Physics 8 Work, Energy, and Power (6 of 37) Inclined Plane (Frictionless)

[@yirmeyahuM]

One of the best professors ever! I've just watched a few videos and I can't believe just how much I understand now.Things are not fuzzy! God bless you!

[@jonathankim6013]

Third method: KE = (Fnx) (d) Fnx = (F - mgsintheta) KE = (F - mgsintheta) (d) I was in your PH2AW class last semester and I'm taking it again this summer and these videos are helping me so much to study and prepare for PH1A next semester. Thank you so much for these videos professor!

[@MichelvanBiezen]

Nur Husnina Since there is no friction, we don't need to know the normal force and therefore we don't need to know the cosine component of the weight. Since the force acts parallel to the incline, the direction of the force and the displacement is the same. Remember that W = F * d * cos (theta) where theta is the angle between the direction of the force and the direction of the displacement.

[@STRANGER-_007]

Thanx a lot. i have an exam in mechanics after three hours! I watched all your videos related. You really did an amazing "work" and give us a scientific "energy" with your "powerful" ideas and skills!

[@historyisthebest5831]

Work, Energy, and Power.

[@b..3543]

Sir, without a doubt; this was the most helpful video i've ever watched on this subject ! Thanks for illuminating our path, sincerely from Turkey

[@MichelvanBiezen]

Glad it was helpful! Welcome to the channel!

[@b..3543]

@@MichelvanBiezen got a BA sir; like A- in U.S system. It was my 7th time trying. Now i'll be learning Phy 102 by your videos, thanks in advance

[@photon2724]

omg. i just realised. This man goes through the trouble of printing out and sticking on the writing that is below the whiteboard. This man is amazing!

[@MichelvanBiezen]

That is done by my wife who films, edits, and produces the videos as well as making all the thumb nails.

[@photon2724]

@@MichelvanBiezen You've helped me and several other people a lot with these quality videos, and i just want you and your wife to know that!

[@lee3951]

i had been searching through books for explaination about why mgh in incline planes have to be distance times sin theta for almost half an hour! Glad that i came across your video, you explained it very well!! Thank you!!

[@alexdev404]

Awesome stuff! I really enjoyed this video on *Work, Energy, and Power* ! Keep up the amazing work!

[@MichelvanBiezen]

Glad you enjoyed it!

[@MisterBinx]

This is why in class lectures are useless to me. I can go on the internet and find a clear explanation and watch it 100 times. Anything confuses me I don't have to interupt the class.

[@SteveRelentless]

This guy is a beast.

[@studywithsingh5065]

i hope you had tried to write "BEST"

[@rygalindo]

It is slang to mean that the professor really knowns his stuff or is very skilled. Its a compliment.

[@lukascho2472]

He is real beast.

[@MichelvanBiezen]

Yagmur Erhan Think about it this way: Work done by the 100N force: W = F* d Work done by gravity: W = -mgsin(theta) * d Net work done: F * d - mgsin(theta) PE gained + KE gained = F * d - mgsin(theta)

[@a.j.5747]

What if the force is not given? And the height is 5 m?

[@MichelvanBiezen]

Ammar Junaid Ammar, Not sure what you are asking? What is the question in the problem?

[@deepakgauniyal3799]

so sir PE gained = 490Jand from above equation(including work due to gravity)W(net)=f*d - mgsin(theta)d = 100*20 - 5*9.8*0.5*20implies that W(net)= 2000 - 490 = 1510Jnow if PE gained is 490J then from above equation KE gained would have to be 1020J....

[@abhisheksamyal585]

I am indian and i like your teaching style👍

[@PhotoBomb2000]

THANK YOU SO MUCH! I was totally confused as to how acceleration could be calculated in these problems! You are awesome!

[@ingridsolaina8508]

Wow. You just made it look easy! Thank you, Sir! ☺️

[@egytian100]

I wish you were my teacher

[@yousefsarhan3011]

me too man

[@fatimamahmoud519]

I’ve begun love physics because of you ...thank you so much God bless you ❤️

[@MichelvanBiezen]

That's great. Thanks for sharing.

