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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 kx2/2
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= kx2/2
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