i wish there was a higher education reformation where students were capable of taking courses from tutorial services such as yourself and still count as a credit towards our degree. My issue with "weed out" classes is you pay a lot of money for a service that isn't provided because of its deliberate design to discourage students. If a business tried doing this in the private sector, it would quickly go out of business. thank you for your work and your many examples which, not only provides set-up examples, but provides them in such that it reinforces all the concepts we are learning in school.
Yes, we hope that in the future, it will be possible to earn degrees without having to go to college. It would improve the education because they would have to compete by improving the quality of the education. Thank you for your comment.
physics3a- why is there no initial kinetic energy at the start when we let the block go? I would assume that when we let the block go the spring would immediately propel the block.
You can either count the potential energy of the spring before the block moves, or the kinetic energy once the spring is extended and the block is moving, but not both at the same time. (That would be double counting).
Very Nice. I forget we can skip all the inbetweens for these multistage problems. The way you showed how you can backtrack helped clear up a few of my other misconceptions as well. I keep wanting to treat each stage as its own PE-final or KE-final and it gets tedious, not to mention unnecessary >.>
Sir, why didn`t we assume that energy loss is 0 as we are considering the mechanical energy after(up the incline)- just so we consider the fact that there is no friction up the incline
This video makes sense except for one little detail, since there is an instantaneous moment which the surface switches from flat to sloped, won’t that moment act as a collision? We would lose even more heat that way
We are also ignoring wind resistance, heating effect of the spring, etc. We are simply concentrating on the principles, and ignoring these types of details. Every physics problem in every text book will have these types of "things" in them.
For the equation to work: W + PEi + KEi = PEf + KEf + E lost the energy lost due to friction must be positive. If you move the E lost to the left side of the equation, then it would be negative.
I don't see ( -0.47m), but the block does not have enough initial KE to overcome the friction of the flat portion and therefore does not reach the ramp.
Hello! I have a question. For some reason I was expecting kinetic energy to be an intergral element in this problem. I say this because from the moment the block departs from the force of the spring to the top of the ramp, wouldn't kinetic energy be part of the equation, particularly, Energy Final?
If you only look at the beginning and the end, the block is not moving at t = 0 and it is not moving when reaching the maximum height ) at the end. (Therefore no KE)
I was pulling my hair for half an hour on why d was negative and then when I watched, I did love the video as there is indepth analysis and a new way to think of the problem. Thank you.
me rn lmao. i like to solve the problem and then see if i got it correct, and i was going absolutely crazy because i didnt know how to interpret the negative value.
I'm confused. Shouldn't the heat lost become positive because the work done by friction is negative? Since friction acts opposite of the box's displacement, shouldn't work by friction equal to friction force times distance times cos(180) which would make it negative?
It depends on how the equation is defined. For the initial equation to be correct as shown in the video, the heat lost term placed on the right side of the equation must be positive.
Also I am doing a rube Goldberg rn and in my energy box project we are supposed to show how our calculations show transformation of energy. In mine I have marbles going down ramp and hitting another marble. We must use energy not momentum and impulse and I'm confused. For future students can u do a vid on collisions calculations using ENERGY not impulse and momentum and elastic stuff?? Ty
Hi there prof! On 7:34 why can't i acumulate the term (m*g*sinΘ) to number 2 divisor on the left? Then it would be (k*x^2 - mgμD)/(2*m*g*sinΘ)... Why can't i put like this? The answer seems to be obvious to me, but when i did like how i said, i got positive number to "d" and all the numbers matched. Sorry for the dumb question, and it has been a pleasure watching your videos!
If you multiply both sides by 2, you must multiply m g mu D on the left side as well. Also what you wrote does not include an equal sign, so I can't tell what you did. (always write everything in the form of an equation).
Thanks prof! Sorry about my miss!! Here's what i did, in form of equation(on 7:34): I did: (k*x^2 - mgμD)/(2*m*g*sinΘ) = d Instead of: (1/2*(k*x^2) - mgμD)/(m*g*sinΘ) = d
Yes!! I just got my error and came back here to tell this! Thank you very much, you're the best professor ever seen! Sorry for wasting your time! Your patience and knowledge as a teacher is a life goal to me!
