Physics 8 Work, Energy, and Power (7 of 37) Inclined Plane (Friction)

[@jimmyduca]

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 :)

[@MichelvanBiezen]

Jimmy. Good for you. Work hard. It will be worth it.

[@azizkash286]

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!

[@MichelvanBiezen]

You are very welcome. Thanks for sharing. We like to hear from students around the world. Keep at your studies and good luck.

[@digedanken]

Amazing humble comment!

[@MichelvanBiezen]

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.

[@MichelvanBiezen]

I just took a look. Video # 6 in this set is an example without friction

[@lonewolf3706]

+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

[@triptabhattacharjee7004]

+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

[@viralcat-w4x]

@@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

[@SupermanDee]

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!!

[@MichelvanBiezen]

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.

[@crez12002]

You have a very pleasing teaching style. Thank you brother. Much appreciated

[@znzunid2061]

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.

[@insevanhouts]

"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

[@getdroppedson7799]

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)

[@MichelvanBiezen]

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.

[@arnaskazakevicius8955]

Michel, I would like to ask a quick question: why perpendicular component is with cosine and adjacent component is with sinus?

[@jethior]

hello sir, in this particular problem you didn't use exactly the weight component in the direction of the force, mgsin(theta) but you used only force of the friction wich opposises the applied force. so, why didn't you use mgsin(theta) in this case?

[@MichelvanBiezen]

The definition of the friction force = (normal force) x ( coefficient of friction) = mg cos (theta) x (coefficient of friction)

[@jethior]

@@MichelvanBiezen sir, here we have frictional force = (normal force) x ( coefficient of friction) = mg cos (theta) x (coefficient of friction) is one of the opponent forces to the applied force, and also we have mgsin(theta) as the another resistance forces. so, Fnet=Fapplied-(mgcos(theta) +mgsin(theta)) and work done should be WORKDONE=mgcos(d) + (Fnet)(d) in wich Fnet is ☝. thank

[@tiaan308]

All the way from South Africa and loving the videos. It is helping me a lot for my test tomorrow

[@MichelvanBiezen]

Morne, Good luck on your test.

[@Edwin-su7nq]

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!

[@MichelvanBiezen]

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).

[@Edwin-su7nq]

@@MichelvanBiezen Thanks.

[@gauravbc8226]

wow ..physic is not so Hard as in used to think .....it's like a drinking water tq ..😅for your effort sir

[@starboii568]

Correction H2O

[@geoffphillips8183]

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!!!

[@MichelvanBiezen]

Glad we were able to help straighten that out.

[@debrahm9536]

Can't Thank you Enough Sir.. may GOD bless you👏💖

[@MichelvanBiezen]

Thank you

[@derrickrose5964]

I have a terrible physics teacher so this really helps me thanks so much!

[@neilf659]

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.

[@chrisblack1557]

Thank you, this is highly appreciated, this is a Test and Exam cheat code.

[@KikiShuxteau]

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?

[@MichelvanBiezen]

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.

[@anilrai1582]

Thats so quick.. and i was wasting like hours for solving these type of sums.. thank u sir❤❤

[@kinjimundenda6858]

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 🙏

[@MichelvanBiezen]

We are working on summary videos, now that will clarify that more.

[@nandiniparmar3818]

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?

[@MichelvanBiezen]

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)

[@asiyaamazon6585]

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?

[@MichelvanBiezen]

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

[@amandapamela4516]

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

[@MichelvanBiezen]

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.

[@amandapamela4516]

THANKS

[@josephmontemayor494]

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

[@Godotcoffeestorm]

Great job on the video. This is really helping me review the material.

[@MichelvanBiezen]

Thanks for letting me know. Glad they are helping.

[@yoyovidsv2382]

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 ?

[@MichelvanBiezen]

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.

[@demetriusondieki9560]

i now understand the concept of frictional force ......thank you

[@MuneebIKazi-el9it]

Dear sir, In this particular example, why have we not taken the resultant force instead of the force being applied. Since the object is on an inclined slope, we know that the mg.sin(theta) component of the weight of the object will be acting downwards the slope. So we know the resultant force acting on the object is: 100 N - mg.sin(theta). Therefore, shouldnt work done be equal to = (100 - mg.sin(theta) ) times 20meters ?

[@MichelvanBiezen]

The work done by what? It is necessary to indicate the work done by what force.

[@MuneebIKazi-el9it]

+Michel van Biezen I meant to say that whatever work will be done, it will be done by a resultant force, right? Out of all the force acting on the body, the net force is the one that will cause any work to be done. So, sir, in this case, the NET force acting on the body (upwards the slope) is : 100N - mg.sin (theta) { where mg sin theta is the component of weight opposing the applied force.} My question is. Given that the resultant force is :100N - mg.sin (theta), wont the work done be = (100N - mg.sin theta) × 20 meters. In other words, Workdone = Resultant force (acting upwards the slope) × 20 meters. Thanks. I hope you get my query.

