- #1

Jorgen1224

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## Homework Statement

There are two blocks that have equal mass m, they are connected by a spring (with a coefficient of elasticity k) that has length l0.

Instantly on the left block a force F begins to act. What will be the min. and max. length of this spring during this movement.

There is no friction

## Homework Equations

Newton's second law

F=kx

Eventually kx

^{2}/2

And that's all i figured out

## The Attempt at a Solution

I know that answer is l0-F/k and l0 (confirmed answer), but i don't know how it actually happened. (I actually know where does l0 in the elongation comes from :P)

These are 3 ideas on how to solve it.

(N = Tension)

1st. From the 2nd law (let's call these m1 and m2 to make it easier, for me at least)

m1⋅a=F-N and m2⋅a = N ---> a=N/m2

m1⋅N/m2 = F - N

m1⋅N= F⋅m2-m2⋅N ---> N=F⋅m2/(m1+m2)

Now because N = F·m2/(m1 + m2)

Then shortened spring will be x = N/k = F·m2 / (k·(m1 + m2))

Therefore length L = l0 - x =

**l0 - F·m2 / (k·(m1 + m2))**so it's actually l0 - Fm / k⋅2m therefore it's l0 - F/km

2nd (It's actually not mine idea)

F= kx

^{2}/2

F1=ma

kx=m2a2

N=m2a2

m1a1-N = F, a=N/m

F=2N

x = N/k = 2F/k

l=l0 - x = L0 - 2F/k

3rd

I thought something about centre of mass, but it's not a very highly developed idea, i don't even know how to describe it.

Forgive me for any language mistakes