I have to ask, I've seen most of your videos regarding Mechanics and I don't understand one thing. If the Work-Energy Theorem is W= E_0 - E_f, how come you use W + E_0 + E_0 = E_f + E_f + Heat lost in every example?
+NC Quadraxis It is just a general equation. There is only heat lost when there is a resistance force like friction of wind resistance. There may not be any W (work put into the system). It is just there as a place holder and when you look at the problem you have to decide what term you have and what term does not apply.
+Michel van Biezen Oh ok thanks and thank you for naking these videos, you really make us understand how and why we calculate things without complicating it so uneccessary like most teachers do :)
Should I always multiply the force friction with the distance because that friction played a role for the whole distance? Assuming, it goes the whole way, of course.
Michel van Biezen not the incline, I am just talking about regular friction, do we always multiply the distance and force of friction to find how much energy was lost?
There was a similar question that I have a problem with. Can you look into it and provide me the solution? The problem is from Kahn, Yoni; Anderson, Adam. Conquering the Physics GRE (Kindle Locations 1009-1012). Cambridge University Press. Kindle Edition., Section 1.3.5 question 1 The following three questions refer to the diagram: a pinball machine launch ramp consisting of a spring of force constant k and a 30 ° ramp of length L. 1. You want to launch the pinball (a sphere of mass m and radius r) so that it just barely reaches the top of the ramp without rolling back. What distance should the spring be compressed? You may assume friction is sufficient that the ball begins rolling without slipping immediately after launch. The answer given in the answer sheet is (mgL/k)^0.5. My answer came out to be (2mgL/k)^0.5.
Umm..what about the work done by friction which will be umgLcos30? Why this hasn't been incorporated even though it is given that frictional force is acting on the surface?
Since that term is placed on the right side of the equation, it must be positive in order to balance the equation. If you move it to the left side, it will be negative.
it say's that the incline is frictionless. wouldn't that imply that the friction effects the block before the incline? PEs - friction = MGH?! I'm totally lost lol
Either you let W be the work done to compress the spring or you account for it by using the PE stored in the spring, but you cannot count both at the same time. (that would be "double dipping")
+mahim .nirob That is already accounted for by the stored energy in the spring. W is used when there is an outside force applied to the system doing work.
This is an equation hardly anyone else uses tho I am getting confused cuz teacher says use wext + kei + pei = kef + pef. I wish u would do it this way cuz it's hard to relate
An excellent lecture... Thank you ❤
Lismary,
I took a note of your request.
It takes a little while before we can get a video through our filming cycle and production cycle.
What a great way to take a problem that had a small error in it and turn it into something special. Very cool.
Thanks 👍
Thank you for sharing solutions for these types of questions. We usually encounter these questions during exams and it is not fun lol.
i wish there was a higher education reformation where students were capable of taking courses from tutorial services such as yourself and still count as a credit towards our degree. My issue with "weed out" classes is you pay a lot of money for a service that isn't provided because of its deliberate design to discourage students. If a business tried doing this in the private sector, it would quickly go out of business.
thank you for your work and your many examples which, not only provides set-up examples, but provides them in such that it reinforces all the concepts we are learning in school.
Yes, we hope that in the future, it will be possible to earn degrees without having to go to college. It would improve the education because they would have to compete by improving the quality of the education. Thank you for your comment.
Thanks to this video I aced my physics quiz today. Thank you so much!
physics3a- why is there no initial kinetic energy at the start when we let the block go? I would assume that when we let the block go the spring would immediately propel the block.
You can either count the potential energy of the spring before the block moves, or the kinetic energy once the spring is extended and the block is moving, but not both at the same time. (That would be double counting).
Very Nice. I forget we can skip all the inbetweens for these multistage problems. The way you showed how you can backtrack helped clear up a few of my other misconceptions as well. I keep wanting to treat each stage as its own PE-final or KE-final and it gets tedious, not to mention unnecessary >.>
Yes, only considering the beginning and end points makes the problem much easier.
Sir, why didn`t we assume that energy loss is 0 as we are considering the mechanical energy after(up the incline)- just so we consider the fact that there is no friction up the incline
No energy is lost going up the incline, but energy is lost sliding over the horizontal surface.
