ParthKohli
  • ParthKohli
hmm
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
ParthKohli
  • ParthKohli
|dw:1443892871026:dw|
ParthKohli
  • ParthKohli
\[dW = \vec F \cdot \vec {dr} = (F \hat i ) \cdot ??? \]
ganeshie8
  • ganeshie8
let \(F=\langle f(x),0 \rangle \) \[W = \int\limits_{P1}^{P2} F.dr = \int\limits_{P1}^{P2} \langle f(x),0 \rangle.\langle dx,dy \rangle = \int\limits_{P1}^{P2} f(x)dx \]

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

ParthKohli
  • ParthKohli
\[|dr| = l|d\theta| \]
ganeshie8
  • ganeshie8
Once you setup the work integral, it is just a normal integral, you can evaluate it using any of the tricks that you're familiar with
ParthKohli
  • ParthKohli
you know, it contains \(mg\) in it. I took a good look at the choices and the only one where I'm getting \(W = 0\) putting \(\theta =0\) is \(mg \ell (1 - \cos \theta)\), so some clever work needed right there.
ganeshie8
  • ganeshie8
whats the value of F ?
ParthKohli
  • ParthKohli
it can vary.
ParthKohli
  • ParthKohli
honestly it doesn't even matter because I think they've calculated it using initial and final potential energies.
ParthKohli
  • ParthKohli
and all these users visit this thread only when I create my dumbest thread...
ParthKohli
  • ParthKohli
I don't see why all the credit for changing the potential energy is going to \(F\) when some of the contribution is made by \(T\)
ganeshie8
  • ganeshie8
then simply use that, its just a geometry problem
ganeshie8
  • ganeshie8
the force T is always there, but i think we ignore it because it is always perpendicular to the direction of bob
ParthKohli
  • ParthKohli
still has a horizontal component no?
ganeshie8
  • ganeshie8
|dw:1443893772125:dw|
ganeshie8
  • ganeshie8
it haas no component along the direction in which the bob is moving
ParthKohli
  • ParthKohli
hmm, of course, yeah. that's true. in absence of other forces, if we see a bob undertaking pendulum motion, we can say that the potential energy which is increased is actually a consequence of previously possessed kinetic energy and not a force. cool.
ParthKohli
  • ParthKohli
nice. alright. good. great. understood.
ganeshie8
  • ganeshie8
looks good to me too initially when you release it from an angle, it has highest potentil energy, which transforms to kinetic energy when the bob goes to its lowerst point, then transforms back to potential energy when the bob goes goes back to its highest point..

Looking for something else?

Not the answer you are looking for? Search for more explanations.