@@yoprofmatt Hey, thank you so much for having an intrinsic motivation to teach not only your students, but me as well. I'm at U of Utah, and not all of us are as fortunate to have a physics professor as thorough and thoughtful as you!
The concept of work in Physics is completely erroneous or wrong and work is not really force times displacement(F•d). Work is a vector quantity(i.e. it's not a scalar) and is ,basically, defined as a "physical-displacement(ms)" by or due to a force, an impulse, or a kinetic force.
Professor Anderson, At the beginning of the course until the entire lesson why didnt you considered Gravity Force (mg), as main factor of work??? And you didnt prove of how this "W=Fdcos(A) formula appeared.
Riley Thomas If we consider displacement as horizontal, it might make more sense? If we apply force F directly horizontally, our angle between the displacement vector and force vector would be 0 So, W=F•d•cos(0)=F•d In the first example, we are taking the angle between the horizontal displacement and vertical force, so the angle is PI W(horizontal)=F•cos(PI)•d=0 If we study the work in vertical axis, we get W(vertical)=F•h•cos(0)=F•h Where h is displacement in y axis(height) F here is mass•gravity So W(vertical)=mass•gravity•h
Hi sir, could tell me when multiply two vectors i.e force and displacement we are getting scalar , don't tell me that we need to use dot product I didn't understand why we are involving it here , I am vexed with all answers .... I don't understand when we apply force on body and body gets displaced in certain direction why it will be scalar ....please clarify me about using of dot product and why it is scalar still we have both vectors
this topic has been bugging me until now, i could not understand this energy (work) formula because it is seriously not making sense to me.... can someone help me ?
When you left the pen and move it to another place you don't feel you do a work ! but if you consider the time here you will feel you are really tired because we don't consider the time in the work ! Like example if I left a pen with like 10 gram 4 one hour without moving my hand I will feel I do a lot of work
At this point, I should be paying this guy a thousand dollars for my education.
Thank you so much for these videos Mr. Anderson! they are extremely helpful!
best professor ever.
Physics with Benedict Cumberbatch!
Thanks, I'll take it.
Now if only I could act.
Cheers,
Dr. A
@@yoprofmatt Hey, thank you so much for having an intrinsic motivation to teach not only your students, but me as well. I'm at U of Utah, and not all of us are as fortunate to have a physics professor as thorough and thoughtful as you!
BEST PHYSICS LECTURES EVER !!!
The concept of work in Physics is completely erroneous or wrong and work is not really force times displacement(F•d). Work is a vector quantity(i.e. it's not a scalar) and is ,basically, defined as a "physical-displacement(ms)" by or due to a force, an impulse, or a kinetic force.
Very nice sir
Professor Anderson, At the beginning of the course until the entire lesson why didnt you considered Gravity Force (mg), as main factor of work??? And you didnt prove of how this "W=Fdcos(A) formula appeared.
Riley Thomas
If we consider displacement as horizontal, it might make more sense?
If we apply force F directly horizontally, our angle between the displacement vector and force vector would be 0
So,
W=F•d•cos(0)=F•d
In the first example, we are taking the angle between the horizontal displacement and vertical force, so the angle is PI
W(horizontal)=F•cos(PI)•d=0
If we study the work in vertical axis, we get
W(vertical)=F•h•cos(0)=F•h
Where h is displacement in y axis(height)
F here is mass•gravity
So W(vertical)=mass•gravity•h
Hi sir, could tell me when multiply two vectors i.e force and displacement we are getting scalar , don't tell me that we need to use dot product I didn't understand why we are involving it here , I am vexed with all answers .... I don't understand when we apply force on body and body gets displaced in certain direction why it will be scalar ....please clarify me about using of dot product and why it is scalar still we have both vectors
because we use dot product!!!!!!!!!, DOT PRODUCT.
6:11 I believe that should be "component of the force," not "component of the work."
I agree. Nice catch.
Cheers,
Dr. A
Does component of work is vector or scalar please clarify ?
@@yoprofmatt Does component of work is vector or scalar please clarify ?
this topic has been bugging me until now, i could not understand this energy (work) formula because it is seriously not making sense to me.... can someone help me ?
Did this video help you?
Dr. A
Thank you!
When you left the pen and move it to another place you don't feel you do a work ! but if you consider the time here you will feel you are really tired because we don't consider the time in the work !
Like example if I left a pen with like 10 gram 4 one hour without moving my hand I will feel I do a lot of work
Yes, but that is what we call "physiological" work, not physical work. In physics, no movement means no work.
Cheers,
Dr. A
Sir use hindi language because we are from India, bihar