Principle of Work and Energy (Learn to solve any problem)

Learn about work, the equation of work and energy and how to solve problems you face with questions involving these concepts. Animated examples, step by step, completely solved.
The crate, which has a mass of 100 kg, is subjected to the action...
The force F, acting in a constant direction on the 20-kg block...
The 8-kg cylinder A and 3-kg cylinder B are released from rest...
The 12-kg block has an initial speed of...
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Book used: R. C. Hibbeler and K. B. Yap, Mechanics for engineers - dynamics. Singapore: Pearson Education, 2014.

[@QuestionSolutions]

Some viewers asked how the 2 equations at 10:31 were solved, please see: kzbin.info4euH1289_Kg

[@DirkdeZwijger]

I've never seen a video on dynamics or statics with this level of finesse. Actually incredible that this even exists

[@QuestionSolutions]

Thank you so much, I hope they were helpful to you!

[@Itisjafferi]

Then you haven't watched Jeff henson yet.

[@DirkdeZwijger]

@@Itisjafferi I haven't, but I don't seem to need it either. Question Solutions uses the same books and similar examples as the one I'm using: the Hibbeler series. This content feels like a premium upgrade to an already solid series

[@alienx097]

The animations , the way u explain everything just awesome .. Thanks for Ur hard work

[@QuestionSolutions]

Thank you very much, I appreciate your comment :)

[@Daschum]

Very clear, intuitive and concise explanation Shared it with my Mechanical Engineering class :)

[@QuestionSolutions]

Many many thanks for the share, I truly appreciate it!

[@AlanaClaassen]

Your videos are such a great help! Thank you so much for taking the time and effort to make your videos so easily understandable and entertaining. Keep up the good work!

[@QuestionSolutions]

You're very welcome and thank you for taking the time to write such a nice comment. I wish you the best with your studies, keep up the great work!

[@fargo8246]

You are truly fantastic at explaining Dynamics topics and how to solve problems! I am eternally grateful for your videos!

[@QuestionSolutions]

Thank you very much for your kind comment. Really appreciate it and I am very glad to hear they are helpful to you. Best wishes with your studies!

[@nahommerk9493]

This channel is underrated! Thanks for the simple and clean explanations, man!

[@QuestionSolutions]

Thank you very much and you are very welcome. I wish you the best with your studies.

[@annar1652]

I am currently studying dynamics in Chinese because I am living in Taiwan. You have no idea how much your video has helped me. Thank you so much!

[@QuestionSolutions]

I am really glad it helped. I wish you the best with your studies!

[@jor2416]

I wish i found this chanel earlier, the explenations and animations are crystal clear. I thank you for your hard work!

[@QuestionSolutions]

Thank you for your kind comment! I wish you the best with your studies. :)

[@bonkgaming3646]

You are just incredible, i never thought that the concept of work could be defined that easily, i just love this channel, life saver

[@QuestionSolutions]

Thank you very much. Happy to hear the channel is helpful :)

[@shahzaibraja1650]

Thank You Man! Your lectures are short and to the point,very helpful in other words.

[@QuestionSolutions]

Thank you very much! :)

[@PKKAAAAA]

It was a perfect , clear explanation. Thank you very much.

[@QuestionSolutions]

You are welcome! Best of luck with your studies.

[@jimmydripps4011]

Excellent lecturing skills. I listen to this video slowed down for even better understanding. Also I pause it to take screenshots. Thanks for making this.

[@QuestionSolutions]

Thank you very much! I hope all the videos help you out and wish you the best in your future endeavors. Keep up the awesome work.

[@BODYBUILDERS_AGAINST_FEMINISM]

This class has been kind of a drag so far but this is actually a very interesting concept. I've always wondered how to calculate things like this, I'm happy I finally get to learn these things. Thank you for the video, as always

[@QuestionSolutions]

Glad to hear you are trying to relate these topics to real life scenarios. I hope you learn a lot, and these videos help you out in your class. Best wishes with your studies!

[@adeptusxiao6473]

YOUR CHANNEL IS UNDERRATED SIR. VERY CLEAR AND HELPFUL EXPLANATIONS

[@QuestionSolutions]

Thank you very much!

[@sweetdana81]

Thank you so much your videos are always clear and concise I appreciate it

[@QuestionSolutions]

I am glad to hear that! Best of luck with your studies.

