Work and Kinetic Energy Playlist ▶: kzbin.info/aero/PLQgL7RKnNg_zXge310jzf_CAxuqk3Bm_o Access solved numericals & downloadable PDFs: ▶ thesciencecube.com/ Join our Telegram Channel: ▶t.me/TheScienceCube_Community
@JioHotstarTelegu2 ай бұрын
Wow ...nice explanation sir ❤
@TheScienceCube2 ай бұрын
So glad you liked it :) I'm always interested in knowing what specific things students like so that I can do more of it in my next video. Can you tell me one or two things that you really liked.
@uttamvashishtha12723 ай бұрын
Thank you so much sir 👍👍👍👍👍
@TheScienceCube3 ай бұрын
So glad you liked it! Consider joining our Telegram Channel: ▶t.me/TheScienceCube_Community
@Sharanya3868 ай бұрын
3:36 sir could you please explain how and why w= u1-u2?
@TheScienceCube8 ай бұрын
Sharanya, you may need to read this properly and watch this lesson as well kzbin.info/www/bejne/r3enqHmEgd6Niq8 The Work-Kinetic Energy Theorem tells us that the work done on an object equals the change in its kinetic energy or W = ΔKE. According to the Law of Conservation of Energy, the total energy (kinetic + potential) in a system remains constant. So, if we express this law, it looks like: K₁ + U₁ = K₂ + U₂, where K represents kinetic energy and U represents potential energy at initial (1) and final (2) states. If we rearrange the equation focusing on the change in kinetic energy, we get: K₂ - K₁ = U₁ - U₂. This equation shows the relationship between the change in kinetic and potential energies. Now, replacing ΔK (which is K₂ - K₁ ) with (U₁ - U₂) in the Work-Kinetic Energy Theorem (W = ΔKE), we get: W = -ΔU = U₁ - U₂. This essentially means the work done is equal to the negative change in potential energy, illustrating how energy transitions between kinetic and potential forms within a system.
@Sharanya3868 ай бұрын
yes sir got it. thank you very much. Your animated videos are quite informative please keep making more of it!@@TheScienceCube
@TheScienceCube8 ай бұрын
@@Sharanya386 Glad to know! Please do share this channel with your friends, that will be great help :) Best wishes