Springs in series question

[onedirection]

I thought springs in series were

ktot = 1/k1 + 1/k2 + ...

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[sps27]

I think the total extension will be the sum of individual extensions of each spring so like kx1+kx2+kx3 = kx. So T(total) = NT. Not too terribly sure but that's what I think.....

 

[monkeyvokes]

I thought springs in series were

ktot = 1/k1 + 1/k2 + ...

this is an interesting problem.

So we know a few facts: when springs connect in series, their x values add. sum of Xs = NX
Also, 1/ktotal = sum of 1/kindividual = N*1/kindividual

Tindividual = x * kindividual

Ttotal = (NX) * (N*1/kindividual)
Ttotal = N * (X*1/kindividual)
Ttotal = N * (X/kindividual)

So Ttotal can not be A.) NTindividual, because Tindividual does not equal (X/kindividual).
I don't think B.) sqrt(N)*Tindividual makes sense because there would be no reason to use squares.
D.) Ttotal =Tindividual/N would be saying that (N * (X/kindividual)) = (kindividual*x)N, when it does not.

That only leaves C.) Ttotal=Tindividual
I think it makes conceptual sense that their tensions would be additive..

I thought about this problem way too much and wrote out an example problem with 2 springs where ki=3 and x=4. I did the math and got the same force of tension for an individual spring as I did for the total force.

Yeah I spent too much time evaluating this one haha

 

[monkeyvokes]

I think the total extension will be the sum of individual extensions of each spring so like kx1+kx2+kx3 = kx. So T(total) = NT. Not too terribly sure but that's what I think.....

Think about this in the extreme sense, imagine connecting an infinite amount of strings that all have the same original tension. Rather than all of their tensions adding and getting infinitely tight, I think they'd maintain the same original level of tension

 

[onedirection]

The answer was actually A. NT for this one