Physics 8 Work, Energy, and Power (7 of 37) Inclined Plane (Friction)
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- เผยแพร่เมื่อ 9 ก.ย. 2013
- Visit ilectureonline.com for more math and science lectures!
In this video I will show you how to find the kinetic needed to pushing an object up an inclined plane with friction..
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This is awesome! I've spent 2 years learning work power and energy and still am struggling. These videos make me understand a lot more about the topic! Thanks!!
Hi Michel, I go back to school after many years of working, and my subjects are maths and physics (grade 12). By watching you video, i learn more about physics that helps me through most of my home works and tests . You are my physics teacher, Michel! Thanks :)
Jimmy.
Good for you.
Work hard. It will be worth it.
Tripta,
The friction force between an object and a flat surface of an incline is defined as the normal force time the coefficient of friction. On an incline that should be N * mu = mg cos(theta) * mu
I would have to look at the other video you are referring to see what is happening there.
I just took a look. Video # 6 in this set is an example without friction
+Michel van Biezen These videos make the work that I've done for almost a year seems like child's play... so easy to understand! Thank you
+Michel van Biezen thank you sir I get it.I think I was too stressed last time due to the examination.but i got another problem. I cant calculate the kinetic energy using net force like u showed in last video. i think i am missing something in calculating net force.I calculated it,
net F= 100 - mg*0.2(meu)*sin30 = 95.1 and
K.E.= net F* x= 95.1 * 20= 1902
BTW ur videos in physics are my favorite
@@triptabhattacharjee7004 the x-component of weight (w) is missing in your net F and also your friction force should be multiplied to cos(30) not sin(30). Therefore net F = 100N - mg*0.2(meu)*cos(30) - mgsin(30) = 67.01N
KE = net F * x = 67.01N * 20m = 1340J
Michel van Biezen,you are my hero,you saved me from failing physics.I love you soo much sir! it would mean the world if you could reply to me sir!
You are very welcome. Thanks for sharing. We like to hear from students around the world. Keep at your studies and good luck.
Amazing humble comment!
Shane,
In this example, part of the work done is converted to potential energy and part is converted to kinetic energy. The acceleration will not be constant and the equation that you are using only works when the acceleration is constant.
thanks Michel Sir.
most of the time i watch Indian physics great teachers videos . this is first time i watch you and find you are not less then any body .
thanks for such a great help to all student.
"most of the time i watch Indian physics great teachers" "I find you are not less then any body". So you assumed white teachers are inferior to indian ones? Sounds racist
Thats so quick.. and i was wasting like hours for solving these type of sums.. thank u sir❤❤
You have a very pleasing teaching style. Thank you brother. Much appreciated
Thank you, this is highly appreciated, this is a Test and Exam cheat code.
Michel, I would like to ask a quick question: why perpendicular component is with cosine and adjacent component is with sinus?
I have a terrible physics teacher so this really helps me thanks so much!
Great video thanks!! Before I was always confused with the - or + signs of WD against Friction, Work done by driving force, PE & KE. you explained it perfectly clearly!!!
Glad we were able to help straighten that out.
Could you calculate the KE in the same way At previous video? That calculating a from 2nd Newton Law and V from V=v0+ at?
Hi sir i like your method of teaching you define every thing clearly sir keep on doing this
Thanks a lot. Your way of explaning is good. ! got a lot
thank you very much your are a great person and you've helped me alot , thank you again
Please consider this or help clear this up if I am wrong, but calculating the Vf from the KE alone only works if there is no change in height. The reason being is that some of the energy in the system is converted to potential energy with the increase in height, therefore sqrt(2(KE-PE)/5) is the correct way to calculate Vf because the total energy of the system is KE-PE not just KE.
thank you so much for this amazing video! it helped me a lot !
thank you so much for this video you helped me finally understand this concept
Great video! It almost felt like you read my mind when you accidentally said 170N instead of work and then changed it just as I thought about it haha.
Great job on the video. This is really helping me review the material.
Thanks for letting me know. Glad they are helping.
All the way from South Africa and loving the videos. It is helping me a lot for my test tomorrow
Morne,
Good luck on your test.
Thank you so much for this, I understand it sooo much better now.