@@MichelvanBiezen Thank you sir I now get it
This video makes sense except for one little detail, since there is an instantaneous moment which the surface switches from flat to sloped, won’t that moment act as a collision? We would lose even more heat that way
We are also ignoring wind resistance, heating effect of the spring, etc. We are simply concentrating on the principles, and ignoring these types of details. Every physics problem in every text book will have these types of "things" in them.
What about the sign convention? The friction causes negative work??
For the equation to work: W + PEi + KEi = PEf + KEf + E lost the energy lost due to friction must be positive. If you move the E lost to the left side of the equation, then it would be negative.
The number (-0.47m) what does it tell us about the position of the block besides the fact that it doesn't go up the incline.
I don't see ( -0.47m), but the block does not have enough initial KE to overcome the friction of the flat portion and therefore does not reach the ramp.
@@MichelvanBiezen Thanks I was referring to the value of d
I appreciate your support and time, thanks
Hello! I have a question. For some reason I was expecting kinetic energy to be an intergral element in this problem. I say this because from the moment the block departs from the force of the spring to the top of the ramp, wouldn't kinetic energy be part of the equation, particularly, Energy Final?
If you only look at the beginning and the end, the block is not moving at t = 0 and it is not moving when reaching the maximum height ) at the end. (Therefore no KE)
What would change in case the incline had friction in it?
Then you need to add the energy lost by friction term = W = F d = mg cos(theta) mu d
I was pulling my hair for half an hour on why d was negative and then when I watched, I did love the video as there is indepth analysis and a new way to think of the problem. Thank you.
me rn lmao. i like to solve the problem and then see if i got it correct, and i was going absolutely crazy because i didnt know how to interpret the negative value.
I'm confused. Shouldn't the heat lost become positive because the work done by friction is negative? Since friction acts opposite of the box's displacement, shouldn't work by friction equal to friction force times distance times cos(180) which would make it negative?
It depends on how the equation is defined. For the initial equation to be correct as shown in the video, the heat lost term placed on the right side of the equation must be positive.
Also I am doing a rube Goldberg rn and in my energy box project we are supposed to show how our calculations show transformation of energy. In mine I have marbles going down ramp and hitting another marble. We must use energy not momentum and impulse and I'm confused. For future students can u do a vid on collisions calculations using ENERGY not impulse and momentum and elastic stuff?? Ty
Are there any examples of where this equation is used?
Are you referring to the energy conservation equation? We use it in many of the examples in the energy playlists.
Hi there prof! On 7:34 why can't i acumulate the term (m*g*sinΘ) to number 2 divisor on the left?
Then it would be (k*x^2 - mgμD)/(2*m*g*sinΘ)... Why can't i put like this?
The answer seems to be obvious to me, but when i did like how i said, i got positive number to "d" and all the numbers matched.
Sorry for the dumb question, and it has been a pleasure watching your videos!
If you multiply both sides by 2, you must multiply m g mu D on the left side as well. Also what you wrote does not include an equal sign, so I can't tell what you did. (always write everything in the form of an equation).
Thanks prof! Sorry about my miss!! Here's what i did, in form of equation(on 7:34):
I did: (k*x^2 - mgμD)/(2*m*g*sinΘ) = d
Instead of: (1/2*(k*x^2) - mgμD)/(m*g*sinΘ) = d
There is still an algebraic error. Note that the 1/2 is only part of kx^2
Yes!! I just got my error and came back here to tell this! Thank you very much, you're the best professor ever seen! Sorry for wasting your time! Your patience and knowledge as a teacher is a life goal to me!
No waste at all. This is how we learn. Glad to be of help.
Gotta say I might have been stumped by the negative result, but it's what gives this problem an extra element of learning capacity, in my opinion.
Not sure which negative result you are referring to.
@@MichelvanBiezen Oh....the first answer was a negative 0.47 m.
If the block is moving, it has velocity.....why is KEf = 0?
When maximum height is reached, the block will come to a stop.
I have to ask, I've seen most of your videos regarding Mechanics and I don't understand one thing.
If the Work-Energy Theorem is W= E_0 - E_f, how come you use W + E_0 + E_0 = E_f + E_f + Heat lost in every example?
+NC Quadraxis It is just a general equation. There is only heat lost when there is a resistance force like friction of wind resistance. There may not be any W (work put into the system). It is just there as a place holder and when you look at the problem you have to decide what term you have and what term does not apply.
