Work Problems - Calculus

This calculus video tutorial explains how to solve work problems. It explains how to calculate the work required to lift an object against gravity or the work required to push a car with a constant force to a certain displacement. This video also explains how to calculate the work done by a variable using calculus. The definite integral of a force function with respect to x is equal to the work done. This lesson explains how to calculate the work required to pull a rope to the top of the building as well as the work done in pulling half of the rope to the top of the building. In addition, it explains how to calculate the work required to stretch a spring beyond its natural length using hooke's law and by calculating the value of the spring constant. Finally, this tutorial contains an example problem that explains how to calculate the work required to pump all of the water inside an inverted conical tank to the top of the tank. You need to know the density of water and the gravity acceleration for that problem.
Arc Length Problems: • Arc Length Calculus Pr...
Surface Area Problems:
• Surface Area of Revolu...
Work Problems - Calculus:
• Work Problems - Calculus
Integration By Parts:
• Integration By Parts
Tabular Method:
• Integration By Parts -...
_______________________________
Trigonometric Integrals:
• Trigonometric Integrals
Trigonometric Substitution:
• Trigonometric Substitu...
Integration By Partial Fractions:
• Integration By Partial...
Trapezoidal Rule:
• Trapezoidal Rule
Simpson's Rule:
• Simpson's Rule & Numer...
________________________________
Improper Integrals:
• Improper Integrals - C...
Integration Into Inverse Trig:
• Integration into Inver...
Integration of Rational Functions:
• Integration of Rationa...
Integral of Logarithmic Functions:
• Integral of Logarithmi...
Integrating Exponential Functions:
• Integrating Exponentia...
__________________________________
Calculus Final Exam and Video Playlists:
www.video-tuto...
Full-Length Videos and Worksheets:
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[@TheOrganicChemistryTutor]

Next Video: kzbin.info/www/bejne/qYi2fX-FaKmGZ6s Final Exams and Video Playlists: www.video-tutor.net/

[@brianbc4]

Not sure how I'm going to pass Cal 2 without this channel and the tuition for nonstate is costing me $1400 per credit. Therefore, I owe so much to this channel.

[@redpanda4102]

fr lol if my prof doesn't make sense im not sitting around waiting for office hours or anything like that im coming right here

[@Luke-gp1gz]

Ah, my favorite. Every draining a tank problem in our textbook or given to us for homework involved cones or triangular troughs, but when it comes to exam prep, the practice exams have trapezoidal troughs just to make the process different and hard to find help on solving... Great video, though! Very clear and informative!

[@EmpyreanLightASMR]

Why can't engineers stop using inverted cones for their water tanks 😂

[@BioHazarDeath1]

I have learned so much in 30 min then I have in 2 classes of calc ...... thank you so much

[@YozoraYue]

BioHazarDeath1 my teacher just basically gave us an essay explanation of all the steps (literally, typed and printed out; and it is a bad idea because it’s a math class) and told us that work is this or that without telling us how they got that information from the problem given. So I was totally confused how it went from kg m^2 / s^2 to Joules and other stuff

[@Rihiihiiihiiihii]

Keep up the good work. Helping a lot of people.

[@gardyion2869]

Heh, you said the thing.😏

[@Sadkiller100]

Srsly helping me 3 years later

[@threeoutchange669]

If I wasn't so irresponsible and hadn't left to study the day before the exam, I would get a good grade. I'll probably fail Calculus 2. Good video though

[@joseangelesrojas6406]

sooo how did it go?

[@threeoutchange669]

@@joseangelesrojas6406 I passed it hehe. My final grade was much better than I expected. The next bosses are Data Structures and Linear Algebra

[@icono__7136]

Good luck on those! Data Structures is a b*tch so don't loaf on the class or you'll not know what hit you when you guys come around to Trees and Queues

[@threeoutchange669]

@@icono__7136 my man, thank you. Thing is, I'm already suffering and haven't even hit these ones yet. Currently I'm on Linear Sorting, which I already find it difficult. But let's go anyway!

[@icono__7136]

@@threeoutchange669 it's okay man. Have you tried GeeksForGeeks videos or Abdul Bari?

[@vsalukir7019]

I think for problem 6, the correct equation is "The integral from 0 to 80 of 240 - 3X dx". Then for part B just change the limits to be from 0 to 40. No breaking it into parts is necessary. It makes the computation simpler. And simply reverses the spring method where the force increases to one where the force decreases.