At the last video u calculated the frictional force using mgsin(theta) bt in this one u used N=mgcos(theta) as reactional force. I can understand that in the last one only mgsin(theta) worked against force. but shouldnt in this case also be N=mgsin(theta)? please reply soon...
while we are finding W shouldnt we use (F-mg.sin(theta)) . d ?
Can't Thank you Enough Sir.. may GOD bless you👏💖
Thank you
i now understand the concept of frictional force ......thank you
Thank you! This is super helpful! Would the energy lost due to friction change if the box was being pulled up the incline by a pulley when a hanging mass is dropped? If so, how would the calculation procedure for this be different?
The calculation for the sliding block would be similar, but you would have to include the other object in the calculations as well. We have examples of that if you look at the other videos in the playlists on work, energy, and power.
Thank you! Very helpful
Very useful, thank you!!
that was really helpful .... thanks a ton .... i would surly prefer this videos to my friends aswell :) god bless you
Thanks Muhammad, much appreciated.
Hi Michel, why do you not take the x-component of the gravity force acting on the object into account? Just like friction opposes the object's motion (which leads to energy loss), the x-component of gravity does too because it has the same direction as the friction. This must mean that both friction force and x-component perform work on the object in the opposite direction of motion. Hence the equation would be:
W(Drag force) + W(friction) + W(x-component) = E(kinetic) + E(potential)
where W(friction) + W(x-component) < 0 (takes energy away from the object)
The force of gravity is alrady taken into account by accounting for the potential energy. It is either one or the other, but not both.
Hey Michael thank you for the response, here is my confusion: when we determine work done by friction down a ramp (of a constant angle like an inclined plane or a varying angle like a curving ramp), we find that the work comes out to depend on the "horizontal length" of the ramp and not the length of the ramp itself (-umgl, where l = horizontal length if the rough ramp). In such a scenario, when work by friction seems to be path independent, how do we justify it as a non conservative force?
It is better to apply the definition of work: W = F d cos(theta) If the force is directed along the incline, then the work DOES depend on the distance traveled along the incline.
very well explanation! Thanks
So can we say that total work = | KE + PE +E(lost) |
I thought the total work = KE - PE - E(lost) because work done by friction is opposing the force and when we are pushing it "up" we are opposing gravity ??
very simple explain , very good examples , big thanks to you from Iraq prof.
Welcome to the channel!
wow ..physic is not so Hard as in used to think .....it's like a drinking water tq ..😅for your effort sir
Correction H2O
Great video! Please can you explain when a cyclist pedals from a height h!on an alert incline plane the work done by him will be gain by KE and work done against friction, but GPE is being lost as height is decreasing. So what will be total work done equation?
Assuming the cyclist is standing up. Eo = Ef W + PEo + KEo = PEf + KEf + E lost Thus 0 + mgh1o (cyclist relative to bicycle) +mgh20 (bicycle + cyclist) + KEo = mgh1f (cyclist relative to bicycle) + mgh2f (bicycle + cyclist) + KEf mg cos(theta) d mu
I loved ur explanation
Hello Michel thank you for helping me with physics and linear algebra. i have a question wouldnt the F in the y direction (F sin theta) also need to be substracted from N? or added? thank you very much and i would appreciate a reply. Thank you
F is directed along the plane of the incline, so there are no components of F to consider. The direction of F is in the same direction of the displacement (d).
Quick question - when calculating the work done (W = Ff*d), why is the angle cos180 instead of cos30? Is it because we already took into consideration of the angle 30 degrees when we calculated what the force of friction was?
180 degrees is the angle between the direction of the friction force (pointing down the incline) and the direction of the displacement (pointing up the incline)
Hi sir! I love your videos so much! They really have been a lifesaver. If you wouldn't mind, I have a physics question based off my homework that I could really use some help on. It goes as so:
"A box is currently sliding across the floor at 1m/s. How much work is required for a person to push the 20kg box 10 meters horizontally while maintaining the speed of 1m/s if the coefficient of friction is 0.25?"
W = friction force x distance = m x g x (mu) x d = 20 x 9.8 x 0.25 x 10
Michel van Biezen excuse me but Ihave a doubt .....and how much is the work without friction?
Hello Michel, I wonder if you can show me how did you get cos 180 degrees? When you have calculated E,lost.
Thank you: :)
I'm three years late but basically the friction force is in the opposite direction of the applied force. This is 180 degrees apart, if it was perpendicular it would be 90 degrees apart.
