Not sure what I can really say, but I owe you a lot. I'm in university, and I can't understand a single word from my professor. Let's say that I'm mainly looking to get into computer science engineering which makes it quite obvious that I'm not into physics. Without you I would have failed my course. I can know understand stuff, and find them very reasonable even that I'm pretty weak at physics. Thank you.
dr is a infinitessimally small piece of the displacement. If displacement is in the x direction, it could be dx. In the y direction, it could be dy. In some non-constant direction, dr.
Question: If force varies but in only one dimension, then there is no dot product in the integral. If force varies as well as displacement, then there is a dot product(between F and dr) in the integral. Is that correct? @Dan Fullerton
Hi Shreya. That first point is incorrect. If force varies in only one dimension, there can still be a dot product if the force and the displacement are not perpendicular. As long as some component of force is in the direction of the displacement, you will have a non-zero dot product.
How is it that you can cancel the dt in your derivation of the work energy theorem? My calculus teacher said that you can't treat variables of integration like variables.
+Bobby Physics I have a differential of t in the numerator, and a differential of t in the denominator. Item A over Item A is equal to one. You don't treat variables of integration like variables (it's actually a differential we're talking about), but what we did here is OK. Short answer -- ask your math teacher. ;-)
Not sure what I can really say, but I owe you a lot. I'm in university, and I can't understand a single word from my professor. Let's say that I'm mainly looking to get into computer science engineering which makes it quite obvious that I'm not into physics.
Without you I would have failed my course. I can know understand stuff, and find them very reasonable even that I'm pretty weak at physics. Thank you.
So glad to hear that the videos and the APlusPhysics site are helping you out. Best of luck to you, and thank you for taking the time to write me!
🐶🏠😦
Yo man how's life, by now you've graduated
you made this so much easier than it would've been otherwise. thank you sir
You're very welcome!
dr is a infinitessimally small piece of the displacement. If displacement is in the x direction, it could be dx. In the y direction, it could be dy. In some non-constant direction, dr.
In the integral equation for , how do you manipulate the "dr"?
Even after 10 years you’re helping so many people! Thank u so much!
Thrilled to hear it -- good luck!
Would this still apply to today”s ap physics c mech exam?
i studied work energy theorem in class today and it didn't make any sense to me, but this video helped me a lot! i really appreciate.
Glad to hear it helped!
Question: If force varies but in only one dimension, then there is no dot product in the integral. If force varies as well as displacement, then there is a dot product(between F and dr) in the integral. Is that correct? @Dan Fullerton
Hi Shreya. That first point is incorrect. If force varies in only one dimension, there can still be a dot product if the force and the displacement are not perpendicular. As long as some component of force is in the direction of the displacement, you will have a non-zero dot product.
How is it that you can cancel the dt in your derivation of the work energy theorem? My calculus teacher said that you can't treat variables of integration like variables.
+Bobby Physics I have a differential of t in the numerator, and a differential of t in the denominator. Item A over Item A is equal to one. You don't treat variables of integration like variables (it's actually a differential we're talking about), but what we did here is OK. Short answer -- ask your math teacher. ;-)