[@gawdzamn]

Thank you very much for making this video. I feel much more confident going into my last midterm for dynamics! Immediately subscribed

[@QuestionSolutions]

Best of luck with your midterm! Do you best and thank you for the subscription :)

[@marcelplchEng]

Best dynamics channel out there, thank you !

[@QuestionSolutions]

Thank you very much! I hope it's helpful to you :)

[@vikramnagarjuna3549]

Hi this is vikram from India, Thanks for increasing video length... Animations are good...

[@QuestionSolutions]

Hi again Vikram, thanks for your comment. :)

[@camerongillespie870]

Great video my friend. This has helped my understanding greatly.

[@QuestionSolutions]

Happy to hear that :)

[@syedwasi2304]

Man I just want to meet you and just want to tell you how great full i am, your way of teaching is unexceptional , fantastic , keep the work going ❤

[@QuestionSolutions]

Thank you so much! I really appreciate it and I was very happy to read your comment. So thank you for taking the time to write it. I wish you the best with your studies! ❤

[@YazanSY]

this is some super helpful content, thank you so much for the effort sir.

[@QuestionSolutions]

Really glad to hear it helped. You're very welcome!

[@YazanSY]

@@QuestionSolutions it absolutely did alot to explain the chapter to me, I have a test tomorrow and your videos gave me a wide vision on the core of this topic! And I am really looking forward to see more content on mechanics soon on your channel. Have an amazing day sir.

[@QuestionSolutions]

@@YazanSY That's great! I wish you the best on your test and thanks so much for taking the time to comment. It's appreciated.

[@azash_omar]

very clear explanation thank you!

[@QuestionSolutions]

Thank you and you're very welcome!

[@user-po3dz1ib8h]

Really amazing, it's very useful for me and I realize principle of work and energy.

[@QuestionSolutions]

Wonderful and I am glad it was useful! Keep up the great work.

[@emza.the.robustdrexlermad_3905]

speed and accuracy is perfectly demonstrated ..Sir

[@QuestionSolutions]

Thank you very much :)

[@wickoab3142]

Awesome man... i love all your videos

[@QuestionSolutions]

Thank you so much!

[@sirbillybobjoe8021]

Thanks SO much for all the help

[@QuestionSolutions]

Wow, this is the first time someone used the "thanks" button!!! I am really grateful and thank you so much! I really appreciate it :)

[@sirbillybobjoe8021]

@@QuestionSolutions no worries u deserve it for all ur hard work uve saved me so many times I rlly appreciate it

[@QuestionSolutions]

@@sirbillybobjoe8021 I am really glad to hear that! Best wishes with your studies and future endeavors.

[@twilight3902]

How come this channel has reached a million subscribers yet? God bless you boss for this work and explanation

[@QuestionSolutions]

Thank you and I am glad you enjoyed the video :)

[@smbudk9341]

This is amazing! Thank u very very much!

[@QuestionSolutions]

You're very welcome!

[@kingzzzmen7242]

You saved my ass for my quiz tomorrow lol very concise and clear!

[@QuestionSolutions]

Glad to hear! Best wishes on your quiz.

[@anonymoususer4356]

Thank you for these videos

[@QuestionSolutions]

You're very welcome!

[@farazhaider4123]

Very cleanest video you ever made 🤩

[@QuestionSolutions]

Thank you very much!

[@antiquarian1773]

This is amazing work! (put intended) Thanks ! :)

[@QuestionSolutions]

haha! Thanks, keep up the good work 😅

[@mgthat8670]

Thank you sir. May Lord bless you always.

[@QuestionSolutions]

Thank you very much!

[@benjaminrockson8579]

that was a very nice explanation

[@QuestionSolutions]

Thank you very much!

[@rod1217]

Thank you for the help!!!

[@QuestionSolutions]

You're very welcome!

[@ruihong7912]

thank you! this is awesome

[@QuestionSolutions]

Glad to hear :)

[@guillaume6373]

It all makes sense now!!

[@QuestionSolutions]

Glad to hear :)

[@karem3523]

Amazing video!

[@QuestionSolutions]

Thank you very much!

[@MisterBinx]

For a second I thought you were integrating the frictional force too and that confused me (they are both blue). I see now. I did this example on my own and did not integrated the varying force. Got an answer like 14.42 which is definitely wrong. Good practice problem to remember you have to integrate if the force isn't constant.