Please if we push the object and leave it go up the incline and rising ‘. How can we solve this problem ‘ Than you very much
WOW 😍, so impressive work, but my suggestion is, you should be adding all the questions you are answering, from question to question, so that we understand how questions comes and how to tackle them in such a time, rest my case Sir 🙏
We are working on summary videos, now that will clarify that more.
how is the friction or energy lost affected if the force is applies horizontally?
do i add the y component of force to normal before multiplying by mu?
+randy perez
If the force is applied horizontally, then you'll have to find the perpendicular and parallel components of the force. (The perpendicular component would have the effect of making the object seem heavier and it would increase the friction force). The parallel component would push the object up the incline.
Hey Michael, thank yuo for posting such amazing videos. I make use of some of the questions while teaching them. I had a question that has been bothering me so much and haven't been able to find out what's wrong there. Please let me know how I can contact or ask you that the same.
The best way to contact me is through these comments. You can try my e-mail address at El Camino College, but my busy schedule doesn't allow me to get to that all the time. If you do write a question, start up a new comment, because the old comments move down quickly as new comments come in.
Now I can easily solve it in exam. Thanks sir from Bangladesh😘
Welcome to the channel. Glad we can be of help.
Sir you are my best teacher
For the first part of the equation, when you show that W= FxD = (100N)(20m) why isn't the component of gravity that is parallel to the crate included as well? Then F= (100N) - (5*9.8*sin22). The applied force is not the net force is it?
Olga,
For the first part we are only calculating the work done by the 100 N force which is 2000 J
Remember, that is the work done BY the 100 N force.
This work is then attributed to
1) PE gained
2) KE gained
3) E lost due to overcoming friction.
The work required to overcome m g sin (30) = 490 J
The work remaining after accounting for m g sin(30) = 1510 J
Quick question, why didn’t we consider mgsin(theta) work done? It is the force done by gravity and it will be opposite of the motion, much like the friction, so there is an energy lost. Or was the initial force (100N) originally 125N? i.e (125-25)=100N ?
You can pick each force (including mg sin(theta) and determine the work done) separately and calculate the work done by each force. (In this case mg sin(theta) will do negative work because it acts in the opposite direction of the displacement.
question that is quite related to the subject if we wanted to know the maximum velocity in this case should we consider the potential energy is 0 ?
Yes, and in this case you also need to consider the energy lost due to friction.
Hello sir ,I am a student from Bangladesh .Thank you so much for your lectures which helped me tremendously.I hope to meet you soon.
You are most welcome
sir can you please tell me what are the forces acting on a body if it is moving on a horizontal road with const speed and why is the frictional force equal to the applied force in that case and also if the velocity is constant acceleration should be zero and net force should also be taken to zero ?????but why is it not so
+Rahul Tiwari Remember Newton's first law. A body in motion will continue in motion unless a force comes along to change that.Once an object is in motion, it will continue at a constant speed as long as there is no net force (Force of friction equals the applied force). If the net force is zero, there is no acceleration. (F = ma)Note that you first have to get the object moving at constant speed before the applied force can be equal to the friction force.
mike id love to know why you dont subtract work done by component of the weight parallel to the surface from the work done by applied because it does negative work on the object just like friction
Don't think about negative or positive work. It is simply an equation. On the left side of the equation you have the work added to the system ( = Fd). On the right side you have what the work did: 1) Added PE 2) Added KE 3) Overcame friction The 2 sides should be equal to one another.
THANKS
sir i think that the work done by the weight component of the system must be added (-Work) to the left side of the equation. Just like the +100 force that acts on the object the F(weight) is negative. that contributes resistance to the system just like friction
thank you sir
When you are finding the work at the beginning, why don't you subtract the parallel force from the applied force to get F x d?
The definition of work is: W = F x d x cos(angle). Since the force pushes in the same direction as the displacement, the angle is 0.
Can we found the KE by finding the acceleration first then plug it in the formula v^2=2ad ...??
Not recommended. Look at the other examples in this playlist.
I have a question, then what about an friction inclined plane with a string pulling the mass up instead? Would it apply the same or different way? Like for example, finding the amount of work in the string?
+ShanaFlame15
There are 34 videos in the playlist: PHYSICS 8 WORK, ENERGY, AND POWER
Take a look at those, they should be able to answer you question.
