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ParthKohli
 one year ago
hmm
ParthKohli
 one year ago
hmm

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ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.1dw:1443892871026:dw

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.1\[dW = \vec F \cdot \vec {dr} = (F \hat i ) \cdot ??? \]

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1let \(F=\langle f(x),0 \rangle \) \[W = \int\limits_{P1}^{P2} F.dr = \int\limits_{P1}^{P2} \langle f(x),0 \rangle.\langle dx,dy \rangle = \int\limits_{P1}^{P2} f(x)dx \]

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.1\[dr = ld\theta \]

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1Once 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
 one year ago
Best ResponseYou've already chosen the best response.1you 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
 one year ago
Best ResponseYou've already chosen the best response.1whats the value of F ?

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.1honestly it doesn't even matter because I think they've calculated it using initial and final potential energies.

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.1and all these users visit this thread only when I create my dumbest thread...

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.1I 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
 one year ago
Best ResponseYou've already chosen the best response.1then simply use that, its just a geometry problem

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1the force T is always there, but i think we ignore it because it is always perpendicular to the direction of bob

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.1still has a horizontal component no?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1dw:1443893772125:dw

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1it haas no component along the direction in which the bob is moving

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.1hmm, 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
 one year ago
Best ResponseYou've already chosen the best response.1nice. alright. good. great. understood.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1looks 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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