[@QuestionSolutions]

Glad to hear you got it :) The blue was there to show that it was to work. 😅

[@kamsydraws]

Thank you very much for the excellent video. For the last question, I was wandering why the ACC is 0

[@QuestionSolutions]

Please provide me with a timestamp to the place you're referring to. Thank you :)

[@johnlestermendoza118]

Thank you!

[@QuestionSolutions]

You're very welcome!

[@changjunyang7471]

most underrated utube channel

[@QuestionSolutions]

😅

[@ifyouwantanythinginthislif7338]

Thank you❤❤❤

[@QuestionSolutions]

You're very welcome! ❤

[@catrinacraft]

Thank you 😊

[@QuestionSolutions]

You're very welcome! :)

[@riyadsalehhassen2056]

thank you 🙏

[@QuestionSolutions]

You're welcome 👍

[@TheMounten1]

Hi, at 12:35, how do you know which value of s_1 is correct as it is a quadratic equation with two solutions?

[@QuestionSolutions]

Did you figure out the two values? If yes, you can tell which one makes the most sense since one is positive and the other is negative.

[@ruandre5619]

Do you have any videos on power and efficiency? I looked through the playlist :) Thank you for your time

[@QuestionSolutions]

Hmm, I don't believe I do, sorry :(

[@sharoonaftab8894]

@@QuestionSolutions you have a great future on youtube keep going.

[@QuestionSolutions]

@@sharoonaftab8894 Thank you very much!

[@Airriii]

Hi your videos have always been helpful but I wonder when the block bounces on the spring B, does the size of the block matter to reach the 0.6m distance spring A?

[@QuestionSolutions]

Thank you! As per your question, I apologize, but I don't understand what you're asking. Also, could you give me a timestamp so I know where you're referring to?

[@iamsobadatthis7947]

Hey quick question. For the second example how do you solve for s when one of the terms is s to the power of 3/2?

[@QuestionSolutions]

You can graph the equation, or you can use symbolab or wolfram alpha.

[@tirthahalder]

Great Explanation, could you please explain again, why, 2 times Change of Sa + 2 times change in Sb = 0? at 8:34

[@QuestionSolutions]

Thank you! Please watch this video first: kzbin.info/www/bejne/f6bHgYN9frtoh68 In simple terms, all we are doing is taking the derivative. Any constant will turn to 0. So "l" is the constant, and when we take the derivative, we get a zero.

[@genarorivera9750]

At 7:52, how did you solve for s? I am having a hard time solving for s. Other than that, great video, definitely deserves a like!

[@QuestionSolutions]

Easiest way is to graph it, otherwise, you can use something like symbolab or wolfram alpha to get an answer. :)

[@michaelmalgioglio1532]

@@QuestionSolutions so you don't know how to do it by hand?

[@QuestionSolutions]

@@michaelmalgioglio1532 You should know how to do it by hand, but that doesn't fall into a dynamics course. That's for a calculus course. If I were doing a series on calculus, I would show how to do an integral, but that's not how university courses work. You learn the integration from calculus, then you apply that to other courses, like dynamics, statics, thermo, etc. I just want to emphasize that the goal of this set of videos isn't to teach how to solve for a variable in an equation, the goal is to show how t apply new equations learned in dynamics.

[@firasal-ghamdi5004]

Thanks alot

[@QuestionSolutions]

You're very welcome!

[@katlynriggs-mills5517]

Really love your explanations, super clear and concise! Please consider making a Thermodynamics video series!!!

[@QuestionSolutions]

Really glad to hear that, and Thermodynamics is on my list to do since this has been requested quite a bit. :)

[@roonyyoouness7115]

Hello, please can you tell me how can i know if the work done by the spring is positive or negative. And thank you for all your kind tutorials🤚🤚🙂

[@QuestionSolutions]

Generally speaking, and this is a nutshell answer, if the spring is doing what you want it to, you will have positive work. In other words, you are considering the position of the item from the datum and the spring gives it more energy in the direction you want it to go. That can be considered to be positive work. If however, the spring is reducing the velocity of the object or something similar to that, then it would be negative work. I hope that makes sense.

[@roonyyoouness7115]

@@QuestionSolutions got it, thanks.

[@francoprins8928]

stellar

[@QuestionSolutions]

Thank you!