Oh I see. Thank you professor.
Hey man i have a question (my energy test is two days from now lol), why is the displacement in the equation for the work lost to overcome friction 20m? Didnt it move 0m because static friction prevents it from moving? Or am i missing something?
+Kaiwen Yu
Static friction is only considered when the object is not moving. Once you overcome the static friction and the object moves, then there is only kinetic friction between the object and the floor
Sir, i have a doubt about when i should use work energy theorem and where to use conservation of energy.
I m also confuse in that...
Just like friction mgsintheta also opposes motion right? so somework is also lost against overcoming this force?so then how do we find pe and ke? thanks. is the work done against mgsintheta being stored as PE?
The work done pushing the object up the incline will give it both potential and kinetic energy.
Hello Michel. Thanks so much for posting all of these videos; however, I have a question at 6:35. Will the work of friction always be added in that equation? Is there any scenario where it would be subtracted? Thanks!
It depends on how the equation is written. But if on the left side you add up all the initial energies and work put into the system, then on the right you must add up all the final energies and the energy lost due to friction. (In this way it is always added).
@@MichelvanBiezen Thanks.
If we would want to calculate only work done by net force, would work then be equal to pe+ke?
That is correct.
you're a life saver thank you so much
Thank you. Glad you found our videos! 🙂
can you help me explain this sir? A 200 kg cart is pushed slowly up an inclined plane. How much work does the pushing force do in moving the object up along the incline to a platform 1.5 m above the starting point if friction is negligible
Since the car is being pushed slowly, we can assume that it does not accelerate and therefore does not increase its kinetic energy. We are told we can ignore any energy losses due to friction. Then all of the work done results in the gain in potential energy. Therefore the work done = mgh = (mass) (acceleration due to gravity) (height gained)
Hi, may i confirm the KE included in your first energy equation will be lost slowly after 20m and it will be converted into GPE and overcoming the friction?And when it stopped after awhile,the KE will be 0, GPE gained and energy lost will increase right?Thank you
+Jesper Sim That is correct.
Just as you said in the first video, Work done to overcome friction is positive bc Force to overcome friction is opposite to the direction of frictional force and here Energy loss is the work done to overcome the friction so its positive?
If you use the definition of work: W = dot product between F and d (Force and Distance). Therefore work is related to a force and the distance and direction the object travels with respect to the direction of the force. If the direction of the force and the direction of travel is the same, the work done is positive.
Hi sir, I was wondering why the angle you used changed each time? At first it was cos0 then cos30 and then cos180. Thanks!
+Julia Alaimo
30 degrees is the angle of the incline
0 degrees is the angle between the force pushing the mass up the hill and the direction of motion (dot product application)
180 degrees is the angle between the weight component parallel to the incline and the direction of motion
what if the questiom is what is the work of firction in 20m should we use positive sign or negative sign? zjust clarifying thank uou so muh
That depends on how the question is asked. It the video here the sign is positive because you want to ADD all the the sources of energy that was converted from work.
hello! can I ask, could you also use this formula of kinetic energy = 1/2 mv^2 instead?
Yes, (1/2) mv^2 can always be used for the KE
I need help in this sum:
find the work done by frictional force as the block of mass 10kg slides down an inclined plane of angle 37 degree to horizontal. coefficient of friction is 0.5. and ht of incl plane is 3m.
The friction force on the incline is: F fr = mgcos(theta) * mu Work done is force x distance along the incline.
Michel van Biezen sir, the answer is supposed to be -200J. The closest value I've got was 278J when I added mgcos37 and Nu
love the bow tie haha!
Sir,
when i started to find KE first and did the same thing like a=F-mgsin(theta)/m
I found the acceleration to be 15.1 and the velocity to be 24.57(approx)
Why do you think i got a wrong velocity even though i should be getting something close to 23 m/s
Since there is friction, you must take into account the friction force. Look at the videos on inclined planes in this playlist: PHYSICS 5 APPLICATIONS OF NEWTON'S LAWS th-cam.com/play/PLX2gX-ftPVXV0Hc1CGnYKaPq19P3RX0Lx.html
Does the horizontal component of the weight oppose motion and if so why haven't you subtracted the work done against it from the total work done?