[@MichelvanBiezen]

No, in this example we are calculating the work done by the 100 N force pushing the block up the incline. (and the video correctly shows you how it is done).

[@MuneebIKazi-el9it]

+Michel van Biezen Yes. But wont the 100 N force have to "overcome" the mg.sin(theta) component to move the object up the slope and do work. This is the core of what i want to ask sir. Thanks !

[@MichelvanBiezen]

OK now we are in the correct place to clarify this. Work can be defined in 2 ways. 1) W = F x d or 2) Work done on an object give it kinetic or potential energy (or both). Thus work done BY A FORCE can be calculated by multiplying (dot product) the force time the distance over which the force acts OR it can be calculated by finding the energy gained by the object. Work can also be done to overcome friction. In this case the work done by the 100 N force over a distance of 20 m is 2000 N which imparts 1340 J of KE, 590 J of potential energy and 170 J of it is used to overcome friction.

[@Arhm]

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?

[@MichelvanBiezen]

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.

[@goravm1365]

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

[@MichelvanBiezen]

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).

[@maya-pena-lobel]

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?

[@MichelvanBiezen]

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.

[@leonoreellenrose]

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!

[@MichelvanBiezen]

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.

[@leonoreellenrose]

@@MichelvanBiezen Thank you Sir!

[@temurbekmukhammadiev6909]

Thanks a lot. Your way of explaning is good. ! got a lot

[@giovannifrrri5495]

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?

[@MichelvanBiezen]

Try it and see if you get the same answer.

[@giovannifrrri5495]

@@MichelvanBiezen I did and it's different, I don't think I made a mistake anywhere

[@rahimhussain8511]

Now I can easily solve it in exam. Thanks sir from Bangladesh😘

[@MichelvanBiezen]

Welcome to the channel. Glad we can be of help.

[@ASIFIQBALSHAKIR]

Hi sir i like your method of teaching you define every thing clearly sir keep on doing this

[@Feynman_Fries]

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?

[@MichelvanBiezen]

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.

[@vachan-maker]

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?

[@MichelvanBiezen]

The work done pushing the object up the incline will give it both potential and kinetic energy.

[@3zoabdullah333]

hello professor i have a question, does the work to overcome friction force only affect the knietic energy and not the potential energy?

[@MichelvanBiezen]

It affects total energy which means it can affect both kinetic and potential energy indirectly.

[@vincentmontalban97]

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!

[@MichelvanBiezen]

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.

[@vincentmontalban97]

@@MichelvanBiezen thank you so much sir!

[@iniaku6640]

Why you do not calculate the (mgsin(tetha)) as Flost

[@iniaku6640]

Ohh i see. Because we have include it in potensial energy calculation

[@PrakashPandey-xj8co]

Sir, i have a doubt about when i should use work energy theorem and where to use conservation of energy.

[@sayana9308]

I m also confuse in that...

[@mclio78]

Hello Michel, I wonder if you can show me how did you get cos 180 degrees? When you have calculated E,lost. Thank you: :)

[@vg789-d9b]

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.

[@Mushtaq_kut]

very simple explain , very good examples , big thanks to you from Iraq prof.

[@MichelvanBiezen]

Welcome to the channel!

[@edison16171]

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

[@MichelvanBiezen]

+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.

[@edison16171]

+Michel van Biezen thanks very much professor.

[@albertopoli8896]

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?

[@mariellelegaspi1809]

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

[@MichelvanBiezen]

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)

[@potatoguy5559]

Hey Michel! Quick question. When youre calculating the work used to overcome counteracting forces, why do you not include gravity along with friction? Intuition tells me that both are resisting the upwards push force, so you'd set it up as W(total) = W(push)-(W(g)+W(friction)) Thanks!!

[@adienaarshad7686]

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?

[@MichelvanBiezen]

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.

[@akmalizzat8663]

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 ..

[@cc-to2jn]

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?

[@MichelvanBiezen]

Did you get the correct answer?

[@cc-to2jn]

@@MichelvanBiezen No, it would not work simply because PE only works vertically, and does not follow the path of the object.

[@olgarademan6864]

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?

[@MichelvanBiezen]

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

[@kingonion2102]

Doesn't the X-component of G (the weight of the object) perform work on the object as well?

[@MichelvanBiezen]

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.

[@Yah11]

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?"

[@MichelvanBiezen]

W = friction force x distance = m x g x (mu) x d = 20 x 9.8 x 0.25 x 10

[@albertopoli8896]

Michel van Biezen excuse me but Ihave a doubt .....and how much is the work without friction?

[@深夜-l9f]

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 ?

[@MichelvanBiezen]

+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.

