# Extension in a Spring: Which Mass Causes Greater Extension?

• Ameya Darshan
In summary, the 5 kg block would experience greater acceleration if it was hanging from the spring from the ideal pulley than if it was hanging from the spring from the other side supported by a mass of 10 kg.
Ameya Darshan

## Homework Statement

Scenario: a block of mass 5 kg is hanging from a spring from an ideal pulley from one side, the other side supporting a mass of 10 kg through a STRING.
Now, the 10 kg block replaced with a 5 kg block.

In which case would the extension in spring be greater, assuming constant acceleration.

## Homework Equations

10(g-a)=5(g+a) - kx1
and
5(g+a)-kx2= 5(g-a)

## The Attempt at a Solution

I know the answer is the first case, visualising it(and according to the solution). But I'm not able to prove this mathematically. From my equations I'm getting x1 proportional to (15a-5g) and x2 proportional to 10a and i don't know how these are related.
Thanks for the help :-)

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Ameya Darshan said:
In which case would the extension in spring be greater, assuming constant acceleration.
What spring? Your scenario says nothing about a spring.

kuruman said:
What spring? Your scenario says nothing about a spring.

my bad I'm so sorry. the spring is connected to the block.

Ameya Darshan said:
Scenario: a block of mass 5 kg is hanging from a spring from an ideal pulley from one side, the other side supporting a mass of 10 kg through a STRING.

That description doesn't make any sense to me. Any chance of a diagram?

this is the second case.

Can you find the acceleration in each of the two cases?

kuruman said:
Can you find the acceleration in each of the two cases?

how can i? i mean, it's related to k and extension in the spring. sorry if I'm missing something very common.

Let's look at the second case first with the two 5-kg masses. What do you think the acceleration is in this case?

i use the equation 5(g+a)-kx = 5(g-a) to get a = kx/10.

Wherever you got that equation, it does not apply here. Suppose I hid the spring with a piece of cardboard (see figure) so that you are not able to see it. All you see is two equal masses on either side of the pulley. You don't know the spring is there and you assume that the string is continuous behind the cardboard. Which way do you think the pulley will be accelerating under these circumstances? Clockwise or anticlockwise?

wouldn't a be 0 if we assume no spring?

Exactly. It is zero. What is the tension in the left string? What about the right string?

5g in both.

Very good. Now suppose I remove the cardboard and you discover that there is an extended spring behind it. Would anything change? Would the masses all of a sudden accelerate?

Ameya Darshan
okay, so the extension will be kx=mg so x =mg/k. am i right?
and was the equation for my first case right?

Ameya Darshan said:
okay, so the extension will be kx=mg so x =mg/k. am i right?
That is correct.
Ameya Darshan said:
and was the equation for my first case right?
One thing at a time. I hope you have seen by now that the tension in the string is the force that causes the extension. So now you need to find the tension in the first case.

Ameya Darshan
Ameya Darshan said:
i use the equation 5(g+a)-kx = 5(g-a) to get a = kx/10.
A massless spring exerts the same force at each end. You seem to have taken it as only exerting a force at one end.

l don't understand, could you please explain further?

kuruman said:
So now you need to find the tension in the first case.

I get T = 10(g-a) = kx. is that right? but this has 'a' in it's equation, so how would i compare it to the second case?

Ameya Darshan said:
l don't understand, could you please explain further?
Ok, but it will be easier if you first explain how you get
Ameya Darshan said:
5(g+a)-kx2= 5(g-a)

the fact that I'm not able to explain how i got it certainly tells me i was wrong. :-/

i understood the second case thanks to kuruman but I'm still not able to work out the first case.

Ameya Darshan said:
i understood the second case thanks to kuruman but I'm still not able to work out the first case.
Can you draw a force diagram of the left, 10 kg, mass with just a small piece of the string attached to it? Just include part of the string but not the spring. Include, that is, what is showing below the cardboard on the left side of the picture I posted, but with the 10 kg instead of the 5 kg mass. Pretend that you are seeing the mass accelerating down. Can you tell me what forces are acting on it? Remember, you have no idea the spring is there.

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## What is extension in a spring?

Extension in a spring refers to the change in length of a spring when a force is applied to it. This can either be an increase or decrease in length, depending on the direction of the force and the properties of the spring.

## What causes a spring to extend?

A spring extends when a force is applied to it. This force can be applied through stretching, twisting, or compressing the spring. The amount of extension is directly proportional to the amount of force applied.

## What factors affect the extension of a spring?

The extension of a spring is affected by several factors, including the type of material the spring is made of, the diameter and length of the spring, and the magnitude and direction of the force applied to it.

## How is the extension of a spring measured?

The extension of a spring is typically measured using a ruler or measuring tape. The original length of the spring is measured and then the new length is measured after a force is applied. The difference between the two lengths is the extension of the spring.

## Why is extension important in a spring?

Extension in a spring is important because it allows the spring to store potential energy, which can be released when the force is removed. This makes springs useful in a variety of applications, such as in car suspension systems, mechanical toys, and door hinges.

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