[@joanneciupak209]

Same

[@3bdul-3ziz38]

Hi Michel , l have a question: μ=0 W(net)=F(net)•d F(net)=-mgsin(θ)+F W(net)=(-5(10)(1/2)+100)•20=1500J?

[@MichelvanBiezen]

Fnet does have a specific meaning. It means the vector sum of all the forces acting at a point or acting on an object. Wnet does not have that same meaning. Work is associated with a particular force acting over a paticular distance. We can talk about the work done by a force pushing an object. The work done to overcome friction. The work done by the force of gravity. The work done to increase the PE, etc. They are all independent of one another.

[@3bdul-3ziz38]

@@MichelvanBiezen In this video we talked about the work pushing the object?

[@kiduseyob4402]

I have a question! why didn't you use the net force while calculating the work done. You used the applied force (100N) which didn't include into consideration the component of weight that opposes the applied force i.e. mgsin(theta)

[@MichelvanBiezen]

When you calculate the work done it is usually associated with a particular force. (e.g. the work done by the force pushing, the work done by the gravitational force, the work done by the friction force, etc.)

[@kiduseyob4402]

@@MichelvanBiezen so had the question asked the net work done, does that mean we would have used the net force instead of the applied one?

[@MichelvanBiezen]

That is correct. Also note that in the energy balancing equation, the potential and kinetic energy are accounted for separately.

[@kiduseyob4402]

@@MichelvanBiezen Thank you so much for taking the time to answer my questions

[@abhigyansrivastava4247]

Doubt- At 5:49, Why did we subtract mgsin theta while calculating acceleration but not for calculating total work done?

[@schoolofscienceglobal]

Very impressive sir!!. But sir, please I need you to recommend a very good physics text book where I can see and solve questions like the ones you usually solve in your videos... Especially for mechanics

[@MichelvanBiezen]

All the major physics text books are good, with plenty of good exercise problems. Choosing one becomes a personal preference.

[@raspberry765]

I had a similar question except the object was moving down the frictionless ramp, therefore changing its PE and KE. Q was asking for the distance travelled down the ramp. However, when I did Wg= deltaKE + deltaU, I get the calculated value for deltaKE+deltaU equal zero. Wg=mgsin theta *d=delta KE + delta U. If I set delta U equal to zero, then I am able to find d and then plug it back in to get my Wnet=Wg= a numerical value (955 J). When I individually calculate deltaU and deltaKE, they are equal to -955 and 955, respectively. I am not clear about why the net Work isn't equal to a combination of PE and KE in my case. Thank you so much in advance!

[@MichelvanBiezen]

You cannot account for the PE and the work due to gravity at the same time. You can only pick one or the other. That is why the initial PE = the final KE, or the negative work done by gravity = KE final.

[@raspberry765]

Michel van Biezen is this because PE is converted to KE aka they are the “same” energy that is a result of the work being done by a force (gravity in my case)? Thank you sooo much for answering my question!

[@DualStixGaming]

Second time watching this vid... I am truly greatful for your videos

[@srii.abhigyan2938]

Thanku respected sir, love and respect from india.

[@MichelvanBiezen]

Thank you and welcome to the channel! 🙂

[@tuktak835]

A very good mental arithmetic!

[@MichelvanBiezen]

Thank you.

[@historyisthebest5831]

Sorry sir it may seem like a stupid question. Actually I understand half of it but not all of the process W = Fd in this case. Why can we multiply F and d directly without resolving them into vector components? And why do we need to resolve them later when you're finding the velocity?

[@jaelly853]

thank you sir ur explaination is very2 clear.. i finally understand the logic of it now

[@kellyhighmore2982]

Thank you for this great video. Can i ask a question can we use Work=the change in kinetic energy in this problem?

[@MichelvanBiezen]

Work can indeed be defined as the change in energy, but you have to account for both the kinetic energy and the potential energy.

[@vachan-maker]

Can we use work energy theorem to say that the total work done by all forces is equal to the change in kinetic energy

[@MichelvanBiezen]

No, because there can also be a change in the potential energy as is the case here.

