21.5 Work-Kinetic Energy Theorem in 2D and 3D

MIT 8.01 Classical Mechanics, Fall 2016
Instructor: Dr. Peter Dourmashkin

License: Creative Commons BY-NC-SA

[qiushizhang]

thank you professor dourmashkin, and all the other faculties in charge of MIT 8.01 for presenting such lucid course, you really modified the way in which I view motion in nature and have aided my research substantially. cannot stress how great work MIT and people like you are doing!

[alihajib]

we really need such good professors as you in Morocco, great demonstration !!

[abhinavgupta]

GREAT EXPANATION DEAR SIR. LOVE FROM INDIA BY AN IIT ASPIRANT (GRADE 12TH)

[wanik]

And I'm over here trying to relearn polynomial factorization. This is impressive.

[PawanKumar-ptqi]

great work for learners of physics by mit

[titosystems]

this may seem as a stupid question, but if kinetic energy is the sum of the squared components of the velocity multiplied by mass/2, can you see the kinetic energy as a sum of the "components" of the "total" kinetic energy? (i say "components" because kinetic energy is a scalar, not a vector)

[Samiya-]

Sir for 3D we just have to put z axis tooo

[samanthaknepp]

This has been very helpful. However, I cannot figure out for the life of me how you are writing backwards. It must be a mind trick!

[BTWPhysics]

The Work-Energy Theorem is incorrect to begin with and the change in kinetic energy does not represent the net work done. Work is not a function of displacement neither and W ≠ F ⋅ ds. Newton's second law is also incorrect and F≠ ma. The substitution ds = vdt into the integral is incorrect in this case too and it would instead be ds = 1/2vdt. Kinetic energy is not 1/2mv^2 neither.