A Level Physics Revision: All of Work, Energy and Power (in 18 minutes)

[@zhelyo_physics]

Hi guys! I have also filmed some past paper questions on Work Energy Power to help you with your revision: kzbin.info/www/bejne/horUipmDhMaFaJI Hope this is helpful!

[@augustineshako1540]

very helpful. thanks big man

[@BTWPhysics]

Why are you censoring me with my one comment? I know that it was deleted.

[@amaansiddique2875]

3:45 just shocked me back to life

[@ritwikgossain]

Nice

[@daniyalhussain4174]

Im revisiting for my mechanics resit in my foundation year and finding out how to get the Kinetic energy equation just took me. Thats so bloody cool and I dont know why. Thank you so much please keep doing what your doing

[@zhelyo_physics]

Thank you so much! And I agree, the kinetic energy derivation is great!

[@dnishal]

hey i commented on your motion video earlier, well im on ocr spec a: with your videos revisions been a BREEEEEZEEE i swear!. i can easily juggle in enough time for my other subjects which also require alot of time, im doing physics, maths, and computer science alevels. i finished motion, forces in action, materials and today im doing laws of motion and momentum and work energy and power. i just wanna say without these videos i would have honestly gave up doing my revision, i would have known i was failing and i couldnt be asked. but these are so damn helpful, thank you so much Zphysics

[@zhelyo_physics]

so amazing to hear!! Thank you so much for commenting and good luck with revision! Drop a comment if something doesn't make sense.

[@williamedwards7881]

Wow this really helped me understand, thank you! Just starting to understand the basics for physics.

[@zhelyo_physics]

Thank you so much for the kind words!! Comments like these really make my day!

[@mishalfatima2858]

Thank you, this was so helpful 💗

[@zhelyo_physics]

Glad it was helpful! : ) Thanks!

[@marcoaltamura7512]

14:30 , I think a more understandable way of writing it would be 1/2mv²-mgh=-Fx, since the angle between F and x is actually 180°, and the cosine of that angle is -1 (furthermore, we expected the work to be negative since it's a resistive force). It's obvious that writing it the same way you doesn't change anything, but isn't 100% clear. Good job with the video!

[@zhelyo_physics]

Excellent point, thanks a lot!

[@Miftahul_786]

My favourite way to do it is by starting with 1/2mu^2 and adding and subtracting all the energies which leads to 1/2mv^2. To decide if you’re adding or subtracting energies, just look at the direction of the force. g acts downwards, and since the ball is going downwards, it is technically gaining mgh. (From a very physical point of view, if you are infinite distance away from a body, you have 0 potential energy. So the closer you get to it the more potential energy that object has, which is quite counterintuitive to how it is taught but it’s true.) Since the friction acts opposite to the direction of motion down the slope, we take the energy away due to friction across a distance. So our final equation: 1/2mu^2 + mgh - Fd = 1/2mv^2 You can now rearrange for Fd pretty simply. Can’t go wrong with this method

[@eirana1368]

thanks a lot! i was really confused T-T

[@ivanflitcroft1408]

Really cleared things up! Thank you

[@zhelyo_physics]

I am glad!

[@Ansariz-y51]

I have to say, U r just amazing Thank you so much👊!

[@zhelyo_physics]

Glad you find it useful! : )

[@vaithiesh.jjayasankar7910]

Thank you !! , it really helped me out!

[@zhelyo_physics]

Glad it helped!

[@vaithiesh.jjayasankar7910]

@@zhelyo_physics your content has never failed to help !!!!

[@luddy20]

Hi sir I know ive been asking so many questions on all your vids😂! but at 13:58 why do we do PE-KE wouldn't I be the other way round since KE is the final energy? I could be wrong but why is it that? thanks sir!

[@zhelyo_physics]

that's okay! So we are really looking for how much energy we have lost - to get a positive answer we would take away the first one take away the 2nd one. Doing it the other way around is fine too, but we would get a negative which may need to be ignored later in the question. Hope this helps!

[@PatelHarnisha]

These video's are soo useful!!!!

[@zhelyo_physics]

thank you so much!

[@AmL-L-L]

great video! just a quick question, is gpe = ke only valid for when there is no resistive force present? so then when resistive forces are considered it becomes gpe - ke = fx , since not all energy is converted to the other form due to the resistive force? thanks

[@khalidelra]

Best content creator ever

[@zhelyo_physics]

Thanks a lot for the comment! Much appreciated !

[@mathsphysics702]

Sir I am Indian English is note coming to me and you are very good explain to the derivation Love you sir ✍️✍️✍️❣️❣️❣️

[@zhelyo_physics]

Thanks a lot for the comment! Glad this is useful!

[@azmanfaruqui2861]

Very well explained

[@zhelyo_physics]

Glad it was helpful!

[@TheObloINATOR777]

thanks 👍 really informative

[@zhelyo_physics]

Anytime! : )

[@annadavletshina1432]

Thank you!

[@zhelyo_physics]

anytime!

[@mamun7911]

Helpful

[@zhelyo_physics]

thank you so much for the comment!

[@azilius5302]

Why did you subtract the P.E from the K.E at 16:

[@zhelyo_physics]

So there is a rule in physics that the change in energy = work done, and energy goes from PE to KE, if there was no work done, the change would be 0. There is work done though, so we find the difference and we can set that equal to Force x Distance. Hope this makes sense!