[@nkosinathintuli3610]

sir, i would just like to know why are we not going to take into consideration the directions, where one would choose a certain direction to be positive, preferably upwards, since it's the original motion of the block, and in that case downwards would be negative and that would make the gravitational acceleration, g, to be a -9.8m/s^2...

[@MichelvanBiezen]

The video is correct as is. When dealing with vectors, the sign indicates the direction, but the direction is uaually expressed in terms of the x and y axis, and therefore you would need to find the vector components. Better just to calculate the magnitude and determine the direction separately. (Note that the magnitude of vectors cannot be negative).

[@vachan-maker]

A doubt, why do we not consider the work done against mgsintheta downwards? so net work would be, work done by external force - work done by mgsintheta and this work gets converted to energy?

[@vachan-maker]

ive just made myself more confused

[@MichelvanBiezen]

You need to take each force separately. If you find the worlk done by the force pushing the object up the hill, the work done is simply the dot product of the force and the displacement.

[@JosieC991]

Thank you so much sir! You're such a fantastic teacher!

[@tanzimahmed2065]

If mgsin(theta) is acting opposite of F, then why we don't use (F-mgsin(theta))*d when calculating work done? Thanks

[@MichelvanBiezen]

We have to ask the question: "who or what is doing the work?" If we want to find the work done by F we use F in the equation. If we want to find the work done by mg sin(theta) then use that force.

[@rajaswalavalkar6712]

I have a doubt, The total work done calculated is 2000J, then 490J is the PE, so you said remaining is 1510J is KE, but there is mg force on the block which have some work done to overcome this. So there will be some energy loss

[@MichelvanBiezen]

The work to overcome mg is equal to the gain in PE. PE gained = mgh Work done to overcome gravity = mgh

[@saimakhankhan2517]

When finding work done, Shouldnt F be 100N minus the component of the weight acting down the slope.

[@MichelvanBiezen]

Not if you are finding the work done by the 100 N force.

[@mormar8608]

Michel u r amazing!!thank u for all the videos❤️

[@natashaevans6379]

Thank you so much! Your videos are clear and very concise

[@MichelvanBiezen]

Natasha. Thank you for the feedback.

[@yankeefan1280]

Great video! One question: Are we assuming that the force required to push the block up the hill is constant?

[@petsworld7846]

if I may ask Sir, through your calculations, Work done by Fg was not considered though, we had (vertical displacement... ) that part confuses me

[@MichelvanBiezen]

Since the force is pushing against the component of Fg which is mg sin(theta), it is actually considered in this problem.

[@ChristineCasul]

Good day Sir! May I know why you used cosine theta in the y-axis? and sine theta in the x-axis? I

[@MichelvanBiezen]

On an inclined plane we prefer to express the components of the weight (mg) of the object on the incline in terms of the perpendicular (to the incline) and parallel (to the incline) components. Then if you look at the triangle, the perpendicular component is adjacent to the angle and the adjacent component is opposite to the angle.

[@travisl7960]

Very helpful, thank you for taking the time to make these...

[@earyrockwell8030]

sir is it doable to use W=Fnet x d? Just finding the unbalanced force/s and multiply with the displacement covered?

[@MichelvanBiezen]

That will give you the work done by the unbalanced ( or net) force

[@kunalkumar-qh7jd]

and the work done by resultant force is is equals to change in kinetic energy

[@christabelnneoma3158]

There's something I don't understand. Can someone explain why the work done is force × distance and not net force × distance. Why didn't we remove mgsinetheta from 100N before calculating the workdone

[@MichelvanBiezen]

The work done by the force is always going to be FORCE x DISTANCE x COS (angle between the direction of the force and the direction of the displacement). This work done will give the block kinetic energy, or potential energy, or both.

[@ThaboJoykieGurl]

Shouldn't the work done include the force that is applied & the weight's component that is parallel to the incline?

[@MichelvanBiezen]

It depends on the question (what is asked). Typically the question will state: "what is the work done by the force applied?"

