Net 48N Force Stretches .5m Spring to .78m

[c4iscool]

how would I set this up: a spring that is .5m in length elongates by .03m when a 4 Newton force is exerted to stretch the spring. what is the total length of the spring(in m) when a net 48 Newton force is used to stretch the spring?

 

[al_201314]

F=kx where x is the length of extension, k is the spring constant. From here you can find k right? With the k you can find for the extended length at 48N and not forgetting the original length.

 

[c4iscool]

k = F/L. that would make k=8?

 

[al_201314]

I'm not sure how you got 8 but it's not correct. To find k, F/x => 4/0.03 and so the spring constant is 133.3N/m. From here are you able to continue?

 

[c4iscool]

o, i c. i was using .5m as x.

 

[c4iscool]

now that I have the constant, it should look like this:
f=k*x
48=133.3*x and I solve for x?

 

[al_201314]

You canot use 5m to find the spring constant, because at 5m, it is the natural length of th spring, and there is no force acting on it to provide the 5m.

Yes, and this x you get is only the length of extension, and the question wants the total length so I guess it's pretty easy from here!

 

[c4iscool]

thanks for your help