Spring-block system: why doesn't M1 move when spring force acts on it?

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tellmesomething said:
So work done in bringing the block from the x position to mean position is -0.5kx^2
Why would this work be negative?
 
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TSny said:
Why would this work be negative?
Oh. That was really stupid. Thankyou very much. Also thankyou @PeroK for enduring this throughout and giving me new insights.
 
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tellmesomething said:
Oh. That was really stupid. Thankyou very much. Also thankyou @PeroK for enduring this throughout and giving me new insights.
First, the problem explicitly says that ##x## and ##y## are distances. They are both positive.

Second, the system starts with some PE, which we take to be positive (relative to zero at equilibrium) and ends with PE, which must also be positive (relative to zero at equilibrium). It's not the case that one PE is positive and the other negative. The mass has positive PE at the start position and positive PE at the final position.

That second observation comes from a physical understanding of the problem. And, that physical understanding should determine how you set up your equations.

I would have simply equated:
$$PE_1 = PE_2$$
$$\implies \frac 1 2 k_1 x^2 = \frac 1 2 k_2 y^2$$And not given it a second thought. That must be the correct equation. If you asked me to fully justify that by applying the work-energy theorem rigorously at every step, then I'd have to be careful about negatives. But, I can't see any reason to complicate this problem.
 
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PeroK said:
I can't see any reason to complicate this problem
I have a nagging feeling, nevertheless. Block B is released and S1 starts to decompress, transferring spring energy that is converted to kinetic energy of block B plus spring energy from the compression of S2 plus kinetic energy of block M2.

At the moment when block B is passing its original position ('##y=0##'), there is also a part of the energy converted to energy to stretch spring S1 (which we know to be at its natural length at that point), plus kinetic energy of M1.

## \frac 1 2 k_1 x^2 = \frac 1 2 k_2 y^2## holds if S1 is at its natural length when block B is at its leftmost position and block M1 is not moving. Neither is credible.

We don't know the ratio of the mass of block B wrt blocks M1 and M2, so I am inclined to claim the exercise as stated in post #1 isn't just complicated: it's unsolvable

🤔

##\ ##
 
BvU said:
I have a nagging feeling, nevertheless. Block B is released and S1 starts to decompress, transferring spring energy that is converted to kinetic energy of block B plus spring energy from the compression of S2 plus kinetic energy of block M2.

At the moment when block B is passing its original position ('##y=0##'), there is also a part of the energy converted to energy to stretch spring S1 (which we know to be at its natural length at that point), plus kinetic energy of M1.

## \frac 1 2 k_1 x^2 = \frac 1 2 k_2 y^2## holds if S1 is at its natural length when block B is at its leftmost position and block M1 is not moving. Neither is credible.

We don't know the ratio of the mass of block B wrt blocks M1 and M2, so I am inclined to claim the exercise as stated in post #1 isn't just complicated: it's unsolvable

🤔

##\ ##
M1 has "negligible mass".
 
Ah ! Overlooked that. Thanks !