[@sreebommineni5123]

Sir when you were explaining vectors you told that Fx= Fcos(theta) and Fy= Fsin(theta) But when you divided mg into its x and y component saying mgsin(theta)on its x axis and mgcos(theta) on its y axis. I know I messed it somewhere , but could you please clarify this doubt And your videos are really helpful

[@MichelvanBiezen]

Fx is not always equal to F*cos(theta) and Fy is not always equal to F*sin(theta). It depends on which angle you are referencing, so every problem must be evaluated separately. Remember the definition of cos. Cos(theta) = adjacent side / hypothenuse and sin(theta) = opposite side / hypothenus

[@bsreepaul]

+Michel van Biezen sir , Thank you for your quick replies and for clarifying my doubts

[@imajahan188]

THANK YOU SO MUCH SIR. I understood it perfectly

[@MichelvanBiezen]

Great 👍

[@hazimamery6878]

Hi! Thank you and i love your videos! So much actually. I have a question, for the net work dont we need to take in consideration work done by the gravity?

[@MichelvanBiezen]

When calculating the work done it is typically referenced to a particular force. (i.e. the work done by the force pushing the object up the incline. Or the work done by the force of gravity. etc.)

[@hazimamery6878]

@@MichelvanBiezen thank you!

[@danielcordeiro6003]

​@@MichelvanBiezen If you can do that why is it possible to know the kinetic energy of an object that way? If you used the net force to calculate the work you would have a different result for the KE.

[@asheeldatharam2932]

What about the parallel component of gravity? The work due to F - work due to parallel component of gravity = Wnet

[@kaylachoi4791]

How can there be constant velocity if there is acceleration? when you find the velocity through the kinetic energy, what does that symbolize? The velocity when the force was moving the block? What I mean is how can kinetic energy be proportional to the square of velocity when work has force and therefore acceleration?

[@mendozanailajasmined.720]

Hi. Is total work = PE + KE + work to counteract the component of the weight of block? I'm a little confused, please enlighten me. Thanks.

[@stevenboncolmo8809]

Is it always w=PE+KE? Thanks! :) I really understand all your examples. More power.

[@MichelvanBiezen]

Unless energy is lost due to friction etc.

[@hallobiyo]

@@MichelvanBiezen what about if the energy is lost due to friction

[@nurrizzatimohammadzambree9553]

May I know why the work done by the gravity (mg sin @)is not put Into account,, sorry for the typos

[@MichelvanBiezen]

By the what? (I think you left that out of your question). (but all was taken into account).

[@nurrizzatimohammadzambree9553]

@@MichelvanBiezen here sir☝️, I've edited my question...

[@MichelvanBiezen]

Ah, thank you for clarifying. The work done by the force of gravity is taken into account by calculating the gain (or loss) in potential energy. You can account for one or the other, but not both. That would be double dipping.

[@nurrizzatimohammadzambree9553]

@@MichelvanBiezen I see,, thanks a lot sir

[@jorriesmith3152]

Can you add what you are trying to solve (most is done in videos but I am trying to study with these and I cant solve the problem if it isnt indicated.

[@historyisthebest5831]

Hi sir sorry I have a question. If you have only one force on the object when you draw a free-body diagram and the acceleration is zero, how do I write the equation?

[@MichelvanBiezen]

If there is only one force, that means there must be a net force and therefore there must be acceleration.

[@historyisthebest5831]

@@MichelvanBiezen Oh ok thank you!

[@josephmendez29]

I have a question Is it correct if I calculate the kinetic energy first calculating the net forces in the x-direction and then calculating the work done by the net force? ex: I did 100N - (5*9.8*sin30) = 75.5N. Then w = f*d = 75.5 * 20 = 1510J. Help me, please. Thank you for your videos !!!!

[@MichelvanBiezen]

I suggest you use the method used in the video.

[@adienaarshad7686]

What is the angle you are taking while finding the components of mg? Why is it 30?

[@MichelvanBiezen]

Because the incline is 30 degrees.

[@hassankhan2808]

than k you sir you helped me a lot our teacher spent a week on this and still couldnt explain this to me,

[@pierregarcon3664]

Sorry you felt this way jason. I will work harder next time.

[@saimaarshad8959]

Is there any work done against the component of the weight.

