# How Does a Force Affect Spring Length Between Two Equal Mass Blocks?

• Jorgen1224
In summary: 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.
Jorgen1224

## Homework Statement

[/B]
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

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Jorgen1224 said:
m1⋅a=F-N and m2⋅a = N
The masses will not have the same acceleration.
But they have the same mass, so you can make the algebra easier by just using m instead of m1, m2.

Jorgen1224 said:

## The Attempt at a Solution

[/B]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 I don't know what 'a' is, but if it is the acceleration of m1, it won't be the acceleration of m2.
They move at different rates, as implied by the question, when it asks about the variation in spring length.

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 Why?
F1=ma F1? Isn't that just F? ma? Is that m1.a1 ? But, F = m1.a1 ignores the spring force.
kx=m2a2
N=m2a2
m1a1-N = F, a=N/m Which a is this?
F=2N Why?
x = N/k = 2F/k
l=l0 - x = L0 - 2F/k

## 1. What is a block connected by a spring?

A block connected by a spring is a simple mechanical system where a block is attached to one end of a spring, while the other end of the spring is attached to a fixed point. When the block moves, the spring stretches or compresses, creating a force that acts on the block.

## 2. How does a spring affect the motion of the block?

The spring affects the motion of the block by exerting a force on the block that is directly proportional to the displacement of the block from its equilibrium position. This force is known as the spring force or the Hooke's law force.

## 3. What is the relationship between the spring constant and the stiffness of the spring?

The spring constant is a measure of the stiffness of the spring. It is defined as the amount of force required to stretch or compress the spring by one unit of length. The higher the spring constant, the stiffer the spring is and the more force it takes to stretch or compress it.

## 4. How does the mass of the block affect the motion of the system?

The mass of the block affects the motion of the system by changing the inertia of the block. The more massive the block is, the more inertia it has and the harder it is to change its state of motion. This means that a heavier block will have a slower acceleration compared to a lighter block when connected to the same spring.

## 5. What factors affect the frequency of the block-spring system?

The frequency of the block-spring system is affected by the mass of the block, the spring constant, and the amplitude of the oscillation. A heavier block or a stiffer spring will result in a lower frequency, while a larger amplitude will result in a higher frequency. The frequency is also inversely proportional to the square root of the mass, meaning that a heavier block will have a lower frequency compared to a lighter block.

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