The gravitational energy at the start is calculated as mgh. The mass is unknown so we leave it as an unknown and just use the letter "m" to represent it. We also chose to use 10 for "g" instead of 9.8. The height at the start is 70m. The gravitational energy is then mgh or m x 10 x 70 = 700m. The "m" is the unknown mass and NOT the unit for meters. I hope this helps.
Conservation of energy was a working hypothesis developed by Leibnitz, as it gave much simpler mathematical solutions. His hypothesis later became a Law of Physics. Einstein explains how important Leibnitz' hypothesis became for physicists to recognise Heat as a form of Energy. He goes on to explain that it was also important for the development of his own famous equation equating Mass and Energy
At 2:40 you can cancel the m. At 2:50 you can bring the gha over to the other side so you get 700 - gha = 1/2 Va^2 The left side works out to 700 - 9.8*40 which is 308 on the left = 1/2 Va^2 on the right. Then you multiply both sides by 2 so the 1/2 is cancelled and you get 308*2 = Va^2 Then you get 616=Va^2 then take the square root of both sides you get 24.8 = Va
@Rockification09 it's not 700m as in meters. It's 700 x m. As in mass. And it's 700 because its mgh where m is unknown so we left it, g is 10m/s^2 ,and h is 70m. So multiply h by g and you get 700!
mmmm, my main issue with understanding conservation of energy part of the fact that mgh + 1/2mv^2. Isn't mgh a vertical component? Is velocity an omnidirectional component because from what I can recall, it's not possible to add two different vectors....
Hi, could you make a video on how to calculate the speed of the car 1) when it reaches the bottom of the loop for the first time, 2) when it reaches the top of the loop for the first time, 3) when it reaches the bottom of the loop for the second time? This is for GCSE exam, and we are only given the mass of the car (1500 kg, the height of the loop (8.5 metres), and starting position at 16 m above ground. Thanks!
If you're travelling through a series of hills in an automobile will you conserve fuel by accelerating through the downhill then coasting through the uphill or by maintaining a constant speed limit? I'm trying to convince my roommate that the latter will use more fuel... but I'm not a Physician...
You're just saying sir why don't you explain and discuss it clearly because some of students watching this videos doesn't understand you where did you get the 24 and 25? Just saying I hope you understand me♥️🤗
GetEmAndy It's not an estimation of mass. Since acceleartion due to gravity is 10 m/s^2, and the given height is 70 m, we will have: m(10)(70) = 700m where "m" is the mass. This basically means that for every kilogram (SI unit) of mass, there will be 700J of energy.
He has Eg there right? Well Eg = Potential Energy Potential Energy = mass * gravity * height The mass =1 The gravity in his example = 10 (it should be 9.81 but he used 10 to simplify) The height in his example = 70 Potential Energy = 10 * 70 = 700
You are correct it does say 700m, but he uses the -10m/s^2 approx. for accel. of gravity so simplified you get => mgh which is (m)(10)(70) which equals 700m [m for mass not bringing in units for this]
Wow, what a well explained video! He went so clearly between all the steps and explained his process so well! I totally understand now! (This is all sarcasm, I don’t understand any of what just happened. Thanks for nothing.)
I think that as with most of the videos in physics topics, this video is missing to address an issue that every effective presentation in physics should; To give the viewer a general approach that will help him/her to solve any problem or to answer any question in the topic. Teaching a topic with an example only, might be pleasant and easier to follow but doesn't give the student the broader and deeper knowledge and methodology required to tackle any problem and master physics for the long run.
THANK YOU BEARDED GUY!! YOU TAUGHT ME MORE IN 3 MINS THAN MY TEACHER COULD IN AN HOUR!!
The gravitational energy at the start is calculated as mgh. The mass is unknown so we leave it as an unknown and just use the letter "m" to represent it. We also chose to use 10 for "g" instead of 9.8. The height at the start is 70m.
The gravitational energy is then mgh or m x 10 x 70 = 700m. The "m" is the unknown mass and NOT the unit for meters.
I hope this helps.
@JoannaPotts Glad to help. We love to hear that it made a difference. Tell you friends about us.
Thank you so much! I now know how to do this problem for my test tomorrow. You taught me more than my teacher has taught me all year long. Thanks! (:
just got my grade thank you bearded man :)
Very good. I am a physics tutor and just wanted a quick refresher! This was awesome.
appreciate the help but that's a lot of skipping of contents there...
damn you really skipped like 20 steps, but still got the answer! Nice job!
@Candorly Hi. That's great! We don't get to hear the results so it was nice of you to take the time to let us know. Best of luck with your course.
11 years later ... Thank you so much sir
how would you find the velocity if this was an opened system where energy escapes as heat due to friction?
