I Miss Krista King and her Videos!!! :) I've been a fan for 10 years.. I enjoy Learning, then forgetting, then RELEARNING Calculus!! :) Love your videos... thankyou
Thank you so much for explaining that weight is the same as the force! I was wondering why a problem I did recently didn't take gravity into account, but it did because we were given the weight of the bucket. Thank you very much!! Your descriptions of everything are very clear! Keep up the good work!
Why isnt the distance 500-x? the length of rope will shorten as the rope gets hoisted up thus making it do less work as more part of the rope gets hoisted. At the ground level, it will have 500x4 . at the top of the building, it will have done 1x4 work done.
Very Interesting Krista! Imagine that my son have simple problems on 5 Grade then I could say that their answer is wrong because they did not add the work done on the rope. Right! We Math Lovers Love to give or get the precise answer. Anyway, well Done!
Question, how do we know when to label the x sub i as x or in other cases (500-x). What are the different situations where both of those would take place?
I'm having some trouble understanding a Calculus 2 Problem involving Force and Work, and I was wondering if anyone could help me. The problem asks “How much work is done in lifting a 1.2 kg book off the floor to put it on a desk that is 0.7 m high? Use the fact that the acceleration due to gravity is g = 9.8 m/s^2.” It says that the force is equal to the mass times acceleration, which is just mass times gravity, but wouldn't there be a moment of acceleration upwards? I'm having trouble wrapping my head around the force needed to lift the book being exactly equal to the books weight at all times...
Omg you remind me of my high school alg2/calc1 teacher, she was the best teacher. These college professors can't explain anything lol thanks so much for taking the time to do all these videos.
Great video! Work is always a confusing topic for me. In the case of the rope's work, do we get the SI units (ft-lbs) due to the product of the rope's force, the distance to dx and dx itself; lb/ft x ft x ft?
You know, I understood integration, I really did, but that very last bit of this video was what I needed to become comfortable with it. But yeah, my midterm study guide was like "yo did you know that even though you thought that force unit was really easy, you didn't actually know how to do physics with calculus at all? Yeah you totally just took highschool physics and knew your formulas, hah, well just try figuring out this sand problem!" Now these aren't carbon copies of each other, but I think my practice in non-calc based physics can carry me the rest of the way through this.
If it's expressed in lbs., then the weight has already factored in the gravitational constant, so you would never need to use the 32ft/sec^2 in this context. She stated only the 9.8 m/s^2 because the only time you would need the gravitational constant is if you were working with meters and kg.
Love you and your videos. Just would point out a slight error (you know everyone loves to correct the instructor ~ sorry), however you state at 1:34 to multiply by 9.8 (which is Meters per second, correct?) But you are using Lbs. not Grams so you should be using 32 ft/sec^2. Which would give you the correct Force. Right? ~ Still love your videos, keep them coming....very helpful.
Your the Only person on youtube that actually explains this with a Weight. Thank you!
I Miss Krista King and her Videos!!! :) I've been a fan for 10 years.. I enjoy Learning, then forgetting, then RELEARNING Calculus!! :) Love your videos... thankyou
Thank you so much for explaining that weight is the same as the force! I was wondering why a problem I did recently didn't take gravity into account, but it did because we were given the weight of the bucket. Thank you very much!! Your descriptions of everything are very clear! Keep up the good work!
I'm so glad it helped you! :)
damn 10 years and still helping people
Glad you like them! :)
Can I chat with you. I want to be your friend.
@@johntyler7155cringe
You're essentially saving me in my Calc 2 class. Thanks for the videos!
thanks!! i'm so glad you like the videos!! :D
Love u Mam ❤️
Form India ❤️🇮🇳
*INDIA*
The capital of Mathematical knowledge
awesome!! you're welcome!! :D
Love your lessons
thank you so much
I just wish that you post video examples that you shows the tricky questions that most people get confused.
+TayRoo qz You're welcome, I'll try to post harder examples. :)
You're welcome! :)
Why isnt the distance 500-x? the length of rope will shorten as the rope gets hoisted up thus making it do less work as more part of the rope gets hoisted. At the ground level, it will have 500x4 . at the top of the building, it will have done 1x4 work done.
Yes u are right
Very Interesting Krista! Imagine that my son have simple problems on 5 Grade then I could say that their answer is wrong because they did not add the work done on the rope.
Right! We Math Lovers Love to give or get the precise answer. Anyway, well Done!
Question, how do we know when to label the x sub i as x or in other cases (500-x). What are the different situations where both of those would take place?
Thanks, and I know what you mean!! :D
essentially this seems to be incorrect
Anyway you could do a course in Astronomy applications. Haven't found one anywhere.
Thank you. Your videos are very helpful :)
Can the distance be (500-x) ?
Yes
The function for the distahce is 500-x, not just x
thanks!! helped soo much!
If the rope was being lowered, would the work be the same as this except negative?
I'm having some trouble understanding a Calculus 2 Problem involving Force and Work, and I was wondering if anyone could help me. The problem asks “How much work is done in lifting a 1.2 kg book off the floor to put it on a desk that is 0.7 m high? Use the fact that the acceleration due to gravity is g = 9.8 m/s^2.”
It says that the force is equal to the mass times acceleration, which is just mass times gravity, but wouldn't there be a moment of acceleration upwards? I'm having trouble wrapping my head around the force needed to lift the book being exactly equal to the books weight at all times...
can this be an arc-length * weight/ft problem?
got it, thanks
Omg you remind me of my high school alg2/calc1 teacher, she was the best teacher. These college professors can't explain anything lol thanks so much for taking the time to do all these videos.
I'm honored! Thank you! :)
Great video! Work is always a confusing topic for me. In the case of the rope's work, do we get the SI units (ft-lbs) due to the product of the rope's force, the distance to dx and dx itself; lb/ft x ft x ft?
calc two final tomorrow and here i am binge-watching your videos lol
Lol, hope the final went great and that you passed the class! :D
Why r u so good in math? can u make some video about shear stress... :D ty..
Thank you! :D
seems like a way oversized rope?
how do you find work/job? lol
can you please do another work video? maybe about pumping water out of a tank?
I've got that video in my online school, I just haven't published it to TH-cam yet... sorry about that!
You know, I understood integration, I really did, but that very last bit of this video was what I needed to become comfortable with it. But yeah, my midterm study guide was like "yo did you know that even though you thought that force unit was really easy, you didn't actually know how to do physics with calculus at all? Yeah you totally just took highschool physics and knew your formulas, hah, well just try figuring out this sand problem!" Now these aren't carbon copies of each other, but I think my practice in non-calc based physics can carry me the rest of the way through this.
I feel you. I'm just about to pass Integral Calc.
cute and smart
If it's expressed in lbs., then the weight has already factored in the gravitational constant, so you would never need to use the 32ft/sec^2 in this context. She stated only the 9.8 m/s^2 because the only time you would need the gravitational constant is if you were working with meters and kg.
Not using Si-UNITS leads to confusion and mistakes for Christ' sake.
Why can't we just do (4*500) *(500) = 1000000 then divide by two to get average work done I.e 500000 + work done on weight.
Love you and your videos. Just would point out a slight error (you know everyone loves to correct the instructor ~ sorry), however you state at 1:34 to multiply by 9.8 (which is Meters per second, correct?) But you are using Lbs. not Grams so you should be using 32 ft/sec^2. Which would give you the correct Force. Right? ~ Still love your videos, keep them coming....very helpful.
Love your voice haha
Ever heard for metric system? :)
500000+15000 is not 65K its 75k
300/4
LOL use SI