[@jaycivandehey14]

In the last problem why is frictional force doing negative work after it touches spring b. Wouldn't the frictional force be doing positive work when the block moves left, since the frictional force will be pointing right

[@QuestionSolutions]

Good question. So when the block is moving left, all forces in the left direction does positive work. In other words, any force that helps the block move to the left does positive work. Friction does negative work since it's pointing to the right. So when we write the work equation, think of it as picking the movement side to be positive. And since friction is always opposite to the direction of movement, it's almost always negative work. I hope that helps.

[@flee3695]

sir thankyou! but i have a question, in example #3 using 1st condition, using the formula WOE equation. T1-47.09(0.3+s1)-300 * 1/2 (s)^2= 0 why do we have to multiply 1/2 to 300? what is the explanation for that sir? and why the s is squared?

[@QuestionSolutions]

This is the work done by a spring. Please take a look at 2:12 and watch for a few minutes. 👍

[@flee3695]

@@QuestionSolutions oh right, thankyou for always answering questions!

[@QuestionSolutions]

@@flee3695 You're very welcome!

[@dannyoBruhh]

I am doing the same question for my course. Except i have different numbers in the problem. The box in my problem weighs 25 lbs. But on her answer key, to get the normal force she doesn't multiply by G value. Her normal force is just 25. And also for T1 when plugging in for m, her m value is 25/32.2. Can you tell me why her steps are different.

[@QuestionSolutions]

You can solve these problems in many ways and arrive at the same answer. I only showcase a single method, but you should do whatever method you're more comfortable with (or whatever method your professor wants you to use). :)

[@matheresahalog5771]

Can you elaborate on the algebra in problem 2 in how you got S? Where the force is variable

[@QuestionSolutions]

Are you asking about the integral? Like how to solve that?

[@5-vcnvzjjcc430]

Peace be upon you. I am from Iraq. Thank you for scratching my treasures. How did you extract it? Speed ​​values?👍✍️🌺

[@QuestionSolutions]

Really sorry but I am not sure what you are asking :(

[@kiakyonduai5164]

Thank sir , I am now ready to do my exam. This video really helps me.

[@QuestionSolutions]

Awesome! I wish you the best with your exam 👍

[@producer5078]

thank u soooo much !

[@QuestionSolutions]

You're very welcome!

[@typhoon9896]

How was the value for x in the second question found when the powers of s were different?

[@QuestionSolutions]

Please use timestamps, I don't know where you're referring to. Thanks!

[@GxR-yt]

Do have any reference book suggestions

[@QuestionSolutions]

Please check the description, I always list the books used :)

[@michaelbitz9127]

couldnt you solve the last problem using only W(friction) = 0.5(mass)(delta speed)^2 since springs do work in both directions, thus cancelling out. This would give -mg(uk) s = -0.5(12)(4^2) ==> s = 2.039 m

[@QuestionSolutions]

There are so many ways to solve these problems, I just show one method using the equations pertaining to the selected chapter. 👍

[@sgpark5056]

After some thinking, I came to the same conclusion. In this problem, friction is the only nonconservative force.

[@stephanie-gp8uq]

correct me if i'm wrong but isn't the normal force at 11:29 supposed to be 129.72N? i think the gravitational acceleration value used was wrong. great video nevertheless!!!

[@QuestionSolutions]

No, it's correct. Acceleration due to gravity is 9.81 m/s^2, if you want it really precise it's 9.80665 m/s^2.

[@muhaminatora2163]

at 7:55, can you go through a thorough explanation on how you solved for s

[@QuestionSolutions]

So I don't think I can explain that over a comment but I am assuming you're stuck at the integral? If yes, please see: www.integral-calculator.com/#expr=50x%5E%281%2F2%29&lbound=0&ubound=x and press go, it will show you all the steps. After that, you can plot it on a graph to figure out where the intercepts occur.

[@andrewkatongo8028]

The value of s in example #2 how was it evaluated I'm a bit behind because on my end it's giving me a different answer?

[@QuestionSolutions]

Please see: www.symbolab.com/solver/step-by-step/%5Cleft(%5Cleft(50x%5E%7B%5Cleft(3%2F2%5Cright)%7D%5Cright)%2F%5Cleft(3%2F2%5Cright)%5Cright)-58.86x%3D1890?or=input It's easier to just graph it and see where it intersects the x-axis though. See: www.desmos.com/calculator/6umwtsvuqk

[@dominicdillon9458]

this guy is good

[@QuestionSolutions]

👍

[@koladek.p8166]

thanks, but I'm having issues finding the final values for s(distance) in the first two questions you solved

[@QuestionSolutions]

For the first question: bit.ly/2OhO6Rz For the second question: bit.ly/2CrNiXC It's simply solving for one variable, don't overthink :)

[@koladek.p8166]

@@QuestionSolutions thanks

[@QuestionSolutions]

You're welcome! Best of luck with your studies.

