You are Here: Home

# M3 elastic strings Tweet

Maths and statistics discussion, revision, exam and homework help.

Announcements Posted on
TSR launches Learn Together! - Our new subscription to help improve your learning 16-05-2013
IMPORTANT: You must wait until midnight (morning exams)/4.30AM (afternoon exams) to discuss Edexcel exams and until 1pm/6pm the following day for STEP and IB exams. Please read before posting, including for rules for practical and oral exams. 28-04-2013
1. M3 elastic strings
Spoiler:
Show

After the string exceeds a length of 1m I would have thought tension acts and it would contribute to the 'work done' along with friction. According to the answers, only friction has been considered (μ=0.11), and I was wondering why you ignore tension in the work-energy principle?
2. Re: M3 elastic strings
You don't ignore the work that tension does here, as you have to use EPE. Remember that work done = force x distance. Here the force is tension, and the distance is how much you extend the string with that force, but both these things are variable as the string extends - they're not constant so you can't just multiply final tension and extension. You have to integrate tension with respect to distance (extension), and this gives you elastic potential energy (i.e. work done just in extending the string). EPE = [lambda.x^2]/[2l], all of this is given in the book

For this question you consider that mgh lost - work done against friction = EPE gained, and mu does come out as 0.11
3. Re: M3 elastic strings
(Original post by scherzi)
You don't ignore the work that tension does here, as you have to use EPE. Remember that work done = force x distance. Here the force is tension, and the distance is how much you extend the string with that force, but both these things are variable as the string extends - they're not constant so you can't just multiply final tension and extension. You have to integrate tension with respect to distance (extension), and this gives you elastic potential energy (i.e. work done just in extending the string). EPE = [lambda.x^2]/[2l], all of this is given in the book

For this question you consider that mgh lost - work done against friction = EPE gained, and mu does come out as 0.11
So tension has already been considered through the use of EPE. I think I get it now, thanks