[@azilius5302]

@@zhelyo_physics it did. Thanks a lot for both the video and the explanation. Amazing channel 💚

[@jasperano]

Why does this legend sound so close to Joe from the Captain Joe yt

[@zhelyo_physics]

hmmm is this the pilot Joe? : )

[@PatelHarnisha]

Please make videos on aqa spec

[@zhelyo_physics]

thanks for the comment! My videos are fully applicable to the AQA spec too. Several ones need special ones, e.g. particle physics, I have it here: kzbin.info/www/bejne/mH7YopWgfcuhhs0si=ZAlycUdgoTS91oj2 Also the paper 3 topics and I have a separate playlist for this: kzbin.info/aero/PLSygKZqfTjPB7h0saqGd1tJvDw6_r5BHb&si=0w7j16GQLuLpV90e

[@thavaneshn07]

What chapters of the text book do these cover

[@zhelyo_physics]

the cover the content of A level Physics basically fully. If you spot anything missing please drop a comment!

[@LutfunNaher-t2b]

10:33

[@PJ-er1nk]

is this only AS level?

[@zhelyo_physics]

Hello, this contains AS material only, however if you are taking A2 it is examined there as well. If you are taking the AS exam, still examinable.

[@BTWPhysics]

The equation ΔE = W is incorrect and the change in energy does not equal the work done. The equation KE = 1/2mv^2 for kinetic energy is incorrect because the other equations Wnet = ΔKE and Wnet = fnet * s you're really referring to that are used to derive the equation for kinetic energy are incorrect. Let's take a look at the net work equation in deriving the equation for kinetic energy like shown in the video. Wnet = Fnet * s Wnet = ma * s Wnet = m * as a = (V - U)/t s = 1/2( U + V )t U = 0 a = V/t s= 1/2Vt as = 1/2vt * v/t as = vt/2 * v/t as = v^2t/2t as = v^2/2 V^2 = 2as ΔKE = m ⋅ as ΔKE = m ⋅ V^2/2 ΔKE = m V^2/2 We can see that the equation Wnet = Fnet ⋅ s causes us to multiply two kinematic equations ,thereby, deriving a third kinematic equation V^2 = 2as shown in the video which has time factored out. We can refer to Occam's razor which ,basically, states that entities should not be multiplied unnecessarily. It is not necessary to multiply two kinematic equations in order to derive a third kinematic equation with time omitted. We can show this using the other kinematic equations. a = V - U/t s = 1/2 at^2 + Ut U = 0 a = V /t s = 1/2 at^2 2s = at^2 2s/a = t^2 t = √2s/a at = V a√2s/a = V V = √2as We can see that we have arrived derived the equation V = √2as through substitution. We can refer to Occam's razor again which tells us that when we have two competing theories or explanations for something that the simplest of the two is preferred to one that is more complex. In this case, we have two competing equations that have time factored out. The simplest equation of the two is V = √2as. We did not have to multiply two kinematic equations unnecessarily like with the equation V^2 = 2as and there also was no need to square the velocities. It is easy to believe that work is a function of displacement because this aligns with our everyday experience. We see that we had moved an object a certain distance and we also feel like we did work when we expend energy. The same goes with lifting an object up against gravity. But, ask yourself, is displacement the only possible factor other than net force to determine the net work? We focus on the distance we moved an object, but we lose track of time. It is easy to lose track of the time that passes when doing work and ,so, we don't factor time in with how much work is done. Yes, the displacement increases with work, but so does the time increase. Let's take a look at the two possible factors starting with the net work equation. a = V/t s = 1/2Vt Wnet = Fnet * s Wnet = ma * s Wnet = m * as Wnet = m * (V /t) * 1/2Vt We can see that the equation Wnet = Fnet ⋅ s causes us to multiply two kinematic equations like shown before. Let's now take a different approach using time as the factor. s = 1/2 ⋅at^2 + Ut U = 0 s = 1/2 ⋅at^2 ms = m * 1/2(at^2) ms = 1/2 * mat^2 ms= 1/2 * Fnet * t^2 Wnet = 1/2 * Fnet * t^2 We can refer to Occam's razor which basically states that entities should not be multiplied unnecessarily. It is not necessary to multiply two kinematic equations considering that we naturally get net force as shown and that we have time which is a possible factor that increases with work done. Occam's razor tells us that when we have two competing theories or explanations for something that the simplest of the two is preferred to one that is more complex. In this case, we have two competing factors of displacement and time when considering the amount of net work done. The equation Wnet = 1/2 * Fnet * t^2 is the simplest out of the two equations and ,hence, time is the correct factor. We can see that we did not multiply two kinematic equations with this approach and that we naturally get Fnet. Sure, we had to multiply by mass, but that is in both cases. We can see that with the product ms that displacement increases with the independent factor of time. So, this equation has net work as a function of time as opposed to displacement. We can also see that the unit for net work is kilogram-meter, as reflected in the product ms, which is not a unit of energy. This means that the net work done is not equal to the change in kinetic energy and the equation W = ΔE is incorrect. This also means that the equation KE = 1/2mv^2 for kinetic energy is incorrect and this also means that the equation PE = mgh for potential energy is incorrect. The equation for gravitational potential energy PE = mgh is not "true" or correct for another reason. Gravitational potential energy is energy that is supposedly stored in an object as the result of the gravitational attraction of the Earth for the object. The fact is that the gravitational attraction for an object decreases with height even when near the Earth's surface. This simple fact means that you cannot look at nor treat gravitational potential energy ,if at all, in the same way as you do with elastic potential energy in terms of being stored energy. This also means that the work done in lifting an object does not increase the gravitational potential energy as it pertains to stored energy and that the energy transferred to an object being lifted does not get converted into gravitational potential energy. Instead, the energy transferred to an object being lifted simply goes into balancing out gravity allowing the object to move straight up at a constant speed. The equation P = W/t for power cannot down to P = Fx/t or P = Fv/t for the reason given which is that Wnet ≠ Fnet * s . This means that power is not the rate that energy is transferred since W ≠ ΔE.

[@basset5889]

this really helps thx 🦾

[@zhelyo_physics]

Anytime!