[@undefined.infinity3106]

@@QuestionSolutions can you give the link again, please? the link isnt working

[@QuestionSolutions]

@@undefined.infinity3106 www.wolframalpha.com/input?i=%28800cos30%29x%2B%281000*%284%2F5%29%29x-156.2x%3D%2850*6%5E2%29 www.wolframalpha.com/input?i=%28%2850x%5E%283%2F2%29%29%2F%283%2F2%29%29-58.86x%3D1890 👍

[@alifnajmi2543]

at 7.54 how do you solve for s because i didn t know how to solve for s^3/2

[@QuestionSolutions]

Easiest would be to graph it and see where it intersects the axis. www.desmos.com/calculator/a7f36rzhmb or the conventional way: www.symbolab.com/solver/step-by-step/%5Cfrac%7B%5Cleft(50x%5E%7B%5Cfrac%7B3%7D%7B2%7D%7D%5Cright)%7D%7B%5Cfrac%7B3%7D%7B2%7D%7D-58.86x%3D1890?or=input

[@darrylcarter3691]

This second quest is kind of hard to solve. I seem to solve the 50(2/3)s^(3/2) - 58.86s = 1890. I tried the wolfarm website to find the step by step solutions but it won’t show me unless I pay for it. It there anyway else I can go. Is there a section that help me solve these kinds of problems.

[@QuestionSolutions]

Hmm, wolfram alpha is usually my go to website if I need to check an answer. If you are a student, you might be able to get a better price as well. Other than that, if you search for equation solvers, a lot of websites do come, but I don't know if they can solve every equation. Other than that, to get an answer, you can always graph it. Sorry I couldn't be more helpful with your question.

[@darrylcarter3691]

@@QuestionSolutions do you have to pay for it

[@QuestionSolutions]

@@darrylcarter3691 Yes, I believe for students, they have a plan for 4.75 per month. If you are a student at an institution, they might already give it for free.

[@elenaserranomarin9844]

hey, how did u solve the equation in 7:54?

[@QuestionSolutions]

Graph it or use wolfram alpha. If you want to do it by hand, see: mathforyou.net/en/online/equation/arbitrary/?e0=50x%5E%283%2F2%29-88.29x%3D2835&v0=x&o0=1&from=google

[@antonioandressantacruz9610]

hi someone knows how obtain the solution of question 2 at 7:55

[@QuestionSolutions]

Graph it or use wolfram alpha. If you want to do it by hand, see: mathforyou.net/en/online/equation/arbitrary/?e0=50x%5E%283%2F2%29-88.29x%3D2835&v0=x&o0=1&from=google

[@joyedwards6230]

Hi where did you get the 81.8736 from?

[@QuestionSolutions]

Please give me a timestamp so I know where to look.

[@joyedwards6230]

@@QuestionSolutions 12:36

[@QuestionSolutions]

@@joyedwards6230 it's the numerical value you get when you isolate for the 2 variables.

[@su3ad885]

how did you solve for vb and va in 10:35

[@QuestionSolutions]

You can plot the 2 graphs, use substitution, or whatever method you're comfortable with. 👍

[@e-nes4042]

why we did not receive directly normal force (50^1/2) ? can you explain simply at 7.54

[@QuestionSolutions]

The force F is based on the distance travelled (s), so we can't just plug values in, it has to be integrated. It's not a fixed number.

[@GxR-yt]

Hey! How you write 2Sa=Sb in the video 8:22 time

[@QuestionSolutions]

Please see: kzbin.info/www/bejne/f6bHgYN9frtoh68

[@AdamWatts-or9vg]

NICE

[@QuestionSolutions]

Thanks!

[@elizabethclarawilliams816]

@8:05 what is he calling the line that he draws? a datum?

