Introduction to Newton’s Second Law of Motion with Example Problem

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Күн бұрын

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The application of Newton’s Second Law is when you really understand what the net force equals mass times acceleration where both force and acceleration are vectors really means. Therefore, we introduce Newton’s Second Law and then do an example problem.
0:00 Intro
0:11 Defining Newton’s Second Law
1:00 The example problem
1:51 Drawing the Free Body Diagram
2:48 The Force of Gravity
3:42 The net force in the y-direction
5:28 The acceleration of the book in the y-direction
6:38 The net force in the x-direction
7:59 Solving for the dimensions of acceleration
8:54 Constant net force means constant acceleration
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@Uncertaintycat 7 жыл бұрын

You helped me work through my lab problems which weren't explained by my teacher or book! You, sir, are a living contradiction to charges that videos can't really teach. I stop the video and work the problem ahead of your student-clones, and it helped me watch for pitfalls and truly understand what I was doing, not just plug numbers into formulas. Thank you!

@FlippingPhysics 7 жыл бұрын

Videos can teach. It just takes a lot of work to make videos that actually will! www.flippingphysics.com/making-a-video.html Thanks for all your positive comments; they really help.

@FlippingPhysics 9 жыл бұрын

Newton’s Second Law! We understand it best by doing an example problem. Therefore we draw a free body diagram, sum the forces, sum the forces and find acceleration. #PhysicsED #flipclass

@andrewfreeman5165 3 жыл бұрын

tekram

@rachelprince1970 6 жыл бұрын

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@FlippingPhysics 6 жыл бұрын

You are absolutely welcome. Keep breathing.

@MonkeyOfLies22 3 жыл бұрын

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@noorh9471 8 жыл бұрын

You have amazing videos :-) Thank you so much, because physics doesn't come easy to me.

@FlippingPhysics 8 жыл бұрын

+Noor H You are welcome. I hope I am making it easier to understand. :)

@nikhilmish001 8 жыл бұрын

very very very very very good making physics more understand able

@FlippingPhysics 8 жыл бұрын

you are very very very very very welcome

@parmmanoo3691 8 жыл бұрын

You deserve so many more subscribers

@FlippingPhysics 8 жыл бұрын

+Parm Manoo Thanks. Sadly, educational KZfaq videos don't draw the number of subscribers of many other topics. I'll keep on working on it though.

@CH-fh2lh 8 жыл бұрын

Keep up the good work. Your attention to detail is excellent. Initially your videos come off as just another of many physics sequences. However as I have watched more I am realizing that they are thorough and reliable and cover many of the details that people miss.

@FlippingPhysics 8 жыл бұрын

+Clayton Handleman Thanks. Awesome comment.

@toptutorsga4258 2 жыл бұрын

Constantly amazed by your wonderful animations, experiments, descriptions!!!!! THANK YOU!!!!

@FlippingPhysics 2 жыл бұрын

Glad you like them!

@kittykat248100 8 жыл бұрын

much appreciation!

@johncelorio6023 2 жыл бұрын

This is by far the TOP physics explainer channel I have found in KZfaq. Thanks a lot!

@FlippingPhysics 2 жыл бұрын

Thanks!

@arhantbagde8037 3 жыл бұрын

Thanks it helped a lot

@mohammedasif2081 8 жыл бұрын

thanx alot

@rocks813 2 жыл бұрын

Best lightbulb moment here: finding out the overlap between forces and the kinematic equations. I learned so much from this video; thank you!

@FlippingPhysics 2 жыл бұрын

You are welcome!

@muhammadusman-pc3on 7 жыл бұрын

nice idea

@zubaidamir2758 7 жыл бұрын

I wish I found these videos earlier! You're amazing!

@FlippingPhysics 7 жыл бұрын

Thanks. Tell all your friends so they can find them earlier!