[@深夜-l9f]

Michel van Biezen I see, professor. Thank you.

[@kaiwenyu8354]

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?

[@MichelvanBiezen]

+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

[@randyperez5465]

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?

[@MichelvanBiezen]

+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.

[@anastasiatimofeeva8184]

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)

[@MichelvanBiezen]

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

[@anastasiatimofeeva8184]

Thank you for a quick answer, now I got it!

[@prosenj1002]

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?

[@ogardiga6240]

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

[@MichelvanBiezen]

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)

[@zh-vu7zc]

If a car is moving at uniform velocity on a frictional horizontal plane? then is the applied force by car is equal to frictional force? Is the power of the car 's engine = Force applied × Uniform Velocity?

[@MichelvanBiezen]

You also have to consider efficiency (part of the power the engine puts out, is lost due to heat (thermodynamic inefficiency)), and part is lost due to wind resistance, part is lost due to internal moving parts, and as you indicated part is lost due to friction between the tires and the road. 🙂

[@zh-vu7zc]

@@MichelvanBiezen Thank you very much! You're the best! 💖

[@somayashikder9073]

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.

[@MichelvanBiezen]

You are most welcome

[@sayajinppl417]

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 ?

[@MichelvanBiezen]

Yes, and in this case you also need to consider the energy lost due to friction.

[@rojodemesa8496]

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

[@MichelvanBiezen]

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)

[@rojodemesa8496]

@@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.

[@MichelvanBiezen]

That is correct.

[@mybangtanbiaseskeepchangin7607]

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.

[@MichelvanBiezen]

The friction force on the incline is: F fr = mgcos(theta) * mu Work done is force x distance along the incline.

[@mybangtanbiaseskeepchangin7607]

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

[@alghamdigh8478]

The force component of the weight of the box in the x-axis direction does work opposite to the thrust force, why is it ignored?

[@MichelvanBiezen]

Technically that is not the x-direction, but the direction parallel to the incline. You are correct that the mg sin(theta) force acts against the force pushing the block, but the work done to overcome that is accounted for by the increase in potential energy = mg d sin(theta). We can account for it as work of as potential energy but not both at the same time.

[@kinjimundenda6858]

@@MichelvanBiezen please sir you are really helping us much, now my question is,y don't you write all the questions in order, so that we see as you are moving from answering, question to question, I suggest you must do so 🙏

[@muhammadhasnain2257]

Sir you are my best teacher

[@raamishiqbal3869]

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?

[@raamishiqbal3869]

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

[@MichelvanBiezen]

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.

[@raamishiqbal3869]

Thank you, wasn't expecting a reply to be honest as you seem like a busy person hahahaha. Keep up the great work! :)

[@Mascutting]

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.

[@furkan3592]

Hi Professor , why don't we use Fnet = (100N - m.g.sin(30°)) Wtot = Fnet.displacement ?

[@MichelvanBiezen]

It depends on what you are trying to calculate. If you are trying to calculate the work done by the force that is doing the pushing, your equation will not give you that result.

[@furkan3592]

@@MichelvanBiezen Thank you professor!

[@BlackMambaKO]

If we would want to calculate only work done by net force, would work then be equal to pe+ke?

[@MichelvanBiezen]

That is correct.

[@MimarOd]

Can we found the KE by finding the acceleration first then plug it in the formula v^2=2ad ...??

[@MichelvanBiezen]

Not recommended. Look at the other examples in this playlist.

[@lalistubiyyaa]

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

[@MichelvanBiezen]

That is accounted for by calculating the change in potential energy (which was done in the video).

[@darkeyeballs]

Thank you so much for this, I understand it sooo much better now.

[@ShanaFlame15]

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?

[@MichelvanBiezen]

+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.

[@ShanaFlame15]

Oh I see. Thank you professor.

[@rahultiwari9003]

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

[@MichelvanBiezen]

+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.

[@yousefabuali1961]

thank you very much your are a great person and you've helped me alot , thank you again

[@aryoseno11]

Hi, i have another question. Isnt the force friction should be added by mg sin theta? Because it is the total force that pushes the block....

[@MichelvanBiezen]

Aryo SenoIt is the net force that accelerates the block. There are three forces acting on the block parallel to the incline. Each force performs work

[@aryoseno11]

Michel van Biezen Oh, but what I meant was that why didn't you add mg sin theta with the force friction. That would increase the force needed to push the block up since it is inclined right?

[@MichelvanBiezen]

Aryo Seno That is accounted for by the PE gained. (490J) which is subtracted from the 2000 J

[@aryoseno11]

Michel van Biezen oooooo ok thanks

[@danielyoworld1745]

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

[@MichelvanBiezen]

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.

[@paulucc95videos]

Can you also do some lesson about the integrals and derivate in applicated to the physics?