[@QuestionSolutions]

That is correct. They are fully covered in this video: kzbin.info/www/bejne/f6bHgYN9frtoh68

[@willgggg900]

from the last question how can s1 be 0.598 if s1 one is the distance of the spring being compressed because if you add it with 0.3 with how much the box moves it will be 0.898 while the whole system is 0.6 m long

[@QuestionSolutions]

I think maybe you're misunderstanding the question. The 0.6 m is only the distance from the edge of the red spring to the blue spring, it doesn't represent the total system. Look very carefully at the start of the question and where the 0.6m is and what it's showing. We don't know how long the springs are, or the total distance from the ending edge of the spring to the ending edge of the other spring. All we know is that it's 0.6 m between the start of each spring. In addition, keep in mind that these diagrams are only starting steps of what it could be, not what it really is. It's not to scale or anything, and when you get the real values, you will have to redraw it, or use your imagination to get a better idea of what it actually looks like. I hope that helps. 👍

[@willgggg900]

@@QuestionSolutions ooh just realised thanks for the help is much more clearer now

[@QuestionSolutions]

@@willgggg900 You're very welcome! Best wishes with your studies :)

[@panacea488]

Hi sir, how did you get the S1=0.598 in the last problem?

[@QuestionSolutions]

It's a quadratic equation so you can solve it in a multitude of ways. You can graph it, use the quadratic equation, etc, whatever is easy for you. www.mathsisfun.com/algebra/quadratic-equation.html

[@panacea488]

Thanks sir, I get it now

[@QuestionSolutions]

@@panacea488 Awesome! :)

[@abrw2342]

10:30 how did you get VA and VB

[@QuestionSolutions]

Isolate for one variable in the first equation and then plug it into the 2nd equation (substitution method). Please see: flexbooks.ck12.org/cbook/ck-12-cbse-math-class-10/section/3.5/primary/lesson/solving-simultaneous-linear-equations-by-substitution/ An easier method is to just graph the 2 equations and see where they intersect. Use desmos for that. 👍

[@QuestionSolutions]

This is a bit late, but here is a short video on how those 2 equations were solved: kzbin.info4euH1289_Kg

[@andrewgrimes1771]

Can you give greater detail to how S simplifies without becoming negative and the exponents being different at 7:54

[@QuestionSolutions]

Are you referring to the integral portion or where the "-58.86s" comes from? If it's the latter, it's the frictional force multiplied by the distance travelled. If it's the first, please see: www.symbolab.com/solver/definite-integral-calculator/%5Cint_%7B0%7D%5E%7Bx%7D50x%5E%7B%5Cfrac%7B1%7D%7B2%7D%7D%20dx as that is more of calculus than dynamics. 😅

[@andrewgrimes1771]

@@QuestionSolutions no i'm referring as to how that (100(s^3/2)/3)-58.86s =1890 simplifies to s=20.52

[@QuestionSolutions]

@@andrewgrimes1771 Oh, you're just solving for "s", it's a single equation with a single variable. Easiest it to just graph it: www.desmos.com/calculator/3bfofpeifb 👍

[@willgggg900]

@@QuestionSolutions wait sorry how would you solve for s im slighty confused ?

[@QuestionSolutions]

@@willgggg900 You're just looking to see where the graph intersects the x-axis, or you can solve it algebraically (faster to graph it): www.symbolab.com/solver/step-by-step/%5Cleft(%5Cleft(50x%5E%7B%5Cleft(3%2F2%5Cright)%7D%5Cright)%2F%5Cleft(3%2F2%5Cright)%5Cright)-58.86x%3D1890?or=input

[@trueislam2170]

perfect

[@QuestionSolutions]

👍

[@kunnskapshushet4411]

In the pulley question why didn't we consider the work done by tension

[@QuestionSolutions]

Work done by which tension? Ropes? If so, it's negligible unless otherwise stated.

[@e-nes4042]

ı have another question. How do we decide which work is negative? at 9,57

[@QuestionSolutions]

It depends on your assumption. Here, we assumed down to be positive, but the cylinder is going up, which makes it negative.

[@3boody777]

Thank you very much ! Can you help me with this question, please : When s=0.45m, the spring is unstretched and the 11-kg block has a speed of 6 m/s down the smooth plane, if the block stops when the distance s is 2.6m, the force F equal to : Note: the smooth plate makes 30 degrees with the negative x-axis You can see the figure of the question if you type it in google. Thanks for everything ❤️

[@QuestionSolutions]

You're very welcome. I am sorry but I don't really solve questions from comments because it opens a floodgate of other questions that I would have to solve. 😅

[@clearflow7925]

9:29 we ignore the work of tension cause it doesn't have mass right?

[@QuestionSolutions]

Yes since we are told to neglect the mass of the cords and pulleys.

[@mizashazwina7943]

for 8:38 will the length always be zero?