[@MichelvanBiezen]

Yes, that is accounted for by using this method.

[@j0mezzy]

I'm just wondering why total work= KE + PE . I thought total work = KE - PE because wouldn't the Potential Energy be -mgh since gravity is pushing down but we are pushing the block "up" opposing gravity?

[@MichelvanBiezen]

+Jomel Sagsagat It depends on how you want to set up the equation. If an outside force performs work on an object, it will increase its energy. That can be PE or KE or both.

[@MichelvanBiezen]

+Jomel Sagsagat It depends on how you want to set up the equation. If an outside force performs work on an object, it will increase its energy. That can be PE or KE or both.

[@Peter_1986]

+Jomel Sagsagat I like to use the formula W0 + P0 + K0 = P + K for these kinds of problems; the terms W0, P0 and K0 are the initial work, initial potential energy and initial kinetic energy, and P and K are the final potential energy and final kinetic energy. In this problem, you have an initial work W0 being done, which is 2,000 J, and you also have a final potential energy P and final kinetic energy K at the top of the ramp, and the final potential energy P is 490 J. The formula will therefore end up like this: W0 = P + K Solve for the final kinetic energy and you get K = W0 - P = 2,000 J - 490 J = 1,510 J.

[@kalisteticss]

Your videos are so helpful! Thank you so much!

[@ronjayzapanta9887]

Great explanations!

[@Yah11]

What is the definition of work? I understand the equation but I'm just trying to think about what it means conceptialy.

[@MichelvanBiezen]

When you perform work on an object, you will give an equivalent amount of energy to the object (kinetic and/or potential) or the work overcomes the friction forces.

[@xmadness6658]

Hi how are you I get confused Isn't work=the change in kinetic energy? so from where you bring potential energy

[@MichelvanBiezen]

Work is change in total energy (not just KE)

[@NinaMakeAstory]

Sir, why do you cut the cos30 and = 1 ?

[@antoniomantovani3147]

this was of 20 lessons, now of 37, but good lessons

[@aryoseno11]

Mr michel where is the relative angle of cos and sin in the part where you explained mg cos theta etc....

[@MichelvanBiezen]

Aryo Seno The angle between mg and the component mg cos(theta) is the same as the angle between the horizontal and the incline.

[@aryoseno11]

ooohh since theyre perpendicular ok ok thanks...

[@jummyfal8584]

Why are you finding the Fnet? And why is it 100N - mgSin(theta)?

[@MichelvanBiezen]

because a = Fnet / m total

[@temurbekmukhammadiev6909]

Thanks a lot for yuor helpful lectures. It really helped mee

[@Ephilia-gs3zy]

why would we not use the net force to find the overall work?

[@fowz2073]

for the very first . shouldnt we suppose to use w=fdcos30 instead w=fd only . am i wrong?

[@MichelvanBiezen]

The definition of work is: W = F * d (or the dot product of F and d if they are not in the same direction).

[@nadilashafira97]

is it because they have same direction. So their theta is 0, right?

[@saimaarshad8959]

Nadila Shafira yes the force is in the direction of motion so theta is 0

[@nadilashafira97]

saima arshad thank you so much

[@kunalkumar-qh7jd]

sir we are doing work done against mg sintheta so cant we write (f-mgsin0)*d=work done

[@DualStixGaming]

How would you solve this problem if you were only given distance the object travelled, work done, and force.. Basically how do you find the angle

[@MichelvanBiezen]

blkbear h is a function of the angle. h = d sine (theta)

[@lamnguyen-bi8jm]

I have an urgent question! why didn't you account the force of gravity( m*g*sin(delta)), which is pulling the object in the opposite direction, to calculate work. I always thought that we should use the net force to calculate the work. Please help me out! Thank you so much.

[@MichelvanBiezen]

+lam nguyen That depends on what you are trying to calculate.In this example , we are calculating the work done by the 100 N force.You can also calculate the work done by mgsin(theta) which would be negative work in relation to the work done by the 100 N force.

[@lamnguyen-bi8jm]

Thank you so much for responding Sir! Your video is really helpful!