[@vsalukir7019]

Alex DulZ Exactly. My equation works for all intervals without breaking the problem into multiple pieces.

[@pkt1e735]

Why would it be 240 - 3x?

[@ethanlane9873]

@@pkt1e735 Because as you pull the rope up, the weight of the rope hanging over the edge decreases by 3 pounds per foot (x).

[@ruhaanvasudeva9120]

@@vsalukir7019 i think so you dont have to subtract 3x

[@fflspectrum5833]

he meant to say 2400 - 3x@@pkt1e735

[@christianarmandocruz]

Your the best man, you’re the reason I understand calculus

[@whitlow4291]

Thank you so much for this! My calculus teacher at the moment is a piece of garbage so I was excited when I saw that you had made a good, comprehensive video on how to do this stuff. You sir are a lifesaver!

[@Awai_quotes]

Why school or college teachers dont teach like that?🤔

[@haruharudude]

For the integral of the last problem, it's a lot easier if you just factor out the constant and do integration by parts if you can't use calculators

[@DoctrinaMathVideos]

Good point, but in a lot of the calculus textbooks applications including this one comes before the introduction of integration by part (aka product rule for integration).

[@redpanda4102]

@@DoctrinaMathVideos yeah my class is doing this before integration by parts

[@BarbaraRademacher]

Your videos are the BEST and most helpful for me as I review Calc-2 after - wait for it - 39 years!

[@georgesadler7830]

Professor Organic Chemistry Tutor, thank you for a deep explanation and analysis of Work Problems in Calculus Two. This material is also presented in Engineering Physics One, Dynamics and Engineering Electromagnetics for all students in Science and Applied Engineering. This is an error free video/lecture on KZbin TV with the Organic Chemistry Tutor.

[@smalltamm8264]

0:00 how much work is required to lift a 5kg book 2m high above the ground? 2:05 Find the work done in pushing a car a distance of 15m with a constant force of 500N 2:50 A variable force ( in Newtons) acting on an object is represented by the function F(x) = x^2 - 4x + 7 where x is the distance in meters from the origin. How much work is required to move the object from x = 1 to x = 4 5:45 A force of 50lb is required to hold a spring stretched 5 inches beyond its natural length. How much work is required to stretch it from its natural length to 10 inches beyond its natural length in ft-lbs? Too lazy to do the rest, add on the comment below for people searching for specific problems ;)

[@sighbrows]

5:28 Spring problem

[@flyin5952]

Thank you so so so much!!! I could not have learned all of this without your awesome help in teaching us slowly and with so many examples!! Thank you so much for taking tiem out of your day to help us!!!

[@andybabay9879]

that last problem hurt my brain

[@elysel9424]

I agree 😭

[@luqmanshah7942]

Tell me about it.....

[@pianoman2247]

that’s considered the “simple” problem for my high school multi var test this week

[@redpanda4102]

missed my lecture due to covid, literally im up to speed off of this video, truly he explains everything so well

[@liammcbride9210]

Awesome video, taught me a lot more than a college course could do!

[@yamnational9461]

John Green and his brother (Crash Course) and this guy is all you need to pass any class. straight up OCT deserves the nobel peace prize. at least alot more than Obama does.

[@cocopowderkid]

You are a life saver sir. I will donate as soon as i can

[@aryanprakash1048]

I'm ecstatic that I'm finally understanding this stuff after so long trying to

[@davidarnoldstevenson3007]

For the last problem, could you explain why you integrated from 3 to 12 in a better way. I understood most of your work, but I still have some questions.

[@notkupid]

Did you get your question answered?

[@lucasdwyer9018]

For Number 6 I don't think you had to integrate part B as the bottom half of the rope is not varying you can simply take the (Length)*(Weight per foot)*(Distance) so 40ft of rope at 3lbs per foot is 120lbs and it will travel 40ft = 4800.

[@melodyearley1247]

Your the reason i wont fail my calc 2 test tomorrow! THANK YOU

[@braeling3677]

Dude you saved my life thank you

[@Kanra2981]

This video is a LIFESAVER. Keep up the good work!

[@yingyihuang7316]

Please please please do videos for C++!!!

[@haasofdetroit9828]

#5 can be done using the cm values then converting to m at the very end. It was much more manageable that way. You'll get 330 then just divide by 100 to get 3.3

[@BM-_]

My calc teacher put this video up on his notes for reference. Little does he know, everything I know for calc is from the Organic chem tutor and not him xD

[@sergiosanchez5137]

26:42 OCT: "If you wanted to know that." Us: "WRITE THAT DOWN WRITE THAT DOWN"

[@triparadox.c]

I am thankful to KZbin.