@vikkykumar3565 8 жыл бұрын

VERY NICE

@fatimatalpur1427 7 жыл бұрын

Your videos are so amazing you are very best teacher of physics really you give very nice examples i easily understand all of your videos thank you so much with your videos help i got full marks in my school test i am so happy you make physics easy and fun God bless you

@FlippingPhysics 7 жыл бұрын

Wow. This makes me very happy. I am very glad to know I have been able to help you learn!

@cassieknight6602 2 жыл бұрын

Wow! Although I easily aced physics, your videos have brought a whole new dimension to my understanding of everything I have reviewed thus far. Than you, again.

@FlippingPhysics 2 жыл бұрын

Wow! Wonderful to hear!

@tar_las 5 жыл бұрын

I appreciate these videos so much

@FlippingPhysics 5 жыл бұрын

Great!

@shashijayasuriya7830 5 жыл бұрын

Hi I am from Sri Lanka. This is the best physics lessons at KZfaq

@FlippingPhysics 5 жыл бұрын

Thank you for the compliment!

@aryensujjan 5 жыл бұрын

Dedication is excellent

@FlippingPhysics 5 жыл бұрын

Thank you!

@isaacong 3 жыл бұрын

Would the coefficient of friction be 0.23?

@gj9665 6 жыл бұрын

Love the way u teach

@FlippingPhysics 6 жыл бұрын

Great! Glad to know I am able to help you learn.

@thepun5vrv831 4 жыл бұрын

Best way of solving problems 😍😍

@FlippingPhysics 4 жыл бұрын

thanks!

@Devanshgupta 6 жыл бұрын

According to my physics teacher, Newton's second law is "the force applied is directly proportional to the rate of change of momentum of a body". Is it wrong Mr.p?

@FlippingPhysics 6 жыл бұрын

Assuming you meant "net force", then he is correct. See: www.flippingphysics.com/impact-force.html

@Raage. 4 жыл бұрын

*F* is directly proportional to d *P* /dt *p* = m *v* By differentiation F = mdv/dt + vdm/dt dv/dt = *a* and since we consider mass to be constant. *V*/dt will be 0 And so *F* = m*a* Here *F*is force *a*is acceleration *v*is velocity m is mass t is time *p*is momentum

@diegozablah6697 7 жыл бұрын

Hi! This video is great as the others! I have a question though. Is speed ever included in this equation? I mean when you cant use kinematics or conservation of energy?

@FlippingPhysics 7 жыл бұрын

I am sorry, however, I do not understand your question. If you would like to clarify the question, I will do my best to answer it.

@diegozablah6697 7 жыл бұрын

Never mind what i thought was, wasnt. Thanks anyway you are great!

@hazemelyamany5551 5 жыл бұрын

I'm from Egypt, I have been searching too much and I watched a lot of videos in this concern but when I found this one I realized that there is no need to search in such topic anymore you made it easier more than I could imagine ..Thanks is not even enough.. keep going bro I have just a query why did you mention "V i" in the givens as it's worthless? and what does "i" indicate?

@FlippingPhysics 5 жыл бұрын

I am very glad to have helped you learn! The "i" stands for "initial". So "Vi" is the velocity initial which is zero because the book starts out at rest. You would need the velocity initial if you were to use any of the uniformly accelerated motion equations at the end to find out more information about the book, like how far it moves over a given change in time.

@hazemelyamany5551 5 жыл бұрын

@@FlippingPhysics Now it's clear Appreciate your efforts

@qsapolo1449 5 жыл бұрын

In the newton second law, can 9.81 m/s^2 be always substitute by acceleration?

@nathanmontoya439 Жыл бұрын

No, it can only if objects are in free fall near the surface of the planet.

@codosacho5924 9 жыл бұрын

according to newtons second law the net forces equal m*a but what if the body moves and had no acceleration that means that f is zero but it actually move with uniform velocity ??