[@beckybuck5031]

Sir, can you explain why it's -mg "sin theta" that counters the 100 N force? I'm confused on the x- and y-components of Fg. I look at that and think x=cos, x runs left to right... I'm a new subscriber to your channel and find your videos very insightful and helpful! Thanks for all you do!

[@MichelvanBiezen]

Becky Landis Becky, Welcome to the channel. Take a look at the playlist: PHYSICS 5 APPLICATIONS OF NEWTON'S LAWS and the videos starting with: Physics - Mechanics: Applications of Newton's Second Law (2 of 20) That is where I explain how the angle used in the components of the weight of the object on the incline is the same as the angle of the incline.

[@jianguili8203]

Hi Michel, Sorry for asking the same question again. I still don't understand when to use sin and cos. Those videos u recommended to me, There is too many of them could u (if u have time) tell me which ones to watch? ps: great videos!:D

[@MichelvanBiezen]

Jian Gui Li Take a look at the last 2 videos in the playlist: PHYSICS 5 APPLICATIONS OF NEWTON'S SECOND LAW

[@reduildcarpe3399]

Shouldn't we have to calculate Fnet first? As far as I know we have to substract mgcos30 before we find W?

[@MichelvanBiezen]

The method shown in the video is correct. But there are multiple ways in which you can solve the problem. Try it your way and see if you get the same answer.

[@reduildcarpe3399]

@@MichelvanBiezen Thank you sir, I guess I made it correct in my exam :)

[@tamarplotzker5231]

can you find the velocity by doing KE=1510j = 1/2(5)v^2 = 1510/2.5=v^2 > v=squareroot of (1510/2.5)= 24.576.....

[@tamarplotzker5231]

oh i just saw that that is the correct answer sorry for bothering u

[@MichelvanBiezen]

No bother at all. Any time.

[@ersinsener4626]

hmm i will ask something you did w = 100x20 but i think 100 should be 75 because of mgsin30 direction of mass to south west ??

[@MichelvanBiezen]

W = (100 N) x (20 m) is correct

[@ersinsener4626]

Michel van Biezen thank you professor

[@parisaaoladi3997]

great teacher

[@jiyavarma4847]

is 100N force that is applied is a non conservative force?

[@faizasikandar6050]

Sir what is the useful work done .input and out put to calculate efficiency

[@oussamabaadj9355]

sir, The potential energy when t = 0 is it PE = 0 J ?

[@MichelvanBiezen]

It doesn't have to be. It can be any value. The work done will cause an increase in PE and KE. (You can set the reference PE to any value)

[@danielcordeiro6003]

If we used the net force to calculate the work we would not have the same kinetic energy in both cases. And that's confusing me a lot.

[@MichelvanBiezen]

You cannot use the "net" force to find the work done. Calculate the work done by a force lifting an object up vertically. If done at a constant speed, the net force would be zero, but yet you lift the object higher.

[@anilsharma-ev2my]

Use enligted black board so we see it clearly?

[@grahamflowers]

There is no kinetic energy in a moving mass there is force Mv squared , kinetic energy is the energy of consistent work from a consistent force regards Graham Flowers

[@MichelvanBiezen]

I think you will find a lot of physicists who do not agree with that statement.

[@grahamflowers]

@@MichelvanBiezen yes I know but can they prove me wrong regards Graham Flowers

[@sarahlolo2100]

how did you determine the direction mg is pulled down ( right or left)

[@nashidixit5]

Can someone explain the part where he finds the net force on the block by dividing it into its components? I think it is around 4:25

[@MichelvanBiezen]

+Ishan Dixit There are lots of examples of forces and acceleration on the inclined plane in this playlist: PHYSICS 5 APPLICATIONS OF NEWTON'S LAWS

[@mihribank2026]

I dont understand why we calculate friction force to calculate net force?Is not it frictionless?And why we calculate it like this?Why dont we use N.μ ?? Thank you so much......

[@MichelvanBiezen]

Even if there is no friction force, you will want to calculate the net force for a number of reasons, both in dynamic and static type problems. In this example, there are two forces (the force pushing and the force due to gravity). If there is a net force the object will accelerate. (Fnet = ma).