+Michel van Biezen Oh ok thanks and thank you for naking these videos, you really make us understand how and why we calculate things without complicating it so uneccessary like most teachers do :)
Why no initial friction work . 50 cm surface also have a friction .Because 50 cm into the D=5m surface
W0=mgk50cm
The problem states that the block slides 5 m over the horizontal surface.
Should I always multiply the force friction with the distance because that friction played a role for the whole distance? Assuming, it goes the whole way, of course.
The incline is frictionless.
Michel van Biezen not the incline, I am just talking about regular friction, do we always multiply the distance and force of friction to find how much energy was lost?
Thank you very much sir - - - Can we put. (0٠1). instead. (0 - 2 ) then the body can go on the inclide.cerfase
That would work. We have other videos that show similar examples.
Thank you again sir
There was a similar question that I have a problem with. Can you look into it and provide me the solution?
The problem is from Kahn, Yoni; Anderson, Adam. Conquering the Physics GRE (Kindle Locations 1009-1012). Cambridge University Press. Kindle Edition.,
Section 1.3.5 question 1
The following three questions refer to the diagram: a pinball machine launch ramp consisting of a spring of force constant k and a 30 ° ramp of length L. 1. You want to launch the pinball (a sphere of mass m and radius r) so that it just barely reaches the top of the ramp without rolling back. What distance should the spring be compressed? You may assume friction is sufficient that the ball begins rolling without slipping immediately after launch.
The answer given in the answer sheet is (mgL/k)^0.5. My answer came out to be (2mgL/k)^0.5.
You need to use an energy balancing equation: Eo = Ef (1/2) kx^2 = mgh where h = L/2 Therefore (1/2)kx^2 = (1/2) mgL
Umm..what about the work done by friction which will be umgLcos30? Why this hasn't been incorporated even though it is given that frictional force is acting on the surface?
Instead of potentional energy with height i used mgsin theta * distance and got the same negative result. Seems like it would work too
It would be better to change the compression of spring to 1 meter so the block have enough energy to make it up incline.
Very tricky problem but I'm glad I learned how to work problems like this. I'm kind of tired of easy problems
This is an amazing detailed problem! Thanks!
Why you have taken work done by friction (heat lost) positive ?
Since that term is placed on the right side of the equation, it must be positive in order to balance the equation. If you move it to the left side, it will be negative.
it say's that the incline is frictionless. wouldn't that imply that the friction effects the block before the incline? PEs - friction = MGH?! I'm totally lost lol
If you move the term for the friction loss to the left side of the equation it would become negative and it would equal the equation you wrote above.
thanks a lot, your videos are the best.
Why does W become 0? I don't get that part.
Either you let W be the work done to compress the spring or you account for it by using the PE stored in the spring, but you cannot count both at the same time. (that would be "double dipping")
Why the W is zero? didn't the spring do any work?
+mahim .nirob
That is already accounted for by the stored energy in the spring.
W is used when there is an outside force applied to the system doing work.
He explained that we're starting after the spring is already compressed
The heat will be also lost on inclined plane, why you didnt took it into account.
According to the problem there is no friction on the incline.
50 CM =.50 METERS NO? NOT 5.0?
+Sandra Rivas
I wrote 0.5 not 5.0. Are you referring to the distance of 5.0 m along which the block slides?
50 cm is half a meter so 0.5m
50 cm = 5 deci-meters = 0.5 meters = 0.05 decameters = 0.005 hectometers = 0.0005 kilometers , don't forget mili->centi->deci-> normal meter -> deca -> hecto -> kilo , and so on..
In India high school kids are solving this problem for fun ..
Yes the schools in India are very demanding!
This is an equation hardly anyone else uses tho I am getting confused cuz teacher says use wext + kei + pei = kef + pef. I wish u would do it this way cuz it's hard to relate
The equation that I use is correct. Depending on how your teacher interprets the equation he/she uses it could be wrong.
so the small d=1.33m
I see what you did there :)
(150-19.6)/98=1.33............mguD=19.6.......suppose to be 19.6 not 196
(20)*(9.8)*(0.2)*(5) = 196 (not 19.6)
distance of 50 m and not 5m
maybe, I plugged in the wrong values.
for my work done by friction I got 1950j. (.5)(1200)(50m)= 1950j
50 cm = 0.5 m