 m/s where it started. Anytime during free fall, the gravity/acceleration is 10 m/s^2 and this is constant!!!! We solve the distance in this free fall exercise exactly like we do in the previous one. Remember, we can only use our equations from the top of the path.
Things began to get difficult when we started to study falling at angle because we had to mix vertical velocity with horizontal velocity. The equations for vertical velocity are d=1/2gt^2 and v=gt. The equation for horizontal velocity is v=d/t. The horizontal velocity is always constant. If asked to draw a picture of a man jumping off of a 45m high cliff with a horizontal velocity of 10 m/s, it would look something like this,
                                        10 m/s -->     
                                         |
                                         |
                                         |
                           45m       |
                                         |
                                         |
                                         |___________________________________________
Immediately we would look at the equation d=1/2gt^2 to figure out how long the man traveled.
45=1/2(10)t^2. 45=5(t)^2. 45/5 = 5/5 (t)^2. 9=t^2. Square root of 9 = 3. t=3. So the ball traveled for 3 seconds. To find the vertical velocity of the man, we use v=gt. v=10(3). v=30 m/s.
                                        
                                          1 second = 10m/s    2 seconds = 10m/s     3 seconds = 10 m/s
                                         |  
                                         |  10 m/s
                                         |
                                         |
                                         |
                                         |  20 m/s
                            45 m     |
                                         |
                                         |
                                         |  30 m/s
                                         |__________________________________________
Now that we have our velocities and speeds, we need to find out how far down field the man got. We solve this by the Pythagorean theorem, a^2 + b^2 = c^2.

                                          1 second = 10 m/s           2 seconds = 20 m/s           3seconds = 30 m/s
                                        | 10 m/s 
                                        |                 _10_____
                                        |             10|\
                                        |                 |  \   square root of 10over 2 = 14.1m/s
                                        | 20 m/s
                    45m             |
                                        |
                                        | 30 m/s
                                        |_________________________________________________________


When we learned throwing at an angle, I got super confused! But when I started understanding falling at an angle, it got better. The Pythagorean theorem is also used in this as the 5,4,3 problem. If given the speed of a ball at a certain angle, it is easy to fill in the remaining numbers.
Example. A ball is thrown up at a 45 degree angle with a speed of 50 m/s

                         |    /50m/s
               30m/s|  /
                         |/_________
                            40 m/s              The horizontal velocity is a constant. So, if the horizontal velocity was 40 m/s, it will always be like that. So when asked to find how far downfield the ball got, we use v=d/t because v=d/t is the horizontal equation and we are finding out how far down field (horizontally) the ball got. So, v=d/t. 40=d/6. 40(6)=d. 240m =d. So the ball got 240 m down field. This example is also in free fall so the acceleration/gravity is still 10 m/s^2 which is always constant!



What I have found difficult about what I have studied is incorporating horizontal and vertical velocity into the same problems. The light bulb finally dinged a little bit just today when going over throwing things up at an angle and falling at an angle. It clicked the more Mrs. Lawrence drew the examples out, and she was going through the process slowly.

I am usually very involved in physics. I am not afraid to ask questions when I am confused and I am not afraid to help others when one doesn't understand something that I do. I am very consistent on my homework completion every night whether I truly understand or I just attempt the problems and come in an ask questions. My blogs I feel have improved over the past weeks. Especially this one. I truly feel like I put everything I know into this reflection.

My self confidence in physics took a deep plunge today when we were reviewing today in class for a test tomorrow and I had noooo idea what was going on with angles. But, when I got home I looked back at my notes from todays class and it just clicked.

My goal for the next unit is to go a step further in my learning ability in physics. I can do this by asking questions even when I think I understand things and reviewing once a week with Mrs. Lawrence just to be sure I understand what's going on in class.


While studying this unit, I kept picturing me throwing up a ball and actually knowing what was going on with the balls speed, velocity, and acceleration. It is a very cool experience.

Free Fall Video

This video is exactly like the presentation Mrs. Lawrence did in class when she had the tube and the feathers and penny's. In this video a man first shows a tube that has air in it and contains feathers and penny's. When there is air in the tube the penny's hit the bottom before the feathers. Because there is air in the tube, there is air resistance. Air resistance is a force that pulls an object up. The heavier the object, the harder it is for the air resistance to catch up with its gravity pulling it to the bottom. The lighter the object, the easier it is for the air resistance to resist the object from falling as fast. This is exactly like the penny's and feathers. The penny has a greater mass so the air resistance is not as fast to try and stop the object. The feathers are lighter so the air resistance is greater. When the tube has no air in it, there is no air resistance acting on the object, only gravity. Because there is no air force, the objects fall at the same time because in free fall, gravity is always constant no matter the mass of an object. The gravity will always be 9.8 or 10 in labs.

This video was very useful because it clearly shows the differences of air resistance and free fall. Free fall NEVER has air resistance so no matter where the object is, it always has a gravitational force of 10 and an acceleration of 10 m/s2 because acceleration is proportional to gravity in free fall. I hope this video helped!

Newton's Second Law Video Resource

Obviously, this video is recommended for children so that they understand Newton's Second Law. I realize that we are not 5 years old, but I also realize that science is science no matter the age. I chose this video because it clearly states the definition of Newton's Second Law which is that acceleration depends on the mass of an object and the amount of force applied. This was shown when we did the lab in class with the carts and the extra mass that we added or took away from the cart or kept constant and the acceleration increased when the mass decreased and the acceleration decreased when mass increased. Remember, mass is inversely proportional to mass which means when acceleration does one thing, mass does another. The other equation in Newton's Second Law is that acceleration is directly proportional to force. So when acceleration increases, force increases and when acceleration decreases, force decreases.

This video helps me learn Newton's Second Law because of the visual examples. When the man pushes a child with a certain amount of force and then pushes a man with a greater mass than a child with the SAME amount of force, the child goes faster and higher which means the acceleration is greater for the child and less for the man who has a greater mass. This example goes right back to the equation acceleration is inversely proportional to mass. Since the boy had a lower mass, the acceleration was higher. When the man had a greater mass, the acceleration was lower.

I found this video useful for me because sometimes what helps me understand things is when I am spoken to like a kid. Slowly and using examples that even a chimp would understand. Less is more!

I hope this video helps you!