[@supermanley101]

Around the 15 minute mark you said weight density. Weight density is also called specific weight which is lb/ft^3, I'm not sure if lb/ft has a unit name. In engineering statics we're just calling it weight.

[@samuelpachoochoo5114]

30:49 can you explain why the displacement would just be x?

[@CJ-nd2nd]

Love all our videos! they are so helpful. Qst - in the first problem. How can a problem in kg and m can be different than a similar problem in ft and Lbs? They are both the same type of mass and distance? I don't understand? Can you explain why when we see kg and m we have to use the formula W = mgd and when we see lbs and ft we have to use W = Fd Thank you!

[@joshuas5310]

kg is a unit that measures mass, while lbs is a unit that measures weight, which already accounts for the effects of gravity.

[@ckosmic]

Cant wait to use this every day in my computer science job 😊😊

[@younghoonyoungseo1405]

For question 7, a more efficient way could be this: W = F * d = Weight * distance. Weight (Total) = Weight (Rope) + Weight (crate) = 4*(200-x) + 300 = 1100 - 4x If you do integral with bounds [0,200], you would get 140,00 (without having to divide the process into two steps)

[@DanielInfrangible]

You're a hero, and you have my gratitude.

[@markosuchy825]

Just saved my ass the night before a test. 💕 much love

[@edidiongedemeka2096]

shouldn't the interval on the integral in number 8 have its limits from 9 to 12?

[@jeroen3648]

I really learned alot from your videos. Thank you

[@sherlockai937]

Thank you, this helped me.

[@darkknight5314]

In example number six. What happened to the height of building 200ft

[@sunnyliu1038]

Nice lecture! so clear explanation, thank you very much!

[@danieljohnsopardenilla997]

The force equation at 22:37 implies that the force required varies as the crate is being lifted up. Why can't it just be the overall lbs of the rope (800 lbs)? Does it get lighter as the crate is being lifted? If so, shouldn't there be more context indicated in the problem (e.g. mentioning a pulley or something?

[@Awai_quotes]

I was finding this video from past 1 week

[@cilva7able]

I would like to see more examples per video. How about when it comes down to the power rule you skip the math and just state the answer and then you can use that time to do more examples.

[@arshiaa8474]

quick question : how can we use F = KX for the rope problem. Isnt F = KX only used for a spring?

[@mitchellcarlson1371]

You are seriously awesome man. Thank you!!

[@phetolomojela3014]

you the best man, you very helpful

[@CrAzY6tothe6]

For the last problem how come you can't just take the slope and get r=5y/12 which would then be pi×(5y/12)^2. I know you have x but it can be confusing if the limits of integration are vertical therefore wouldn't it best be everything respect to y?

[@luisarmenta2619]

This came out 3 days after my test exactly on this lol

[@matthewm3]

lol that sucks

[@AK-gb9rp]

Came out about 2 years before my test haha

[@georgebrian9273]

Watching from Africa

[@nickd.2449]

We need more of this stuff, great work!

[@edwardhope4166]

I am completely confused on the second part of pulling up half of the rope. We know that half of 80 is 40, so why wouldn't the answer be integrating from 0 to 40 3x? Why does take 3 fourths the work to pull do half the work? I just seems logical that if it took me 100J to pull an entire rope then it would only take me 50J to pull half of it.

[@cameronguilbeau5888]

Think about it in small chunks. If you were pulling a rope with a small weight attached to the end up from over a building, it would be a little heavy and require some work. Now imagine You got through the hard part and you have about half a foot left to pull over the edge. That last bit is going to be way easier seeing as you already pulled up most of the load.

[@edwardhope4166]

@@cameronguilbeau5888 Your post doesn't answer my question. The concept behind the rope being lighter as the ropes length getting shorter is not my problem. I understand how to calculate the work pulling up the entire rope, but I am lost on how he went about calculating the work for pulling up half of the rope. Can you explain that to me.

[@Chrisymcmb]

@@edwardhope4166 So I was confused by this as well. Basically, when we calculate the force needed to pull the first half of the rope, there is still a piece of rope dangling under the first half of the rope we pulled, therefore we use the variable force function f(x)=3x. However, if we are going to pull the rope all-together, then there is no unknown part of the rope dangling beyond the the end of the part of the rope that we are pulling, we are pulling up the whole rope. Therefore, right there we don't need the unknown x in the f(x)=3x, instead, f(x)=3(40). So I think, the "x" is basically when there is something still dangling beyond the the point at which we finish pulling the rope.

