Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

An object is placed on a frictionless table with its one end attached to a cord which is connected to a pulley and the tension is maintained constant at 25 N. what is the change in kinetic energy of the object from its position 3m away to 1m away from the raised platform?

poorly drawn, so not at all to scale...just shows the setup.

I think that we should integrate work with respect to theta here, over the interval $\tan^{-1}(1.2/3)$ to $\tan^{-1}(1.2/1)$. But for that, we need to differentiate work with respect to $\theta$, right? So doesn't it end up the same if I differentiate it and then integrate it? Or do I integrate the expression $F s \cos(\theta)$ with $\mathrm{d}\theta$ without differentiating it first?

share|cite|improve this question
Thanks a ton, David! – Saurabh Raje Apr 6 '13 at 5:19
Are u trying for JEE? :) – Sreekanth Karumanaghat Apr 7 '13 at 11:43
Yes...although my classes haven't started yet. But tell me, is it one of the tough questions in that category, or easier? Btw, I found this in Halliday walker book – Saurabh Raje Apr 7 '13 at 14:20
this isnt one of the toughest problems as far as JEE is concerned I would rate this as medium complexity. The tough questions will be a combination of mechanics,electro magnetism and themodynamics.. :) JEE is the toughest engineering entrance in the whole world.. even MIT entrance test is easier, all the best keep a back up there is no guarantee that u will succeed even if u study 10 hrs a day for 3 years.. dont expect too much... make ur fundas strong.. – Sreekanth Karumanaghat Apr 10 '13 at 12:27
I havent heard about vidyamandir... yeah HC Verma is physics.. sorry I forgot chemistry books names. – Sreekanth Karumanaghat Apr 11 '13 at 17:23
up vote 0 down vote accepted

As the work $W=\int\vec F. {d\vec s}$ The work will be only dependent on the component of force along the displacement , or otherwise the length of rope moved along the force.

I hope we are done!.

Now you can try it.

share|cite|improve this answer
so is it basically displacement times sec(theta) times force? as displacement times sec(theta) lies along the force, ie the hypothenuse?? – Saurabh Raje Apr 11 '13 at 10:10
@SaurabhRaje Join – ABC Apr 11 '13 at 10:17

First of all, you pretty much never need to integrate work. There is no formula I can think of which involves the integral of work.

Secondly, just because you are going to be integrating something over $\theta$ doesn't mean you have to differentiate something with respect to $\theta$. That is one way to do the problem, but it may not be the most straightforward way. I would suggest going back to the formula $W = \int \vec{F}\cdot\mathrm{d}\vec{s}$ and trying to express it in terms of $\theta$. Drawing a diagram and doing some geometry should help.

share|cite|improve this answer
Won't just calculating the length of rope pulled out be enough to get work done? – ABC Apr 6 '13 at 4:51
Well that was integral of dW...sorry about the typo, – Saurabh Raje Apr 6 '13 at 5:21
How would I get f in terms of dS? Isn't fcos(theta) along displacement? So fcos(theta) is changing with a change in shouldn't I integrate it with theta? – Saurabh Raje Apr 6 '13 at 5:28
@exploringnet sure, but then you bypass all the practice you would get doing integrals ;-) – David Z Apr 6 '13 at 6:32
@SaurabhRaje You will need to choose one variable to parametrize the motion. It could be $\theta$ or $s$ for example; any variable that takes a different value at every point along the path. Then you need to express everything else in terms of that variable. – David Z Apr 6 '13 at 6:33

We should note that theta varies here... As far as the kinetic energy change is concerned.. Kinetic energy change is nothing but work done by a varying force... At any point work done is W = F s $\cos(\theta)$ here it has been mentioned that F is maintained a constant we are only concerned about Fcos theta wer theta varies from $\tan^{-1}(1.2/3)$ to $\tan^{-1}(1.2)$

so the ultimate formula is W = $\int \vec{F}\cdot\mathrm{d}\vec{s}$

We have to find a relation between ds and d$\theta$ by some geometric method and approximation finding the relation between a small change in linear and angular displacements in the triangle

d$\theta$ is $\theta$2 - $\theta$1 and ds is S2- S1 we need to establish a relation between ds and d$\theta$ in the process we can use approximation that d$\theta$ is very small and ds too is very small.. I have not been able to do that yet.. if any one is able to do that lemme know.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.