Finding Young's Modulus and Energy Stored in an Elastic Cord

[ElBell]

1. Two part question.

An elastic cord with cross sectional area 4.00mm^2 needs a force of 3.6N to increase its length by 12%

Part One- Find the Youngs Modulus of Elasticity

Part Two- The unstretched cord is 0.80m long. Find the energy stored in this strain.

2. Homework Equations :
Y= Stress/ Strain
Stress= Force/ cross sectional area
Strain= change in length/ original length


Part One Attempt-
Stress= 3.6N/ 4* 10^-2
Stress= 90

Strain= 9.6/ 80
Strain= 0.12

Y= 90/ 0.12
Y= 750

Now I don't know what to do! What are the Units? Is this correct?

Part Two Attempt-

I believe this has something to do with Potential Energy...is this correct?

Any help would be much appreciated! Thanks so much for your time

 

[pat666]

your first part is wrong. You HAVE to change to si units if you want to do the rest of the problem... also what is the deal with "3.6N/ 4* 10^-2" what that is effectively saying is that stress is 3.6 multiplied by 400! do you understand that the the ^2 given in the question is part of the unit and that you DONT actually square the number! You need to work on your exponents and what they mean, there is nothing wrong with the actual physics you've done. (sort of)

 

[pat666]

you also need to post the entire question you need help with. where did the lengths you have come from??

 

[ElBell]

Thanks for your help so far!

I understand from your help on my last question that you don't square your numbers, but I havent. I thought I had to change my units to metres, so I changed 4 o (4*10^-2)? Maybe I should have just written 0.004 to save confusion.

Am I right in even converting this to metres?

Ive just attempted again:

Stress= 3.6N/ 0.004
Stress= 900

Strain= 900/ 80
Strain= 11.25

Y= 90/ 11.25
Y= 8

Is this more like an answer I should be looking for?

 

[ElBell]

sorry, also forgot to mention that for some reason we are never given the lengths! I assume because we are using percentages I could just input the easiest figure and the result will always be the same??

 

[pat666]

kk go to google type in (4mm^2) to m^2 its 4*10^-6 not 0.004 also what i was saying before.. you had x/x^-2 what hapens when you divide by a number with a negative index?

 

[ElBell]

Ooooh OK I see what you are saying.

Stress= 3.6N/ (4* 10^-6)
Stress= 900000

Strain= 900000/ .80
Strain= 1125000

Y= 90/ 1125000
Y= 8^-5

:( Now I am confused...and hopeless!

 

[pat666]

no DO NOT PUT the - index in and your good.

 

[pat666]

ok you were actually correct for part a sorry...thought you were making the same mistake as before...

 

[pat666]

and yes for deleta L on L it will be 0.12... were did you get 0.8??

 

[ElBell]

OK, sorry for this, but now I am totally confused.

Is the answer I had originally correct? 750? Or is my new answer correct? haha!?

 

[pat666]

new your 1st one was way off

 

[pat666]

your 1st stress, your youngs modulus is still wrong check you change in length!

 

[ElBell]

I got .8 because it was .8 of a metre? previously i had 80cm written in but i guessed because we are using metres, .8 would have been the correct input?

 

[ElBell]

Stress= 3.6N/ (4* 10^-6)
Stress= 900000

Strain= .096/ .80
Strain= 0.12

Y= 900000/ 0.12
Y= 75* 10^5 ?

 

[pat666]

ok but delta L would be .8*0.12

 

[ElBell]

Thanks for your help so far:

Can anyone help with part two?

So far I have equations:

W= 1/2kx^2
F=kx (so k= f/w)??

So:

k= 3.6/ 0.096= 37.5

W= (1/2)(37.5)(0.096^2)
W= 0.1728J ----does anyone know if this appears correct?

 

[pat666]

looks good to me

 

[ElBell]

pat666, you are a champion :) Thankyou.

 

[pat666]

i just had a proper look at it then and its actually not correct sorry. k doesn't equal f/w since w doesn't = x
your actual procedure and answer is correct but this " k= f/w)" isnt. your answer is correct just that one error. (which you didnt use in your calculations)
3.6=k*.096
k=37.5Nm
w=.5*37.5*.096^2
w=0.1728J
where are you getting these questions that don't have answers?