[@GerbilIV]

Where is the power video relating to this?

[@johanniral8116]

greetings, dear teacher. i hope you are alright. teacher, i have a question, when you are calculating the net force, why is the algebraic sign of the component of the force [m*g*sin(theta)] negative? thank you dear teacher.

[@MichelvanBiezen]

+Johann Iral Since the direction of the acceleration is up the incline, make any force in the direction of the acceleration positive and every force acting in the opposite direction relative to the acceleration is negative.

[@johanniral8116]

Great, teacher. I have understood. :D I really thank you for your help, and now I am going to do the exam of physics today pretty good. May God bless you. Have a good day. (y)

[@kimtaeyeonismyeverything]

at the very beginning why is W not F*d cos 30isn't the angle between them 30?thank you.

[@asaadalalawi3561]

thank you professor!

[@yunusemreuzmez8308]

Why dont we calculate net force then times with d in this question ? I think ı can solve this problem like that (F-mgsin@) x d - PE

[@MichelvanBiezen]

There are indeed multiple ways in which you can solve this problem.

[@kushraj6768]

Sir please keep making videos

[@MichelvanBiezen]

We will

[@danielyoworld1745]

just wondering why cos tetha = 1 tho cos 30 = squareroot of 3 over 2

[@danielyoworld1745]

our exam is on the next day after tommorow please reply thank u so much

[@MichelvanBiezen]

The cosine of an angle is the ration of the adjacent side to the hypotenuse.

[@danielyoworld1745]

+Michel van Biezen thank u so much for the fast reply. ur video helps a lot.

[@MichelvanBiezen]

Here is a good playlist for you if you want to brush up on trigonometry. TRIGONOMETRY BASICS - PRECALCULUS 6 kzbin.info/aero/PLX2gX-ftPVXWiPVqdVFdCvLDGEtCj8iyy

[@danielyoworld1745]

Thank u so much.

[@quamos99]

mass 15000 kg. X 500 m theta 2.5• Resistance 800N(constant) work done by driving force????

[@MichelvanBiezen]

Assuming that the resistance force is parallel to the incline and assuming that the 500m is the distance along the incline. Work done to overcome resistance = F * d = 800 N * 500 m = 400,000 J. Work done to gain height = mgh = (15,000 kg) * (9.8 m/sec^2) * (500 m x sin 2.5) = 3,206,000 J Total work done is the sum = 3,606,000 J

[@quamos99]

Michel van Biezen thank you sir. i got one more problem- power 30kw time 14 sec initial v 20 m/s final v 30 m/s distance travelled??

[@katyushah.beilschmidt6705]

I dont get 4:58..whats with sine theta?

[@katyushah.beilschmidt6705]

nvm I got it thx!!

[@Nothing_serious]

I don't understand why it works. The potential energy and kinetic energy have different directions so I don't understand why the equation works

[@MichelvanBiezen]

Energy does not have a direction because they are scalar quantities they are not vectors.

[@janayool1001]

sir in this incline we have a degree of 30 so Wp=mgycos30 right it is not flat right ?

[@MichelvanBiezen]

Jana, The force is directed along the incline (in the same direction as the displacement) Therefore the angle between the force and the incline is 0 degrees and cos (0) = 1 That is why I put that example up so you can see the difference.

[@janayool1001]

thank you

[@logancouture4257]

what would be the net work done on the box?????

[@MichelvanBiezen]

Logan, We typically don't calculate the "net work". But the total work done will result in the change of energy for the box and if there is friction, the work will also be responsible for overcoming the friction.

[@logancouture4257]

Okay because I have a homework problem at is asking for the network done on the box and its an extra credit problem so I watched your video to see if you awnsered it.

[@MichelvanBiezen]

Logan Couture My guess would be: Net work = total work - work required to overcome friction In other words Net work = energy gained by the box.

[@logancouture4257]

Okay .. I will watch your video again to see what my awnser will be. That helps. Thank you...:)

[@danielyakubov3677]

What level is this lesson on? By level I mean either high school or college.

[@MichelvanBiezen]

Daniel Yakubov High school and college (1st year)