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Question: A block of Mass m is connected to another block of mass M by a massless spring of spring constant k . The blockes are kept on a smooth horizontal plane and are at rest. The spring is unstretched when a constant force F starts acting on the block of mass M(horizontally) to pull it. Find the maximum extension of the spring. figure I solved it by two different methods:

Method 1: here I assumed that max. extension(x) will be produced when both the blocks would be moving with constant acceleration. Therefore constant acceleration is

$$a=F/(m+M)$$ 1

and by considering free body diagram of block with mass m

$$kx=ma$$ 2

From 1 and 2 $$x=mF/(k(m+M))$$

Method 2: by conservation of energy In the reference frame of center of mass.

For block 'm', we have two forces acting $mF/(m+M)$ and $kx$, in opposite directions.

For block 'M' we have three forces acting $MF/(m+M)$ and $kx$, and $F$ in the opposite direction.

Assuming m moves a max distance x_1 from CM M moves a distance x_2 from CM. Then work done by external force will be

$$W=mF(x_1+x_2)/(m+M)$$

This will be stored as the internal energy, therefore

$$1/2·k·(x_1+x_2)^2=W$$

On solving this we get

$$x_1+x_2=2mF/(k(m+M))$$ Which one is the wrong and why?

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closed as off-topic by John Rennie, Mark Eichenlaub, ZeroTheHero, Kyle Kanos, Jon Custer Apr 26 '18 at 14:02

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Method 1 is correct. The error in Method 2 enters with your expression for "work done by external force". This should be a sum of the the work done on $m,$ which is $x_1mF/(m+M)$ and that done on $M,$ which is $x_2(F-MF/(m+M))$. To solve using this method, you will also need an additional relation: the ratio of $x_1$ to $x_2$, which is just $M/m.$ In your expression you are assuming that the work done on the spring is the total length by which it is stretched multiplied by the ultimate tension. But the tension in the spring only reaches that value at the end of the stretching phase, so the work is only half as much.

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