hmm

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|dw:1443892871026:dw|
\[dW = \vec F \cdot \vec {dr} = (F \hat i ) \cdot ??? \]
let \(F=\langle f(x),0 \rangle \) \[W = \int\limits_{P1}^{P2} F.dr = \int\limits_{P1}^{P2} \langle f(x),0 \rangle.\langle dx,dy \rangle = \int\limits_{P1}^{P2} f(x)dx \]

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\[|dr| = l|d\theta| \]
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
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.
whats the value of F ?
it can vary.
honestly it doesn't even matter because I think they've calculated it using initial and final potential energies.
and all these users visit this thread only when I create my dumbest thread...
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 \(T\)
then simply use that, its just a geometry problem
the force T is always there, but i think we ignore it because it is always perpendicular to the direction of bob
still has a horizontal component no?
|dw:1443893772125:dw|
it haas no component along the direction in which the bob is moving
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.
nice. alright. good. great. understood.
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..

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