FROM 2009 WHEN THIS VEDEIO WAS POSTED AND AT THAT TIME I WAS 1 YEAR OLD TO 2024 AND AT THIS TIME I AS 15 YEARS OLD >THANK YOU FROM IRAQ
Conservation of energy was a working hypothesis developed by Leibnitz, as it gave much simpler mathematical solutions. His hypothesis later became a Law of Physics. Einstein explains how important Leibnitz' hypothesis became for physicists to recognise Heat as a form of Energy. He goes on to explain that it was also important for the development of his own
famous equation equating Mass and Energy
At 2:40 you can cancel the m. At 2:50 you can bring the gha over to the other side so you get 700 - gha = 1/2 Va^2 The left side works out to 700 - 9.8*40 which is 308 on the left = 1/2 Va^2 on the right. Then you multiply both sides by 2 so the 1/2 is cancelled and you get 308*2 = Va^2 Then you get 616=Va^2 then take the square root of both sides you get 24.8 = Va
@Rockification09 it's not 700m as in meters. It's 700 x m. As in mass. And it's 700 because its mgh where m is unknown so we left it, g is 10m/s^2 ,and h is 70m. So multiply h by g and you get 700!
mmmm, my main issue with understanding conservation of energy part of the fact that mgh + 1/2mv^2. Isn't mgh a vertical component? Is velocity an omnidirectional component because from what I can recall, it's not possible to add two different vectors....
@PhysicsEH Little off-topic but I have done only five pages of force practice problems and I got 90% on my Forces (No vectors) test.
@PhysicsEH at 2:40 yiu have one mass on one side and two on the other, how did they all cancel out?
Hi, could you make a video on how to calculate the speed of the car 1) when it reaches the bottom of the loop for the first time, 2) when it reaches the top of the loop for the first time, 3) when it reaches the bottom of the loop for the second time? This is for GCSE exam, and we are only given the mass of the car (1500 kg, the height of the loop (8.5 metres), and starting position at 16 m above ground. Thanks!
How do the m’s cancel out?
when your finding Va.Why did u square both sides?Rather than square rooting, both sides
If you're travelling through a series of hills in an automobile will you conserve fuel by accelerating through the downhill then coasting through the uphill or by maintaining a constant speed limit?
I'm trying to convince my roommate that the latter will use more fuel... but I'm not a Physician...
Why were the three masses cancelled out at 2:43? owo
can you explain how you rearranged the equation?
Thank you, helped a lot
Don't give up. You need to know a bit of physics to understand this problem. Best of Luck.
how did you get 25m
may i rest these nuts on your chin?
Thanks for the video Mr Morin!
how did the height in b became 25? is it not 42.something?
@Candorly Glad to help. Best of luck on your quiz.
LOVE YOU MY GLORIOUS KING
You're just saying sir why don't you explain and discuss it clearly because some of students watching this videos doesn't understand you where did you get the 24 and 25? Just saying I hope you understand me♥️🤗
thank u .. very helpful for me and now I solved the problem
that was a good question and was explained clearly, thanks :)
@flagellajello Glad to help!
@nguyen25520 Thanks, glad to help. Tell your friends about us.
@xoizzzyyyox Thanks for letting us know. Best of luck on your test!
Very easy to understand. Thank you :]
How did you get the mass? (700g)
Frank Liuu that's not mass
it's an estimate.
GetEmAndy It's not an estimation of mass.
Since acceleartion due to gravity is 10 m/s^2, and the given height is 70 m, we will have:
m(10)(70) = 700m where "m" is the mass.
This basically means that for every kilogram (SI unit) of mass, there will be 700J of energy.
i love you bearded man
very helpful... thank you
how did u go from 70m to 700m explain things to its potential of understanding
He has Eg there right?
Well Eg = Potential Energy
Potential Energy = mass * gravity * height
The mass =1
The gravity in his example = 10 (it should be 9.81 but he used 10 to simplify)
The height in his example = 70
Potential Energy = 10 * 70
= 700
I have the same question
How are you canceling the masses out?
When do people usually learn this? I'm pretty sure learning physics in 8th grade is normal, right?
Fushimi Saruhiko most kids in America never learn it
Why is it 700m when the diagram says 70m?
+Britany Cheung that m is not for metre. Its for mass.
You are correct it does say 700m, but he uses the -10m/s^2 approx. for accel. of gravity so simplified you get => mgh which is (m)(10)(70) which equals 700m [m for mass not bringing in units for this]
@@OCReefer i thought that g would equal 9.8?
Wow, what a well explained video! He went so clearly between all the steps and explained his process so well! I totally understand now! (This is all sarcasm, I don’t understand any of what just happened. Thanks for nothing.)
Thanks
thanks heaps
what about energy loss due to friction?
@lyttylgrenadier One of them was me.
THANK YOU!!! :D
I think that as with most of the videos in physics topics, this video is missing to address an issue that every effective presentation in physics should; To give the viewer a general approach that will help him/her to solve any problem or to answer any question in the topic. Teaching a topic with an example only, might be pleasant and easier to follow but doesn't give the student the broader and deeper knowledge and methodology required to tackle any problem and master physics for the long run.
you look like Zack Galifiniakis !
Thats a big loop
hi
Mr.Walker???
very helpful
do i know you from somewhere...?? hmmm aren't u that painter?
？？