[@Born2Losenot2win]

11:52 Wait, why did u subtract the domain of the wanted work by the domain of the given work to get the answer? Wasn’t it enough to put 15 and 20 cm for the domain in meters?

[@Born2Losenot2win]

Oh I get it nvm

[@jayall2155]

this is so so so so helpful!!!!

[@jackschmell2186]

The formula for calculating the work done in a cone is incorrect. The tiny dx contributes to the Volume. So instead of V(x) it should be A(x) inside in the integral.

[@audryssayuanwie6107]

For no.6, so does that mean the value 200m does not matter at all?

[@merbun3170]

do u have more videos on the inverted cylinder problem?

[@zachmoir]

Your awesome dude. Thank you so much!

[@potchi3086]

Are the limits of 3 and 12 correct for the last problem?

[@Junker_1]

On the first problem, with the book that weighs 5 kg. Doesn't he confuse weight with mass? The book weighs 5 kg but the mass isn't 5kg.

[@copernicus99]

This is great, thank you! At 3:27, to calculate the work done in lifting an object above the surface of the Earth over a distance, would you substitute the gravitational force equation, F=Gm1m2/r^2, in the integral?

[@seoinchoi4595]

can anyone tell me what (around 14:50) X is for question 6? It says that F=kx, and according to F=ma shouldn't X be acceleration? But the unit is ft meaning X is the length, so I am a bit confused here🥲🥲

[@JKMath]

They are two different formulas for force. F=ma is for an object with mass m that is in motion, but F=kx is for the force required to stretch or compress a spring a distance of x given the spring constant k (Hooke's Law). They are not the same formula. Does this help? Let me know! I know my response is a little late, but maybe it will help someone else out who is reading the comments. Have a great day! :)

[@nelsoncoop3774]

the spring problem is wrong apparently, you have to use some kind of spring formula. following your steps gave me drastically different answers.

[@loganlauck2093]

In example 4, before you converted to ft*lbs, why were the units lbs*in and not lbs/in ?

[@aliduaneabdi7464]

number 8. I thought the height of the small section should be dy instead of dx!!!!!! Clarify please

[@broskie1]

Doesn't the k in kx have a negative sign that is carried through the integration?

[@lailanesanchez4181]

thank you

[@tyront83]

For problem number 5 was there a need to convert cm to m? I'm not saying it's wrong but I guess I'm lost as to why he did it?

[@EricZhao-n2d]

Why do we want k in newtons/meters? 11:00

[@Se7enMile-Slider]

why did you have to calculate the bottom half as well, if you were only required to find the work of pulling up half the rope?

[@vangmountain]

You have to do the bottom also because work has to been done to it to get it to move up. How else did it move up? You may be thinking that work was done on the top half and because it was connected to the bottom half, all you have to do is apply work to the first half, move it up, and the bottom automatically comes up because they are connected, but that's not the case. Work is done to any and all pieces of the rope that has moved any distance. The only difference with the bottom half is that all, dx parts of the rope, move the same distance so the same amount of work is done to all the dx pieces of the rope, hence, summing all those infinitely small work at the bottom will give the total work done on the bottom half. Hope this helps. The key is to remember that if no work is done on the bottom half, it would not move up. Cheers :).

[@pedrooreosg9180]

My brain just goes blank when I see a word problem

[@brianpk7206]

A statement has never been more true

[@superligitguy]

wow dude txs so much this is great

[@vangmountain]