It depends on how you solve the problem. Note that there is an equation: W = KE + PE + E lost. That means that the work put into the system equals the sum of the KE gained, the PE gained, and the energy lost due to friction. That is the correct equation.
In the equation w= ke + pe + energy lost, why is the work done to overcome mgsin30 is not also accounted as energy lost?
Ex: w= ke + pe + e friction + e mgsin30 ..
what will happen to kinetic energy if work done is equal to mgsintheta times d neglecting friction
If the force applied is equal to the mgsin(theta) component, the net force would be zero (Ignoring friction), then the acceleration would be zero, and the KE would remain the same.
Hoping that my question be answered, why is the Ffriction being multiplied to cos 180 and not cos 30? Thank you.
Your videos are really helpful in making us, students, understand things over our subjects. I aspire! Thank you SIr!
When calculating the work done use this equation: W = F . d (where F and d are vectors) = W = F d cos (theta) where theta is the angle between the direction of the force and the direction of the displacement.
@@MichelvanBiezen Thank you Sir!
thanks really help.
Could you please tell me why do you use sin along plane and cos in direct 90 to the plane
Look at the right triangle made when using the force of gravity, (mg), and the components making up that force. If you want to determine the magnitude of those components you need to use the sin and cos. sin = opposite / hypotenuse cos = adjacent / hypotenuse
Hi, why isn’t the friction work (Ff + Mgx)d cos(180)? Because both mgx and the friction is working against the pushing force. I would appreciate an answer asap (have a test in 1 day) //Sweden
Use the definition: Friction force = normal force x coefficient of friction. The normal force is the perpendicular force between the object and the surface which in this case is mg cos(theta)
I want to ask a question.. why the work of mgsin(a) isn't included? I'm a little bit confused since i thought mg sin(a)will slow down the force and has a work by multiplying it with displacement. Please help me
That is already accounted for with the gain of potential energy (490J). Note that the PE calculation contains the sin(theta) component.
hello professor i have a question, does the work to overcome friction force only affect the knietic energy and not the potential energy?
It affects total energy which means it can affect both kinetic and potential energy indirectly.
Wouldn't there also be energy lost due to gravity, which is mg sin(theta)*d?
I'm confused because we apply mg sin(theta) when we do dynamics.
Please help!
Take it one step at a time. Word done is simply force x distance (and time the cos of the angle between the force and the displacement if not zero). Work will overcome friction, and add energy to the object (or both). In this case work is used to overcome friction, add potential energy due to height gained, and add kinetic energy due to velocity gained.
@@MichelvanBiezen Understood. Thank you!
Can I ask you sir. Where did 180 deg came from? Thank you so much. You've really help me a lot. No words can explain how much I'm thankful of your so very crystal clear explanation lecture videos. God bless you sir!
If the force is acting in one direction and the displacement is in the opposite direction then the angle between the force vector and the displacement vector is 180 degrees.
@@MichelvanBiezen thank you so much sir!
Hi Michel! Great videos, thank you so much, it really helps. But could I ask you to explain one more time (or in the other way): when we calculate W=PE+KE+E*lost, why is friction with a plus sign? (time: 6:50)
Anastasia Timofeeva
Think of it this way:
On the left side of the equation you have all the energy that is already there (the initial energy) + any additional energy created by doing work.
This energy is then used to give potential energy and kinetic energy to the final state AND is used to overcome any friction.
So you have to add up all the energy on the right side of the equation
Thank you for a quick answer, now I got it!
Sir in this part you took
W(total) = K:E+P:E + E(loss)
But if I want to put the value of E (loss) which is (-170 J) then is it will be in + or I will put that as it is with -ve sign?
thank you :)
Hi Professor Michel I would just like to clarify if the cos theta in the formula W = FDcos(theta) means that the angle of the work done? Because my intuition at first to solve for the work on here is simply by plugging W = (100N)(20m)cos 30 because it is inclined 30 degrees. However in the video shown you applied it as W = 100(20)cos0
Work = force x displacement x cos (angle between the directon of the force and the direction of the displacement) So the work done by the foce pushig the mass is: F x d = 100N x 20 m But the work done by the frtiction for is: Ffr x d cos (180) = mg cos(30) x 20 x (-1)
@@MichelvanBiezen I already understand the way you approach this problem. Very clear to be honest. I assume the W=Fd that resulted to 2000J is you imagined that the box is on a straight surface causing cos0. Then you used the 30 degree to find the height by using trigonometry. and so on.
