M3 elastic strings

Spoiler

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?

Reply 1

scherzi

4

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

Reply 2

snow leopard

OP

1

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 :smile:

M3 elastic strings

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