[@QuestionSolutions]

Yes, the constants will always turn to zero. Please see: kzbin.info/www/bejne/f6bHgYN9frtoh68 This is fundamental to any pulley problem you face.

[@leandroneconoelgara8620]

How about the work of a force that is applied instantaneously? Like hitting a baseball with a bat, the force is instantaneous but will make the ball move

[@QuestionSolutions]

Those are covered in future videos. Please check the dynamics playlist.

[@leandroneconoelgara8620]

@Question Solutions is work done only when the force is applied all throughout the distance traveled?

[@QuestionSolutions]

@@leandroneconoelgara8620 Please see: pressbooks.uiowa.edu/clonedbook/chapter/work-the-scientific-definition/ I think it will help clear a lot of the questions you have. :) If you need to account for time, so how long a force is applied to an object, you have to get into impulse as well. There is a video on that too.

[@saiprasadsatya3677]

How to resolve components towards the head or away from the tail?

[@QuestionSolutions]

Please provide a timestamp so I know where to look. Thanks!

[@saiprasadsatya3677]

From 4:51

[@QuestionSolutions]

@@saiprasadsatya3677 If you are asking how to break forces into components, please watch this video: kzbin.info/www/bejne/hKOvZpdjZ6iUmLM

[@jamess3493]

can you refer me to something to help me remember how to solve the equation at 7:52 ?

[@QuestionSolutions]

If you want to solve this step by step, see: mathforyou.net/en/online/equation/arbitrary/?e0=50x%5E(3%2F2)-88.29x%3D2835&v0=x&o0=1&from=google

[@jamess3493]

@@QuestionSolutions thanks

[@joebethtechmarkcreatives]

Why do we have to integrate because in the first example we didn’t

[@QuestionSolutions]

The force is given as a function of distance. It's not a constant force, whenever something is not constant, you will most likely need to integrate.

[@johnlestermendoza118]

labyu daw sabi ni andrae!! papasa na siya

[@QuestionSolutions]

Not sure what this says, sorry!

[@saiprasadsatya3677]

Friction is force which resists motion how we considered it as work

[@QuestionSolutions]

Work is force times distance, so if friction is a force, and it's applied for a certain distance, wouldn't that create work? It would just be negative work.

[@saiprasadsatya3677]

@@QuestionSolutions thank u

[@QuestionSolutions]

@@saiprasadsatya3677 You're welcome :)

[@sasukekianhoong6130]

if it is frictionless surface , will the block still come to rest ?

[@QuestionSolutions]

No, unless you account for air resistance. An opposite force has to stop the block.

[@sasukekianhoong6130]

@@QuestionSolutions let's said the block is initially at rest, then I push the block to the right on frictionless surface and the air resistance is neglected. if I want to count the distance travel by the block, it will go infinity? Cuz the block will not come to rest.

[@QuestionSolutions]

@@sasukekianhoong6130 That is correct, nothing will stop the block unless you apply another force from the opposite direction.

[@sasukekianhoong6130]

@@QuestionSolutions If the block is pushed to the right until spring B is compressed 50 mm and released,where at first it has travel distance (0.3+0.05)=0.35m then 0.5k'x'^2=0.5kx^2 , 0.5 (100)(x'^2)= 0.5(300)(0.05)^2, x=0.0866m , so my total distance travel by block is 0.35 + (0.6+0.0866)+(0.6+0.05)+(0.6+0.0866)+... and so on ? it is correct ?

[@sasukekianhoong6130]

As my spring B is compressed 50 mm and released , my block initial speed is 0 and i know that the block will compress spring A as well and my final velocity of block is 0 again so the eq is 0.5k'x'^2=0.5kx^2 am i right ?

[@sharoonaftab8894]

why is 2sa+sb=0 ( why is the change in dispalcement equal to 0)?

[@QuestionSolutions]

Because the total length of the rope does not change, it's a constant. So there is no displacement for the total length of the cable. Please take a look at this video: kzbin.info/www/bejne/f6bHgYN9frtoh68 👍

[@mayraloredo3251]

in the first example, I keep getting 1781 for N

[@QuestionSolutions]

Please see: www.symbolab.com/solver/step-by-step/x-100%5Cleft(9.81%5Cright)-800sin%5Cleft(30%5Cright)%2B1000%5Cleft(3%2F5%5Cright)%3D100%5Cleft(0%5Cright)?or=input