@FlippingPhysics 9 жыл бұрын

If the net force acting on an object equals zero (all the forces balance one another out and add up to zero) then, according to Newton's Second Law, the product of the mass and acceleration of the object will also be zero. Because the mass of the object will not be zero, the acceleration of the object must be zero. Acceleration equals change in velocity divided by change in time. Therefore an acceleration of zero means the change in velocity will be zero and therefore the velocity will not be changing. If the object is at rest, then it will remain at rest with a velocity of zero. If the object is in motion, then it will maintain a constant velocity. A common misconception here is that no acceleration means no forces acting on the object. This is not correct. No acceleration means no NET force.

@ShawnDypxz 7 жыл бұрын

Please Sir, Reply me. I always get confused when it's stated that a body is acted upon by a force (Say 5 N). Is that a constant force that continues it's application on that body and causes a constant acceleration? Or couldn't it be just a 5 N push only to give a certain acceleration ? and After it's acceleration , it just continues it's motion with the final velocity. And also , What if the body is acted upon by a variable force during it's change of momentum ... There's so much confusion I have with variable force . If variable force is applied on the body for a certain period of time , What will be the acceleration ?

@FlippingPhysics 7 жыл бұрын

Okay, I will do my best to answer your question. 1) Understand it is the net force which causes an object to accelerate. www.flippingphysics.com/second-law.html 2) It an object has no net force acting on it, it will be in equilibrium. www.flippingphysics.com/equilibrium.html 3) Typically, unless the class requires calculus, the effect caused by a variable net force is not addressed. If the net force does vary as a function of time, then calculus would be required to determine what happens. Here is a link to an in-class lecture I did which includes a variable force demonstration. kzfaq.info/get/bejne/Y7pkldqjptmbknk.html I would suggest you also use the lecture notes posted here: www.flippingphysics.com/calculus.html Hope that helps!

@umesh83013 6 жыл бұрын

At 9:24 did u said nine point o metres ?

@FlippingPhysics 6 жыл бұрын

Nope. 1.0 meters. FYI: You can always turn on close captioning to read what I am saying in addition to hearing.

@jasonearls1821 5 жыл бұрын

no he said 1.0 meters

@drishtysharma876 6 жыл бұрын

i want yu to teach laws of motion

@jimkadel3003 7 жыл бұрын

An otherwise excellent presentation, let me offer a suggestion for IMHO an improvement. In previous video/s the "Force Normal(Fn) and Gravitational force(Fg)" have already been described, and shown, in the case of a book lying on a table, as equal and opposite forces. Therefore if one is following "lesson x lesson", the general presentation on forces, it is needless and somewhat confusing to go further into ΣFy = Fn - Fg = m(ay) other than to say IT must be zero and therefore (ay) = 0.

@FlippingPhysics 7 жыл бұрын

I am going to respectfully disagree with your suggested approach. Here is why. In my experience, the more steps students skip, the less they understand. They may grasp what you suggest in this simple problem, however, when it comes to more complicated problems, the simplified approach of just "knowing Fn = Fg" by seeing it in the free body diagram, fails the students. Take an example of a roller coaster car going over the top of a hill. Fn is up and Fg is down, just like in this problem, however, Fn does _not_ equal Fg. If students are not in the habit of walking all the way through Newton's Second Law, they will miss that fact.

@arsenal78910 5 жыл бұрын

I thought there was always acceleration due to gravity in the y-direction, as I remember in kinematics. How come in this case it's not -9.80m/s^2?

@FlippingPhysics 5 жыл бұрын

The object is not in free fall. When an object is in free fall, it has an acceleration in the y-direciotn of -9.81 m/s^2. www.flippingphysics.com/introduction-to-free-fall.html

@unnamedexodus3902 Жыл бұрын

It helps when they answer the questions incorrectly. Shows me why the answer is wrong. Thank you.

@Phoenix-ly5yz 3 жыл бұрын

By dimensions do you mean units?

@FlippingPhysics 3 жыл бұрын

Yes, I do.

@marioandultrachap 8 жыл бұрын

if it says magnitude of all force. why did you only do the y direction and not the x direction. such as force applied and force of friction.

