This guy is simple, my physics professor makes it seem like we need to do way more than what's actually needed. This guy just helped release a lot of stress the day before the exam.
For my later use 0:11 :(part A) horizontal force and friction example (finding net horizontal force) 2:53 : (part B) acceleration, time, initial speed=0 (finding distance) 4:22 : (part A) frictional force (N), force horizontal (calc net horizontal force) 6:50 : (part B) calc acceleration 7:45 : (part C) final speed after a distance 8:52 : initial speed, final speed, time value (finding acceleration)
This really helped me, thank you so much, my Fiancé has started redoing some school subjects and she’s been really struggling with this and so I wanted to help. You have helped me understand and now I can help her 🙌🏻
Yo for reAl like right now am trying to solve a problem with only two weights (50kg and 80kg) it has no friction ..... calculate the speed .... me ....THE FUCK
He's skipping the steps of wiring out the formulas and units. If you do that, it's more tedious, but much easier to follow. It's also the correct way to do it. He's also ignoring sigfigs, etc.
I understand this but i just dont know when and how to use the formula You really exerted plenty of effort here. Even though I am still 11 years old, I now have a little information about this. Thank you The Organic Chemistry Tutor
This channel is the best in the world I have been trying to understand this from my lectures for two days with no hope but here in few minutes I could understand the picture please carry on , and give us some examples when there is more forces on y axes as well so we can understand how it will work altogether thanks for your support
I learned something in like 12 minutes and actually learned something than sitting in a classroom where the teacher just lectures for half an hour, saying things that doesnt make sense. Thank you!
Force does not exist physically like an object with mass. Force, as we know it, turns out to be nothing more than an expression to express an idea like one would use the word "Love" to express once feelings. Force or Net Force is not the initial cause of motion. Energy (applied energy) is the origin of motion and not Force. Example: Choose an object of your choices and without applying the energy from within you, try to push and pull by applying "only" the Force or Net Force. Meaning that Energy (applied energy) is the origin of motion and not "Force". Once Energy (E) is applied, it creates what is known as Momentum (p). When this Momentum (p), which is mass and velocity, comes in contact with an object(s), it makes a surface contact that will enable you to push and pull. Energy (E) and Momentum (p) or "Ep", are not to be confused with Kinetic Energy in any way. Without Energy, there is no Momentum. Without Momentum, there is no surface contact on an object(s) to push, pull, work, shaping objects, motion, gravity, magnetism, radiation, etc. Momentum does not and cannot exist without the applied Energy that creates it. Energy and Momentum or "Ep", is the only common denominator that links all fundamental forces of nature. Without Ep, all fundamental forces of nature would be inert or non-existence. ~Guadalupe Guerra
I hope this videos may help me to understand cause I was the only dumb one in my class while everyone finished the work in 5 mins saying that it SO EASY!!!!!
To the people who can’t understand, Write with him! Write what he’s writing and it might make sense, I do this with all his videos and I understand it quite alot
Sir ma'am, for Qn2 ur answer is wrong part b). The tensional force exerts a vertical force upwards of 175N. Since the weight of the 12kg block is 117.6N. Thus there would be a net vertical force exerted onto the block of 57.4N. With this and the net horizontal force acting on the block, there would be a resultant force of 308.51N (3s.f). Acceleration is therefore 25.7m/s^2 (3s.f).
You are so smart. I'm good at Physics but AP Physics is a little challenge for me. I need a 5 on the exam and I believe that you and my prep book will allow me to get a 5. 😃
Question 2. the 12kg box actually feels an upward force exerted by the surface, the value is - 57.4 Newton (yes, minus 57.4 N) if a 350 Newton force acting on it at an angle of 30 degrees to the horizontal level. It means, the box has been actually lifted from the surface, therefore all the subsequent calculations of the horizontal net force, the acceleration, and the final speed after 200 meters will be invalid.
yes but actually no. You are right if you're only talking about the picture he drew, in that some force would have to hold the box to the surface for it to keep traveling only horizontally (rather then a "negative upwards force" It's probably easier to call it as a positive downwards force, eg gravity, similar to how friction is a positive backwards force, not a "negative forwards force" [even though mathematically they are the same thing of course]). Notice though that the actual question includes no mention of a surface, or definition of the frictional force as a RESULT of drag on a surface. Therefor, for the purposes of solving the question, the box is floating in liquid, air, or space and the frictional force will be continuously working on the box from all sides, not just from a surface. So in respect to the net horizontal force, in the video is absolutely correct. You're also right however that the acceleration and final speed are incorrect unless we assume that there is some force (eg gravity) that is stopping the box from moving upwards. Otherwise the acceleration and traveling would happen diagonally; the frictional force would be the same (assumed from the word "constant" and no mention of it only being horizontal horizontal) but we'd need to calculate with the original 350 N. I get a = 19.17 m/s(squared) and Vf = 87.57m/s for this
The problem with the 12 kg box pulled to the right by a 350 N force at 30 degrees above the horizontal contains a contradiction in the initial conditions. The scalar component of the downward normal force (i.e., the force that is supposed to be pressing the box against the surface) in this case is (12 kg)(9.8 m/s^2)sin(-90 deg) + (350 N)sin(30 deg) = 57.4 N, which is a positive quantity. It means that there will not be a sliding motion along the surface, since the latter requires that the scalar component of the downward normal force be negative. In other words, if you pull a 12 kg box with a force of 350 N at 30 degree angle, the box will take off the surface. Thus, the problem, as stated has no physical meaning, and the solution provided is not correct, I'm afraid.
Isn't it 8400N for the force exerted by the car and the net force is 8400N - 3500N (frictional force) which is 4900N? So, I think the average force would be equal to 5950N. Please let me know your thoughts about this. Thanks! BTW, I really appreciate your vids. :)
can you answer this plss..the velocity of an object at a time is given by v=4.0+2.0t. find the magnitude of force acting on the object of mass=4.0kg find the kinetic energy of an object of a mass-4.0kg and time 3.0t setting the initial position x=0.0m of an object at time t=0.0s calculate the time (t>0.0s) at the position x=21m of the object?
Great video but so many missing links. Like I didn't understand when the acceleration was squared and when it wasn't. And when initial velocity was also squared and it wasnt
So draw a line pointing up at a angle that looks like this, this is the force vector. / / / / Now turn it into a right triangle /| / | / | /θ_| Remember Soh cah toa Sin θ is opposite/force vector Cos θ is adjacent/force vector Solve for the opposite and adjacent to find fx and fy
I don't know if you will see this but @6:33 can you please tell me what the net force would be because I worked it out but I am not sure if I am right.
just a question about number 2, the force of 350 N at an angle of 30 degrees is enough force to lift the box from the ground, so the acceleration should have a horizontal and vertical component; am I missing something?
This author doesn't pay attention to things like that so he often makes mistakes. If it flew in the air, that friction would cease, yes. He doesn't even track sigfigs. Not a good teacher after all, other than the general concepts
Very nice explanation...just you should use 's' as a symbol for displacement...you are using 'd' for distance...this is not wrong but in equations of motion ...'s' is used...so not for confusion it should used...☺️🤗😊
for the acceleration of the second problem you divided the newton by 12 kg, wouldnt you have to convert the kg to 120 newtons? Just asking, kind of confused.
Newton's Second Law of Motion needs a slight correction since they did not know about cubing 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.