That is correct.
What would've happened if we calculated the net force subtracting friction there? And then we find the acceleration, then final velocity and then KE and subsequently GPE, and adding them we get a different result. So that way is wrong?
Try it and see if you get the same answer.
@@MichelvanBiezen I did and it's different, I don't think I made a mistake anywhere
thanks so much!!
I didn't see any comments regarding significant figures. I'm guessing in this example significant figures was not your priority? Thank you!
I tend not to push the significant figures and concentrate on the principles and how to solve the problems. We have separate videos on the topic of significant figures.
Sir, I have a doubt??
Why can't we use F-mgsin(theta) - mu mgcos(theta) = ma to calculate acceleration(a)
Then we can easily calculate KE by v^2 = u^2 + 2as. So, KE = 1/2 m v^2 .
Ever realized your name is an anagram of Stupido? Guess not /s
Michel Van Biezen.. Sir Can you please explain how I will find the net force when there is friction.. Like in pervious vedio U have shown how to find out KE by finding out the velocity with the help of net force. If I want to find the KE in this case with the help of net force... Can you please tell me how I will find the net force in this case...
We have lots of examples like that in the playlist.
Doesn't the X-component of G (the weight of the object) perform work on the object as well?
When an object is on an incline, it makes more sense to take the weight of the object and subdivide it into the parallel and perpendicular components.
Hey, I did this problem slightly differently. Not sure if I'm correct; Instead of substituting h=d*sin0, I substituted d=h/sin0 into (W=Fd) Wfric= mu*mgcos0*(h/sin0). Is this still correct?
Did you get the correct answer?
@@MichelvanBiezen No, it would not work simply because PE only works vertically, and does not follow the path of the object.
In the work-energy principle, i always take the kinetic and potential energy as negative, if the outcome is negative and vice versa. Havent came across a question with negative friction though. Im confused? So we only have to change the sign for friction?
I use this, Work done by driving force= Kinetic energy+Potential+ any work done by resistance. So if my kinetic
energy comes out as negative, i use it as negative in the principle. Why not for friction. Would highly appreciate a reply. Thank you
Rather than worrying about the sign, think of it like this: If you do work on an object you will add energy to it. If the friction acts in the opposite direction of the force doing the work, it will take energy away from the object.
Thank you, wasn't expecting a reply to be honest as you seem like a busy person hahahaha. Keep up the great work! :)
Energy lost by friction, work done by friction is converted into thermal and sound energy, 170 J was converted to thermal and sound energy, right ?
+The Scientist Technically you are correct, however, sound energy tends to be very small or negligible. We typically only consider energy converted into thermal energy.
Michel van Biezen I see, professor. Thank you.
Are you not supposed to find the energy lost due to mgsin(theta) cause is still a force that is pulling the box down even if we negelect the frictional force
That is accounted for by calculating the change in potential energy (which was done in the video).
Hello professor, when calculating the PE why do we use mg.h and not mgsin(theta).h, isn't mgcos(theta) cancelled by the normal force in that equation? thanks
+edison ortiz 歐帝森
PE is always mgh regardless of how the object gets there.
If traveling along the incline h = mgd sin(theta) with d being the distance along the incline.
+Michel van Biezen thanks very much professor.
Proffesor I have a question, so if the net work is the sum of all the forces, shouldn't it be: (2000J)+(-170J)+(490J) ? which = 2320J
Because you wrote that 2000J =.... but isn't that 2000J just the work done on the object by the 100N?
I'm confuse how you got the net work and the total K
Thanks for the videos and the website, when I saw your website I saw heaven hahahah thank you with my very deep gratitude for the videos :)
+Arturo Fernandez
The problem is correct as worked out in the video.
Ef = W - E lost due to friction. where Ef = KEf + PEf E = energy KE = kinetic energy PE = potential energy
why are we not using the net force to calculate the work done? I solved the problem first before I watched the entire video and what I did was finding the net force that is pushing the block on the ramp. I found 1830 N. my PE and KE are still the same answer you found.
Edem,
There are usually multiple ways to do a physics problem, thus sometimes it comes down to preference.
My purpose in most of my videos is to enhance the understanding of the concepts.