@FlippingPhysics 8 жыл бұрын

+marioandultrachap I am not sure I understand your question. I summed the forces in the y-direction at 3:42 and in the x-direction at 6:38. Perhaps you are confused as to why all the forces are not included when you sum the forces. Because force is a vector, when you sum the forces you need to pick a direction and sum the forces in that direction, which is why we have to sum the forces in the x and y-directions separately.

@marioandultrachap 8 жыл бұрын

+Flipping Physics like where it says the magnitude of all the forces acting upon the book seem like you just did the y direction but not the included the x direction that's where I'm confused at. On part A of the question

@marioandultrachap 8 жыл бұрын

Like what I'm asking is why is there no x direction including in part A of the problem

@FlippingPhysics 8 жыл бұрын

+marioandultrachap The answers to the x-direction forces of Part A of the problem are in the problem statement itself. In other words, the magnitudes of the Force Applied and the Force of Friction are given in the problem statement, therefore, there is no reason to solve for them.

@marioandultrachap 8 жыл бұрын

Oh ok sorry I didn't catch that at first so you don't need to add those forces that you did calculate to the overall magnitude?

@GuerrasLaws 3 жыл бұрын

Physics F=ma test: “Without” applying the Energy from within you, choose an object of your choice, and apply only the Force or Net Force needed to push it away from you or pull it towards you. Please let me know if you’re able to. This is only a test. Thank you.

@learningisecstatic9348 5 жыл бұрын

Sir there is a tress on the fact that g is positive 9.81 si units. Mass is always positive, then why is force of gravity negative? If the force of gravity is negative due to the fact that the direction toward the center of the earth is chosen as the negative direction then g also acts along that direction, therefore it should also be negative 9.81si units. Sir please help. Thank you for your mighty effort and enthusiasm and love and affection that result in such accomplished videos. Love you sir.

@FlippingPhysics 5 жыл бұрын

The acceleration due to gravity, "little g", is a scalar and does not have direction, therefore you _cannot_ say the direction of "g" is _anything_ because it does not have a direction. The force of gravity is a vector and therefore does have a direction. Hope that helps!

@learningisecstatic9348 5 жыл бұрын

@@FlippingPhysics Sir thank you for your quick reply. But my confusion is increased. I am unable to get the why behind the reason of little g being a scalar. If it is a scalar why is mg, the product of two scalar a vector? Sir please help. Or you may suggest me a link of your video that can clarify my understanding. Sir please. Love you sir.

@learningisecstatic9348 5 жыл бұрын

Sir I have just found in wikipedia that gravitational acceleration is a vector. And in some other sites it is said to be negative 9.81 taking direction towards the centre of the earth to be negative. Sir please help.

@FlippingPhysics 5 жыл бұрын

@@learningisecstatic9348 Wow, I am sorry about this. I think I totally forgot that g is a vector. So, here goes: You are correct that g is a vector. You are correct that the gravitational field on a planet points towards the center of the planet which we consider down and negative. Therefore g is down and negative. _However_ , the vast majority of the time we work with g we are using the magnitude of the g which means we are using the positive value of g. If you look through all of the videos I have made which use g, we have always been using the magnitude of g: free fall, range equation, gravitational potential energy, force of gravity, Newton's Second Law, etc. However, you can solve Newton's Second Law problems using unit vectors, and in that case, you do include the direction of g. However (again), I do not typically solve problems that way. Sorry for the confusion!!

@learningisecstatic9348 5 жыл бұрын

@@FlippingPhysics Sir thank you for your reply. It is clear to me. May the almighty God bless you with all sorts of happiness and achievements. Let us continue teaching learning, teaching from your end and learning from our end. love you sir.

@ronerickson8083 5 жыл бұрын

Newton's Second Law of Motion needs a slight correction since they did not know about squaring a number in Newton's era we can assume that this may have been overlooked. As such (F=ma^2) is a natural correction that can be easily proofed through calculation, formulation, and experimentation. This correction is 33% more accurate if you do not accept it proof it yourself.