I was looking at the work problem for the rope and it's actually mistaken. The answer is correct, but the explanation is incorrect. The total work done is figured by summing all the infinitely small amounts of work(dw). dw(change in work) is simply, FD. In the example, he has F = 3X. This is incorrect. F is in fact, NOT 3x. F = dx ft *3 lbs/ft. The "ft" units cancel leaving you with a force of 3dx lbs. F = 3dx NOT "3X" as he shows. The reason why force is" 3dx", is because you divide the rope into dx sizes. As a result, the weight per dx size of the rope, is simply, dx ft * 3lbs/ft which is 3dx lbs. To get the dw(changing weight) for each of these dx lengths of the rope, we simply use F*D. Substituting "3dx" for F and "X" for D gives us "3dxX" which can be rewritten as "3Xdx". As such, even with this error in explanation, the problem still works. For students learning this material, they need to understand that the total work done is the sum of all these small changes in work. They need to know where everything in the integral comes from. "dx" are the infinitely small pieces of rope that the rope is divided up into. We can figure it's weight(force due to gravity), by simply multiply this dx length by how much the rope weighs per lbs. What we see is that we are moving infinitely small pieces of rope which all have the same weight (force), but each is moving a different distance X, hence, to get the total work done on the rope, we have to add all these little tiny bits of changing work. This is the understanding that needs to happen to understand these rope problems. A note about the 2nd half of this problem is that you do not need to integrate it to get the work done on the bottom half. Because the same amount of work is done on ALL dx lengths of rope (all dx lengths of rope move the same distance), the work done is simply the total of all the dx lengths, which in this case is 120 lbs(force), times the distance traveled, which is 40ft.

[@Chrisymcmb]

You should make a video explanation please

[@vangmountain]

@@Chrisymcmb are you saying you don't quite understand and would like to see it in video form to help you better understand? Yes, in video form it would make it easier to understand. The problem with a lot of these is that he shows you the steps but the explanations are not that great. Sometimes there's just no explanation. If all you are doing is following directions you will never understand. You must understand the concept behind the methods. Anyone can follow directions...but what happens when the problem changes...then the directions themselves have to be changed. Only someone who understands the concepts can come up with a new set of directions. If you are only following directions with no understanding, when those directions are noy applicable, what do you do? That's the problem students fail to understand. If you are in need of help and would like to understand how these rope work problems are done, i guess i could do one, but I was really waiting to do it on an educational channel I'm trying to get started on.

[@Chrisymcmb]

@@vangmountain it’s be great if you could do one on this topic man

[@vangmountain]

@@Chrisymcmb Sure...I can put up a quick one...nothing special from vid standpoint but for sure it will help you to grasp these rope problems

[@Chrisymcmb]

@@vangmountain that’s awesome, thanks man!

[@kanem8751]

For the cone problem, wouldn't it be the integral from 9 to 12, not from 3 to 12? The water was filled up to 9m.

[@dev5224]

I think that's because you want to empty water that is filled with 9 m on the bottom of cone by simply subtracting 3 from 12. If you would do the integral from 9 to 12, you empty water that is filled with 3 m. I also have a question on this video, how did he get 1000 for density?

[@kanem8751]

@@dev5224 The density of water is 1g/mL. One cubic meter of water is 10^6 mL, and 1000kg is 10^6 grams. Therefore, 1000kg/(m^3).

[@karabokhumalo3285]

I don't understand why the 200ft was not used at all in n.o6

[@ahmedsalim5370]

How can the height of the disk be equal to dx instead of dy? isn't the height equal to dy when the cross section is perpendicular to y-axis? I got R = (5/12) y

[@JensenPlaysMC]

it doesnt matter in this context. but yes, use dy

[@vsalukir7019]

He represented the vertical axis as X and used it consistently that way.

[@hugobarahona5552]

(Everything in context of y) Your way: R = (5/12)y now let y = 12. R = 5/12 (12) = 5 His way: R = (5/12) (12 - y) now let y = 0. R = 5/12 (12) = 5 You were looking at the problem from the bottom (zero) to the top (12). Whereas he was looking at the problem from top (zero) to bottom (12). It's like if you were standing at the bottom of a house looking up. You would see 12 meters above you. And he was standing on the roof of the house. He would see 12 meters bellow him.

[@andrewsalazar98]

On problem 7, why is F=mg not used since the crate is affected by it? Can we assume that the weight includes this?

[@pianodan1608]

Weight itself is a force, which means weight = mg. This means that the the quantity of the weight already takes into account gravity.

[@johnreyhunat243]

For no.8 how come the outcome had come to that?

[@sharjeelhayat6787]

if problem 8 why are you using the formula of cylinder for volume of cone ?? should you not use the formula 1/3*pi*r^2*h. I think you missed 1/3.

[@vangmountain]

He is correct. The volume displaced in the cone is the sum of an infinite number of cylinders that are infinitely thin. It's so thin, the volume of each cylindrical slice is essentially just the surface area of the top of the cylinder. The only thing is that from one cylinder to the next, the radius changes. It gets so small that it becomes a point for the very bottom slice. It's for this reason why you use the volume of a cylinder not a cone. The cone is not made of cones but rather, cylinders with varying radius. ⁰

[@maanpatel9716]

god bless your soul

[@kellyrivera7938]

wouldn't it be 0.02 instead of 0.03??? please explain that on why its 0.03

[@colanseivanp.3157]

In no.4 how do you get a 125 times 4 over 3 times 4??