[@MichelvanBiezen]

Paolo, That is a good suggestion. It will work on that when I get the opportunity.

[@paulucc95videos]

wao! You make me very happy, because you're awesome and you can teach very good =)

[@rosenrot32]

What about the horizontal component of the weight, doesn't it do any work?

[@MichelvanBiezen]

That is accounted for with the potential energy gain = mgh

[@rosenrot32]

@@MichelvanBiezen ooooh. Ok. I understand. Thanks!

[@Hoopsjvl]

I didn't see any comments regarding significant figures. I'm guessing in this example significant figures was not your priority? Thank you!

[@MichelvanBiezen]

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.

[@triptabhattacharjee7004]

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...

[@abinabraham1153]

Sir which frictional force will be there when a bicycle moves up a slope? Is it sliding frictional force or rolling frictional force or both? If rolling frictional force , how to do this problem ?

[@MichelvanBiezen]

Examples of rolling friction can be found in the mechanical engineering videos on friction. MECHANICAL ENGINEERING 11 FRICTION

[@abinabraham1153]

Michel van Biezen can you please give a link for that

[@MichelvanBiezen]

All videos can be easily found by going to the home page.

[@suryakathan]

Could you please tell me why do you use sin along plane and cos in direct 90 to the plane

[@MichelvanBiezen]

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

[@abdirashiidabdullahi7987]

hellow sir, why is the work done overcoming the friction is lost?

[@MichelvanBiezen]

If you slide an object over a flat surface it will eventually stop. That means that the initial kinetic energy the object had is removed by the friction force (the work done by the friction force). KEo = E (lost)

[@Feynman_Fries]

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.

[@MichelvanBiezen]

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.

[@jespersim8668]

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

[@MichelvanBiezen]

+Jesper Sim That is correct.

[@TheGamingWattsit]

Would energy lost take into account all resistive forces (ie. The component of its weight going down the slope mgsin(30)) not just friction?

[@MichelvanBiezen]

Energy lost is the mechanical energy changed into thermal energy (heat). This is caused by friction, wind resistance, etc. Potential energy changing into kinetic energy would not fall into this category.

[@TheGamingWattsit]

@@MichelvanBiezen but surely some of the total work done would be against the force of mgsin30N

[@MichelvanBiezen]

Yes, but that is accounted for with the potential energy and kinetic energy, it does not get converted to heat energy.

[@TheGamingWattsit]

@@MichelvanBiezenok thankyou very much

[@mudassirgulzar5030]

what will happen to kinetic energy if work done is equal to mgsintheta times d neglecting friction

[@MichelvanBiezen]

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.

[@farhanm5021]

Yahoo!! Thank youuu!!

[@MichelvanBiezen]

You're welcome!!

[@monsieurali8484]

Won't work also need to be done against mgsin θ since that force is also opposite to the direction of motion?

[@MichelvanBiezen]

The video is correct as shown. Remember that work done results in an increase of either kinetic or potential energy, (or to overcome friction)

[@monsieurali8484]

@@MichelvanBiezen Thank You

[@mimidimiri2839]

If the applied force is not given how will we find the work done

[@MichelvanBiezen]

You can only find the work done by a force if the force is given. Depending on what is given, we can sometimes surmise the force from the change is speed, the amount of friction, etc.

[@Jirayaaaaaa]

Why you didn't multiply the distance at dcostheta?

[@MichelvanBiezen]

Because the direction of the force and the direction of the displacement are the same direction. (the angle is zero)

[@eon7860]

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!

[@MichelvanBiezen]

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.

[@eon7860]

@@MichelvanBiezen Understood. Thank you!

[@annwr2a2]

hi sir , why you didn't calculate the WORK done by gravity ( component of weight ) witch is ( Wx = mg sin ( theta)) ?

[@annwr2a2]

I meant ( Work Done by Gravity )

[@MichelvanBiezen]

That is already accounted for in the Potential Energy term. (PE = mgh)

[@jonathaniniguez2897]

thank you so much for this video you helped me finally understand this concept

[@nel233]

you're a life saver thank you so much

[@MichelvanBiezen]

Thank you. Glad you found our videos! 🙂

[@tayebaadiba9312]

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...

[@MichelvanBiezen]

We have lots of examples like that in the playlist.

[@anggadamanik8506]

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

[@MichelvanBiezen]

That is already accounted for with the gain of potential energy (490J). Note that the PE calculation contains the sin(theta) component.

[@sudiptoghosh5757]

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 .

[@insevanhouts]

Ever realized your name is an anagram of Stupido? Guess not /s

[@kayanna9843]

in the problem I have the force is horizontal and nor parallel to the plane. im not sure hot to solve this problem

[@MichelvanBiezen]

We have examples just like that in this playlist: PHYSICS 4.7 FRICTION & FORCES AT ANGLES