GCSE Physics - Elastic Potential Energy and F = ke Equations #45

[@Cognitoedu]

If you’d like to practise the material covered in this video, check out our platform at www.cognitoedu.org - it's totally free, and has been built to make learning and revision as easy as possible. The main features are: - Lessons organised by topic, only the lessons relevant to your specific exam board and tier are shown. - Automatic progress tracking. Progress bars tell you what you’re doing well at, and what you need to spend some time on. - Practise quizzes so you can test your knowledge. You can quiz yourself on any combination of topics you like. - A huge number of fully-hinted questions that take you step-by-step through some of the trickiest calculations & concepts. - A comprehensive bank of past exam papers, organised both by year, and also by topic. Amadeus & Tom

[@PreshW-vc7qi]

Hi btw elastic potential energy is in paper 1 not two and it is under energy not forces as shown on your website. Thank you👍🏾

[@nityaramanathan1523]

thank you! your channel is the only one that has gotten me through all my phys, chem and bio tests!

[@Cognitoedu]

Thanks Nitya, glad we could help!

[@ig3043]

this is amazing honestly am just going to year 10 and these videos are the only thing that help me revise easily with good examples thankyou so much

[@Cognitoedu]

Glad we could help out Iram! Best of luck!

[@syedbasimafzal3789]

amazing video i hope ur happy with the work u have done so far as it helped so many people. Scrolling thru the comments are just so suprisingly amazingly good. I think i speak for everyone when i say this.... THANK YOU COGNITO

[@suxela360]

It was so helpful that I even watched the ad lol

[@dekucuredmydepression5273]

thank you!! my phy mock is tomorrow tytyty

[@IbraheemSiddiqui-l6q]

This was so helpful it taught me more in four minutes than my physics teacher could teach me in an hour

[@strangerthingsfan5166]

Thank you so much! Your channel has helped me a lot with physics ^_^

[@skyjake4042]

This was really helpful with my physics homework. Thanks!

[@thedarkknight646]

watching this because I have an online physics class and my professor sucks at explaining this crap.

[@umaysmohammad8154]

Hope you are getting more subscribers and likes.

[@coolieboi252]

This video was extremely helpful

[@Cognitoedu]

Thanks Coolieboi, glad you enjoyed it 😊

[@yemchikwa1]

This video is sooo accurate 😊😅

[@learnnewconcept_everyday932]

most helpful channel❤❤❤

[@Sponley]

Great video, very informative

[@ethanwinters2096]

Would agree

[@headfreeze9302]

Thank you sirrr

[@straw_bennry]

you saved us thank you very much💙💙💙

[@Cognitoedu]

Thanks Soobin's Twin!! 😊

[@violetcooper1307]

I hate Phisics but this just makes it so much more fun😂

[@Adrianalyna]

I really appreciate.

[@anxhelo8649]

thank you!

[@qtjps2739]

Great!

[@LilithLorenz]

Bruh the ad i got at the beginning called me out for looking this video up i-

[@lulubee9959]

Was very helpful thanks!

[@natu.d9546]

My physics teacher never mentioned the graph(3:45) before, weird. Anyway thank you

[@samueloommen1066]

Very helpful!!!!!

[@danielbalhatchet-adams8958]

Hello class

[@fedhaamour]

Hi

[@visual_dorito]

Hi

[@aisyahnurirdina]

Thankyou it is really helpful

[@faizanrasool6156]

Can you find mass with this formula

[@harsangovinath4833]

In a certain circumstance, yes. When you stretch a spring, kinetic energy is converted into elastic potential energy. Remember that stretching a spring requires force which essentially increases the extension. Therefore, if you know the spring constant, and the speed at which it was stretched, you can do: Elastic potential energy = kinetic energy 1/2ke^2 = 1/2mv^2 (just plug in the values)

[@nagendrakumar9578]

Sir for spring force formula their must be negative sign but in your formula their is no negative sign why? I'm confused can you give clarity about that sir

[@Cognitoedu]

Hi, what makes you think there should be a negative sign?

[@shaaravabrol384]

Negative direction shows that restoring force is acting in the opposite direction of the force applied on the object

[@nagendrakumar9578]

Thanks for your response but my exams had completed 1 month ago

[@APower2004]

very cool

[@jjolatunji2481]

Great Video!

[@MilkesaMelka]

2:28 min k=700

[@mellow_swirl5406]

Where did they get 0.5 from all of a sudden

[@MI-lo2hj]

you mean at 3:18 ? its because in the equation for Ee there is 1/2 at the begining which is the same as 0,5

[@dublestufedoreo2606]

do rubber bands follow hooke's law? I see many sources showing graphs that it does. But there are also graphs showing otherwise!!!! Help!!!

[@adeolaadenusi1387]

Why are there 2 equations, and how do we know which one to use?

[@Fatima-zf5ym]

The first one is Hooke’s law.. it would be helpful if you google it to know more, basically it is a spring that exert force direct proportional to the distance,, and you will use the other equation just if you were asked about the potential energy ( I hope all of my informations are right 🤣)

[@Fatima-zf5ym]

+ the second equation is used to measure the area under the graph

[@adeolaadenusi1387]

@@Fatima-zf5ym thank you

[@mariloujagonal6613]

Please someone elaborate why 1/2 is in the formula?

[@okenough2124]

I think if you plot out every result on a graph, it will form a triangle. The answer is just the area of that triangle on the graph. So you use 1/2 * b * h. The values on the base and height of the triangle somehow involve spring constant and extension on it.

[@sitakwasitakwa9798]

Hope there more like this one too

[@syedmohidtaqipro724]

BANKAI MINAZUKI

[@H12rh]

It’s Hayden what u lot saying

[@antonroy415]

wsg bro

[@ethanwinters2096]

“Cheeky”

[@chita0310]

how do we get the extension with only the force and the spring constant?

[@harsangovinath4833]

F=ke where F is force(N) , k is spring constant(N/m) and e is extension(m).Rearranging this gives you e(extension)=F/k. Hope this helps!

[@unknown-md4bc]

Please I learnt that an object is more elastic if it has more resistance to the deforming force and also returns to it's original position very fast after the influence of the deforming force have been removed... but you said the exact opposite

[@boost1382]

I think what the video said was right, for example a car has more resistance to the deforming force, thus returning to its original position very fast, which means it is *less* elastic that something like a spring, which has less resistance to the deforming force, so returns to its original position slower

[@yumna4414]

Can anybody solve this question? A 100g mass is suspended from a spring next to a vertical meter rule. The top of the spring is level with the 0.0cm mark. The bottom of the spring is level with the 27.2cm mark. The 100g mass is replaced with a 600g mass. The length of the spring is now 89.7 cm. The spring has not reached the limit of proportionality. The 600g mass is replaced with 200g mass. What is the length of the spring?

[@jrplayss4536]

Posay🤩🤩🤩

[@boost1382]

25cm?