[@AboudSawatt-dz5pb]

Is this video for Calc 2 or Calc 1?

[@Eng586]

2

[@LafiajiSamuel-g3z]

but why is the final integral 3 to 12 ?

[@twiththecaps]

13:24 How come some teachers/textbooks use x (sub i) star when calculating the work? What does it mean and how come you don’t use it here

[@fip2554]

Is also because is a sort of index, is kind of when the limit begins since to calculate the work we use infinite cross sections, xi is the beginning

[@mikemeberg2852]

Where did you get the 1,000 from in the last example?

[@Giblet12]

Density of water

[@colanseivanp.3157]

Or where do you get that?

[@ale4462]

in 23:25 how come the weight of the rope is more than a 300lb create?

[@lmdemps]

Someone reply to this in like a year so I can reminisce on the past grind of calc 2

[@megavirus3305]

Not a year but here’s 6 months

[@mrmansir3734]

at 6:56, how are we now able to write F as a function of x???

[@calebfoxschool]

Because force is defined to be a function of x. In cases where the force is already given and you have to solve for x or k, it's given as the f(x) of the function. For example, consider the equation 50=k0.05. In this case, we know that f(0.05)=50. That's as good as I understand it, if I fully understood it I wouldn't be watching this video, but I hope that helps.

[@fookiebear1]

why isn't the displacement 5-x for the last problem?

[@nick425ful]

5 is the radius

[@dawoodjarrar8110]

Hey guys my test is tomorrow wish me luck! xD

[@cardinaljelly174]

i love you

[@joshguevara9570]

For number 4 I used the same integral to check that the work to get from 0 - 5 inches really was 50lbs, and I instead got 125ft *lb, not 50. Does anyone know why that is??? Please help I'm going crazy

[@Prism019]

ft*lbs is a measure of Work, whereas lbs is a measure of Force. The Force applied by the spring when held constant at 5 inches past the natural length is 50lbs. The Work required to stretch the spring from 0 to 5 inches would be 125 ft*lbs. You calculated something (Work) different than what you were trying to find (Force).

[@YoutubeChannel-oq6ss]

Should not the work equation for the last question be W = pg * integral from a to b of v(x)*delta x..... the v(x) has the dx within it.

[@triparadox.c]

Vx has a dx in it. So what happen when you take that dx out from Vx?

[@YoutubeChannel-oq6ss]

@@triparadox.c You dont take it out... you just replace v(x) with PI*R^2* h where the height would be delta x

[@triparadox.c]

@@KZbinChannel-oq6ss You said V(x) has dx in it. After you substitute the value of V(x) in place of V(x), you can just take the dx from V(x) out and it matches the original form. The reason why the dx is NECESSARY to be written is that on this particular problem, it just happen that dx is inside V(x). A general form must consider all possibility.

[@YoutubeChannel-oq6ss]

@@triparadox.c ooh that's interesting. Ok I stand corrected. Thanks for letting me know.

[@triparadox.c]

@@KZbinChannel-oq6ss No worries. It's a do or die for me on 18th..

[@rewrew5762]

"lenth"

[@tomaffatigato1498]

I think for the purpose of doing the math and passing a test, your vids are unsurpassed. However, it would be nice, at least to myself, if there was a bit more of the explanation of the actual physical concepts. For example, in the last example, there was no explanation of why you did what you did as it relates to what you are trying to achieve. instead just a rote manipulation of terms.

[@Dina-he1uc]

there are other videos for that. he makes things simple by not overcomplicating things which suits many people

[@toxin_d]

this pound thing confuses me so much...

[@haruharudude]

It only really makes sense if you have a slight physics background or know it from general knowledge. Basically, pounds are a measure of force which is basically just weight within the scope of calc classes. Weight is just mass times acceleration due to gravity. For work problems, pounds don't mean mass

[@toxin_d]

@@haruharudude when I look up for the dimensions of pound in Google it tells me it's mass

[@haruharudude]

@@toxin_d Different contexts. Force is proportional to mass so lbs being measured for mass is perfectly fine. That being said, it still is a measure of force due to it being mass multiplied by acceleration due to gravity. That's all you need to know for these problems

[@sigma7208]

17:00 is wrong