Extension in a Spring: Which Mass Causes Greater Extension?

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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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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?
 
s4.gif

this is the second case.
 
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.
 
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?
Ameya_3.png
 
wouldn't a be 0 if we assume no spring?
 
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.
 
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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?
 
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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